Graphs with Eigenvalues of High Multiplicity - Semantic Scholar
Washington University in St. Louis
Washington University Open Scholarship Arts & Sciences Electronic Theses and Dissertations
Arts & Sciences
Summer 8-15-2015
Graphs with Eigenvalues of High Multiplicity Casey Boyett Washington University in St. Louis
Follow this and additional works at: http://openscholarship.wustl.edu/art_sci_etds Recommended Citation Boyett, Casey, "Graphs with Eigenvalues of High Multiplicity" (2015). Arts & Sciences Electronic Theses and Dissertations. Paper 636.
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WASHINGTON UNIVERSITY IN ST. LOUIS Department of Mathematics
Dissertation Examination Committee: John Shareshian, Chair Renato Feres Robert Pless Ari Stern David Wright
Graphs with Eigenvalues of High Multiplicity by Casey Boyett A dissertation presented to the Graduate School of Arts & Sciences of Washington University in partial fulfillment of the requirements for the degree of Doctor of Philosophy August 2015 St. Louis, Missouri
c 2015, Casey Boyett
TABLE OF CONTENTS List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Abstract of the Dissertation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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1 Introduction . . . . . . . . . 1.1 Background . . . . . . 1.2 Definitions . . . . . . . 1.3 Summary of Chapters
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1 1 2 5
2 Key Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Classic Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 New Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3 Summary of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4 λ = −1 . . . . . . . . . . . . 4.1 P1 , P2 , and P3 . . . . 4.2 P3 (a, b, c) + K1 (d) . . . 4.3 P4 (a, b, c, d) . . . . . . 4.4 K1,3 (a, b, c, d) . . . . . 4.5 C4 (a, b, c, d) . . . . . . 4.6 P5 (a, b, c, d, e) . . . . . 4.7 C6 (a, b, c, d, e, f ) . . . . 4.8 Coefficients . . . . . . 4.9 Bounds on eigenvalues
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5 λ=0. . . . . . . . . . . . . . 5.1 Q0 , Q2 , and Q3 . . . . . 5.2 2P2 (a, b, c, d; z) . . . . . 5.3 P4 (a, b, c, d; z) . . . . . . 5.4 P5 (a, b, c, d, e; z) . . . . . 5.5 Kite(a, b, c, d; z) . . . . . 5.6 Bull(a, b, c, d, e; z) . . . . 5.7 House(a, b, c, d, e; z) . . . 5.8 Broom(a, b, c, d, e, f ; z) . 5.9 P2 × K3 (a, b, c, d, e, f ; z) 5.10 Coefficients . . . . . . . 5.11 Bounds on eigenvalues .
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100 100 102 104 107 113 118 123 127 133 135 137
6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 ii
LIST OF FIGURES 1.1
C4 + K1 and K1,4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1 2.2 2.3 2.4 2.5
Bow tie and subgraphs . . Blowup and split example P4 . . . . . . . . . . . . . Bipartite Q4 . . . . . . . . Non-bipartite Q4 . . . . .
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ACKNOWLEDGEMENTS
I would first like to thank my advisor, John Shareshian, for allowing me to pursue a topic of my own choosing. While it was more work on me, the satisfaction of working on my own problem as opposed to trudging through one handed to me more than made up for it. Much effort was expended on independent research, but I do not believe I would have enjoyed the alternative. Next, I would like to thank all of the wonderful friends I have made over the last few years as well as those lifelong friends I can never get rid of. Without them I would have easily lost my sanity (what little I still retain). First is my local gaming friends: Sarah Morris, Meredith Sargent, Lauren Katz, and Joseph Colton. I am fortunate to have met each and every one of them, even if occasionally I make a tree fall on them. Next is my “down south” group: Jared Palmer, Lisa Palmer, and Victoria Porter. Even though there are hundreds of miles between us, we still finds ways to keep each other entertained. Many late hours have been spent with them chatting away online. Lastly is Morgan Harrison. Above all others she just gets me. Last, and most important, I would like to thank my family. Without their love and support I would not be where I am today. My mom, my dad, and my niece mean the world to me, and I hope I have made them proud. My only regret is that my dad is not here to see all that I have accomplished as he passed away before I could finish my work. Unfortunately it is not possible to list every single friend I have made along my journey. Even if I am not able to name them all here, they are still very important to me, and I am grateful for all they have done for me.
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Dedicated to my dad.
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ABSTRACT OF THE DISSERTATION Graphs with Eigenvalues of High Multiplicity by Casey Boyett Doctor of Philosophy in Mathematics Washington University in St. Louis, 2015 Professor John Shareshian, Chair Given a graph G we can form a matrix AG indexed by the vertices of G and which encodes the edges of G. AG is called the adjacency matrix of G. From the adjacency matrix we may find the eigenvalues. We would now like to know what information we may garner from the eigenvalues. It turns out quite a bit may be determined from the eigenvalues, collectively called the spectrum.
One big question is to ask whether or not a graph can be uniquely determined by its spectrum. Much research has been done in this area, and it is conjectured that almost all graphs may in fact be determined by their spectra. This is however a difficult task. In this dissertation we look at a subset of all graphs, namely those with either −1 or 0 in their spectrum with a given multiplicity. We first show that any such graph must either be primitive in a sense, or that it is obtained from a primitive graph by an elementary operation of blowing up or splitting vertices. We then show that the set of primitive graphs is finite, for a fixed multiplicity. Lastly, we analyze graphs with −1 or 0 in their spectra with multiplicities up to 4, and show many which are uniquely determined by their spectra.
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Chapter 1 Introduction We assume a basic understanding of graph theory and linear algebra. Beyond these basics we will keep the contents as self-contained as possible. Throughout we will use the notation [n] to mean the set of integers {1, 2, . . . , n}.
1.1
Background
Graph theory is a subject with a myriad of uses. Algebraic, and more specifically spectral, graph theory is no exception. Using the eigenvalues of a graph to study properties of that graph has advantages in that the numbers give quantifiable data, and moreover these eigenvalues can be quickly computed. For example, the spectrum of a graph can give information on structures of the graph such as acyclic digraphs, spanning trees, Hamiltonian cycles, independent sets, cliques, Eulerian orientations, vertex colourings, and so on. Spectral graph theory also has more concrete uses in other areas of study such as information theory, coding theory, chemistry, geographic studies, social sciences, epidemiology, and game theory. Let us look at a couple of these uses in more detail. In information theory one often asks how much information can be transmitted over a channel or a network of channels. The Shannon capacity gives a precise number to how much information can be transmitted, and bounds
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on the Shannon capacity can be given in terms of the spectrum of the graph representing the channel network. Chemistry is a natural subject in which to use graph theory. The structure of a molecule can easily be modeled by a graph, and information about the spectrum of the graph can quickly give information about the structure of the molecule. One very powerful use of spectral graph theory in chemistry is in determining whether two molecules are the same. For large molecules this can be a difficult task, but by calculating the spectra of the associated graphs and comparing them, it is often possible to determine whether or not the molecules are the same. This is a topic we explore further in this thesis. That is, we look at whether or not certain graphs are uniquely determined by their spectra. Finally, social science is yet another subject in which graph theory has natural uses. With the wealth of social networking sites, a graph representing connections between various individuals is a simple way in which to examine connections through the site. Sites such as Facebook take advantage of these network graphs in order to continually improve their functionality. Spectral graph theory has many theoretical and practical applications. This thesis focuses on the theoretical aspects, and primarily we try to approach the Graph Isomorphism problem through spectral techniques. This approach is not new, but we take a look at a subset of graphs that as of yet have gone untouched.
1.2
Definitions
Definition 1.2.1. A graph G is a pair (V (G), E(G)) where V (G) is a finite set and E(G) is a subset of V (G) × V (G). The elements of V (G) are called the vertices of G, and the elements of E(G) are the edges of G. If e = (v, w) is an edge, then v is called the tail of e and w is the head. An edge of the form (v, v) is called a loop. If both (v, w) and (w, v) are edges, then we call {v, w} an undirected edge. Edges that are not undirected are directed. Similarly, a graph with only undirected edges is called undirected. Unless otherwise noted all graphs are loopless and undirected. Where no confusion will be incurred we will refer to the vertex set V and the edge set E without reference to G. We 2
will commonly label the vertices V (G) = {v1 , . . . , vn }. By abuse of notation an edge {v, w} will often be written vw. A few key properties of graphs will be needed. Definition 1.2.2. The complement of a graph G is the graph G with the same vertex set V (G) and E(G) = {vw : vw 6∈ E(G)}. Definition 1.2.3. A path in a graph is a sequence of distinct vertices v1 v2 · · · vk+1 such that vi vi+1 is an edge for all i ∈ [k]. Such a path is said to have length k. Definition 1.2.4. A cycle in a graph is a sequence of distinct vertices v1 v2 · · · vk such that vi vi+1 is an edge for all i ∈ [k − 1], and v1 vk is also an edge. Such a cycle is said to have length k. Definition 1.2.5. A graph is connected if for every two distinct vertices v and w there is a path containing v and w, otherwise the graph is disconnected. Definition 1.2.6. A forest is an acyclic graph. A tree is a connected acyclic graph. Definition 1.2.7. The neighborhood of v ∈ V (G) is the set N (v) = {w : vw ∈ E(G)}. The closed neighborhood of v is N [v] = N (v) ∪ {v}. Definition 1.2.8. The degree of a vertex v is the number of edges containing v and is denoted by d(v). The degree sequence of G is (d(v1 ), . . . , d(vn )). Definition 1.2.9. The maximum degree of G is ∆(G) = max{d(vi )}. The minimum degree of G is δ(G) = min{d(vi )}. Definition 1.2.10. If S ⊆ V is such that no two vertices in S form an edge, then S is called an independent set. The size of the largest independent set is called the independence number of G and is denoted by α(G). Definition 1.2.11. If S ⊆ V is such that every two distinct vertices in S form an edge, then S is called a clique. The size of the largest clique called the clique number of G and is denoted by ω(G). Definition 1.2.12. Graphs G and H are isomorphic if there exists a bijection φ : V (G) → V (H) such that xy ∈ E(G) iff φ(x)φ(y) ∈ E(H). 3
An easy observation is that α(G) = ω(G) and ω(G) = α(G). There are also a few special classes of graphs and operations on graphs that will be important. Definition 1.2.13. The complete graph Kn has vertex set V (Kn ) = {v1 , . . . , vn } and edge set E(Kn ) = {vi vj : i 6= j}. Definition 1.2.14. A graph is called k-partite if V can be partitioned into k disjoint independent sets. If k = 2 we will call the graph bipartite. Definition 1.2.15. The complete multipartite graph Kn1 ,...,nk has vertex set V (Kn1 ,...,nk ) = V1 ∪· · · · ∪· Vk with |Vi | = ni for all i and edge set E(Kn1 ,...,nk ) = {vi vj : vi ∈ Vi , vj ∈ Vj , i 6= j}. Definition 1.2.16. The cycle graph Cn has vertex set V (Cn ) = {v1 , . . . , vn } and edge set E(Cn ) = {vi vi+1 : 1 ≤ i < n} ∪ {v1 vn }. Definition 1.2.17. The path graph Pn has vertex set V (Pn ) = {v1 , . . . , vn } and edge set E(Pn ) = {vi vi+1 : 1 ≤ i < n}. Definition 1.2.18. Given a subset S ⊆ V , the induced subgraph G[S] has vertex set S and edge set E(G[S]) = {vw : vw ∈ E(G) and v, w ∈ S}. If S = V \T then we will write G − T for G[S] or G − v if T = {v}. Definition 1.2.19. Given two graphs G and H on disjoint vertex sets, we can form the graph G + H with vertex set V (G + H) = V (G) ∪· V (H) and edge set E(G + H) = E(G) ∪· E(H). Since this thesis is primarily focused on the spectrum of a graph we need to define the adjacency matrix of G. Definition 1.2.20. For a graph G we define the adjacency matrix AG as the (0, 1)-matrix indexed by the vertices V (G) = {v1 , . . . vn } with entry aij = 1 when vi vj is an edge of G and aij = 0 otherwise. This is not the only matrix naturally associated with a graph, but it is the only one we will study. Given a matrix it is natural to consider the eigenvalues. This leads to our next definitions. Definition 1.2.21. Given a matrix MG associated with a graph G we define the characteristic polynomial of G with respect to MG as PMG (x) = det (xI − MG ) where I is identity matrix. 4
Definition 1.2.22. Given a matrix MG we call its eigenvalues, treated as a multiset, the M-spectrum of G. We denote this spectrum by M-Spec(G). For our purposes we will eliminate the symbol AG in the above definitions and look solely at the characteristic polynomial PG (x) and its associated spectrum Spec(G). Since we are only considering undirected graphs, the adjacency matrices are all real and symmetric, and therefore all eigenvalues are real. We will often denote Spec(G) = {λ1 , . . . , λn } with λi ≥ λi+1 for all i ∈ [n−1]. Moreover, when explicitly listing the spectrum, superscripts will be used to denote multiplicity, while multiplicity of a single eigenvalue λ will often be denoted mult(λ). We now arrive at two of the most important definitions. Definition 1.2.23. Two graphs G and H are called M-cospectral if M-Spec(G) = M-Spec(H), treated as mutlisets. Equivalently, G and H are M-cospectral if PMG (x) = PMH (x). We denote G and H being M-cospectral by G ∼M H. If a graph G is not M-cospectral to any non-isomorphic graph H, then we say G is determined by its M-spectrum or that it is M-DS for short. As noted above we will drop the A in these definitions and say two graphs are cospectral or that a graph is DS. Since cospectrality is an equivalence relation, this allows us to partition a set of graphs. Ideally we would hope that all graphs are DS, but unfortunately this is not the case. We can see this by looking at K1,4 and C4 + K1 . Both of these graphs have the spectrum {−2, 03 , 2}.
Figure 1.1: C4 + K1 and K1,4
1.3
Summary of Chapters
In Chapter 2 we will explore the key results for this dissertation, starting with the classical theorems and working our way to the new results. In Chapter 3 we give an overview of the classes of graphs that will be proved to be DS or shown to have cospectral mates. Chapters 4 5
and 5 contain the bulk of the calculations that were summarized in Chapter 3. Chapter 4 focuses on graphs with λ = −1 in their spectra with high multiplicity, while Chapter 5 focuses on graphs with λ = 0 with high multiplicity. Finally, Chapter 6 closes out this dissertation with final conclusions, conjectures, and routes for future work.
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Chapter 2 Key Theorems Results without proofs or specific citations can be found in [1, 2, 4, 5, 6].
2.1
Classic Results
The adjacency matrix of a graph has been studied extensively. Many properties of a graph can been obtained from the spectrum while other properties are lost. For example, the number of vertices and the number of edges can be determined from the spectrum while connectedness cannot. The latter can be seen from the K1,4 and C4 + K1 example in Chapter 1. The number of vertices is clearly equal to the number of eigenvalues, while the number of edges can be determined using the following proposition. Proposition 2.1.1. The number of closed k-walks in a graph G is given by λk1 + · · · + λkn . A walk is similar to a path except that the vertices need not be distinct, and a closed walk is one that starts and ends at the same vertex. Therefore, a closed 2-walk is nothing more than an edge, and so the number of edges is given by 12 (λ21 + · · · + λ2n ). Another consequence of this theorem is that λ1 + · · · + λn = 0. This is true because a closed 1-walk would be a loop, and all of our graphs are assumed to have no loops. For a general graph it is difficult to say much about its spectrum, but there are several classes of graphs for which the spectrum is explicitly known. These include the complete 7
graphs, cycle graphs, and path graphs, among many others. A few are summarized in the next proposition. Proposition 2.1.2. • Spec(Kn ) = {−1n−1 , n − 1} • Spec(Kn ) = {0n } √ √ • Spec(Km,n ) = {− mn, 0m+n−2 , mn} • Spec(Cn ) = 2 cos
2kπ n
• Spec(Pn ) = 2 cos
kπ n+1
, k ∈ [n] , k ∈ [n]
One way of calculating the eigenvalues of a graph is by first calculating eigenvectors. The adjacency matrix gives us a way find eigenvectors using the structure of the graph. Proposition 2.1.3. λ ∈ Spec(G) with v as an associated eigenvector iff vi = λ
X
vj .
vj ∈N (vi )
In particular, what Proposition 2.1.3 says is that the entry of an eigenvector associated with vertex vi is dependent upon the entries of the neighbors of vi . Knowledge of the structure of G is often useful in calculating the eigenvalues and eigenvectors of G. Perhaps the most important theorem needed is the Interlacing Theorem. It gives a way of relating eigenvalues of an induced subgraph G[S] with the eigenvalues of G itself. Theorem 2.1.4 (Interlacing Theorem). If λ1 ≥ . . . ≥ λn and µ1 ≥ . . . ≥ µn−1 are the eigenvalues of G and G − v, respectively, then λi ≥ µi ≥ λi+1 for i ∈ [n − 1]. More generally, if S ⊂ V (G) is a subset of m vertices of G and ρi are the eigenvalues of G − S, then λi ≥ ρi ≥ λn−m+i for i ∈ [m]. Corollary 2.1.5. If H is an induced subgraph of G with |V (H)| = |V (G)| − k, and λ ∈ Spec(H) with multiplicity m, then λ ∈ Spec(G) with multiplicity between m − k and m + k. Similarly, if λ ∈ Spec(G) with multiplicity m0 , then λ ∈ Spec(H) with multiplicity between m0 − k and m0 + k. 8
This gives us a way to study a graph by looking at its induced subgraphs. Moreover, since we are primarily interested in graphs with eigenvalues of multiplicity greater than 1, we know that these subgraphs must have these same high multiplicity eigenvalues. Take √ for example the bow tie graph (see Figure 2.1(a)) with eigenvalues { 12 (1 ± 17), −12 , 1}. Since −1 has multiplicity 2, Corollary 2.1.5 guarantees that each subgraph on 4 vertices must have −1 in its spectrum with multiplicity of 1, 2, or 3.
We can quickly check
that each such subgraph has spectrum either {−12 , 12 } (first graph in Figure 2.1(b)) or {−1.4811 . . . , −1, 0.3111 . . . , 2.1700 . . .} (second graph in Figure 2.1(b)).
(a) Bow tie
(b) Vertex-deleted subgraphs
Figure 2.1: Bow tie and subgraphs Knowing that a graph has an eigenvalue with multiplicity greater than 1 is useful for interlacing, and it does indeed guarantee certain eigenvalues in subgraphs, but it would be nice to be able to work up from a given graph and guarantee eigenvalues with increasing multiplicity. That is, we would like to work our way up from the bottom instead of from the top down. We will see in the next section how we can approach this problem. Before investigating further there are a few more properties we can deduce from the Interlacing Theorem. In particular, we can compute elementary bounds on α(G) and ω(G). Proposition 2.1.6. Let n = |V (G)|, n− = |{i : λi < 0}| and n+ = |{i : λi > 0}|. Then α(G) ≤ min{n − n− , n − n+ }. Proof. Let S ⊆ V (G) be a maximal independent set. Then G[S] is a graph with no edges. In particular, each eigenvalue of G[S] is 0. If |S| = s, then Theorem 2.1.4 gives λn−s+i ≤ 0 ≤ λi . From this we can infer that 0 ≤ λs , and that n− ≤ n − s. Similarly, by looking at −AG , we get n+ ≤ n − s. Therefore s = α(G) ≤ min{n − n− , n − n+ }.
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In a similar fashion, we get the following bounds on ω(G): Proposition 2.1.7. Let n = |V (G)|, N− = |{i : λi ≤ −1}|, and N+ = |{i : λi ≥ −1}|. Then ω(G) ≤ min{N− + 1, N+ + 1, λ1 + 1}. Since the graphs we will be studying will have spectra containing either 0 or −1 with high multiplicity, Propositions 2.1.6 and 2.1.7 tell us that the graphs will contain either large independent sets, in the case of 0, or large cliques, in the case of −1. Another property of a graph we can gather from the spectrum is whether or not the graph is bipartite. This will be useful when examining the λ = 0 case in Chapter 5. Proposition 2.1.8. A graph G is bipartite iff Spec(G) is symmetric about 0. Proof. Let V = S ∪· T be a bipartition of the vertices of G, and let λ ∈ Spec(G). If v is an eigenvector for λ, then v0 defined as having entries vi for each vertex vi ∈ S and −vj for each vertex vj ∈ T is an eigenvector for −λ. For the converse, we need to make use of the Perron-Frobenius theorem for real square matrices, but this is outside of the scope of this thesis. One last key theorem we will utilize relates the characteristic polynomial of a graph to the characteristic polynomial of its vertex-deleted subgraphs. Proposition 2.1.9. If V (G) = {v1 , . . . , vn }, then n
X d PG (x) = PG−vi (x). dx i=1 Proof. The derivative of the characteristic polynomial of an n × n matrix is the sum of the characteristic polynomials of its n principal submatrices obtained by deleting the i-th row and i-th column. The matrix obtained by deleting the i-th row and column of AG is precisely the adjacency matrix of G − vi . Thus, we get n
X d PG (x) = PG−vi (x). dx i=1
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(a) P4
(b) P4 with blowup
(c) P4 with split
Figure 2.2: Blowup and split example
2.2
New Results
Before we move on to the new results we need some definitions. Definition 2.2.1. If v1 , v2 ∈ V (G) are such that N (v1 )4N (v2 ) = {v1 , v2 }, then we call {v1 , v2 } a blowup or blowup partition. More generally, if S ⊆ V (G) with |S| > 1 is such that {vi , vj } is a blowup for all vi , vj ∈ S, then S is a blowup partition. We call G a blowup graph if it contains a blowup partition. Definition 2.2.2. If v1 , v2 ∈ V (G) are such that N (v1 )4N (v2 ) = ∅, then we call {v1 , v2 } a split or split partition. More generally, if S ⊆ V (G) with |S| > 1 is such that {vi , vj } is a split for all vi , vj ∈ S, then S is a split partition. We call G a split graph 1 if it contains a split partition. The idea of a blowup partition may be thought of as replacing a vertex in a graph with a clique while keeping all adjacencies from the replaced vertex with all vertices in the clique. Similarly, a split partition may be thought of as replacing a vertex with an independent set, again keeping all previous adjacencies. Examples are given in Figure 2.2. Throughout we will use a specific notation for blowup graphs and split graphs. Notation 2.2.3. Let G be a graph with V (G) = {v1 , . . . , vn }, and let ai ∈ N for i ∈ [n]. G(a1 , . . . , an ) is used to denote the blowup graph obtained from G by replacing vertex vi with a clique of size ai while keeping all adjacencies between cliques that replace adjacent vertices. Similarly, G(a1 , . . . , an ; z) is used to denote the split graph with vertices replace by independent sets of size ai along with z isolated vertices. If we know that a graph has a blowup or split partition, then we immediately gain information about the graph’s spectrum. 1
The notion of a split graph sometimes refers to a graph whose vertex set can be partitioned into a clique and an independent set. This distinction should not cause confusion here.
11
Proposition 2.2.4. Suppose G has a blowup partition S of size k ≥ 2. Then −1 ∈ Spec(G) with multiplicity at least k − 1. Similarly, if G has a split partition S of size k ≥ 2, then 0 ∈ Spec(G) with multiplicity at least k − 1. Proof. Suppose S = {v1 , . . . , vk }. Define vectors vi , i ∈ [k − 1] as having entries vik = 1, vii = −1, and vij = 0 otherwise. We need to consider three cases to show that vi is an eigenvector for eigenvalues 0 or −1. First, suppose v ∈ / N (vi ). Then clearly vi ∈ / N (v). Also, since N (vi )4N (vk ) = {vi , vk } or ∅, then vk ∈ / N (v). Thus, using Proposition 2.1.3 we get that the sum over the neighbors of v is 0, as we needed. Next, suppose v ∈ N (vi ) and v 6= vk . This time, the sum over the neighbors of v will include both vi and vk , and this sum is clearly 0. Lastly, suppose v = vi or v = vk . If vi vk is not an edge, then in both cases the sum over the neighbors of v is again 0. This gives λ = 0 as an eigenvalue. On the other hand, if vi vk is an edge, then the only contributing vertices in the sum will be precisely vi or vk . In either case we simply get a change in sign since vii = −vik . This gives λ = −1 as an eigenvalue. Finally, since the set of vi are clearly linearly independent, we get that the multiplicity of λ is at least k − 1. This says that blowup graphs always have −1 in their spectrum, but the converse is not true. Take for example C6 . We have Spec(C6 ) = {−2, −12 , 12 , 2}, but C6 clearly does not have a blowup. Similarly, not all graphs with 0 in their spectrum have a split partition. It would be nice if we could characterize which graphs have this property. Definition 2.2.5. A graph G is called 0-primitive if 0 ∈ / Spec(G), or if 0 ∈ Spec(G) and G does not contain a split partition or a vertex of degree zero. G is called -1-primitive if −1 ∈ / Spec(G) or if −1 ∈ Spec(G) and G does not contain a blowup partition. When 0∈ / Spec(G) or −1 ∈ / Spec(G), then G is called trivially primitive. Primitivity may be with respect to either splits or blowups, and when needed the distinction will be explicitly given. In most cases it will be clear from context. With this new language we are attempting to classify all primitive graphs. A full classification is slightly out of reach, 12
but we are able to give an algorithmic approach to determining all primitive graphs with a fixed number of eigenvalues other than 0 or −1. Before we examine this, we will first show that the algorithm is guaranteed to terminate. Definition 2.2.6. Define Pkn to be the set of -1-primitive graphs with n vertices and k S eigenvalues other than -1, and define Pk = Pkn . Similarly, define Qnk and Qk for 0-primitive graphs. The set P4 is shown in Figure 2.3, and Q4 is shown in Figure 2.4 and Figure 2.5. Q4 is partitioned into bipartite and non-bipartite graphs. This will make the examination in Chapter 5 easier. One key thing to note is that P4 and Q4 are both finite sets. This is not a peculiarity. We will show in Theorem 2.2.13 that Pk and Qk are finite sets for all k ≥ 0. Before we prove Theorem 2.2.13 we first need a theorem on the rank of the adjacency matrix of G and then a way to relate the characteristic polynomial of G with G. Theorem 2.2.7. Let G be a 0-primitive graph on n vertices, and let r be the rank of AG . Then r n = O 22 .
Corollary 2.2.8. Fix k ≥ 0. Then there is an upper bound on the number of vertices of a graph G ∈ Qk . Proof. Let G ∈ Qk with n = |V (G)|. Since 0 ∈ Spec(G) with mult(0) = n − k, we know that the rank of AG is n − mult(0) = n − (n − k) = k. Theorem 2.2.7 then tells us that k n = O 22 .
A proof of Theorem 2.2.7 can be found in [7]. Corollary 2.2.8 tells us that Qk is finite. To show that Pk is also finite we need to relate the characteristic polynomial of G to that of G. Lemma 2.2.9. Suppose A is an invertible matrix and u and v are column vectors. Then det (A + uvT ) = 1 + vT A−1 u det(A). 13
Kd
Kb
Ka
Kc
Kb
Ka
(a) P3 (a, b, c) + K1 (d)
(b) P4 (a, b, c, d)
Kc
Ka
Ka
Kb
Kd
Kc
Kd
Kb (c) K1,3 (a, b, c, d)
Ka
Kc
(d) C4 (a, b, c, d)
Kb
Kc
Kd
(e) P5 (a, b, c, d, e)
Ka
Kb
Kf
Kc
Ke
Kd
(f) C6 (a, b, c, d, e, f )
Figure 2.3: P4
14
Ke
Kd
Ka
Kz
Kc Kz
Kb
Kb
Ka
Kd (a) 2P2 (a, b, c, d; z)
Kc
Kd
(b) P4 (a, b, c, d; z)
Kz
Ka
Kb
Kc
Kd
Ke
(c) P5 (a, b, c, d, e)
Figure 2.4: Bipartite Q4 Let {µ1 , . . . , µm } be the distinct eigenvalues of G. For i ∈ [m], let Pi be the n×n matrix that represents the orthogonal projection of Rn , with respect to the standard basis {e1 , . . . , en }, onto E(µi ), where E(µi ) is the eigenspace of µi . Finally, let βi =
kPi jk √ , n
where j is the all-1
vector. One property we will exploit below is the spectral decomposition of A. We can write A as A = µ 1 P1 + µ 2 P2 + · · · + µ m Pm . Note also that
P
Pi = I, Pi2 = Pi = PiT for all i, and Pi Pj = O for all i 6= j. With this
we can now calculate the characteristic polynomial of G using a mutlilinear determinental expansion. Proposition 2.2.10. For a graph G, the characteristic polynomial of G is given by ! m 2 X β i PG (x) = (−1)n PG (−x − 1) 1 − n . x + 1 + µi i=1 Proof. Let J be the all-1 matrix. Then AG = J − I − AG . Note that J = jjT . From this we
15
Kd
Kc
Ke Ka
Kd
Kb
Kz
Kz
Ka
Kc
Kb (a) Kite(a, b, c, d; z)
(b) Bull(a, b, c, d; z)
Kb Ka Ka
Kc
Ke
Kb
Kf
Kz
Ke Kz
Kc
Kd (c) House(a, b, c, d, e; z)
Kd
(d) Broom(a, b, c, d, e, f ; z)
Ka
Kb
Kf
Kc
Ke
Kd
(e) P2 × K3 (a, b, c, d, e, f ; z)
Figure 2.5: Non-bipartite Q4
16
Kz
get PG (x) = det ((x + 1)I + AG − J) = det ((x + 1)I + AG ) 1 − jT ((x + 1)I + AG )−1 j ! m 2 X β i . = (−1)n PG (−x − 1) 1 − n x + 1 + µ i i=1
Corollary 2.2.11. If λ ∈ Spec(G) with multiplicity m, then −λ − 1 ∈ Spec(G) with multiplicity m − 1, m, or m + 1. We need one last proposition before we can prove Theorem 2.2.13. Proposition 2.2.12. A graph G has a blowup partition iff G has a split partition. Proof. Suppose {v, w} is a blowup partition in G. Then v and w share all neighbors, save for v and w themselves. Therefore, in G, v and w still share all neighbors, and thus {v, w} is a split partition in G. The converse is proved similarly. Theorem 2.2.13. The sets Pk and Qk are finite for all k ≥ 0. Proof. Qk was proved finite in Corollary 2.2.8. To show Pk is finite, we note that for each G ∈ Pk , we have G in either Qk−1 , Qk , or Qk+1 , all of which we know to be finite. Now that we know that Pk and Qk are always finite, it would be useful to know which graphs belong in each set. Unfortunately there is no simple description of each set, but it is possible to build each set algorithmically. Using a proof similar to that of Proposition 2.2.10 we will relate the characteristic polynomial of G to that of G with a vertex added. Let S ⊆ V (G) and define GS = G +S x where x is a new vertex with edges added between x and all vertices in S. Let cS be the characteristic vector of S. Using the same notation as in Proposition 2.2.10, we get the following. Lemma 2.2.14. Let D be a 1 × 1 matrix, and let A, B, and C matrices of appropriate sizes. Then det
A B
! = (D − 1) det(A) + det(A − BC).
C D 17
Proposition 2.2.15. The characteristic polynomial of PGS (x) is given by ! r X kPi cS k2 . PGS (x) = PG (x) x − x − µi i=1 Proof. The characteristic polynomial is given by xI − A −c G S det (xI − AGS ) = −cT x S
= (x − 1) det (xI − AG ) + det (xI − AG − cS cTS ) = det (xI − AG ) x − cTS (xI − AG )−1 cS ! r X kPi cS k2 = PG (x) x − . x − µi i=1
We are now set to demonstrate with the next few propositions how we can build the sets Pk and Qk . Proposition 2.2.16. Let µ ∈ Spec(G) with multiplicity m ≥ 1, and S ⊆ V (G). Then µ ∈ Spec(GS ) with multiplicity at least m iff cS ∈ E(µ)⊥ . Q Proof. Without loss of generality assume µ = µ1 . Write mG (x) = ri=1 (x − µi ) and PG (x) = Qr mi i=1 (x − µi ) . Then Proposition 2.2.15 gives ! r X PG (x) kPi cS k2 mG (x) PGS (x) = xmG (x) − mG (x) x − µi i=1 ! r r r Y X Y mi −1 2 . = (x − µi ) xm (x) − kP c k (x − µ ) G j i S i=1
Define q(x) = xmG (x) −
i=1
r X
2
kPi cS k
i=1
r Y
j=1 j6=i
(x − µj ).
j=1 j6=i
Now, if cS ∈ E(µ1 )⊥ , then kP1 cS k2 = 0. From this we get that x − µ1 is a common factor throughout q(x), and thus q(µ1 ) = 0 which give that µ1 is a root of order at least m1 in PGS (x). 18
On the other hand, suppose cS 6∈ E(µ1 )⊥ . Then kP1 cS k2 6= 0. From this it is now clear that Q q(µ1 ) = kP1 cS k2 rj=2 (µ1 − µj ) 6= 0. Therefore µ1 is a root of order m1 − 1 in PGS (x). Proposition 2.2.17. Let G be a blowup graph and let S ⊆ V (G). Then ei − ej ∈ E(−1) for some i 6= j. If cS ∈ E(−1)⊥ , then GS is a blowup. Proof. The first assertion is trivial. For the second assertion, we have that cS · v = 0 for all v ∈ E(−1). In particular cS · (ei − ej ) = 0. This says that (cS )i = (cS )j . That is, the new vertex x in GS is either adjacent to both vi and vj , or it is adjacent to neither. In both cases we arrive at N (vi )4N (vj ) = {vi , vj }, and thus GS is a blowup. Proposition 2.2.18. If G is primitive with −1 ∈ Spec(G) with multiplicity m and G − v is a blowup for some v ∈ V (G), then −1 ∈ Spec(G − v) with multiplicity m + 1. Proof. Let G0 = G − v and S = N (v) ⊆ V (G). By Proposition 2.2.17, since G0 +S v is not a blowup then cS ∈ / EG0 (−1)⊥ . By Proposition 2.2.16, since cS ∈ / EG0 (−1)⊥ , then the multiplicity of −1 goes down by one from G0 to G. Since the multiplicity of −1 in G is m the multiplicity must be m + 1 in G0 . Proposition 2.2.19. Let G be primitive with −1 ∈ Spec(G) with multiplicity m. Then there exists a v ∈ V (G) such that G − v has −1 ∈ Spec(G − v) with multiplicity m − 1; moreover, G − v is primitive. Proof. From Proposition 2.1.9 we know n
X d PG (x) = PG−vi (x). dx i=1 Since −1 is a root of order m in PG (x) it must be a root of order m − 1 in
d P (x). dx G
By
Theorem 2.1.4 we know that −1 has multiplicity between m−1 and m+1 in each Spec(G−vi ), and therefore we must have that at least one of PG−vi (x) must have −1 as a root of order m − 1. From Proposition 2.2.18 this G − vi must also be primitive. Propositions 2.2.17, 2.2.18, and 2.2.19 apply equally well to λ = 0 with blowup replaced with split. Most importantly they tell us that the primitive graphs with n vertices, say, can 19
be built up from the primitive graphs on n − 1 vertices. We will now give the algorithm in detail. Algorithm 2.2.20. Fix λ ∈ {−1, 0} and k ≥ 0. Set n = k. Let Gi denote the set of all graphs on i vertices. Let Si denote Pki or Qik . Step 1: For each graph G ∈ Gn , compute Spec(G). If λ ∈ / Spec(G), then add G to Sn . Step 2: If Sn is empty, go to Step 3. Else, for each G ∈ Sn and for each S ∈ {0, 1}n ∩EG (λ)⊥ , compute Spec(GS ). If λ ∈ Spec(GS ) and GS is primitive, then add GS to Sn+1 . Increase n by 1. Repeat Step 2. Step 3: If λ = −1, then set Pk =
S
Si . Else, set Qk =
S
Si .
Step 1 computes the trivially primitive graphs. Step 2 iteratively takes the primitive graphs on n vertices, adds a new vertex in such a way to increase the multiplicity of λ, and checks whether or not it is primitive. Finally, Step 3 simply combines the sets produced in Step 2 to create Pk or Qk . The validity of Step 2 is given by Propositions 2.2.16 through 2.2.19, and Theorem 2.2.13 shows that the algorithm will terminate. Using this algorithm we are able to calculate the sets for Pk for 0 ≤ k ≤ 7 and Qk for 0 ≤ k ≤ 6. P0 = ∅ P1 = {K1 } P2 = {2K1 } P3 = {3K1 , P3 } P4 = {4K1 , P3 + K1 , P4 , C4 , P5 , C6 } |P5 | = 30 |P6 | = 422 |P7 | = 8226
20
Q0 = {empty graph} Q1 = ∅ Q2 = {P2 } Q3 = {K3 } Q4 = {K4 , 2P2 , P4 , P5 , Kite, Bull, House, Broom, P2 × K3 } |Q5 | = 25 |Q6 | = 1654 The last part we need in order to study these sets in-depth is a way to calculate the spectra, or equivalently the characteristic polynomials, of the blowup and split graphs. This is easily done using a quotient matrix formed from the adjacency matrix and a partition of the vertices of the graph. Definition 2.2.21. A partition π = {V1 , . . . , Vk } of V is called an equitable partition if for every 1 ≤ i, j ≤ k there exists an integer bij ≥ 0 such that every vertex in Vi is adjacent to exactly bij vertices in Vj . Define the quotient matrix AG /π as having (i, j) entry equal to bij . It is straightforward to see that blowup and split partitions naturally lead to equitable partitions. Suppose S ⊆ V is a split partition. Since all vertices in S have the same neighbors, they will be adjacent to the same number of vertices in any subset of V , including S itself. Within S each vertex is adjacent to 0 other vertices. The same argument holds if S is a blowup partition, except in this each vertex is adjacent to |S| − 1 vertices in S. Take for example the bow tie graph in Figure 2.1. It is a blowup of P3 , where both end-vertices have been blown up with partitions of size 2. Then for the quotient matrix we get 1 1 0 2 0 2 . 0 1 1 The usefulness of having an equitable partition is that we can show that the characteristic polynomial of the quotient matrix induced by the partition divides the characteristic polynomial of G. Proposition 2.2.22. Let π be an equitable partition of V . Then the characteristic polynomial of AG /π divides the characteristic polynomial of AG . 21
A proof of this can be found in [6]. Going back to the bow tie graph example, we can compute both its characteristic polynomial and that of its quotient matrix. Doing so gives PG (x) = −(x − 1)(x + 1)2 x2 − x − 4 PAG /π (x) = −(x − 1) x2 − x − 4 As expected these differ only in the factors containing the eigenvalues of −1 coming from the two blowup partitions. Proposition 2.2.22 now offers us an efficient way to study all blowup graphs coming from Pk and all split graphs coming from Qk . Proposition 2.2.23. Let H = G(a1 , . . . , an ) be a blowup graph, and let A be the quotient P matrix obtained from the n blowup partitions of sizes ai . Let a = ni=1 ai . Then PH (x) = (x + 1)a−n PA (x). Similarly, if K = G(a1 , . . . , an ; z) is a split graph, and A and a are defined similarly, then PK (x) = xa+z−n PA (x). The characteristic polynomials for elements of P3 and P4 are given and analyzed in Chapter 4, while those for Q3 and Q4 are in Chapter 5. Before ending this chapter we need to answer one last question. Up to this point we have only considered graphs with λ = −1 or λ = 0 of high multiplicity. We have shown that graphs can have either of these of arbitrarily high multiplicity. For any λ ∈ / {−1, 0} we can show that mult(λ) forces an upper bound on the size of the graph. Theorem 2.2.24. Let G be a graph with n vertices, and let λ ∈ Spec(G) with λ 6∈ {−1, 0}. Let t = dim E(λ)⊥ . Then we have either (a) n ≤ 12 t(t + 1) or (b) λ = 1 and G = K2 or G = 2K2 . A proof of this can be found in [5]. Since dim E(λ)⊥ = n − mult(λ), we see that the multiplicity of an eigenvalue other than −1 or 0 is bounded by the size of the graph. Some work has been in determining graphs with eigenvalues of high multiplicity other than −1 or 0, such as in [3] in which all graphs with dim E(λ)⊥ ≤ 5 are determined. 22
Chapter 3 Summary of Results For graphs arising from P3 and P4 , we get the following results: • nK1 (a1 , . . . , an ) is DS for all ai ≥ 1 • P3 (a, b, c) is DS for all a, b, c ≥ 1 • P3 (k, m, 1) + K1 (1) is DS for all k, m ≥ 1. • P3 (k, 1, m) + K1 (1) is DS for all k, m ≥ 1 except for when k = m. In that case we have P3 (k, 1, k) + K1 (1) ∼ P3 (1, k, 1) + K1 (k). • P3 (k, 1, 1) + K1 (m) is DS for all k, m ≥ 1. • P3 (1, k, 1) + K1 (m) is DS for all k, m ≥ 1 except for when k = m. In that case we have P3 (k, 1, k) + K1 (1) ∼ P3 (1, k, 1) + K1 (k). • P4 (k, m, 1, 1) is DS for all k, m ≥ 1. • P4 (k, 1, m, 1) is DS for all k, m ≥ 1. • P4 (k, 1, 1, m) is DS for all k, m ≥ 1. • P4 (1, k, m, 1) is DS for all k, m ≥ 1. • K1,3 (k, m, 1, 1) is DS for all k, m ≥ 1 except for C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1) and P3 (a, n + ac − a − c + 1, c) + K1 (1) ∼ K1,3 (k, m, 1, 1) where 2a(a − 1)c(c − 1) = n2 is a 23
perfect square, k = n2 , and m = ac + n2 . • K1,3 (1, k, m, 1) is DS for all k, m ≥ 1. • C4 (k, m, 1, 1) is DS for all k, m ≥ 1 except for C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1). • C4 (k, 1, m, 1) is DS for all k, m ≥ 1 except for C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1) and C4 (12, 1, 6, 1) ∼ P3 (9, 4, 6) + K1 (1). • P5 (k, m, 1, 1, 1) is DS for all k, m ≥ 1. • P5 (k, 1, m, 1, 1) is DS for all k, m ≥ 1 except for P5 (k + 1, 1, k, 1, 1) ∼ P5 (2, k, 1, 1, k) for k ≥ 1. • P5 (k, 1, 1, m, 1) is DS for all k, m ≥ 1. • P5 (1, k, m, 1, 1) is DS for all k, m ≥ 1. • P5 (1, k, 1, m, 1) is DS for all k, m ≥ 1. • C6 (k, m, 1, 1, 1, 1) is DS for all k, m ≥ 1. • C6 (k, 1, m, 1, 1, 1) is DS for all k, m ≥ 1. • C6 (k, 1, 1, m, 1, 1) is DS for all k, m ≥ 1. For graphs arising from Q3 and Q4 , we get the following results: • 2P2 (k, 1, 1, 1; z) is DS iff k = 2 and z ≥ 0. • P4 (k, 1, 1, 1; z) is DS iff k = 2 or k ≥ 1 is odd and z ≥ 0. • P4 (1, k, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0. • P5 (k, 1, 1, 1, 1; z) is DS iff k ≥ 2 is even except for k ∈ {4, 10, 12, 16, 58, 94} and z ≥ 0 or k = 1 and z = 0. • P5 (1, k, 1, 1, 1; z) is DS iff k ∈ {1, 2} and z = 0. • P5 (1, 1, k, 1, 1; z) is DS iff k ≥ 2 is even and z ≥ 0. • Kite(k, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0.
24
• Kite(1, k, 1, 1; z) is DS for all odd k ≥ 1 and z ≥ 0. • Kite(k, 1, 1, m; z) is DS for all k, m ≥ 1 and z ≥ 0. • Bull(2, 1, 1, 1, 1; 0) and Bull(k, 1, 1, 1, 1; z) are DS for all odd k ≥ 1 and z ≥ 0. • Bull(1, 1, k, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0. • Bull(1, 1, 1, k, 1; z) is DS for all k ≥ 1 and z ≥ 0. • Bull(1, 1, 1, k, m; 0) ∼ Broom(1, 1, 1, d, e, f ; z) iff d = d1 d2 , e = e1 e2 , and f = de + d + e ± (d1 e1 + d2 e2 ), in which case k = de + d + e ± d1 e1 , m = de + d + e ± d2 e2 , and z = de. • House(k, 1, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0. • House(1, k, 1, 1, 1; z) is DS for all odd k ≥ 1 and z ≥ 0 or when k = 2 and z = 0. • House(1, 1, 1, k, m; z) is DS for all k, m ≥ 1 and z ≥ 0. • Broom(k, 1, m, 1, 1, 1; z) is DS for all k, m ≥ 1 and z ≥ 0 with four exceptions: Broom(3, 1, 1, 1, 1, 1; z + 3), Broom(7, 1, 3, 1, 1, 1; z + 1), Broom(1, 1, 1, 1, 1, 1; z + 1), and Broom(3, 1, 3, 1, 1, 1; z + 5). • Broom(1, k, 1, 1, 1, 1; z) is DS for all k ≥ 1 not a multiple of 4 and z ≥ 0 except when k = 1 and z ≥ 1 or k = 5 and z ≥ 5. • Broom(1, 1, 1, k, m, 1; z) is DS for all k, m ≥ 1 and z ≥ 0 except for k = m = 1 and z ≥ 1. • Broom(1, 1, 1, 1, 1, k; z) is DS for all k ≥ 1 and z ≥ 0 except for k = 1 or k = 5 and z ≥ 1. • P2 × K3 (k, 1, 1, 1, 1, 1; z) is DS for all k ≥ and z ≥ 0 except for k = 1 and z ≥ 2 or k = 3 and z ≥ 4.
25
Chapter 4 λ = −1 In this chapter we will primarily analyze graphs arising from P4 . First we will show that all graphs coming from P1 , P2 , and P3 are DS. Characteristic polynomials for each can be found at the end of this chapter. Each graph in P4 is given its own section. Within each of these sections we analyze the blowup graphs where all but two partitions are fixed at size 1. For each of these, we first list the relevant coefficients and give relationships found between them. These will be used with the other graphs from P4 to find all cospectral pairs, if any exist. P1 , P2 , and P3 are each analyzed in one section.
4.1
P1, P2, and P3
P1 and P2 are easy to analyze since they each contain one graph. The following proposition will show that each graph arising as a blowup from P1 or P2 is DS. Proposition 4.1.1. If Spec(G) = {−1n1 +...+nk −k , n1 − 1, . . . , nk − 1}, then G is isomorphic to the disjoint union of Kn1 , . . . , Knk . Proof. The spectrum of A+I is {0n1 +...+nk −k , n1 , . . . , nk }. So A+I is positive semi-definite of rank k and can be written as V T V , where V has columns consisting of vectors v1 , . . . , vn ∈ Rk . 26
Since the diagonal entries of A+I are all 1 we get that each vi is a unit vector. Now, since all entries in A + I are either 0 or 1, this means all of the vectors vi are either orthogonal or are equal. This means A + I can be reordered so that it is a block diagonal matrix with all-ones matrices on the diagonal, the sizes of which are n1 , . . . , nk . Therefore A is the adjacency matrix of Kn1 + . . . + Knk . Corollary 4.1.2. All graphs arising as a blowup from P1 or P2 are DS. Not only does this show that all graphs arising from P1 and P2 are DS, but it show that nK1 (a1 , . . . , an ) is DS for all n. Therefore we may exclude nK1 from consideration when determining which graphs are DS or cospectral. Using this logic with P3 , we see that we only need to check when P3 (a, b, c) is cospectral with P3 (d, e, f ). Proposition 4.1.3. P3 (a, b, c) is DS for all a, b, c ≥ 1. Proof. From Section 4.8, we see that the coefficients for P3 (a, b, c) are c1 = a + b + c, c2 = ac, and c3 = abc. Suppose P3 (a, b, c) ∼ P3 (d, e, f ) for some d, e, f ≥ 1. Then ac = df and abc = def imply b = e. We are then left with a + c = d + f and ac = df , which implies {a, c} = {d, f }, and therefore P3 (a, b, c) ∼ = P3 (d, e, f ).
4.2
P3(a, b, c) + K1(d)
P3 (k, m, 1) + K1 (1): d1 = k + m + 2 d2 = 2k + m + 1 d3 = km − k d4 = km
c3 < c 4 c1 + c4 = c2 + c3 + 1
27
(4.1) (4.2)
With c1 = k + m + 2 and c3 = km − k we get q 1 2 c1 − 3 ± c1 − 6c1 − 4c3 + 9 , m = c1 − 2 − k. k= 2 With c2 = 2k + m + 1 and c3 = km − k we get q 1 2 k= c2 − 2 ± c2 − 4c2 − 8c3 + 4 , m = c2 − 1 − 2k. 4 Using c4 = km, these give, respectively, q 1 2 c4 = c1 + 2c3 − 3 ± c1 − 6c1 − 4c3 + 9 2 q 1 2 c4 = c2 + 4c3 − 2 ± c1 − 4c1 − 8c3 + 4 4 After rearranging, squaring, and simplifying, we get c1 c3 + c23 + c24 + 3c4 = c1 c4 + 2c3 c4 + 2c3 c2 c3 + 2c23 + 2c24 + 2c4 = c2 c4 + 4c3 c4 + c3 Using both of these, we can finally simplify to the equation 2c1 c3 + c2 c4 + 4c4 = 2c1 c4 + c2 c3 + 3c3
(4.3)
P3 (a, b, c) + K1 (d) : Using equation (4.3) to solve for a we get a=
−b2 d2 + 3b2 d − 2b2 − bcd2 + 3bcd − 2bc + bd2 − 4bd + 3b − cd2 + 2cd + 2d2 − 3d bcd − bc + bd2 − 3bd + 2b + cd + d2 − 2d
As messy as this is, we can maximize a, with the assumption that c > 1, and show that a ≤ 1 with equality iff d = 1. Therefore P (a, b, c) + K1 (d) ∼ P3 (k, m, 1) + K1 (1) iff a = d = 1 (or similarly c = d = 1). Supposing now that c = d = 1, we have the equations km − k = ab − a and km = ab, which are enough to conclude a = k and b = m. P4 (a, b, c, d) : First suppose d = 1. From equation (4.2) we then get c = ac + bc 28
This implies a + b = 1 which cannot happen with a, b ≥ 1. So we must have a, d > 1. Next we can look at equation (4.3). a2 bc2 d + a2 bcd2 + 2a2 bc + ab2 cd2 + 2ab2 c + 2abc2 + 8abcd + 2b2 cd + 2bc2 d + 2bcd2 = a2 bc2 + 3a2 bcd + 3ab2 cd + 3abc2 d + 3abcd2 + 3abc + b2 cd2 + 3bcd Since bc 6= 0 is a common factor throughout we can reduce this to a2 cd + a2 d2 + 2a2 + abd2 + 2ab + 2ac + 8ad + 2bd + 2cd + 2d2 = a2 c + 3a2 d + 3abd + 3acd + 3ad2 + 3a + bd2 + 3d With the aid of Mathematica we calculate the mimimum of a2 cd + a2 d2 + 2a2 + abd2 + 2ab + 2ac + 8ad + 2bd + 2cd + 2d2 −(a2 c + 3a2 d + 3abd + 3acd + 3ad2 + 3a + bd2 + 3d) as 4. Therefore equation (4.3) is never satisified for a, d ≥ 2 and b, c ≥ 1.
K1,3 (a, b, c, d): Equation (4.2) gives 2abcd + bcd + a + b + c + d = acd + abd + acd + bc + bd + cd + 1 Now, (b − 1)(c − 1)(d − 1) = bcd − bc − bd − cd + b + c + d − 1 ≥ 0 for all b, c, d ≥ 1 implies bcd + b + c + d ≥ bc + bd + cd + 1 with equality when one of b, c, d equals 1. Suppose d = 1. Then a(b − 1)(c − 1) = 2abc − abc − ab − ac + a ≥ 0 for all a, b, c ≥ 1 implies 2abc + a ≥ abc + ab + ac for all a, b, c ≥ 1 with equality when one of b, c equals 1. Suppose c = 1. We now look at (4.3) with c = d = 1. This gives 4a2 b + 2a2 + 8ab2 + 18ab + 4a − 2b2 − 4b = 4a2 b + 8ab2 + 18ab + 4a − 2b2 − 4b which gives 2a2 = 0. This cannot happen with a ≥ 1. 29
C4 (a, b, c, d): With equation (4.2) we get 3abcd + a + b + c + d = abc + abd + acd + bcd + ac + bd + 1 Solving for a we get a=
bcd + bd − b − c − d + 1 3bcd − bc − bd − cd − c + 1
Now we can show that a ≤ 1 unless c = 1 or b = d = 1. To see this we will show 3bcd − bc − bd − cd − c + 1 ≥ bcd + bd − b − c − d + 1 with equality when c = 1 or b = d = 1. This inequality simplifies to 2bcd − 2bd − bc − cd + b + d = (c − 1)(2bd − b − d) ≥ 0 which gives the desired result. Now suppose b = d = 1. Equation (4.3) gives 3a2 c2 + 4a2 c + 2a2 + 4ac2 + 27ac + 4a + 2c2 + 4c = 2a2 c2 + 7a2 c + 7ac2 + 20ac + 4a + 4c which reduces to a2 c2 − 3a2 c + 2a2 − 3ac2 + 7ac + 2c2 = 0 Thinking of this as a quadratic in a we get √ 3c2 − 7c ± c c2 − 18c + 33 a= 2(c2 − 3c + 2) We immediately see that a complex when 3 ≤ c ≤ 15. For c = 1 we get a = −0.5 and with c = 2 we get a = −4. For c ≥ 16 we will show that both roots are in (1, 2). In the above, if we plug in a = 1, 1.55, 2 we get, respectively, 4c + 2 > 0 −0.2475c2 + 3.6425c + 4.805 < 0 2c + 8 > 0 The roots of −0.2475c2 +3.6425c+4.805 are c = −1.2183..., 15.9355..., and it is thus negative for c ≥ 16. A symmetric argument works to for when a = c = 1. Since a is never a positive integer here, we are done. 30
P5 (a, b, c, d, e): Equation (4.2) gives abcd+abce+acde+a+bcde+b+c+d+e = abc−ace+ac+ad+ae+bcd+bd+be+cde+ce+1 Solving for a gives a=
−bcde + bcd + bd + be − b + cde + ce − c − d − e + 1 bcd + bce − bc + cde + ce − c − d − e + 1
Similar to the C4 case we can show that a ≤ 1 with equality iff c = e = 1. Observe bcd + bce − bc + cde + ce − c − d − e + 1 −(−bcde + bcd + bd + be − b + cde + ce − c − d − e + 1) = bcde + bce − bc − bd − be + b = b(cde + ce − c − d − e + 1) And b(cde + ce − c − d − e + 1) ≥ 0 with equality iff c = e = 1. Now with a = c = e = 1, equation (4.3) gives 2b2 d2 +5b2 d+3b2 +5bd2 +26bd+11b+3d2 +11d−6 = b2 d2 +6b2 d+3b2 +6bd2 +23bd+11b+3d2 +11d−6 which simplifies to b2 d2 − b2 d − bd2 + 3bd = bd(bd − b − d + 3) = 0 This cannot happen with b, d ≥ 1.
C6 (a, b, c, d, e, f ): Similar to above, if we solve equation (4.2) for a and show that a < 1. The calculations follow the same pattern, but they are a bit too cumbersome to fit here.
Therefore we may conclude that P3 (k, m, 1) + K1 (1) is DS for all k, m ≥ 1.
31
P3 (k, 1, m) + K1 (1): d1 = k + m + 2 d2 = km + k + m + 1 d3 = 0 d4 = km c3 = 0
(4.4)
c1 + c4 = c2 + 1
(4.5)
Similar to the case of P3 (k, m, 1) + K1 (1), equations (4.4) and (4.5) give that c1 + c4 = c2 + c3 + 1. We can use the information from P3 (k, m, 1) + K1 (1) to aid us.
P3 (a, b, c) + K1 (d): From equation (4.4) we get abc − acd = ac(b − d) = 0 which gives b = d. Equation (4.5) now gives ab2 c + a + 2b + c = ab + ac + b2 + bc + 1 which reduces to ab2 c − ab − ac + a − b2 − bc + 2b + c − 1 = (b − 1)(abc + ac − a − b − c + 1) = 0 We are now left with either b = 1 or a = c = 1. The former gives k+m+2 = a+c+2 km = ac which says a = k, c = m or a = m, c = k. The latter gives k + m + 2 = 2b + 2 km = b2 which gives k = m = b. So we get that P3 (b, 1, b) + K1 (1) ∼ P3 (1, b, 1) + K1 (b) for all b ≥ 1. 32
The stronger result of P3 (a, b, a) + K1 (b) ∼ P3 (b, a, b) + K1 (a) for all a, b ≥ 1 can be shown through direct computations.
P4 (a, b, c, d): c3 = abc + bcd > 0.
K1,3 (a, b, c, d): From the P3 (k, m, 1) + K1 (1) case we know that c = d = 1. Now we have with equation (4.4) 2ab + a − b = 0 But this is not possible with a, b ≥ 1.
C4 (a, b, c, d): c3 = abc + abd + acd + bcd > 0.
P5 (a, b, c, d, e): From the P3 (k, m, 1) + K1 (1) case we have a = c = e = 1. Now we have with equation (4.4) bd + b + d − 1 = 0 But this is not possible with b, d ≥ 1.
C6 (a, b, c, d, e, f ): We know from the P3 (k, m, 1) + K1 (1) case above that c1 + c4 6= c2 + c3 + 1 for any a, b, c, d, e, f ≥ 1.
Therefore we may conclude that P3 (k, 1, m) + K1 (1) is DS for all k 6= m with P3 (k, 1, k) + K1 (1) ∼ P3 (1, k, 1) + K1 (k).
33
P3 (k, 1, 1) + K1 (m): d1 = k + m + 2 d2 = km + k + 2m d3 = k − km d4 = km
c3 ≤ 0 2c1 = c2 + c3 + 4
(4.6) (4.7)
It is straightforward to show that k = c3 + c4 m = c1 − c3 − c4 − 2 Since c4 = km we get c4 = (c3 + c4 )(c1 − c3 − c4 − 2)
(4.8)
P3 (a, b, c) + K1 (d): Using equation (4.8) we get ac(ab2 cd2 + 2ab2 cd + ab2 c − 2abcd2 − 2abcd − abd − ab + acd2 + ad −b2 d − b2 − bcd − bc − bd2 + 3bd + 2b + cd + d2 − 2d) = 0 Since ac ≥ 1 we can focus on the rest. Consider it as a polynomial in b. That is f (a, b, c, d) = b2 (acd2 + 2acd + ac − d − 1) +b(−2acd2 − 2acd − ad − a − cd − c − d2 + 3d + 2) +acd2 + ad + cd + d2 − 2d We can show that when f (a, b, c, d) has real roots, then they must be in (0, 2). This effectively shows that b = 1 is the only possibility. To see this, we first calculate f at b = 0, 1, 2, and 34
we also calculate the inflection points of f which will lie in ( 21 , 2). f (a, 0, c, d) = acd2 + ad + cd + d2 − 2d f (a, 1, c, d) = ac − a − c + 1 f (a, 2, c, d) = acd2 + 4acd + 4ac − ad − 2a − cd − 2c − d2
It is now easy to see that f (a, 0, c, d) > 0, f (a, 1, c, d) ≥ 0, and f (a, 2, c, d) > 0 for a, c, d ≥ 1 with f (a, 1, c, d) = 0 iff a = 1 or c = 1. Now, we also have ∂f (a, b, c, d) = 2abcd2 + 4abcd + 2abc − 2acd2 − 2acd − ad − a − 2bd − 2b − cd − c − d2 + 3d + 2 ∂b This implies that the inflection point is at b∗ = We need to show
1 2
2acd2 + 2acd + ad + a + cd + c + d2 − 3d − 2 2(d + 1)(acd + ac − 1)
< b∗ < 2. For this we look at
2acd2 + 2acd + ad + a + cd + c + d2 − 3d − 2 − 2(2(d + 1)(acd + ac − 1)) = −2acd2 − 6acd − 4ac + ad + a + cd + c + d2 + d + 2 < 0 2(2acd2 + 2acd + ad + a + cd + c + d2 − 3d − 2) − 2(d + 1)(acd + ac − 1) = 2acd2 − 2ac + 2ad + 2a + 2cd + 2c + 2d2 − 4d − 2 > 0 And thus
1 2
< b∗ < 2. Now, if f (a, b∗ , c, d) > 0, then f has no real roots. If f (a, b∗ , c, d) ≤ 0,
then the roots of f are in (0, 1). Since the only integer solution is at b = 1 with a = 1 or c = 1, we are left to consider if P3 (a, 1, 1)+K1 (d) ∼ P3 (k, 1, 1)+K1 (m). Using k +m+2 = a+d+2 and km + k + 2m + ad + a + 2d we conclude that a = k and d = m.
P4 (a, b, c, d): c3 = abc + bcd > 0.
35
K1,3 (a, b, c, d): Similar to P3 (a, b, c) + K1 (d) we need to look at equation (4.8). In this case we get f (a, b, c, d) = 4a2 b2 c2 d2 + 4a2 b2 c2 d + a2 b2 c2 + 4a2 b2 cd2 + 2a2 b2 cd + a2 b2 d2 + 4a2 bc2 d2 +2a2 bc2 d + 2a2 bcd2 − 2a2 bcd − a2 bc − a2 bd + a2 c2 d2 − a2 cd − 4ab2 c2 d2 −2ab2 c2 d − 2ab2 cd2 − 2ab2 cd − ab2 c − ab2 d − 2abc2 d2 − 2abc2 d − abc2 −2abcd2 + 4abcd + 2abc − abd2 + 2abd − ac2 d − acd2 + 2acd + b2 c2 d2 +b2 cd + bc2 d + bcd2 − 2bcd = 0 This time we will consider this as a polynomial in a and will show the real roots are in (0, 1), and therefore has no positive integer roots. Looking at f (1, b, c, d) we have f (1, b, c, d) = b2 c2 d2 + 2b2 c2 d + b2 c2 + 2b2 cd2 + b2 cd − b2 c + b2 d2 − b2 d + 2bc2 d2 +bc2 d − bc2 + bcd2 + bc − bd2 + bd + c2 d2 − c2 d − cd2 + cd From this we can see that f (1, b, c, d) > 0 for all b, c, d ≥ 1. We also need to calculate the inflection point a∗ . As a fraction it is too unwieldy to list here, but from it we can calculate in a similar fashion to P3 (a, b, c) + K1 (d) that 0 < a∗
0.
P5 (a, b, c, d, e): As above we use equation (4.8) to resolve this case. For P5 (a, b, c, d, e) we can factor out a c which leaves us with a linear equation in c left to look at. Solving said linear equation for c leads to a rational function with many terms, but it is straightforward to show from it that c < 1. In particular c can never be a positive integer.
36
C6 (a, b, c, d, e, f ): Using equation (4.8) and solving for a as well as the inflection point a∗ leads to a, a∗ < 0 when a is real. Therefore a is never a positive integer.
Therefore we may conclude that P3 (k, 1, 1) + K1 (m) is DS for all k, m ≥ 1.
37
P3 (1, k, 1) + K1 (m): d1 = k + m + 2 d2 = km + 2m + 1 d3 = k − m d4 = km
c3 < c 4 c1 + c4 = c2 + c3 + 1
(4.9) (4.10)
From c1 = k + m + 2, c3 = k − m, and c4 = km we get 4c4 = (c1 + c3 − 2)(c1 − c3 − 1)
(4.11)
P3 (a, b, c) + K1 (d): Equation (4.10) says abcd − abc + acd − ac − ad + a − bd + b − cd + c + d − 1 = (abc + ac − a − b − c + 1)(d − 1) = 0 This happens iff a = c = 1 or d = 1. If a = c = 1, then b + d + 2 = k + m + 2 and bd + 2d + 1 = km + 2m + 1 imply b = k and d = m. If d = 1, then a + b + c + 1 = k + m + 2, abc − ac = k − m, and abc = km give −a2 b2 c2 + 2a2 bc2 − a2 c2 + a2 − 4abc + 2ab + 2ac − 2a + b2 + 2bc − 2b + c2 − 2c + 1 = 0 Treating this as a polynomial in b we get at b = 0, 1, 2, respectively f (a, 0, c) = −a2 c2 + a2 + 2ac − 2a + c2 − 2c + 1 = −(a − 1)(c − 1)(ac + a + c − 1) f (a, 1, c) = a2 − 2ac + c2 = (a − c)2 f (a, 2, c) = −a2 c2 + a2 − 6ac + 2a + c2 + 2c + 1 From these it is immediately clear that f (a, 0, c) < 0 iff a, c > 1 and f (a, 1, c) ≥ 0 for a, c ≥ 1 with equality iff a = c. A little more work will show that f (a, 2, c) ≤ 0 for a, c ≥ 1 with 38
equality iff a = c = 1. Moreover, we actually have f (1, b, 1) = 0 for all b ≥ 1. We have already checked the case a = c = 1 anyway, so we may ignore it. Having either a > 1 or c > 1, we see from f (a, b, c) that b ∈ (0, 2) with b = 1 iff a = c > 1. From here it is easy to check that a = c = k = m. Thus P3 (1, k, 1) + K1 (k) ∼ P3 (k, 1, k) + K1 (1) for all k ≥ 1, which we have already seen above.
P4 (a, b, c, d): Equation (4.11) gives a2 b2 c2 − a2 + 2ab2 c2 d + 4abcd − abc − 2ab − 2ac − 2ad + 3a +b2 c2 d2 − b2 − bcd − 2bc − 2bd + 3b − c2 − 2cd + 3c − d2 + 3d − 2 = 0 It is easy to show that the left side of the above equation is always greater than 0 except when a = b = c = d = 1. However, we know from equation (4.10) that a, d > 1, so the left hand side is strictly greater than 0 for our purposes.
K1,3 (a, b, c, d): Equation (4.10) gives c = d = 1. Equation (4.11) then gives 2a(2ab2 + 2ab − 2b2 + b − 1) > 0 for a, b ≥ 1.
C4 (a, b, c, d): Equation (4.10) gives b = d = 1. Equation (4.11) then gives 4a2 c2 + 4a2 c + 4ac2 + 10ac − 2a − 2c > 0 for a, c ≥ 1.
P5 (a, b, c, d, e): Equation (4.10) gives a = c = e = 1. Equation (4.11) then gives b2 d2 + 2b2 d + 2bd2 + 5bd − 2b − 2d > 0 39
for b, d ≥ 1.
C6 (a, b, c, d, e, f ): Equation (4.10) is never satisfied for a, b, c, d, e, f ≥ 1.
Therefore we may conclude that P3 (1, k, 1) + K1 (m) is DS whenever k 6= m with P3 (1, k, 1) + K1 (k) ∼ P3 (k, 1, k) + K1 (1) for all k ≥ 1.
40
4.3
P4(a, b, c, d)
P4 (k, m, 1, 1): d1 = k + m + 2 d2 = 2k + m d3 = km + m d4 = km
c3 > c 4 2c1 + c4 = c2 + c3 + 4
(4.12) (4.13)
P3 (a, b, c) + K1 (d): c3 = abc − bcd < abcd = c4 for all a, b, c, d ≥ 1.
P4 (a, b, c, d): For c3 > c4 we need c3 − c4 = abc + bcd − abcd = bc(a + d − ad) > 0 Since bc > 0 for b, c ≥ 1 we must have a + d − ad > 0. This happens when a = 1 or d = 1. Assume d = 1. Then equation (4.13) gives abc + 2a + 2b + 2c + 2 = abc + ac + a + bc + b + 4 This simplifies to a + b + 2c − ac − bc − 2 = (2 − a − b)(c − 1) = 0 This says that either a = b = 1 or c = 1. If a = b = 1, then c + 3 = k + m + 2 and c + 2 = 2k + m imply k = 1 and m = c, which is just P4 (1, c, 1, 1) ∼ = P4 (1, 1, c, 1). On other other hand, if c = 1, then a + b + 2 = k + m + 2 and 2a + b = 2k + m imply a = k and b = m, which is just P4 (a, b, 1, 1) ∼ = P4 (k, m, 1, 1). 41
K1,3 (a, b, c, d): c3 > c4 implies abc + abd + acd > 2abcd + bcd Ignore for a moment the bcd term. Then we can factor abc + abd + acd − 2abcd = a(bc + bd + cd − 2bcd). Now 2bcd ≥ bc + bd + cd when at least two of b, c, d are greater than 1. In that case we would have abc + abd + acd < 2abcd + bcd. Therefore we may assume c = d = 1. With that equation (4.13) now gives 2ab + 2a + 2b + 4 = 2ab + a + b + 5 This simplifies to a + b = 1 which is obviously not possible with a, b ≥ 1.
C4 (a, b, c, d): For c3 > c4 we need abc + abd + acd + bcd > 3abcd Suppose c, d > 1. Then abcd ≥ acd, abcd ≥ bcd and abcd ≥ abc + abd = ab(c + d). This last inequality follows from cd ≥ c + d for c, d ≥ 2. Therefore, by the symmetry of C4 , we may assume that at most one of a, b, c, d is greater than 1. Suppose b = c = d = 1. Then equation (4.13) gives 5a + 6 = 4a + 6 This implies that a = 0, which cannot happen.
P5 (a, b, c, d, e): c3 ≤ c4 for all a, b, c, d, e ≥ 1.
C6 (a, b, c, d, e, f ): c3 ≤ c4 for all a, b, c, d, e, f ≥ 1. Therefore we may conclude that P4 (k, m, 1, 1) is DS for all k, m ≥ 1.
42
P4 (k, 1, m, 1): d1 = k + m + 2 d2 = km + k + 1 d3 = km + m d4 = km
c3 > c 4 c1 + 2c4 = c2 + c3 + 1
(4.14) (4.15)
P3 (a, b, c) + K1 (d): Conveniently (4.12) and (4.14) are the same, so we can reuse the information from the P4 (k, m, 1, 1) case. We know that c3 < c4 for all a, b, c, d ≥ 1.
P4 (a, b, c, d): We may assume d = 1 from (4.14). Equation (4.15) then gives 2abc + a + b + c + 1 = abc + ac + a + bc + b + 1 This simplifies to abc − ac − bc + c = (a − 1)(b − 1)c = 0 Either a = 1 or b = 1. First suppose a = 1. Then b+c+2 = k +m+2, b+c+1 = km+k +1, and 2bc = km + m imply k = 1 and either b = 1 and c = m or b = m and c = 1. In either case we are left with isomorphic graphs. Now suppose b = 1. Then we are once again we get isomorphic graphs from a + c + 2 = k + m + 2, ac + a + 1 = km + k + 1, and ac + c = km + m giving a = k and c = m.
K1,3 (a, b, c, d): Inequality (4.14) gives c = d = 1. Equation (4.15) then gives 4ab + a + b + 2 = 2ab + a + b + 2 43
This simplifies to 2ab = 0, which cannot happen.
C4 (a, b, c, d): Inequality (4.14) implies b = c = d = 1. Equation (4.15) then gives 7a + 3 + 4a + 3 This simplifies to 3a = 0, which cannot happen.
P5 (a, b, c, d, e): Inequality (4.14) is never satisfied with a, b, c, d, e ≥ 1.
C6 (a, b, c, d, e, f ): Inquality (4.14) is never satisfied with a, b, c, d, e, f ≥ 1.
Therefore we may conclude that P4 (k, 1, m, 1) is DS for all k, m ≥ 1.
44
P4 (k, 1, 1, m): d1 = k + m + 2 d2 = km + k + m d3 = k + m d4 = km
c1 = c3 + 2
(4.16)
c2 = c3 + c4
(4.17)
P3 (a, b, c) + K1 (d): Equation (4.16) gives a + b + c + d = abc − bcd + 2 Solving for a we get a=
b+c+d−2 bc − cd − 1
The numerator is always positive and the numerator is positive whenever c > 1 or c = 1 and b > d + 1. In either case we can now use this in equation (4.17). Doing so we get bcd(b + c + d − 2) (c + d − bc + cd)(b + c + d − 2) + (bd + cd)(bc − cd − 1) = bc − cd − 1 bc − cd − 1 Assuming c > 1 we can factor out (bc − cd − 1). Doing so we are left with b2 c + bc2 + 2bcd2 + c2 d2 + cd + 2c + 2d = 2bcd + 3bc + c2 d + c2 + cd2 + d2 It can be shown that the left hand side is always strictly greater than the right side for b, c, d ≥ 1. If c = 1, then we have a=
b+d−1 b−d−1
Equation (4.17) then gives b2 + 2bd2 + 2d + 1 = d2 + 2bd + 2b Once again the left side is always greater than the right.
45
P4 (a, b, c, d): Equation (4.16) gives a + b + c + d = abc + bcd + 2 It is easy to now show that b = c = 1. First observe that abc+1 ≥ a+b for all a, b, c ≥ 1 with equality when either a = c = 1 or b = c = 1. In either case c = 1. Similarly, bcd + 1 ≥ c + d forces b = 1. From here a + d + 2 = k + m + 2 and ad = km gives a = k and d = m or a = m and d = k. In both cases we’re left with isomorphic graphs.
K1,3 (a, b, c, d): Equation (4.17) gives 2abcd + abc + abd + acd = bcd + bc + bd + cd The left side factors as a(2bcd + bc + bd + cd), and thus equality above is clearly never possible.
C4 (a, b, c, d): Equation (4.16) gives a + b + c + d = abc + abd + acd + bcd + 2 Equality is clearly never possible.
P5 (a, b, c, d, e): Equality is never possible in equation (4.16).
C6 (a, b, c, d, e, f ): Equality is never possible in equation (4.16).
Therefore we may conclude that P4 (k, 1, 1, m) is DS for all k, m ≥ 1.
46
P4 (1, k, m, 1): d1 = k + m + 2 d2 = k + m + 1 d3 = 2km d4 = km
c3 > c 4
(4.18)
c1 = c2 + 1
(4.19)
c3 = 2c4
(4.20)
P3 (a, b, c) + K1 (d): Conveniently (4.18) is the same as (4.12) and (4.14). c3 < c4 for all a, b, c, d ≥ 1.
P4 (a, b, c, d): Inequality (4.18) gives d = 1. Equation (4.19) then gives a + b + c + 1 = ac + a + b + 1 This simplifies to ac − c = (a − 1)c = 0. Since c ≥ 1 we must have a = 1. Finally, b + c + 2 = k + m + 2 and bc = km give b = k and c = m or b = m and c = k. In either case we have isomorphic graphs.
K1,3 (a, b, c, d): Inequality (4.18) gives c = d = 1. Equation (4.20) then gives 2ab + a − b = 4ab This simplifies to 2ab − a + b = 0, which is not possible with a, b ≥ 1.
47
C4 (a, b, c, d): Inequality (4.18) gives b = c = d = 1. Equation (4.19) then gives a+3=a+2 This implies a = 0, which is not possible.
P5 (a, b, c, d, e): Inequality (4.18) is never satisfied with a, b, c, d, e ≥ 1.
C6 (a, b, c, d, e, f ): Inquality (4.18) is never satisfied with a, b, c, d, e, f ≥ 1.
Therefore we may conclude that P4 (1, k, m, 1) is DS for all k, m ≥ 1.
48
4.4
K1,3(a, b, c, d)
K1,3 (k, m, 1, 1): d1 = k + m + 2 d2 = 2m + 1 d3 = 2km + k − m d4 = 2km c1 + c4 = c2 + c3 + 1 2c4 = (2c1 − c2 − 3)(c2 − 1)
(4.21) (4.22)
P3 (a, b, c) + K1 (d): Equation (4.21) gives abcd + a + b + c + d = abc − acd + ac + ad + bd + cd + 1 This simplifies to abcd − abc + acd − ac − ad + a − bd + b − cd + c + d − 1 = (abc + ac − a − b − c + 1)(d − 1) = 0 This implies either a = c = 1 or d = 1. If a = c = 1, then equation (4.22) gives b2 d2 − 2b2 d + 2bd2 − 2bd = bd(bd − 2b + 2d − 2) = 0 Since bd ≥ 1 we must have bd − 2b + 2d − 2 = 0. If we solve this for b we get b= From this we quickly get that b ≥ 1 iff
4 3
2 − 2d d−2
≤ d < 2, and therefore d is not an integer. On the
other hand, if d = 1 then equation (4.22) gives a2 c2 − a2 + 2abc − 2ab − 2ac + 2a − b2 − 2bc + 2b − c2 + 2c − 1 = 0 Unfortunately from here we run into a problem. If we solve this for b and take the positive answer we get b=
p 2a(a − 1)c(c − 1) + ac − a − c + 1 49
This is an integer when 2a(a − 1)c(c − 1) is a perfect square. This is solved in a lemma below, but the result shows that for almost all a ∈ N there are infinitely many c ∈ N such that 2a(a − 1)c(c − 1) is a perfect square.
P4 (a, b, c, d): We have k = 21 (2c1 − c2 − 3). For this to be positive we obviously need 2c1 > c2 + 3. From this we get that −bd + 2b + 2c + 2d − 3 c+d−2 For b, c ≥ 1 and d ≥ 2 we get that the right side is always at most 3, and as such we get 1≤a
c2 + 3 gives 2a + 2b + 2c + 2d + 2e + 2f > ac + ad + ae + bd + be + bf + ce + cf + df + 3 This is easily shown to never be true with a, b, c, d, e, f ≥ 1.
Therefore we may conclude that K1,3 (k, m, 1, 1) is DS for all k, m ≥ 1 except for C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1) and P3 (a, n + ac − a − c + 1, c) + K1 (1) ∼ K1,3 (k, m, 1, 1) where 2a(a−1)c(c− 1) = n2 is a perfect square, k = n2 , and m = ac + n2 . When is 2a(a − 1)c(c − 1) a perfect square with a, c ≥ 2? If 2a(a − 1) = m2 n, where n is squarefree, then we have the equation m2 nc(c−1) = k 2 . Clearly for this to have a solution we need nc(c−1) = k 02 . Therefore we may first consider when nc(c−1) is a perfect square. Lemma 4.4.1. If (x, y) ∈ Z2 solves the equation x2 − ny 2 = k with k|x, then x2 x2 nxy 2 n −1 = k k k 2 2 is an integral solution to nc(c − 1) = k 02 . If xk < 0 then we may take xk and
x2 ( − 1) k for
positive solutions. Proof. Since x2 − ny 2 = k we get x2 − k y = n Multiplying both sides by n2 x2 gives nx2 (x2 − k) = n2 x2 y 2 . Now, since k|x we can divide 2
both sides by k 2 to get x2 n k
x2 nxy 2 −1 = k k 52
This is clearly an integral solution to nc(c − 1) = k 02 . Lemma 4.4.2. If nc(c − 1) is a perfect square, then c = n1 c21 and c − 1 = n2 c22 where n = n1 n2 . Corollary 4.4.3. Using the notation of the previous lemma, if nc(c − 1) is a perfect square, then (n1 c1 )2 − nc22 = n1 . With these lemmas we have that solving nc(c − 1) = k 02 is equivalent to looking for solutions of x2 − ny 2 = k with the conditions that k|x and k|n. Luckily this is a problem of antiquity. Firstly we can show that knowing two solution (x1 , y1 , k1 ) and (x2 , y2 , k2 ) to x2 − ny 2 = k we can get a third solution, and thus infinitely many. Lemma 4.4.4. If (x1 , y1 , k1 ) and (x2 , y2 , k2 ) both solve x2 −ny 2 = k, then (x1 x2 +ny1 y2 , x1 y2 + x2 y1 , k1 k2 ) is another solution. Proof. This follows immediately from Brahmagupta’s identity which states (x21 − ny12 )(x22 − ny22 ) = (x1 x2 + ny1 y2 )2 − n(x1 y2 + x2 y1 )2 .
What is even better is that we really only need one solution (x1 , y1 , k1 ) of x2 − ny 2 = k to generate the rest through recursion with xi+1 = x1 xi + ny1 yi yi+1 = x1 yi + y1 xi ki+1 = k1 ki The only concern with this now is whether or not ki |xi and ki |n. If k1 = 1, then ki = 1 for all i, and Lagrange showed that x2 − ny 2 = 1 has infinitely many solutions as long as n is not a perfect square. So we are guaranteed solutions, but we need not always have ki = 1. For example we have 22 − 6 · 12 = −2, and −2|2 and −2|6 which gives 6 · 3 · (3 − 1) = 62 . If we use the Lemma 4.4.4 with (2, 1, −2) then we get (10, 4, 4) at the next iteration, and we can check that 102 − 6 · 42 = 4, but 46 | 6. To remedy this we can divide x and y by −2 and 53
k by 4. This gives the triple (−5, −2, 1). We still have (−5)2 − 6 · (−2)2 = 1 and 1|6. To account for this we can modify the lemma above. Lemma 4.4.5. If (x1 , y1 , k1 ) and (x2 , y2 , k1 ) are integral solutions to x2 − ny 2 = k with k1 |n and k1 |xi , then
x1 x2 + ny1 y2 x1 y2 + x2 y1 , ,1 k1 k1
is another integral solution. Proof. This again follows immediately from Brahmagupta’s identity, but this time dividing both sides by k12 . This gives k2 (x2 − ny12 )(x22 − ny22 ) 1 = 12 = 1 = k1 k12
x1 x2 + ny1 y2 k1
2
−n
x1 y2 + x2 y1 k1
2 .
This also gives us a modified recursion that guarantees ki |n and ki |xi : x1 xi + ny1 yi ki x1 y i + y 1 xi = ki k1 = ki
xi+1 = yi+1 ki+1
It can be shown that solutions to x2 − ny 2 = k show up in the study of continued fractions of √ n, but that is not necessary for our purposes. It can also be shown that all solutions can be found using a fundamental solution (x1 , y1 , k1 ) to x2 − ny 2 = k and the recursion above, but again this is not necessary for our purposes. Now that we have solutions to nc(c − 1) = k 02 , we need to consider when 2a(a − 1) = m2 n. In particular we want to know when n = 1. This, however, is just a particular case of our examination above. It is straightforward to get that a fundamental solution for x2 − 2y 2 = k is (1, 1, −1). This leads to the set of solutions for {a, a − 1} as {2Pn2 , 2Pn2 + (−1)n } where Pn is the n-th Pell number which can be defined by Pn =
(1 +
√ n √ 2) − (1 − 2)n √ 2 2
54
These are important because they are the values for a for which 2a(a − 1)c(c − 1) is never a perfect square. The first few values of a in this sequence are {2, 9, 50, 289, . . .}. All other values of a lead back to the equation nc(c − 1) = k 02 .
55
K1,3 (1, k, m, 1): d1 = k + m + 2 d2 = km + k + m d3 = k + m d4 = 2km
c1 = c3 + 2 2c3 + c4 = 2c2
(4.23) (4.24)
Equation (4.23) is the same as (4.16) from P4 (k, 1, 1, m), so we get to reuse that data.
P3 (a, b, c) + K1 (d): We will start by showing that the only possibility is k = m = d + 1. For this we need to look at PK1,3 (1,k,m,1) (d). Since d ∈ Spec(P3 (a, b, c) + K1 (d + 1)) we need PK1,3 (1,k,m,1) (d) = 0. To that effect we can plug d into PK1,3 (1,k,m,1) to get (d + 1)k+m−2 (d4 − d3 k − d3 m + 2d3 + d2 km − 2d2 k − 2d2 m + 2dkm − 2d − km + k + m − 1) For this to vanish we clearly need d4 − d3 k − d3 m + 2d3 + d2 km − 2d2 k − 2d2 m + 2dkm − 2d − km + k + m − 1 = 0 since d ≥ 1. Solving this for k we get k=
d4 − d3 m + 2d3 − 2d2 m − 2d + m − 1 d3 − d2 m + 2d2 − 2dm + m − 1
We will now show that k ∈ (d − 1, d) for m ∈ (1, d − 1); k ≤ −1 for m = d; k = d + 1 for m = d + 1; and k ∈ (d, d + 1) for m ≥ d + 2. The easiest to check are when m = d, d + 1. For m = d it is straightforward to calculate k=
−d − 1 d−1
Similarly for m = d + 1 we get k=
−d(d + 1)2 =d+1 −d(d + 1)
56
These give the desired results for m = d, d + 1. For the other two claims we need to look at d4 − d3 m + 2d3 − 2d2 m − 2d + m − 1 − (d − 1)(d3 − d2 m + 2d2 − 2dm + m − 1) = d3 − d2 m + 2d2 − 3dm − d + 2m − 2 d4 − d3 m + 2d3 − 2d2 m − 2d + m − 1 − (d + 1)(d3 − d2 m + 2d2 − 2dm + m − 1) = −d(d + 1)(d − m + 1) We also need
From
∂k ∂m
∂k d2 (d − 1)2 =− 3 ∂m (d − d2 m + 2d2 − 2dm + m − 1)2
we see that k is decreasing for all m ≥ 1 except when d3 −d2 m+2d2 −2dm+m−1 =
0, which is at m = m
0 when
≤ d − 1 and −d(d + 1)(d − m + 1) > 0 when d ≥ 1 and m ≥ d + 1. This
shows k > d − 1 for 1 ≤ m ≤ d − 1 and k < d + 1 for m > d + 1. To get the bound with d we can look at k when m = 1 and when m = d + 2. We get respectively at m = 1 and m=d+2 k=
(d + 1)(d2 − 2) d 2d2 + 3d − 1 Since k is decreasing we have the desired result. Therefore we must have k = m = d + 1. k=
Now we can use c1 = 2d + 4 and c2 = (d + 1)(d + 3) to get 2c2 − 2cd − 8c + 3 c c2 − cd − 4c + 3 b = c
a = −
Finally, c4 = 2(d + 1)2 reduces to 2c4 d − 4c3 d2 − 16c3 d + 2c2 d3 + 16c2 d2 + 41c2 d − 7cd2 − 32cd + 2c + 9d =0 c But 2c4 d − 4c3 d2 − 16c3 d + 2c2 d3 + 16c2 d2 + 41c2 d − 7cd2 − 32cd + 2c + 9d > 0 for all c, d ≥ 1.
P4 (a, b, c, d): Equation (4.23) implies b = c = 1, and we have already seen that P4 (a, 1, 1, d) is DS. 57
K1,3 (a, b, c, d): Equation (4.24) gives 2abcd + 2(abc + abd + acd − bcd) = 2(bc + bd + cd) This reduces to 2(a − 1)(bcd + bc + bd + cd) = 0, which implies a = 1. Now equation (4.23) gives b + c + d + 1 = −bcd + bc + bd + cd + 2 This reduces to (b − 1)(c − 1)(d − 1) = 0, which implies one of b, c, d is 1. Suppose d = 1. Then k + m + 2 = b + c + 2 and km + k + m = bc + b + c imply b = k and c = m or b = m and c = k. In either case we have isomorphic graphs.
C4 (a, b, c, d): Equality is never possible in equation (4.23).
P5 (a, b, c, d, e): Equality is never possible in equation (4.23).
C6 (a, b, c, d, e, f ): Equality is never possible in equation (4.23).
Therefore we may conclude that K1,3 (1, k, m, 1) is DS for all k, m ≥ 1.
58
4.5
C4(a, b, c, d)
C4 (k, m, 1, 1): d1 = k + m + 2 d2 = k + m d3 = 2km + k + m d4 = 3km
c1 = c2 + 2 3c2 + 2c4 = 3c3
P3 (a, b, c) + K1 (d): Equality is never possible in equation (4.25).
P4 (a, b, c, d): Equality is never possible in equation (4.25).
K1,3 (a, b, c, d): Equation (4.25) gives a = bc + bd + cd − b − c − d + 2 Now, k + m = c2 and 2km = c3 − c2 gives q 1 k, m = c2 ± c22 + 2c2 − 2c3 2 For these to be integers we need c22 + 2c2 − 2c3 ≥ 0. This gives −b2 c2 − 2b2 cd + 2b2 c − b2 d2 + 2b2 d − 2bc2 d + 2bc2 −2bcd2 + 8bcd − 2bc + 2bd2 − 2bd − c2 d2 + 2c2 d + 2cd2 − 2cd ≥ 0
59
(4.25) (4.26)
The only positive integer solutions, with b ≥ c ≥ d ≥ 1, to this are b ≥ 1, c = d = 1; b = c = 2, d = 1; and b = 3, c = 2, d = 1. We can quickly check that these give 4b + 1, 8, and 1, respectively. We can immediately eliminate b = c = 2, d = 1 since 8 is not a perfect square. For b = 3, c = 2, d = 1, we get a = 1, k = 4, m = 1. Finally, for b ≥ 1, c = d = 1 we have 4b + 1 is a perfect square exact when b = r(r + 1) for some positive integer r. This gives a = r(r + 1) + 1, k = (r + 1)2 , and m = r2 . Now for c4 = 3km we have 2(r(r + 1) + 1)r(r + 1) = 3r2 (r + 1)2 which reduces to (r − 1)r(r + 1)(r + 2) = 0. Since r ≥ 1 we must have r = 1, but this once again gives a = 3, b = 2, k = 4, and m = 1.
C4 (a, b, c, d): Equation (4.25) gives c = d = 1. We then get from k + m = a + b and 3km = 3ab that a = k and b = m or a = m and b = k. In either case we have isomorphic graphs.
P5 (a, b, c, d, e): Equality is never possible in equation (4.25).
C6 (a, b, c, d, e, f ): Equality is never possible in equation (4.25).
Therefore we may conclude that C4 (k, m, 1, 1) is DS for all k, m ≥ 1 except for C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1).
60
C4 (k, 1, m, 1): d1 = k + m + 2 d2 = km + 1 d3 = 2km + k + m d4 = 3km c1 + 2c2 = c3 + 4
(4.27)
3c2 = c4 + 3
(4.28)
3c1 + 2c4 = 3c3 + 6
(4.29)
Note that 0 ∈ Spec(C4 (k, 1, m, 1)). Also note that from c1 = k + m + 2 and c2 = km + 1 we get m2 + (2 − c1 )m + c2 − 1 = 0
(4.30)
P3 (a, b, c) + K1 (d): 0 ∈ Spec(P3 (a, b, c) + K1 (d)) iff a = c = 1 or d = 1. If a = c = 1 then equation (4.27) reduces to 2(b + 3)d = 0, which cannot happen. Suppose then that d = 1. Then we can solve equation (4.27) for b=
3(ac + a + c − 1) ac − 3
We can now use this with (4.30) to get (ac − 3)m2 − ac(a + c + 2)m + ac(ac + a + c − 1) =0 ac − 3 Let f (a, c, m) = (ac − 3)m2 − ac(a + c + 2)m + ac(ac + a + c − 1). Then we can show that the only positive integer roots for f are at m = 6, 12. To do this we need to assume a ≥ c and then look at f (a, c, c − 2) = 3a2 c − 3ac2 + 7ac − 3c2 + 12c − 12 f (a, c, c − 1) = 2a2 c − 2ac2 + 2ac − 3c2 + 6c − 3 f (a, c, c) = c(a2 − ac − a − 3c) f (a, c, c + 1) = −2ac − 3c2 − 6c − 3 61
From these we can show that f (a, c, c−2) > 0 for a, c ≥ 1; f (a, c, c−1) > 0 for 1 ≤ c ≤ a−1; and f (a, c, c + 1) > 0 for a, c ≥ 1. This gives m ∈ (c − 1, c + 1) for 1 ≤ c < a and m ∈ (c − 2, c + 1) for a = c. We can do better and show that m ∈ (c − 2, c) for a = c. For the latter case this would imply m = c−1 is the only possible integer root if a = c, but when a = c √ we get f (c, c, c−1) = −c2 +6c−3 which has roots c = 3± 6. Therefore we may eliminate this possibility. For c < a, on the other hand, we must look at f (a, c, c) = c(a2 − ac − a − 3c) = 0. Since c 6= 0 we must have a2 − ac − a − 3c = 0, which gives c=
a(a − 1) a+3
From this we can show that a − 4 < c < a − 3 if a > 9. For a ≤ 9 the only positive integer solutions are a = 3, c = 1 and a = 9, c = 6. For a = 3, c = 1 we would have b=
3(ac+a+c−1) ac−3
undefined. For a = 9, c = 3 we get b = 4, k = 12, and m = 6. That is,
P3 (9, 4, 6) + K1 (1) ∼ C4 (12, 1, 6, 1).
P4 (a, b, c, d): Equation (4.27) gives a + b + c + d + 2(ac + ad + bd) = abc + bcd + 4 We can solve this for a to get a=
−bcd + 2bd + b + c + d − 4 bc − 2c − 2d − 1
We can now use this with equation (4.28) to solve for b as q (c+6)d−6 (c + d − 1) +2 cd b= (c − 2)d + 2 If d = 1 then b = 3 for all c ≥ 1. This gives a = −2. If c ≥ 3 then 1 ≤ b ≤ 3 for all d ≥ 1. Using this we can show that the only integer solutions with d ≥ 4 is b = 1, c = 6, d = 25. This gives a =
24 . 19
Finally we must consider the cases c = 2. With c = 2 we have ! r 3 1 4− +2 b = (d + 1) 2 d
√ From this it is clear that d must equal 3, but even with that we would have b = 4 + 2 3. We have thus exhausted all possibilities. 62
K1,3 (a, b, c, d): Equation (4.27) gives a + b + c + d + 2(bc + bd + cd) = abc + abd + acd − bcd + 4 We can solve this for bc + bd + cd =
a + b + c + d + bcd − 4 a−2
This with equation (4.28) gives a + b + c + d + bcd − 4 3 − 3 = 2abcd a−2 This simplifies to 2a2 bcd − 4abcd − 3bcd − 3b − 3c − 3d + 6 = 0 If we look at the left side as a polynomial f (a, b, c, d), then we can show that the roots in a are in [−1, 0] ∪ [2, 3]. To do this we look at f (−1) = f (3) = 3(bcd − b − c − d + 2) ≥ 0 f (0) = f (2) = −3(bcd + b + c + d − 2) ≤ 0 Equality above happens when at least two of b, c, d are 1. Suppose c = d = 1. Equation (4.27) then reduces to 2b(a − 3) = 0, which implies a = 3. Finally, from here we can solve for b = 2, k = 4, and m = 1. That is, C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1), which we saw above.
C4 (a, b, c, d): Equation (4.28) gives 3ac + 3bd = 3abcd + 3 This simplifies to (ac − 1)(bd − 1) = 0. Thus we have a = c = 1 or b = d = 1, both of which are equivalent. Suppose b = d = 1. We can then use k +m+2 = a+c+2 and km+1 = ac+1 to get a = k and c = m or a = m or c = k. In either case we have isomorphic graphs.
P5 (a, b, c, d, e): Equation (4.29) gives 3(a + b + c + d + e) + 2(abcd + abce + acde + bcde) = 3(abc + bcd + cde − acd) + 6 63
Since 3(a+b+c+d+e) > 6 for all a, b, c, d, e ≥ 1, we must have 2(abcd+abce+acde+bcde) < 3(abc + bcd + cde − acd) for equality above. But now this reduces to 2 (abd + abe + ade + bde) + ae < ab + bd + de 3 This can easily be shown to always be false with a, b, c, d, e ≥ 1.
C6 (a, b, c, d, e, f ): Similar to the P5 (a, b, c, d, e) case above we can show that equation (4.29) is never satisfied.
Therefore, we may conclude that C4 (k, 1, m, 1) is DS for all k, m ≥ except for C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1) and C4 (12, 1, 6, 1) ∼ P3 (9, 4, 6) + K1 (1).
64
4.6
P5(a, b, c, d, e)
P5 (k, m, 1, 1, 1): d1 = k + m + 3 d2 = 3k + 2m + 1 d3 = km − k + m + 1 d4 = 2km + k + m 9c1 + c4 = 4c2 + 2c3 + 21
(4.31)
Using c1 = k + m + 1 and c2 = 3k + 2m + 1 we get k = −2c1 + c2 + 5 and m = 3c1 − c2 − 8. So we also get c3 = −6c21 + 5c1 c2 + 36c1 − c22 − 15c2 − 52
(4.32)
c4 = −12c21 + 10c1 c2 + 63c1 − 2c22 − 26c2 − 83
(4.33)
Since c1 , c2 > 0, we can also show that c3 , c4 > 0 with (4.32) and (4.33).
P3 (a, b, c) + K1 (d): Equation (4.31) gives 9(a + b + c + d) + abcd = 4(ac + ad + bd + cd) + 2(abc − acd) + 21 Solving for a gives 4bd − 9b + 4cd − 9c − 9d + 21 bcd − 2bc + 2cd − 4c − 4d + 9 We can now use (4.32) to show that 1 ≤ b ≤ 3. Let f (b, c, d) be the numerator of (4.32) a=
(with c3 moved to the right side). (f (b, c, d) is a very ugly polynomial that is omitted here.) Then the first thing we can show is that f (3, c, d) ≥ 0 for c, d ≥ 1 with equality iff c = d = 1. Moreover, f (b, c, d) > 0 for b ≥ 4 and c, d ≥ 1.
Now, using c3 > 0, we also get that b > d. Since 1 ≤ b ≤ 3 and d ≥ 1, we can eliminate b = 1. This leaves us with only b = 2 and d = 1, or b = 3 and d = 1 (since b = 3 iff c = d = 1). Now, if b = 2 and d = 1, then we have f (2, c, 1) = −(c2 − 2c + 2)(3c2 − 2c + 1) = 0 65
√ This gives c = 1 ± i, 13 (1 ± i 2). Finally, if b = 3 and c = d = 1, then a is undefined as the denominator above is 0.
P4 (a, b, c, d): Equation (4.31) gives 9(a + b + c + d) + abcd = 4(ac + ad + bd) + 2(abc + bcd) + 21 Solving this for a gives a=
2bcd + 4bd − 9b − 9c − 9d + 21 bcd − 2bc − 4c − 4d + 9
Now we can use this in equation (4.32), just as above, and show that b ≤ 1. Moreover we have b = 1 iff c = 1. This would then give P4 (a, 1, 1, d) which we know is DS.
K1,3 (a, b, c, d): Equation (4.31) gives 9(a + b + c + d) + 2abcd = 2(abc + abd + acd − bcd) + 4(bc + bd + cd) + 21 Solving for a gives a=
−2bcd + 4bc + 4bd − 9b + 4cd − 9c − 9d + 21 2bcd − 2bc − 2bd − 2cd + 9
From this we get that a < 1 if b, c, d ≥ 2. To show this we look at −2bcd + 4bc + 4bd − 9b + 4cd − 9c − 9d + 21 − (2bcd − 2bc − 2bd − 2cd + 9) = −4bcd + 6bc + 6bd − 9b + 6cd − 9c − 9d + 12 This is always less than 0 when b, c, d ≥ 2. Thus we are forced to have at least one of b, c, d equal 1. Suppose d = 1. Now a=−
2bc − 5b − 5c + 12 2b + 2c − 9
With the assumption that b ≥ c, this gives that a ≥ 1 iff b ≥ 4 and c = 1, or b = 3 and c = 2. If c = 1 then we have K1,3 (a, b, 1, 1) which we know is not cospectral with P5 (a, b, c, d, e). If b = 3 and c = 2, then a = 1. This gives K1,3 (1, 3, 2, 1), which we know is DS. 66
C4 (a, b, c, d): Equation (4.31) gives 9(a + b + c + d) + 3abcd = 4(ac + bd) + 2(abc + abd + acd + bcd) + 21 Solving for a gives 2bcd + 4bd − 9b − 9c − 9d + 21 3bcd − 2bc − 2bd − 2cd − 4c + 9 Now, if b, c, d ≥ 2 then a < 2. To see this we look at a=
2bcd+4bd−9b−9c−9d+21−2(3bcd−2bc−2bd−2cd−4c+9) = −4bcd+4bc+8bd−9b+4cd−c−9d+3 This is less than 0 for all b, c, d ≥ 2. With this we may assume that one of, and exactly one of, a, b, c, d equals 1. We may assume any due to the symmetry of C4 and exactly one because C4 (k, m, 1, 1) and C4 (k, 1, m, 1) are never cospectral with P5 (a, b, c, d, e). Suppose d = 1. Then equation (4.32) gives 2b2 − (2ac − 8a − 8c + 14)b + a2 c2 − 5a2 c + 6a2 − 5ac2 + 23ac − 24a + 6c2 − 24c + 22 = 0 If we let b = 0, then the left side becomes a2 (c2 − 5c + 6) + a(−5c2 + 23c − 24) + 6c2 − 24c + 22 This is always greater than 0 for a, c ≥ 2. This implies that b < 0, which cannot happen.
P5 (a, b, c, d, e): Equation (4.31) gives 9(a + b + c + d + e) + abcd + abce + acde + bcde = 4(ac + ad + ae + bd + be + ce) + 2(abc − ace + bcd + cde) + 21 Solving this for c gives c=
4ad + 4ae − 9a + 4bd + 4be − 9b − 9d − 9e + 21 abd + abe − 2ab + ade + 2ae − 4a + bde − 2bd − 2de − 4e + 9
We can narrow down the cases by considering when a, e are 1 or not. If a = e = 1, then c ≤ 0 for b, d ≥ 2. If b or d is also 1 (and the other is not 2), then c = 1. If b = 2 and d = 1, 67
then c can be anything. Now, if a, e ≥ 2 then c < 2, which implies c must be 1. The case for a ≥ 2 and e = 1 is a little more subtle. If a = 2 and d ≥ 3, then c < 7. If a = d = 2, then c = 3b. Finally, if a ≥ 3 and b, d ≥ 1, then c ≤ 3 .
If a = d = e = 1 and b = 2, then equation (4.32) reduces to 2(c − 1) = 0 This implies c = 1. Now, using k + m + 3 = c1 and 3k + 2m + 1 = c2 we get k = 1 and m = 2. These are isomorphic.
If a, e ≥ 2 and c = 1, then we can use equations (4.31) and (4.32) to show that e ≤ 1 if b ≥ 1 and d ≥ 2, which is obviously a contradiction. If d = 1, then e = 1. We are then left with k + m + 3 = a + b + 3 and 3k + 2m + 1 = 3a + 2a + 1 which gives a = k and b = m, which leaves us with isomorphic graphs.
If a = 2 and e = 1, then equation (4.32) reduces to b2 (−d2 + 3d − 2) + b(−2cd + d2 − 5d + 6) + 2d − 2 = 0 If d ≥ 3 and c ≤ 6, then this is never satisfied. With d = 2 and c = 3b, this then becomes 2(6b2 − 1) = 0 This would imply b =
q
1 , 6
which is absurd.
Finally, we can look at a ≥ 3, b, d ≥ 1 and 1 ≤ c ≤ 3. First suppose c = 1. Then we can show that d = 1 is the only integer solution to equation (4.32). We are left with the same situation as above and get a = k and b = m. If c = 2, then we can show again that d < 2. Now, for d = 1 we get b = 2a − a2 , which implies b < 0 for a ≥ 3. This is cannot happen. Lastly, if c = 3 we once again have d < 2 with d = 1 implying 2b = −3a2 + 7a − 2. This again implies b < 0 for a ≥ 3. We have thus exhausted all possibilities.
68
C6 (a, b, c, d, e, f ): We can use equation (4.31) to solve for a and show that a < 1 for all b, c, d, e, f ≥ 1. The fraction for a is omitted for space.
Therefore we may conclude that P5 (k, m, 1, 1, 1) is DS for all k, m ≥ 1.
69
P5 (k, 1, m, 1, 1): d1 = k + m + 3 d2 = km + 2k + m + 2 d3 = 2m d4 = 3km + m
6c1 + c4 = 3c2 + 2c3 + 12 Also, we can easily get that m =
c3 2
c4
and k = c1 − c23 − 3. With these we get 3 3 = c3 c1 − c3 − 4 2 4
(4.34)
(4.35)
P3 (a, b, c) + K1 (d): Equation (4.34) gives 6(a + b + c + d) + abcd = 3(ac + ad + bd + cd) + 2(abc − acd) + 12 Solving for d gives d=
2abc + 3ac − 6a − 6b − 6c + 12 abc + 2ac − 3a − 3b − 3c + 6
Now, if a, c ≥ 2 then d < 2. That would imply d = 1. If d = 1, then 0 ∈ Spec(P3 (a, b, c) + K1 (d)), but we can easily check that 0 ∈ Spec(P5 (k, 1, m, 1, 1)) iff k = m = 1. If a = c = 1 and b ≥ 2, then 2 < d < 3. We will check a = b = c = 1 separately. Finally, if a = 2, c = 1, and b ≥ 2, then 2 > d ≥ 4, with d = 4 iff b = 2 and d = 3 iff b = 3. Similarly we will check a = 2 and b = c = 1 separately.
For (a, b, c, d) = (2, 2, 1, 3) we get k = 8 and m = −2. For (a, b, c, d) = (2, 3, 1, 3) we get k = 6 and m = 0. For (a, b, c, d) = (1, 1, 1, d) equation (4.35) reduces to (3d − 1)2 = 0, which gives d =
1 . 3
For (a, b, c, d) = (2, 1, 1, d) equation (4.35) reduces to 6d2 − 3d − 1 = 0, which gives d = √ 1 (3 ± 33). 12 70
P4 (a, b, c, d): Equation (4.34) gives 6(a + b + c + d) + abcd = 3(ac + ad + bd) + 2(abc + bcd) + 12 Solving for a gives 2bcd + 3bd − 6b − 6c − 6d + 12 bcd − 2bc − 3c − 3d + 6 We can now use this with equation (4.35). That then reduces to a=
(b3 d4 − 8b2 d3 − 8b2 d2 + 21bd2 + 36bd + 12b − 18d − 36)c3 +(4b3 d3 − 8b3 d2 − 8b2 d4 + 8b2 d3 − 4b2 d2 + 24b2 d +24b2 + 42bd3 − 12bd2 − 80bd − 104b − 54d2 + 168)c2 +(3b3 d2 − 12b3 d + 12b3 − 14b2 d3 + 36b2 d2 + 18b2 d −68b2 + 21bd4 − 48bd3 − 2bd2 − 64bd + 184b − 54d3 +108d2 + 120d − 240)c + 6bd3 − 36bd2 + 72bd − 48b −18d4 + 72d3 − 48d2 − 96d + 96 = 0 Call this polynomial f (b, c, d). We can use this to show that if b, d ≥ 3 then c < 2. For this we can calculate f (b, 2, d) > 0 for all b, d ≥ 3 and f (b, c, d) is strictly increasing for b, d ≥ 3 and c > 2. Suppose c = 1. Then we have f (b, 1, d) is (d−1)(b3 d3 +5b3 d2 −12b3 −8b2 d3 −22b2 d2 +2b2 d+44b2 +21bd3 +21bd2 −8bd−44b−18d3 +6d+12) We can now use this to show that d < 2 for all b ≥ 3. But then we would have P4 (a, b, 1, 1), which we know is DS. Therefore, we must have b = 1, 2 and d ≥ 1 or b ≥ 1 and d = 1, 2. Suppose first that b = 1. Then d cannot be 1. If d = 2 then c = 2. This then gives a = 32 . If d ≥ 13 then we can show that 2 < c < 4. If b = 1 and c = 3, then a < 0 for all d ≥ 13. For 3 ≤ d ≤ 12 we can directly show that there are no integer solutions for c.
Now suppose b = 2. If d = 1 then c = 1, which gives P4 (a, 2, 1, 1), which is DS. If d = 2, then c = 4. This gives a = − 23 . For 3 ≤ d ≤ 37 we can directly show there are no integer solutions for c. Finally, if d ≥ 38, then we can show 1 < c < 2.
Next suppose d = 1. Then b ≥ 2. For 2 ≤ c ≤ 27 we can show that b = 1 is the only integer 71
solution. For c ≥ 28 we can show that b = 1 is again the only integer solution with the other solutions being in (5, 6) ∪ (9, 10). Finally, suppose d = 2. If b = 1 then c = 2. This gives a = 23 . If b = 2 then c = 4. This gives a = − 23 . If b = 3, then c = 0. Finally, if b ≥ 4 then c ∈ [0, 1). We have thus exhausted all possibilities.
K1,3 (a, b, c, d): Equation (4.34) gives 6(a + b + c + d) + 2abcd = 2(abc + abd + acd − bcd) + 3(bc + bd + cd) + 12 Solving for a gives a=
−2bcd + 3bc + 3bd − 6b + 3cd − 6c − 6d + 12 2(bcd − bc − bd − cd + 3)
If b ≥ 3 and c, d ≥ 2, then a < 0 except for b = 3 and c = d = 2 in which case a is undefined. If c = d = 1, then 1 < a < 2 for b ≥ 3. a = 0.5 when b = 1 and a is undefined when b = 2. For c = 2 and d = 1 we have a ≤ 1 for b ≥ 2 with equality iff b = 2. This would give K1,3 (1, 2, 2, 1), which we know is DS. Finally, when c = d = 2 we have a = b. With this last case we get from k + m + 3 = 2b + 4 and 2m = 4b2 that k = −2b2 + 2b + 1 and m = 2b2 . This means that k ≤ 0 for b ≥ 2.
C4 (a, b, c, d): Equation (4.34) gives 6(a + b + c + d) + 3abcd = 3(ac + bd) + 2(abc + abd + acd + bcd) + 12 Solving for a gives a=
2bcd + 3bd − 6b − 6c − 6d + 12 3bcd − 2bc − 2bd − 2cd − 3c + 6
72
This with equation (4.35) reduces to 0 = (9b4 d4 − 36b3 d3 − 24b3 d2 − 24b2 d3 + 45b2 d2 +60b2 d + 12b2 + 60bd2 + 6bd − 36b + 12d2 − 36d)c4 +(−24b4 d2 + 24b3 d3 + 24b3 d2 + 72b3 d + 24b3 −24b2 d4 + 24b2 d3 − 48b2 d2 − 88b2 d − 104b2 + 72bd3 −88bd2 − 88bd + 168b + 24d3 − 104d2 + 168d)c3 +(18b4 d4 − 24b4 d3 − 9b4 d2 + 12b4 d + 12b4 − 24b3 d4 −12b3 d3 + 24b3 d2 + 10b3 d − 68b3 − 9b2 d4 + 24b2 d3 +148b2 d2 − 116b2 d + 184b2 + 12bd4 + 10bd3 −116bd2 + 128bd − 240b + 12d4 − 68d3 + 184d2 − 240d)c2 +(−12b4 d2 + 24b4 d + 32b3 d2 − 64b3 d − 12b2 d4 + 32b2 d3 −8b2 d2 + 8b2 d − 48b2 + 24bd4 − 64bd3 + 8bd2 + 96b − 48d2 + 96d)c +9b4 d4 − 24b4 d3 + 12b4 d2 − 24b3 d4 + 46b3 d3 +4b3 d2 + 12b2 d4 + 4b2 d3 − 32b2 d2 − 48b2 d − 48bd2 + 96bd As daunting as this polynomial is, we can use it to quickly narrow our search. Call the polynomial f (b, c, d). First, if c = 0 then f (b, 0, c) = (b − 2)b(d − 2)d(9b2 d2 − 6b2 d − 6bd2 − 14bd + 24) This is easily seen to be strictly positive for b, d ≥ 3. Moreover, we can also show that f (b, c, d) is strictly increasing for b, d ≥ 3 and c ≥ 0. Thus we must have one of b, d equal to 1 or 2. By the symmetry of C4 we may assume that d = 1, 2.
First suppose d = 1. Then f (b, c, 1) = (9b4 − 60b3 + 93b2 + 30b − 24)c4 + (−24b4 + 144b3 − 240b2 + 64b + 88)c3 +(9b4 − 70b3 + 231b2 − 206b − 112)c2 + (12b4 − 32b3 − 28b2 + 64b + 48)c +48b − 3b4 + 26b3 − 64b2 For 1 ≤ b ≤ 8 we can directly calculate that there are no positive integer solutions for c. For b ≥ 9 we can show that c ∈ (−1, 0) ∪ (0, 1) ∪ (1, 2). 73
Now suppose d = 2. Then f (b, c, 2) = 8c2 (18b4 c2 − 12b4 c + 12b4 − 48b3 c2 + 57b3 c − 54b3 +15b2 c2 − 83b2 c + 74b2 + 27bc2 + 27bc − 22b − 3c2 + 14c − 12) Ignoring the 8c2 factor we can show that the rest of f (b, c, 2) is strictly greater than 0 for b ≥ 3 and is also strictly increasing. For b = 2 we get c = 0.
P5 (a, b, c, d, e): Equation (4.34) gives 6(a + b + c + d + e) + abcd + abce + acde + bcde = 3(ac + ad + ae + bd + be + ce) + 2(abc − ace + bcd + cde) + 12 Solving for a gives a=
−bcde + 2bcd + 3bd + 3be − 6b + 2cde + 3ce − 6c − 6d − 6e + 12 bcd + bce − 2bc + cde + 2ce − 3c − 3d − 3e + 6
Using this with equation (4.35) we can show that if c, d ≥ 2 then all real roots b of the resulting polynomial are strictly less than 1 for all e ≥ 1. Thus, we must have either c = 1 or d = 1.
First suppose c = 1. If d ≥ 3 and e ≥ 2, then b < 2 with b = 1 iff e = 2 and d ≥ 3 or d = e = 3. We are left to check when d = 1, 2 or e = 1.
If e = 1 then b ∈ (−1, 2) ∪ (2, 3] with b = 3 iff d = 1. b = 1 is a root for all d ≥ 1.
If d = 1, then the above polynomial becomes (b − 1)e(b − e)(3b2 e − 2b2 − 3be2 − 3be + 2b + e2 + 2e) = 0 From this we can easily show that b = 1 and b = e are the only integer solutions for e ≥ 2 (the other two roots are in (0, 1) ∪ (e + 1, e + 2)), and b = 1 and b = 3 are solutions when e=1
74
Finally, if d = 2, then we can show that for e ≥ 13 the only positive root for b is in ( e+2 , e+3 ). 3 3 For 1 ≤ e ≤ 12 we can directly calculate that b = 1 is the only positive integer root, and this happens iff e = 1 or e = 2.
Now we must check with d = 1. If c = 3 then b ≤ 6. If c = 4 then b ≤ 3. If c = 5 then b ≤ 2 and if c ≥ 6 then b ≤ 1. Finally, if c = 2 then we have the same argument as with d = 2 and c = 1 above. That is, b = 1 is the only solution, and this happens iff e = 1 or e = 2.
Finally, of those possibilities listed above we can check to see that all either give isomorphic graphs or no solutions for k, m except for when a = 2, b = e, c = 1, d = 1, e ≥ 1, in which case k = e + 1 and m = e. This gives P5 (k + 1, 1, k, 1, 1) ∼ P5 (2, k, 1, 1, k) for all k ≥ 1. We can actually do better and show P5 (a(k + 1), a, ak, a, a) ∼ P5 (2a, ak, a, a, ak) for all a, k ≥ 1.
C6 (a, b, c, d, e, f ): Solving equation (4.34) for a we can show that a < 1 for all b, c, d, e, f ≥ 1.
Therefore we may conclude that P5 (k, 1, m, 1, 1) is DS for all k, m ≥ 1 except for P5 (k + 1, 1, k, 1, 1) ∼ P5 (2, k, 1, 1, k) for k ≥ 1.
75
P5 (k, 1, 1, m, 1): d1 = k + m + 3 d2 = km + 2k + m + 2 d3 = 2m d4 = 2km + k + m 3c1 + c4 = 2c2 + c3 + 5 Similar to P5 (k, 1, 1, m, 1) we have m = c4 = −
c3 2
and k = c1 −
c3 2
(4.36) − 3. With these we get
c23 + c1 c3 − 3c3 + c1 − 3 2
(4.37)
P3 (a, b, c) + K1 (d): Equation (4.36) gives 3(a + b + c + d) + abcd = 2(ac + ad + bd + cd) + abc − acd + 5 Solving for d gives d=
abc + 2ac − 3a − 3b − 3c + 5 abc + ac − 2a − 2b − 2c + 3
Now equation (4.37) reduces to (a4 c4 − 6a3 c3 + 12a2 c2 − 8ac)b4 +(−6a4 c3 − 6a3 c4 + 16a3 c3 + 24a3 c2 + 24a2 c3 − 64a2 c2 − 24a2 c − 24ac2 + 68ac − 8)b3 +(−4a4 c4 + 4a4 c3 + 12a4 c2 + 4a3 c4 + 34a3 c3 − 78a3 c2 − 24a3 c + 12a2 c4 − 78a2 c3 +28a2 c2 + 148a2 c − 24ac3 + 148ac2 − 164ac − 24a − 24c + 36)b2 +(12a4 c3 − 14a4 c2 − 8a4 c + 12a3 c4 − 52a3 c3 − 2a3 c2 + 92a3 c − 14a2 c4 − 2a2 c3 +176a2 c2 − 212a2 c − 24a2 − 8ac4 + 92ac3 − 212ac2 + 88ac + 72a − 24c2 + 72c − 52)b +a4 c4 − 8a4 c3 − 5a4 c2 + 12a4 c − 8a3 c4 + 6a3 c3 + 44a3 c2 − 48a3 c − 8a3 −5a2 c4 + 44a2 c3 − 95a2 c2 + 38a2 c + 36a2 + 12ac4 − 48ac3 +38ac2 + 46ac − 52a − 8c3 + 36c2 − 52c + 24 = 0 With this we can show that b < 4 if a = 1 and c ≥ 2 or c = 1 and a ≥ 2; and b < 3 if a, c ≥ 2. From this we immediately get that if b = 3 then one of a, c is 1. Assuming b = 3 76
and c = 1 we can then use this with our equation for d above to show that d < 1 for all a ≥ 3. For a = 1 we get d = 35 , and for a = 2 we get d = 3. Next, if b = 1, then we can show that d < 2 for all a, c ≥ 1. Finally, if b = 2 we can show d ≤ 2 with d = 1 iff a = c = 2 and d = 2 iff a = 2 and c = 1.
With only these three possibilities we can directly calculate that no positive integer solutions for k, m exist.
P4 (a, b, c, d): Equation (4.36) gives 3(a + b + c + d) + 2abcd = 2(ac + ad + bd) + abc + bcd + 5 Solving for a gives a=
bcd + 2bd − 3b − 3c − 3d + 5 bcd − bc − 2c − 2d + 3
Equation (4.37) then reduces to (c4 d4 + 2c3 d3 − 4c3 d2 − 2c2 d + 3c2 )b4 +(−6c4 d3 − 4c4 d2 − 6c3 d4 + 8c3 d3 − 2c3 d2 +12c3 d + 6c3 − 4c2 d3 + 14c2 d2 − 2c2 d − 16c2 )b3 +(12c4 d2 + 14c4 d + 3c4 + 24c3 d3 − 22c3 d2 − 26c3 d − 28c3 + 12c2 d4 −36c2 d3 + 20c2 d2 − 20c2 d + 49c2 − 8cd2 + 24cd − 18c)b2 +(−8c4 d − 12c4 − 24c3 d2 + 20c3 d + 48c3 − 24c2 d3 +76c2 d2 − 20c2 d − 62c2 − 8cd4 + 44cd3 − 68cd2 + 12cd + 26c)b −8c3 − 24c2 d + 36c2 − 24cd2 + 72cd − 52c − 8d3 + 36d2 − 52d + 24 = 0 With this we can now show that b < 3 if d ≥ 2 and c ≥ 1. If d = 1 then b ∈ (−1, 1] ∪ (3, 4) ∪ (5, 6) for all c ≥ 5. For 1 ≤ c ≤ 4 we can directly show that b = 1 is the only integer solution. In fact b = 1 is always an integer solution with d = 1. Regardless, this would give P4 (a, 1, c, 1), which we know is DS. Now using our equation for a above we can show that with b = 1 that a ≤ 7 for all c ≥ 1 and d ≥ 2. With b = 2 we get a ≤ 8 for all c ≥ 1 and d ≥ 2. We can then calculate 77
the only possible combinations for (a, b, c, d) are (2, 1, 4, 8), (2, 1, 8, 4), (3, 1, 3, 11), (4, 1, 6, 4), (5, 1, 4, 5), (7, 1, 3, 7), (2, 2, 3, 8), (2, 2, 8, 3), (3, 2, 2, 8), (8, 2, 2, 3). Finally, with these we can directly compute that no solutions for k, m exist.
K1,3 (a, b, c, d): Equation (4.36) gives 3(a + b + c + d) + 2abcd = abc + abd + acd − bcd + 2(bc + bd + cd) + 5 Solving for a gives a=
−bcd + 2bc + 2bd − 3b + 2cd − 3c − 3d + 5 2bcd − bc − bd − cd + 3
From this we can show that a < 1 for all b, c, d ≥ 1.
C4 (a, b, c, d): Equation (4.36) gives 3(a + b + c + d) + 3abcd = 2(ac + bd) + abc + abd + acd + bcd + 5 Solving for a gives bcd + 2bd − 3b − 3c − 3d + 5 3bcd − bc − bd − cd − 2c + 3 With this we can show a < 2 for b ≥ 2 and c, d ≥ 1. If a = 1, then c = 1. Finally, if b = 1, a=
then a < 1 if d ≥ 2. These give C4 (1, b, 1, d) and C4 (a, 1, c, 1), respectively, which we know are not cospectral with P5 (k, 1, 1, m, 1).
P5 (a, b, c, d, e): Solving equation (4.36) for c gives c=
2ad + 2ae − 3a + 2bd + 2be − 3b − 3d − 3e + 5 abd + abe − ab + ade + ae − 2a + bde − bd − de − 2e + 3
With this we can show that c < 2 for all a, b, d, e ≥ 1 with c = 1 iff a, d ≥ 1 and b = e = 1 or a = d = 1 and b, e ≥ 1. These are isomorphic, so we may assume b = e = 1. Now k + m + 3 = a + d + 3 and 2m = 2d imply that a = k and d = m. These are isomorphic graphs. 78
C6 (a, b, c, d, e, f ): Using equation (4.36), we can show that a < 1 for all b, c, d, e, f ≥ 1.
Therefore we can conclude that P5 (k, 1, 1, m, 1) is DS for all k, m ≥ 1.
79
P5 (k, 1, 1, 1, m): d1 = k + m + 3 d2 = km + 2k + 2m + 1 d3 = −km + k + m + 1 d4 = 2km + k + m
3c1 = c2 + c3 + 7 c2 = c3 + c4
(4.38) (4.39)
P3 (a, b, c) + K1 (d): Equations (4.38) and (4.39) together reduce to (−a2 c2 + 5ac − 3)b2 + (−a2 c2 + 5a2 c + 5ac2 − 13ac − 6a − 6c + 7)b +2a2 c2 − 3a2 c − 3a2 − 3ac2 + 4ac + 7a − 3c2 + 7c = 0 With this we can show that b ∈ (−2, 1) ∪ (1, 2) if a ≥ 3 and c ≥ 2. If we assume a ≥ c (and we can due to symmetry), then we must have either c = 1 and a ≥ 1 or c = 2 and a = 2. If c = 1 then b ∈ (2, 3) for a ≥ 9. For a ≤ 8 we can calculate that the only integer solutions for b are b = 1, 8 at a = 1 and b = 8 at a = 5. These give d = 0, d = 38 , and d = − 35 , respectively. Now, for a = c = 2 we get b = 1 or b = 4, but then we get d = 0 or d = −3.
P4 (a, b, c, d): Equation (4.38) gives 3(a + b + c + d) = abc + ac + ad + bcd + bd + 7 Solving for d gives d=
−abc − ac + 3a + 3b + 3c − 7 a + bc + b − 3
From this we can show that d ≤ 1 with equality iff exactly two of a, b, c are 1. Regardless, we reduce to cases of P4 (a, b, c, d) that we have already shown to be DS.
80
K1,3 (a, b, c, d): In the P4 (k, 1, 1, m) case we already showed that equation (4.39) is never satisfied with a, b, c, d ≥ 1.
C4 (a, b, c, d): Similar to P4 (a, b, c, d) we can solve equation (4.38) for d to get d=
−abc − ac + 3a + 3b + 3c − 7 ab + ac + bc + b − 3
This time we get that d < 1 for all a, b, c ≥ 1.
P5 (a, b, c, d, e): Solving (4.38) for c gives c=
−ad − ae + 3a − bd − be + 3b + 3d + 3e − 7 ab − ae + a + bd + de + e − 3
We can then show that c ≤ 1 with equality iff b = d = 1. Finally, with k + m + 3 = a + e + 3 and km + 2k + 2m + 1 = ae + 2a + 2e + 1 we get k = a and m = e or k = e and m = a.
C6 (a, b, c, d, e, f ): Solving (4.38) for f gives f=
−abc + ace − ac − ad − ae + 3a − bcd − bd − be + 3b − cde − ce + 3c + 3d + 3e − 7 ab + ae − bd + b + c + de + d − 3
With this we have f < 1 for all a, b, c, d, e ≥ 1.
Therefore we may conclude that P5 (k, 1, 1, 1, m) is DS for all k, m ≥ 1.
81
P5 (1, k, m, 1, 1): d1 = k + m + 3 d2 = 2k + 2m + 2 d3 = 2km d4 = 3km + m
2c1 = c2 + 4 9c23 + 6c1 c3 + 4c24 + 12c4 = 4c1 c4 + 12c3 c4 + 16c3
(4.40) (4.41)
P3 (a, b, c) + K1 (d): Equation (4.40) gives 2(a + b + c + d) = ac + ad + bd + cd + 4 Solving for d gives d=
−ac + 2a + 2b + 2c − 4 a+b+c−2
From this we can show that d < 2 for all a, b, c ≥ 1. This would imply that d = 1. More importantly that means 0 ∈ Spec(P3 (a, b, c) + K1 (d)), but we can show that 0 ∈ Spec(P5 (1, k, m, 1, 1)) iff m = 1. For that we observe PP5 (1,k,m,1,1) (0) = −k(m − 1) We can now conclude that P3 (a, b, c) + K1 (d) is not cospectral with P5 (1, k, 1, 1, 1) from our previous results with P5 (k, m, 1, 1, 1).
P4 (a, b, c, d): Equation (4.40) gives 2(a + b + c + d) = ac + ad + bd + 4 Solving for d gives d=
−ac + 2a + 2b + 2c − 4 a+b−2 82
If a ≥ 3 and b, c ≥ 1, then d < 2 which means d = 1 is the only possibility. If a = 2 and b, c ≥ 1, then d = 2. The a = 1 case is symmetric with the d = 1 case from before.
For a = d = 2, equation (4.41) reduces to 2bc + b + c + 2 = 0 This is clearly not possible with b, c ≥ 1. On the other hand, if d = 1, then a=
b + 2c − 2 c−1
With this (4.41) reduces to −12b4 c2 + c(−60c2 + 56c + 13)b3 + c(−72c3 + 128c2 − 14c − 36)b2 +(−12c4 + 37c3 − 14c2 + 19c − 30)b + 18c3 + 42c2 − 138c + 78 = 0 With this we can now show that b < 1 for all c ≥ 1.
K1,3 (a, b, c, d): Equation (4.40) gives 2(a + b + c + d) = bc + bd + cd + 4 Solving for d gives d=
2a − bc + 2b + 2c − 4 b+c−2
Using this with equation (4.41) we can then show that b < 1 for all c ≥ 1.
C4 (a, b, c, d): Equation (4.40) gives 2(a + b + c + d) = ac + bd + 4 Solving for d gives d=
−ac + 2a + 2b + 2c − 4 b−2
Using this with equation (4.41) we can show that b ≤ 2 for a, c ≥ 1, with b = 2 iff a = c = 4 or a = 6 and c = 3. We can then show directly that no positive integer solutions exist for
83
k, m. For b = 1, we can show that the only positive solutions for c are in (0, 1) ∪ (1, 2) for a ≥ 3 and that no positive integer solutions exist for a = 1, 2.
P5 (a, b, c, d, e): Equation (4.40) gives 2(a + b + c + d + e) = ac + ad + ae + bd + be + ce + 4 Solving for a gives a=
−bd − be + 2b − ce + 2c + 2d + 2e − 4 c+d+e−2
From this we can show that a ≤ 1 for all b, c, d, e ≥ 1 with equality iff b = e = 1 or d = e = 1. These are isomorphic, so we may assume that a = d = e = 1. It is then straightforward to show that k = b and m = c.
C6 (a, b, c, d, e, f ): Equation (4.40) gives 2(a + b + c + d + e + f ) = ac + ad + ae + bd + be + bf + ce + cf + df + 4 Solving for a gives a=
−bd − be − bf + 2b − ce − cf + 2c − df + 2d + 2e + 2f − 4 c+d+e−2
From this we can show that a < 1 for all b, c, d, e, f ≥ 1.
Therefore we may conclude that P5 (1, k, m, 1, 1) is DS for all k, m ≥ 1.
84
P5 (1, k, 1, m, 1): d1 = k + m + 3 d2 = km + k + m + 3 d3 = km + k + m − 1 d4 = 2km + k + m c2 = c3 + 4 c1 + c4 + 3 = 2c2 c1 + c4 = 2c3 + 5
(4.42) (4.43) (4.44)
Note that 0 ∈ Spec(P5 (1, k, 1, m, 1)).
P3 (a, b, c) + K1 (d): For 0 ∈ Spec(P3 (a, b, c) + K1 (d)) we need either a = c = 1 or d = 1. If a = c = 1, then we have P3 (1, b, 1) + K1 (d), which we know is not cospectral with any P5 (a, b, c, d, e). Suppose then that d = 1. Equation (4.42) reduces to a + b + c + ac = abc − ac + 4 Solving for a gives b+c−4 bc − 2c − 1 From this we can show that a ≤ 1 for all b, c ≥ 1 unless b = 3 and c = 1. However, a=
in all cases we reduce to P3 (1, b, c) + K1 (1), which we know is not cospectral with any P5 (a, b, c, d, e).
P4 (a, b, c, d): Equations (4.42) and (4.44) together reduce to (b2 d2 − 2bd + b − 1)c2 + (b2 d2 + 2b2 d + b2 − 2bd2 − 4bd − 4b − 2d − 5)c − d2 − 5d − 4 = 0 With this we can show that c ∈ (1, 2) for b = 1 and d ≥ 22 and that c ∈ (0, 1) if b = 2 and d ≥ 5, b = 3 and d ≥ 2, and b ≥ 4 and d ≥ 1. For all other cases missed we can directly check that there are no positive integer roots for c. 85
K1,3 (a, b, c, d): Equations (4.42) and (4.43) reduce to (2c2 d2 + 2c2 d − 2c2 + 2cd2 − 4cd + c − 2d2 + d)b2 +(2c2 d2 − 4c2 d + c2 − 4cd2 − 4cd + 4c + d2 + 4d)b −2c2 d2 + c2 d + cd2 + 4cd − 4 = 0 With this we can show that b ≤ 1 for all c, d ≥ 1 with equality iff c = 1 or d = 1. In either case we have K1,3 (a, 1, 1, d), which we know is not cospectral with any P5 (a, b, c, d, e).
C4 (a, b, c, d): Equation (4.42) gives ac + bd = abc + abd + acd + bcd + 4 Solving for a gives a=
−bcd + bd − 4 bc + bd + cd − c
With this we can show that a < 1 for all b, c, d ≥ 1.
P5 (a, b, c, d, e): Solving equation (4.43) for c gives c=
2ad + 2ae − a + 2bd + 2be − b − d − e − 3 abd + abe + ade − 2a + bde − 2e + 1
With this we can show that c < 3 for all a, b, d, e ≥ 1. Moreover, we can show that c1 + c4 > 2c3 + 5 for all a, e ≥ 2 and b, c, d ≥ 1. Therefore we may assume e = 1. With this we can now use equations (4.42) and (4.43) to reduce to d(3a2 c2 + a2 cd − 2acd − 6ac − c2 d + 3cd − d + 3) = 0 When c = 1 this factors as (a − 1)2 d(d + 3) = 0 which has a = 1 as the only positive root. From here it is easily shown that either k = b and m = d or k = d and m = b. When c = 2 it is d(2a2 d + 12a2 − 4ad − 12a + d + 3) = 0, which only has (a, d) = (1, 3) as a positive integer solution. This then implies b = 0.
86
C6 (a, b, c, d, e, f ): c1 + c4 > 2c3 + 5 for all a, b, c, d, e, f ≥ 1. Therefore we may conclude that P5 (1, k, 1, m, 1) is DS for all k, m ≥ 1
87
4.7
C6(a, b, c, d, e, f )
C6 (k, m, 1, 1, 1, 1): d1 = k + m + 4 d2 = 3k + 3m + 3 d3 = 2km + 2 d4 = 5km + 3k + 3m + 1 3c1 = c2 + 9
(4.45)
5c3 + 2c2 = 2c4 + 14
(4.46)
5c3 + 6c1 = 2c4 + 16
(4.47)
0 is never in the spectrum of C6 (k, m, 1, 1, 1, 1), and 1 is in the spectrum iff k = 1 or m = 1. Finally, using the bounds on the eigenvalues, we can show that 1 ≤ λ3 < 1.5.
P3 (a, b, c) + K1 (d): Solving equation (4.45) for d gives d=
3a + 3b + 3c − ac − 9 a+b+c−3
From this we can show that d < 3 for all a, b, c ≥ 1. We know that 0 is never in the spectrum of C6 (k, m, 1, 1, 1, 1), so we can assume that d = 2 and m = 1. With m = 1, our system of equations simplifies greatly. We can deduce that 2c1 = c3 + 8. With this added relation we get the equation a2 c2 − a2 c − ac2 − ac − 2 = 0 This equation has no positive integer solutions.
P4 (a, b, c, d): Equations (4.45) and (4.46) reduce to a2 (2c2 d − 5c2 + 2cd2 − 11cd + 15c) +a(−11c2 d + 15c2 − 11cd2 + 48cd − 51c) +15c2 d + 15cd2 − 39cd + 6d2 − 32d + 42 = 0 88
From this we can show that if d = 1 then a ∈ (2, 3) for all c ≥ 2 and a = 1 if c = 1; if d = 2 then a ∈ (2, 3) for all c ≥ 1; if d = 3 then a ∈ {3} ∪ (15, 16) for all c ≥ 7, a = 21 if c = 1, a = 18 if c = 2, a = 17 if c = 3, and a = 16 if c = 6; if d = 4 then a ∈ (3, 4) ∪ (6, 7] for all c ≥ 1 with a = 7 iff c = 1; and if d ≥ 5 then a ∈ (3, 5) for all c ≥ 1, and a 6= 4 for any c. (This includes only positive real roots.)
Finally, having narrowed down the possibilities we can directly computer that no positive integer solutions exist.
K1,3 (a, b, c, d): Solving equations (4.45) and (4.47) for a and then setting those equal to one another gives b2 (4c2 d − 5c2 + 4cd2 − 22cd + 15c − 5d2 + 15d) +b(4c2 d2 − 22c2 d + 15c2 − 22cd2 + 96cd − 51c + 15d2 − 51d) −5c2 d2 + 15c2 d + 15cd2 − 51cd − 6 = 0 With this we can show that b ∈ (0, 7) (whenever b is positive and real) for all c ≥ d ≥ 1 with b = 4 iff c = d = 1 and b = 1 iff c = 4 and d = 1. In both cases we get that a = 0.
C4 (a, b, c, d): Assume a ≥ b, c, d and b ≥ d. Then using C4 (c, d, c, d) ≤ C4 (a, b, c, d) ≤ C4 (a, b, a, b) we get d − 1 ≤ λ3 ≤ b − 1. We also know that 1 ≤ λ3 < 1.5. Therefore d − 1 < 1.5 which implies d = 1, 2. With d = 1 equations (4.45) and (4.47) give 2b2 c − 10b2 − 2bc2 − 22bc + 36b − 21c2 + 40c + 6 =0 c−3 On the other hand, using (4.46) and (4.47) we get 2 (2b2 c − 10b2 − 2bc2 − 14bc − 4b − 21c2 − 42) =0 bc − 5b − 7c These together then simplify to bc − 5b − 5c − 6 = 0 which only has {b, c} = {6, 36} as positive integer solutions. With these we can directly compute that there are no solutions
89
for a, k, m. A similar argument for d = 2 leads to 7bc − 10b − 10c − 6 = 0 which has no positive integer solutions.
P5 (a, b, c, d, e): Assuming a ≥ e, similar to C4 (a, b, c, d) we can show that min{c, e}−1 ≤ λ3 ≤ max{a, c}−1. Since λ3 < 1.5 we must one of c, e in {1, 2}. Using these with equations (4.45), (4.46), and (4.47) we get, from c = 1, 2 and e = 1, 2 respectively, the equations 2bd + 2be − 5b + 2de + 5e − 6 = 0 2bd + 2be − 5b + 2de + 5e − 3 = 0 2bcd − 3bc + 2cd + 5c − 6 = 0 2bcd − bc + 4cd + 10c − 6 = 0 Finally, with these we get the possible combinations for (b, c, d, e) the following: (1, 1, 1, 1), (4, 1, 1, 1), (4, 2, 1, 1), (5, 3, 1, 1), (6, 6, 1, 1). From these we can directly compute that no integer solutions then exist for a, k, m.
C6 (a, b, c, d, e, f ): Equation (4.45) gives a=
−bd − be − bf + 3b − ce − cf + 3c − df + 3d + 3e + 3f − 9 c+d+e−3
From this we can show that a ≤ 1 with equality iff only at most two consecutive variables are greater than 1. That is C6 (a, b, c, d, e, f ) ∼ = C6 (k 0 , m0 , 1, 1, 1, 1). From here it is easy to show that k 0 = k and m0 = m.
Therefore we may conclude that C6 (k, m, 1, 1, 1, 1) is DS for all k, m ≥ 1.
90
C6 (k, 1, m, 1, 1, 1): d1 = k + m + 4 d2 = km + 2k + 2m + 4 d3 = 2k + 2m d4 = 5km + 3k + 3m + 1 2c1 = c3 + 8 10c2 = 7c3 + 2c4 + 38 7c1 + c4 = 5c2 + 9
(4.48) (4.49) (4.50)
Note that k ∈ Spec(C6 (k, 1, m, 1, 1, 1)) iff k = m; k + 1 ∈ Spec(C6 (k, 1, m, 1, 1, 1)) iff k = m = 1. We also have the bounds k ≤ λ1 ≤ k + 2, m ≤ λ2 ≤ m + 1, 1 ≤ λ3 < 1.25, and −4 < λn ≤ −2.
P3 (a, b, c) + K1 (d): Since d − 1 ∈ Spec(P3 (a, b, c) + K1 (d)) we must have k = m = d − 1 by the note above. With this we can already compare coefficients to see that there are no integer solutions for a, b, c, d.
P4 (a, b, c, d): From equation (4.48) we get a=
−bcd + 2b + 2c + 2d − 8 bc − 2
This gives that a ≥ 1 iff b = c = 1 and 1 ≤ d ≤ 3. Regardless we get P4 (a, 1, 1, d) which we know is DS.
K1,3 (a, b, c, d): If b ≥ c ≥ d, then we have d − 1 ≤ λ3 ≤ b − 1. Since 1 ≤ λ3 < 1.25, we can conclude that d = 1 or d = 2. Moreover, we can show that if d = 1, then 0 ≤ λ3 < 1. Therefore d = 2 is the only possibility. With this, equation (4.48) gives c=−
2(ab − a − b + 2) ab + 2a − 2b − 2 91
This gives that c 6≥ 2.
C4 (a, b, c, d): We may assume d = 1 or d = 2. With these we get, respectively, from equation (4.48) a=
−bc + 2b + 2c − 6 bc + b + c − 2
a=−
2(bc − b − c + 2) bc + 2b + 2c − 2
In both cases we have a < 1.
P5 (a, b, c, d, e): Assume a ≥ e. Since 1 ≤ λ3 < 1.25, we must have min{c, e} ≤ 2. Using equations (4.48) and (4.49) we can then show that the only possibilities for (b, c, d, e) are (1, 1, d, 1), (1, 1, 2, 3), (2, 1, 4, 10), and (1, 1, 4, 13). With (1, 1, d, 1) we get a = 1, but we know P5 (1, 1, 1, d, 1) is DS. With the other possibilities we have a 6∈ Z. C6 (a, b, c, d, e, f ): Since 1 ≤ λ3 < 1.25 we must have m1 + m2 − 1 < 1.25, where m1 = min{a, c, e} and m2 = min{b, d, f }. In particular this means m1 = m2 = 1. Suppose m2 = f = 1. Then equation (4.50) gives a=
−bcde − bcd − bde + 5bd + 5be − 2b − cde + 5ce − 2c − 2d − 7e + 2 bcd + bce + bc + bd + be + cde + ce − 5c + de − 5d − 5e + 7
With this we can then show that a < 2 for all b, c, d, e ≥ 1, and a = 1 iff b = d = 1 or c = e = 1. That is, we have C6 (1, 1, c, 1, e, 1) or C6 (1, b, 1, d, 1, 1), and from here is it straightforward to show {c, e} = {k, m} or {b, d} = {k, m}.
Therefore we may conclude that C6 (k, 1, m, 1, 1, 1) is DS for all k, m ≥ 1.
92
C6 (k, 1, 1, m, 1, 1): d1 = k + m + 4 d2 = km + 2k + 2m + 4 d3 = 2k + 2m d4 = 4km + 4k + 4m
2c1 = c3 + 8 4c1 + c4 = 4c2 4c2 = 2c3 + c4 + 16
(4.51) (4.52) (4.53)
Assume k ≥ m. We can bound the eigenvalues as k ≤ λ1 ≤ k + 1, m < λ2 < m + 1, λ3 = 1, and −3 < λn ≤ −2.
P3 (a, b, c) + K1 (d): Assume a ≥ c. −1 ∈ Spec(P3 (a, b, c) + K1 (d)) iff d = 2; a = c = 2; a = 5, b = 6, c = 1; a = 6, b = 4, c = 1; or a = 8, b = 3, c = 1. In all cases, for λ3 = 1 we must have d ≥ 2.
If d = 2, then a=
2(b + c − 2) bc − 2c − 2
We can then show that k, m ≥ 2 iff b = 3 and c ≥ 3. Now, if b = 3 then 2 < a ≤ 8 for all c ≥ 3 with a = 8 and c = 3 or a = 5 and c = 4 the only integral possibilities. From these it is easily shown that k, m 6∈ Z. If a = c = 2, then equation (4.51) gives b = 3d. Since b > d, this implies that λ2 = d − 1, but m < λ2 < m + 1 means that λ2 6∈ Z. Thus, this cannot happen. If a = 6, b = 4, and c = 1, then d = 45 . If a = 8, b = 3, and c = 1, then d = 54 . Finally, if a = 5, b = 6, and c = 1, then d = 2. This again gives that λ2 = 1 ∈ Z, which cannot happen.
93
P4 (a, b, c, d): From equation (4.51) we get a=
−bcd + 2b + 2c + 2d − 8 bc − 2
This gives that a ≥ 1 iff b = c = 1 and 1 ≤ d ≤ 3. Regardless we get P4 (a, 1, 1, d) which we know is DS.
K1,3 (a, b, c, d): If b ≥ c ≥ d, then we have d − 1 ≤ λ3 ≤ b − 1. Since λ3 = 1, we can conclude that d = 1 or d = 2. Moreover, we can show that if d = 1, then 0 ≤ λ3 < 1. Therefore d = 2 is the only possibility. With this, equation (4.51) gives c=−
2(ab − a − b + 2) ab + 2a − 2b − 2
This gives that c 6≥ 2.
C4 (a, b, c, d): We may assume d = 1 or d = 2. With these we get, respectively, from equation (4.51) a=
−bc + 2b + 2c − 6 bc + b + c − 2
a=−
2(bc − b − c + 2) bc + 2b + 2c − 2
In both cases we have a < 1.
P5 (a, b, c, d, e): Assume a ≥ e. For λ3 = 1 we must have min{c, e} = 1, 2. If either c or e is 2, then we must have a = c = e = 2. If c = 1 then a ≥ 4 and e ≥ 3. If e = 1 then a or c is at least 3. We can immediately check that with a = c = e = 2 there are no solutions for k, m.
Suppose c = 1. We can use equations (4.51) and (4.52) to show that a ≥ 4 and e ≥ 3 iff one of the following is true: b = 1, d ≥ 14 and e = 3; b = 1, 8 ≤ d ≤ 11, and e = 4; b = 6, d = 1, 94
and e = 3; or b = 7, d = 1, and e = 4. With only these possibilities we can immediately check that equations (4.51) and (4.52) are never simultaneously satisfied.
Now suppose e = 1. Requiring that no three of a, b, c, d, e are 1, we can use equations (4.51) and (4.52) to show that a ≥ 1 iff b = 1 and one of the following is true: c = 2 and d ≥ 2; c = 3 and 2 ≤ d ≤ 4; or d = 2 and 4 ≤ c ≤ 7. With only these possibilities we can immediately check that equations (4.51) and (4.52) are never simultaneously satisfied.
C6 (a, b, c, d, e, f ): For λ3 = 1 we need m1 = m2 = 1. Suppose f = 1. Then equation (4.52) gives a=
−bcde − bcd − bde + 4bd + 4be − cde + 4ce − 4e − 4 bcd + bce + bc + bd + be + cde + ce − 4c + de − 4d − 4e + 4
With this we can show that C6 (a, b, c, d, e, f ) ∼ = C6 (k 0 , 1, 1, m0 , 1, 1) with k 0 ≥ m0 , and then it is immediate that k = k 0 and m = m0 .
Therefore we may conclude that C6 (k, 1, 1, m, 1, 1) is DS for all k, m ≥ 1.
95
4.8
Coefficients
The coefficients presented here are for the characteristic polynomials arising from P3 and P4 , excluding 3K1 and 4K1 . They are factored as (x + 1)c1 −k H(x), where k = 3 or k = 4, depending on whether we are looking at P3 or P4 , and H(x) is x3 + (3 − c1 )x2 + (3 − 2c1 + c2 )x + (1 − c1 + c2 + c3 ) for P3 and x4 + (4 − c1 )x3 + (6 − 3c1 + c2 )x2 + (4 − 3c1 + 2c2 + c3 )x + (1 − c1 + c2 + c3 − c4 ) for P4 . The characteristic polynomials have been factored in this way to facilitate ease in comparing different graphs with one another. P31 (a, b, c) = a + b + c P32 (a, b, c) = ac P33 (a, b, c) = abc P3 + K11 (a, b, c, d) = a + b + c + d P3 + K12 (a, b, c, d) = ac + ad + bd + cd P3 + K13 (a, b, c, d) = abc − acd P3 + K14 (a, b, c, d) = abcd P41 (a, b, c, d) = a + b + c + d P42 (a, b, c, d) = ac + ad + bd P43 (a, b, c, d) = abc + bcd P44 (a, b, c, d) = abcd K1,31 (a, b, c, d) = a + b + c + d K1,32 (a, b, c, d) = bc + bd + cd K1,33 (a, b, c, d) = abc + abd + acd − bcd K1,34 (a, b, c, d) = 2abcd C41 (a, b, c, d) = a + b + c + d C42 (a, b, c, d) = ac + bd C43 (a, b, c, d) = abc + abd + acd + bcd C44 (a, b, c, d) = 3abcd 96
P51 (a, b, c, d, e) = a + b + c + d + e P52 (a, b, c, d, e) = ac + ad + ae + bd + be + ce P53 (a, b, c, d, e) = abc + bcd + cde − ace P54 (a, b, c, d, e) = abcd + abce + acde + bcde C61 (a, b, c, d, e, f ) = a + b + c + d + e + f C62 (a, b, c, d, e, f ) = ac + ad + ae + bd + be + bf + ce + cf + df C63 (a, b, c, d, e, f ) = abc + bcd + cde + def + aef + abf − ace − bdf C64 (a, b, c, d, e, f ) = abcd + abce + abcf + abdf + abef + acde + acef + adef + bcde + bcdf + bdef + cdef
97
4.9
Bounds on eigenvalues
The bounds in this section were found by exploiting symmetries in the graphs and using Theorem 2.1.4.
P3 (a, b, c) + K1 (d): d − 1 is always in Spec(P3 (a, b, c) + K1 (d)). With the assumption that a ≥ c, the other three eigenvalues can be bounded by √ c2 + 6cb + b2 + c + b − 2 ≤ λ01 ≤
1 2
√
1 2
a2 + 6ab + b2 + a + b − 2
c − 1 ≤ λ02 ≤ a − 1 √ √ − a2 + 6ab + b2 + a + b − 2 ≤ λn ≤ 21 − c2 + 6cb + b2 + c + b − 2
1 2
λ01 and λ02 have been used here as we do not necessarily know where d − 1 falls with respect to these.
P4 (a, b, c, d): Assume a ≥ d. Let M = max{b, c} and m = min{b, c}. Then √
1 ( 2
1 (− 2
√
d2 + 4m2 + d + 2m − 2) ≤ λ1 ≤ √ √ 1 d d + 4m + d − 2) ≤ λ2 ≤ ( 2
d2 + 4m2 + d + 2m − 2) ≤ λ3 ≤ √ √ 1 (− a a + 4M + a − 2) ≤ λn ≤ 2
√
1 ( a2 + 4M 2 + a + 2M − 2) 2 √ 1 √ a a + 4M + a − 2) ( 2 √ 1 (− a2 + 4M 2 + a + 2M − 2) 2 √ √ 1 (− d d + 4m + d − 2) 2
K1,3 (a, b, c, d): Assume b ≥ c ≥ d. Then √
1 ( 2
a2 + 10ad + d2 + a + d − 2) ≤ λ1 ≤
√
1 ( 2
a2 + 10ab + b2 + a + b − 2)
d − 1 ≤ λ2 ≤ b − 1 √
1 (− 2
d − 1 ≤ λ3 ≤ b − 1 √ a2 + 10ab + b2 + a + b − 2) ≤ λn ≤ 21 (− a2 + 10ad + d2 + a + d − 2)
98
C4 (a, b, c, d): Assume a ≥ b, c, d and b ≥ d. Then 1 ( 2
√ c2 + 14cd + d2 + c + d − 2) ≤ λ1 ≤
1 ( 2
√ a2 + 14ab + b2 + a + b − 2)
c − 1 ≤ λ2 ≤ a − 1 d − 1 ≤ λ3 ≤ b − 1 √ √ 1 (− a2 + 14ab + b2 + a + b − 2) ≤ λn ≤ 12 (− c2 + 14cd + d2 + c + d − 2) 2
P5 (a, b, c, d, e): Let M1 = max{a, c, e}, M2 = max{b, d}, m1 = min{a, c, e}, and m2 = min{b, d}. Then p f (m1 , m2 ) + m1 + m2 − 2) ≤ λ1 ≤
1 ( 2
p f (M1 , M2 ) + M1 + M2 − 2)
1 ( 2
m1 + m2 − 1 ≤ λ2 ≤ M1 + M2 − 1 m1 − 1 ≤ λ3 ≤ M1 − 1 p p 1 (− f (M1 , M2 ) + M1 + M2 − 2) ≤ λn ≤ 12 (− f (m1 , m2 ) + m1 + m2 − 2) 2 where f (x, y) = x2 + 10xy + y 2 .
C6 (a, b, c, d, e, f ): Let M1 = max{a, c, e}, M2 = max{b, d, f }, m1 = min{a, c, e}, and m2 = min{b, d, f }. Then p f (m1 , m2 ) + m1 + m2 − 2) ≤ λ1 ≤
1 ( 2
p f (M1 , M2 ) + M1 + M2 − 2)
1 ( 2
m1 + m2 − 1 ≤ λ2 ≤ M1 + M2 − 1 1 (− 2
m1 + m2 − 1 ≤ λ3 ≤ M1 + M2 − 1 p p f (M1 , M2 ) + M1 + M2 − 2) ≤ λn ≤ 12 (− f (m1 , m2 ) + m1 + m2 − 2)
where f (x, y) = x2 + 14xy + y 2 .
99
Chapter 5 λ=0 This chapter is broken down similarly to Chapter 4. We will begin with a section dedicated to Q0 , Q2 , and Q3 . The sections following it will focus on Q4 . For these graphs, we fix all but one split partition at size one, with a few exceptions. We also partition and analyze Q4 through bipartite and non-bipartite graphs. We do this because the spectrum of a graph can determine whether or not it is bipartite, and a split graph is bipartite if and on if it comes from a primitive bipartite graph.
5.1
Q0, Q2, and Q3
For Q0 , we get always get nK1 . Since the spectrum determines the number of vertices and edges (see Proposition 2.1.1), it is straightforward to see that all split graphs coming from Q0 are DS as there is exactly one graph with n vertices and 0 edges. Q2 and Q3 are not quite as simple to analyze as Q0 or even P1 , P2 , or P3 . Recall that Q2 = {K2 } and Q3 = {K3 }. Notice that Kn (a1 , . . . , an ; z) is precisely that complete multipartite graph Ka1 ,...,an with z isolated vertices. Unlike the complete graphs and union of complete graphs, we do not get that complete multipartite graphs are DS. The simplest example of this is K1,4 and C4 + K1 , as we saw in Chapter 2. Notice that C4 is K2,2 . In fact, we saw √ √ in Proposition 2.1.2 that Spec(Ka,b ) = {− ab, 0a+b−2 , ab}, which quickly leads us to the following proposition. 100
Proposition 5.1.1. Let a, b ≥ 1 and let n = ab. Then K2 (a, b; z) is DS iff n is prime or a and b are such that they minimize {x + y : x, y ≥ 1 and xy = n}. In the first case we can allow z ≥ 0. In the second case we need z to be less than a + b minus the second smallest element in {x + y : x, y ≥ 1 and xy = n}. Proof. Without loss of generality we may assume a ≥ b. Now, suppose that K2 (a, b; z) is cospectral with K2 (k, m; x) for some k, m ≥ 1 and x ≥ 0. We know that ab = km and a + b + z = k + m + x by equating their spectra. Now, if ab is prime, then clearly we get {a, b} = {k, m}, and thus the two graphs are isomorphic. Thus, K2 (a, b; z) is always DS when ab is prime. On the other hand, if ab is not prime, then from a + b + z = k + m + x we get that z = k + m + x − a − b ≥ 0. Now, if a + b = k + m, then {a, b} = {k, m}. Suppose then that a + b < k + m. Then we see that z is never less than x + y − a − b. If instead a + b > k + m, then z is always 0 or greater. Therefore, in this case K2 (a, b; z) is DS iff a + b < k + m and 0 ≤ z < k + m − a − b. The case for K3 (a, b, c; z) is not quite as simple. Some work has been done in the case of general complete multipartite graphs, but it is difficult to characterize DS graphs as definitively as in Proposition 5.1.1. However, we can show similarly that K3 (a, b, c; z) is DS when abc is prime or when a + b + c is minimal. Perhaps the most useful theorem about complete multipartite graphs is the characterization of their spectra as having only one positive eigenvalue. Theorem 5.1.2. G has exactly one positive eigenvalue iff it is the union of a complete multipartite graph and k isolated vertices with k ≥ 0. This result and more can be found in [8]. Theorem 5.1.2 will allow us to exclude examining K4 (a, b, c, d; z) in the rest of this chapter.
101
5.2
2P2(a, b, c, d; z)
2P2 (k, 1, 1, 1; z): d1 = k + z + 3 d2 = k + 1 d3 = 0 d4 = k
c2 = c4 + 1
(5.1)
2P2 (a, b, c, d; 0) : c2 = c4 + 1 gives ab + cd = abcd + 1. This can only happen if ab = 1 or cd = 1. Suppose c = d = 1. From here we can quickly solve for a, b ≥ 1, ab = k, and z = a + b − ab − 1 = −(a − 1)(b − 1). Since z ≤ 0, we can move the isolated points to 2P2 (a, b, c, d; z). Thus, we have 2P2 (a, b, 1, 1; z + (a − 1)(b − 1)) ∼ 2P2 (ab, 1, 1, 1; z) for all a, b ≥ 1 and z ≥ 0. In particular 2P2 (k, 1, 1, 1; z) has a cospectral mate for all even k ≥ 4.
P4 (a, b, c, d; 0) : We may assume a ≥ d. c2 = c4 + 1 gives ab + bc + cd = abcd + 1. From this we solve for a. a=
bc + cd − 1 b(cd − 1)
With a ≥ d ≥ 1 we can get that either d = 1 and b, c ≥ 1 or d = 2 and b = c = 1. Finally, from these we get P4 (3, 1, 2, 1; z+2) ∼ 2P2 (6, 1, 1, 1; z), P4 (2, 2, 3, 1; z+7) ∼ 2P2 (12, 1, 1, 1; z), and P4 (2, 1, 1, 2; z + 1) ∼ 2P2 (4, 1, 1, 1; z) for z ≥ 0.
P5 (a, b, c, d, e; 0) : Using the bounds on eigenvalues we know that λ2 = 1 and λn−1 = −1. These imply √
em ≤
1
≤
√
aM √ √ − aM ≤ −1 ≤ − em 102
Therefore aM = em = 1, and thus a = b = d = e = 1. Finally we can solve for k = 2c + 1, and thus P5 (1, 1, c, 1, 1; z + c) ∼ 2P2 (2c + 1, 1, 1, 1; z) for c ≥ 1 and z ≥ 0. This says that 2P2 (k, 1, 1, 1; z) has a cospectral mate for all odd k ≥ 3. Therefore we conclude that 2P2 (k, 1, 1, 1; z) is DS iff k = 2 and z ≥ 0.
103
5.3
P4(a, b, c, d; z)
P4 (k, 1, 1, 1; z): d1 = k + z + 3 d2 = k + 2 d3 = 0 d4 = k c2 = c4 + 2
(5.2)
2P2 (a, b, c, d; 0) : c2 = c4 + 2 gives ab + cd = abcd + 2, but for all a, b, c, d ≥ 1 we have ab + cd < abcd + 2.
P4 (a, b, c, d; 0) : We may assume a ≥ d. c2 = c4 + 2 gives ab + bc + cd = abcd + 2. From this we solve for a. bc + cd − 2 b(cd − 1) With a ≥ d ≥ 1 we can only have d = 1. With d = 1 we now have a=
a=
bc + c − 2 b(c − 1)
If we solve for b instead and let c = 1, then b = 1, which just gives P4 (a, 1, 1, 1; z) ∼ P4 (k, 1, 1, 1; z). So we may assume c ≥ 2. If we also assume b ≥ 2, then we 1 < a ≤ 2, with a = 2 iff c = 2. On the other hand, if b = 1, then a = 2 for all c ≥ 2. From these we can then solve for P4 (2, b, 2, 1; z +3b−2) ∼ P4 (4b, 1, 1, 1; z) and P4 (2, 1, c, 1; z +c−1) ∼ P4 (2c, 1, 1, 1; z) for all b, c ≥ 1 and z ≥ 0. In particular P4 (k, 1, 1, 1; z) has a cospectral mate for all even k ≥ 4.
P5 (a, b, c, d, e; 0) : √ Using the bounds on eigenvalues with the fact that λ2 = 2em ≤ k + 2 −
√ k+2− k2 +4 √ 2
√ k 2 + 4 ≤ 2aM
104
we get
√ k 2 + 4 ≤ 2aM , but for k ≥ 2 we have 1 < k + 2 − k 2 + 4 < 2. √ For k = 1 we have 2em < 2 − 3 < 1. Since 2em ≥ 2 we have a contradiction. Therefore 2em ≤ k + 2 −
√
Therefore we conclude that P4 (k, 1, 1, 1; z) is DS iff k = 2 or k ≥ 1 is odd and z ≥ 0.
105
P4 (1, k, 1, 1; z): d1 = k + z + 3 d2 = 2k + 1 d3 = 0 d4 = k
c2 = 2c4 + 1
(5.3)
2P2 (a, b, c, d; 0) : c2 = 2c4 +1 gives ab+cd = 2abcd+1, but for all a, b, c, d ≥ 1 we have ab+cd < 2abcd+1.
P4 (a, b, c, d; 0) : We may assume a ≥ d. Solving (5.3) for a, we get a=
bc + cd − 1 b(2cd − 1)
This implies that a ≤ 1, with a = 1 iff b ≥ 1 and c = d = 1 or c ≥ 1 and b = d = 1. In either case we have P4 (a, b, c, d; z) ∼ = P4 (1, k, 1, 1; z).
P5 (a, b, c, d, e; 0) : √
√ 2k+1− 4k2 +1
√ Similar to P4 (k, 1, 1, 1; z), with the bounds on the eigenvalues and using λ2 = 2 √ √ 2 2 we get that 2em ≤ 2k + 1 − 4k + 1 ≤ 2aM . However, 2k < 4k + 1 < 2k + 1, and thus √ 2em ≤ 2k + 1 − 4k 2 + 1 < 1 for all k ≥ 1, a contradiction.
Therefore we conclude that P4 (1, k, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0.
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5.4
P5(a, b, c, d, e; z)
P5 (k, 1, 1, 1, 1; z): d1 = k + z + 4 d2 = k + 3 d3 = 0 d4 = 2k + 1
2c2 = c4 + 5
(5.4)
2P2 (a, b, c, d; 0) : Equation (5.4) gives 2ab + 2cd = abcd + 5. We get equality iff ab = 3 and cd = 1 or ab = 1 and cd = 3. Assume a = 3 and b = c = d = 1. This gives k = 1, and thus 2P2 (3, 1, 1, 1; z) ∼ P5 (1, 1, 1, 1, 1; z + 1). This agrees with our analysis of 2P2 (k, 1, 1, 1; z).
P4 (a, b, c, d; 0) : From equation (5.4) we can solve for a=
2bc + 2cd − 5 b(cd − 2)
If we assume a ≥ d, then we can use this to show that d ≤ 3. If d = 3, then we must have a = 3 and b = c = 1. This leads to P4 (3, 1, 1, 3; z) ∼ P5 (4, 1, 1, 1, 1; z). If d = 2, then we must have c = 1, 2 and b ≥ 1, or c ≥ 3 and 1 ≤ b ≤ 3. If c ≥ 3, then 3 < a < 4 if b = 1 and 2 < a < 3 if b = 2. We can only have b = 3 if c = 3 in which case a=
13 . 4
If c = 2, then a =
7 2
if b = 1 and 1 < a < 2 if b ≥ 1. If c = 1, then using c2 = k + 3
and c4 = 2k + 1 we get that b = 12 . Finally we consider d = 1. If b = c = 1, then a = 1 and k = 0. If c = 2, then similar to above we can solve for b = 14 . With c ≥ 3 it can be shown that a ∈ (2, 7] with a = 3 when b = 3 and c = 13 or b = 9 and c = 7, a = 5 when b = 1 and c = 5, and a = 7 when 107
b = 1 and c = 3. These lead to P4 (3, 3, 13, 1; z + 42) ∼ P5 (58, 1, 1, 1, 1; z), P4 (3, 9, 7, 1; z + 78) ∼ P5 (94, 1, 1, 1, 1; z), P4 (5, 1, 5, 1; z + 4) ∼ P5 (12, 1, 1, 1, 1; z), and P4 (7, 1, 3, 1; z + 2) ∼ P5 (10, 1, 1, 1, 1; z).
P5 (a, b, c, d, e; 0) : √
√ k+3− k2 −2k+5
√ we get that 2em ≤ With the bounds on the eigenvalues and using λ2 = 2 √ k + 3 − k 2 − 2k + 5 < 4. This implies that 2em = 2, and thus e = m = 1. If b = 1 or
d = 1 we can show that 1 < a ≤ 3. With b = 1 we have a = 3 if d = 1 and c ≥ 1. With d = 1 we have a = 2 with b = c = 3 and a = 3 with b = 1 and c ≥ 1. With these we get P5 (3, 1, c, 1, 1; z+c−1) ∼ P5 (2c+1, 1, 1, 1, 1; z) and P5 (2, 3, 3, 1, 1; z+10) ∼ P5 (16, 1, 1, 1, 1; z). In particular P5 (k, 1, 1, 1, 1; z) has a cospectral mate for all odd k ≥ 3 and z ≥ 0. Therefore we may conclude that P5 (k, 1, 1, 1, 1; z) is DS iff k ≥ 2 is even except for k ∈ {4, 10, 12, 16, 58, 94} and z ≥ 0 or k = 1 and z = 0.
108
P5 (1, k, 1, 1, 1; z): d1 = k + z + 4 d2 = 2k + 2 d3 = 0 d4 = 3k
3c2 = 2c4 + 6
(5.5)
2P2 (a, b, c, d; 0): Equation (5.5) gives 3ab + 3cd = 2abcd + 6. We get equality iff ab = 3 and cd = 1 or ab = 1 and cd = 3. Assume a = 3 and b = c = d = 1. This gives k = 1, and thus 2P2 (3, 1, 1, 1; z) ∼ P5 (1, 1, 1, 1, 1; z + 1). This agrees with our analysis of 2P2 (k, 1, 1, 1; z).
P4 (a, b, c, d; 0): From equation (5.5) we can solve for a=
3(bc + cd − 2) b(2cd − 3)
If we assume a ≥ d, then we can use this to show that d ≤ 2. If d = 2, then we get a = 3, c = 1, b ≥ 1 as the only solution. This leads to P4 (3, b, 1, 2; z + b − 2) ∼ P5 (1, 2b, 1, 1, 1; z) for b ≥ 1 and z ≥ 0, with the one exception of b = 1 and z = 0.
If d = 1, then we get many solutions. The solutions for (a, b, c) are (6, b, 2), (4, 1, 3), (3, 2, 4),
109
(2, 2, 12), (2, 4, 18), (2, 6, 10), (2, 7, 9), (2, 9, 8), and (2, 15, 7). These give P4 (6, b, 2, 1; z + 3b − 5) ∼ P5 (1, 4b, 1, 1, 1; z) P4 (4, 1, 3, 1; z) ∼ P5 (1, 4, 1, 1, 1; z + 1) P4 (3, 2, 4, 1; z + 2) ∼ P5 (1, 8, 1, 1, 1; z) P4 (2, 4, 18, 1; z + 27) ∼ P5 (1, 48, 1, 1, 1; z) P4 (2, 2, 12, 1; z + 3) ∼ P5 (1, 16, 1, 1, 1; z) P4 (2, 6, 10, 1; z + 25) ∼ P5 (1, 40, 1, 1, 1; z) P4 (2, 7, 9, 1; z + 27) ∼ P5 (1, 42, 1, 1, 1; z) P4 (2, 9, 8, 1; z + 32) ∼ P5 (1, 48, 1, 1, 1; z) P4 (2, 15, 7, 1; z + 49) ∼ P5 (1, 70, 1, 1, 1; z) With these we see that P5 (1, k, 1, 1, 1; z) has a cospectral mate for all even k ≥ 2 and z ≥ 0, except when k = 2 and z = 0, and that there is much overlap when k is a multiple of 4.
P5 (a, b, c, d, e; 0): p √ With the bounds on the eigenvalues and using λ2 = k + 1 − k 2 − k + 1 we get that √ em ≤ k + 1 − k 2 − 2 + 1 < 2. This implies that em = 1, and thus e = m = 1. If b = 1 or d = 1 we can show that 1 ≤ a < 2. In either case we arrive at P5 (1, 3, c, 1, 1; z + c − 1) ∼ P5 (1, 2c + 1, 1, 1, 1; z). This is very similar to the P5 (k, 1, 1, 1, 1; z) case. In particular we see that P5 (1, k, 1, 1, 1; z) has a cospectral mate for all odd k ≥ 3 and z ≥ 0.
Therefore we conclude that P5 (1, k, 1, 1, 1; z) is DS iff k = 1, 2 and z = 0.
110
P5 (1, 1, k, 1, 1; z): d1 = k + z + 4 d2 = 2k + 2 d3 = 0 d4 = 2k + 1 c2 = c4 + 1
(5.6)
√ 1,1 Spec(P5 (1, 1, k, 1, 1; z)) = {0k+z , ±11,1 , ± 2k + 1 }.
2P2 (a, b, c, d; 0): Since we know Spec(2P2 (a, b, c, d; z)), we can directly compute c = d = 1 and ab = 2k + 1. Since ab is odd we need both a and b to be odd. Therefore, we have 2P2 (2a + 1, 2b + 1, 1, 1; z + 2ab − a − b) ∼ P5 (1, 1, 2ab + a + b, 1, 1; z) for all a ≥ 1, b ≥ 0, and z ≥ 0. In particular with b = 0, we get that P5 (1, 1, k, 1, 1; z) has a cospectral mate for all odd k ≥ 3 and z ≥ 0.
P4 (a, b, c, d; 0): From the bounds on eigenvalues, and using λ2 = 1, we get 1 √ 4dm + m2 − m ≤ 1 2 This reduces to dm − m − 1 ≤ 0 This is only true if d = 1 and m ≥ 1 or d = 2 and m = 1. We can then use PP4 (a,b,c,d;z) (1) = 0 to get abcd − ab − bc − cd + 1 = 0 This along with d = 1, 2 gives us the only possible solutions as (3, 1, 2, 1), (2, 2, 3, 1), and (2, 1, 1, 2). These give k = 52 , 11 , 3 , respectively, of which none can happen. 2 2 P5 (a, b, c, d, e; 0): 111
With the bounds on the eigenvalues and using λ2 = 1 we get that em ≤ 1, and thus e = m = 1. If d = 1 then PP5 (a,b,c,1,1;z) (1) = 0 gives c(ab − 1) = 0, and thus a = b = 1. On the other hand, if b = 1, then PP5 (a,1,c,d,1;z) (1) = 0 gives acd + ad − a − c − d + 1 = 0. This, too, can be shown to only be true when a = d = 1. In both cases we are led to P5 (1, 1, c, 1, 1; z) ∼ = P5 (1, 1, k, 1, 1; z). Therefore we conclude that P5 (1, 1, k, 1, 1; z) is DS iff k ≥ 2 is even and z ≥ 0.
112
5.5
Kite(a, b, c, d; z)
Kite(k, m, n, 1; z): d1 = k + m + n + z + 4 d2 = km + kn + mn + k d3 = kmn d4 = kmn
c3 = c4
(5.7)
It is easy to see that (5.7) is not satisfied for Bull, House, Broom, or P2 × K3 . Kite(a, b, c, d; z) to satisfy (5.7) then d = 1. Now, with d = 1 we can show that
1 4
For
< λ2 < 21 .
For this we look at 1 PKite(a,b,c,1;z) = 2−2a−2b−2c−2z−2 (128abc − 16ab − 16ac − 16a − 16bc + 1) 4 1 PKite(a,b,c,1;z) = −2−a−b−c−z−1 (4ab + 4ac + 4a + 4bc − 1) 2 From these it is easy to see that PKite(a,b,c,1;z) 41 > 0 and PKite(a,b,c,1;z) 12 < 0 for all a, b, c ≥ 1. Moreover, we can also show that λ2 < 31 iff b = c = 1. For this we look at 1 PKite(a,b,c,1;z) = 3−a−b−c−z−1 (27abc − 9ab − 9ac − 9a − 9bc + 1) 3 This is negative iff b = c = 1. These are enough to show that Kite(k, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0.
On the other hand, if we assume k = 1, then we can solve for p a + ab + ac + bc − abc − 1 ± (a + ab + ac + bc − abc − 1)2 − 4abc m, n = 2 With the assumption that b ≥ c, for these to be real and positive we need a = 1 and 1 ≤ c ≤ b; a ≥ 2, b ≥ 1, and c = 1; or a ≥ 9 and b = c = 2.
With the a = 1 case we get m = b and n = c, and thus Kite(1, b, c, 1; z) ∼ Kite(1, m, n, 1; z). 113
With the a ≥ 2, b ≥ 1, and c = 1 case we get √ 1 2 2 m, n = 2a + b − 1 ± 4a + b − 4a − 2b + 1 2 b 2
for b > 2. More precisely, n ≤ a − 1 b−1 2 for a ≤ ( 2 ) . for b ≤ a(a − 1), with n = a − 1 iff b = a(a − 1). Similarly n ≤ b−1 2 b−1 It is immediately clear that for given m, n, z, there are at most min{a − 1, 2 } possible From this we can quickly see that n < a and n
3c3 .
For Kite(a, b, c, d; 0) we quickly see from (5.11) that d = 3. Now assume m = 1. We can now solve ab + ac + bc + 3a = 3k + 2 abc = k to get k = 2a, b = 2, and c = 1. This shows that Kite(a, 2, 1, 3; z+a−2) ∼ Bull(2a, 1, 1, 1, 1; z) and Kite(1, 2, 1, 3; z) ∼ Bull(2, 1, 1, 1, 1; z + 1) for all a ≥ 2 and z ≥ 0. We can also show the stronger result that for all a, m ≥ 1 and z ≥ 0 we have Kite(a, 2m, 1, 3; z + a − m − 2) ∼ Bull(2a, m, 1, 1, 1; z) or Kite(a, 2m, 1, 3; z) ∼ Bull(2a, m, 1, 1, 1; z − a + m + 2), depending on whether a − m − 2 is positive or negative. Unfortunately this does not encompass all solutions as can be seen from Kite(6, 2, 2, 3; z + 1) ∼ Bull(8, 3, 1, 1, 1; z).
For Bull(a, b, c, d, e; 0) equation (5.11) gives c = d = e = 1 from which we quickly get Bull(a, b, c, d, e; z) ∼ = Bull(k, m, 1, 1, 1; z).
Finally, for House(a, b, c, d, e; 0) we get from equation (5.11) that a = c = 2 and d = e = 1. Solving for k, m, and b from here gives b = 21 (2z + 3), which is never an integer.
118
Therefore we conclude that Bull(2, 1, 1, 1, 1; 0) and Bull(k, 1, 1, 1, 1; z) are DS for all odd k ≥ 1 and z ≥ 0.
119
Bull(1, 1, k, 1, 1; z): d1 = k + z + 4 d2 = 2k + 3 d3 = k d4 = 2k + 1
c2 = 2c3 + 3
(5.12)
c4 = 2c3 + 1
(5.13)
c2 = c4 + 2
(5.14)
We saw previously that with Broom and P2 × K3 that c4 > 3c3 ≥ 2c3 + 1. We can also show that with House that c4 > 2c3 + 1.
For Kite(a, b, c, d; 0) we get abcd = 2abc + 1 from (5.13), which gives a = b = c = 1 and d = 3. However, with these we get c2 = 6 6= 5 = 2c3 + 3.
Finally, for Bull(a, b, c, d, e; 0) (5.13) gives abcd + abce + abde = 2abc + 1 This factors as ab(cd + ce + de − 2c) = 1 which implies ab = 1 and cd + ce + de − 2c = 1. The former implies a = b = 1 while the latter can be seen to give d = e = 1. Therefore c = k, and thus Bull(a, b, c, d, e; z) ∼ = Bull(1, 1, k, 1, 1; z). Therefore we conclude that Bull(1, 1, k, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0.
120
Bull(1, 1, 1, k, m; z): d1 = k + m + z + 3 d2 = k + m + 3 d3 = 1 d4 = km + k + m
c3 = 1
(5.15)
We can immediately rule out P2 × K3 from (5.15). For each of Kite, Bull, House, and Broom (5.15) forces a = b = c = 1.
From c2 = k + m + 3 and c4 = km + k + m we can solve for q 1 2 k, m = c2 − 3 ± c2 − 2c2 − 4c3 − 3 2 From this we get k, m = 0, d with Kite; k, m = d, e with Bull; and k, m = 0, 2de + d + e with House. Broom is explored in full detail below. In particular we will show that if m = 1, then no Broom is cospectral.
Therefore we conclude that Bull(1, 1, 1, k, 1; z) is DS for all k ≥ 1 and z ≥ 0.
From (5.15) we get for Broom(1, 1, 1, d, e, f ; 0) k, m =
p 1 de + d + e + f ± (de + d + e + f )2 − 4(def + de + df + ef ) 2
For this to be an integer it is clear that we need (de + d + e + f )2 − 4(def + de + df + ef ) = n2 for some n ∈ N0 . If we solve this for f we get f = de + d + e ±
√ n2 + 4de
We now need to solve for which n makes n2 + 4de a perfect square. First suppose n = 2r and n2 + 4de = (2s)2 . Then we get de = (r + s)(r − s). On the other hand, if n = 2r + 1 and n2 + 4de = (2s + 1)2 , then de = (r + s + 1)(r − s). We now see that r and s depend on how we can factor de. Suppose we factor de = pq. For n even, if we set r + s = p and r − s = q, 121
then we get r = then we get r =
p+q , and thus n 2 p+q−1 , and once 2
= p + q. For n odd, if we set r + s + 1 = p and r − s = q, again n = p + q. Finally, we arrive at f = de + d + e ± (p + q)
where de = pq. This gives us the following proposition:
Proposition 5.6.1. Bull(1, 1, 1, k, m; 0) is cospectral with Broom(1, 1, 1, d, e, f ; z) iff d = d1 d2 , e = e1 e2 , and f = de + d + e ± (d1 e1 + d2 e2 ) in which case k = de + d + e ± d1 e1 , m = de + d + e ± d2 e2 and z = de. From this proposition we quickly see that k, m ≥ 2 since di ei ≤ de. As an example of this proposition in action consider d = 6 and e = 2. One possible way to factor de = 12 is d1 = 2, d2 = 3, e1 = 2, and e2 = 1. Taking the negatives above we are led to f = 12 + 6 + 2 − (2 · 3 + 2 · 1) = 12, k = 12 + 6 + 2 − 2 · 3 = 14, m = 12 + 6 + 2 − 2 · 1 = 18, and z = 12. The full list for d = 6 and e = 2 is: Bull(1, 1, 1, 19, 8; 0) ∼ Broom(1, 1, 1, 6, 2, 7; 12) Bull(1, 1, 1, 18, 14; 0) ∼ Broom(1, 1, 1, 6, 2, 12; 12) Bull(1, 1, 1, 17, 16; 0) ∼ Broom(1, 1, 1, 6, 2, 13; 12) Bull(1, 1, 1, 24, 23; 0) ∼ Broom(1, 1, 1, 6, 2, 27; 12) Bull(1, 1, 1, 26, 22; 0) ∼ Broom(1, 1, 1, 6, 2, 28; 12) Bull(1, 1, 1, 32, 21; 0) ∼ Broom(1, 1, 1, 6, 2, 33; 12)
122
5.7
House(a, b, c, d, e; z)
House(k, 1, 1, 1, 1; z): d1 = k + z + 4 d2 = 3k + 3 d3 = k d4 = 3k + 1
c4 = 3c3 + 1
(5.16)
We will use (5.16) to quickly show that House(k, 1, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0. Starting with Kite, (5.16) gives abcd = 3abc + 1 This gives abc(d − 3) = 1, and thus a = b = c = 1, and d = 4. We have already shown above that Kite(1, 1, 1, 4; z) is DS for all z ≥ 0. Continuing with Bull, we get abcd + abce + abde = 3abc + 1 This gives ab(cd + ce + de − 3c) = 1, and thus a = b = 1 and cd + ce + de − 3c = 1. If d > 1 or e > 1, then cd + ce ≥ 3c and de > 1, so we require d = e = 1. This then forces c = 0. Next, with House we get abcd + abce + abde + bcde = 3abc + 1 This gives b(acd + ace + ade + cde − 3ac) = 1. Similar to Bull above we are forced to have b = d = e = 1 which leaves a + c − ac = 1. This is true iff a or c is 1. Since we may assume a ≥ c, we can set c = 1, and we are left with a = k, and thus House(a, b, c, d, e; z) ∼ = House(k, 1, 1, 1, 1; z). Lastly we will look at Broom and P2 × K3 . With Broom we have abcd + abce + abcf ≥ 3c for all d, e, f ≥ 1, and thus c4 > 3c3 + 1. We can show this true with P2 × K3 as well with a similar argument. Therefore we conclude that House(k, 1, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0.
123
House(1, k, 1, 1, 1; z): d1 = k + z + 4 d2 = 2k + 4 d3 = k d4 = 4k
c2 = 2c3 + 4
(5.17)
c4 = 4c3
(5.18)
Starting with Kite(a, b, c, d; 0), (5.18) immediately gives d = 4. Using this with (5.17) gives ab + ac + bc + 4a = 2abc + 4 The solutions to this are a = 2, b = 6, and c = 1 or a ≥ 1 and b = c = 2. Thus, Kite(2, 6, 1, 4; z + 3) ∼ House(1, 12, 1, 1, 1; z), Kite(1, 2, 2, 4; z) ∼ House(1, 4, 1, 1, 1; z + 1), and Kite(a, 2, 2, 4; z + 3a − 4) ∼ House(1, 4a, 1, 1, 1; z) for all a ≥ 2 and z ≥ 0.
Next with Bull(a, b, c, d, e; 0), (5.18) gives a, b ≥ 1, c = d = 2, and e = 1. With these (5.17) becomes 3ab − 4a − 3b + 4 = (a − 1)(3b − 4) = 0 Clearly a = 1, and we get Bull(1, 1, 2, 2, 1; z) ∼ House(1, 2, 1, 1, 1; z+1) and Bull(1, b, 2, 2, 1; z+ b − 2) ∼ House(1, 2b, 1, 1, 1; z) for b ≥ 2 and z ≥ 0.
Continuing with House(a, b, c, d, e; 0), (5.18) gives d + e < 4. If we assume d ≥ e, then we have either d = 2 and e = 1 or d = e = 1. With the former (5.18) becomes ac − 2a − 2c = 0 which has solutions a = 3, c = 6; a = c = 4; and a = 6, c = 3. Solving for b in (5.17) gives 31 13 , 27 12
and
28 , 27
respectively. On the other hand, with d = e = 1, (5.18) becomes 2ac − a − c = 0
124
which only has a = c = 1 as a solution. We can then solve for b = k, and thus we get House(a, b, c, d, e; z) ∼ = House(1, k, 1, 1, 1; z).
In a similar fashion as above we get for Broom(a, b, c, d, e, f ; 0) that a = 7, c = 3, d = e = f = 1, and b =
28 . 31
For P2 × K3 we have no simultaneous solutions to (5.17) and (5.18).
Therefore we conclude that House(1, k, 1, 1, 1; z) is DS for all odd k ≥ 1 and z ≥ 0 or when k = 2 and z = 0.
125
House(1, 1, 1, k, m; z): d1 = k + m + z + 3 d2 = km + k + m + 3 d3 = 1 d4 = 2km + k + m
c3 = 1
(5.19)
We saw above that c3 = 1 forces a = b = c = 1 in Kite, Bull, House, and Broom, and that c3 = 1 is not possible with P2 × K3 . Solving for k, m we get p 1 2 2c2 − c4 − 6 ± (2c2 − c4 − 6) + 4(c2 − c4 − 3) k, m = 2 From this we can quickly calculate k = d, m = 0 with Kite(1, 1, 1, d; z) and k = d, m = e with House(1, 1, 1, d, e; 0). With Bull and Broom we get that m < 0, when it is real, for all d, e, f ≥ 1.
Therefore we conclude that House(1, 1, 1, k, m; z) is DS for all k, m ≥ 1 and z ≥ 0.
126
5.8
Broom(a, b, c, d, e, f ; z)
Broom(k, 1, m, 1, 1, 1; z): d1 = k + m + z + 4 d2 = km + 2k + 2m + 2 d3 = km d4 = 3km + 2k + 2m + 1
c2 + 2c3 = c4 + 1
k, m =
(5.20)
p 1 c2 − c3 − 2 ± (c2 − c3 − 2)2 − 16c3 4
Starting with Kite(a, b, c, d; 0), we will first show c = 1. From (5.20) we solve for a a=
bc − 1 bcd − 2bc − b − c − d
For a ≥ 1 and k, m to be real we get that one of c = 1, if we assume b ≥ c. Moreover, for k, m to be real we also get b ≥ 6 and d = 4; b = 4, 5 and d = 5; b = 3 and d = 6; or b = 2 and d = 8. Checking these possibilities, we get the only positive integer solutions for a, k, m are with Kite(1, 3, 1, 6; z) ∼ Broom(3, 1, 1, 1, 1, 1; z + 3) and Kite(3, 7, 1, 4; z) ∼ Broom(7, 1, 3, 1, 1, 1; z + 1).
Next, for Bull(a, b, c, d, e; 0) to satisfy (5.20) we get the possibilities a = b = c = e = 1 and d = 4; a = 3, b = d = 2 and c = e = 1; a = c = 2, b = e = 1, and d = 3; or a = b = 1, d = e = 2, and c ≥ 1. Of these we can only have the last for k, m to be real and positive. Moreover, we have k, m =
√ 1 c + 3 ± c2 − 10c + 9 4
in this case. From here we can show that c2 − 10c + 9 is a perfect square iff c ∈ {1, 9, 10}, and of these c = 10 gives a rational value for m. Thus, we have Bull(1, 1, 1, 2, 2; z) ∼ Broom(1, 1, 1, 1, 1, 1; z + 1) and Bull(1, 1, 9, 2, 2; z) ∼ Broom(3, 1, 3, 1, 1, 1; z + 5).
127
House(a, b, c, d, e; 0) is simpler. (5.20) is only satisfied with a = b = c = e = 1 and d = 4 √ √ or a = b = c = 1 and d = e = 2. These give k, m = 14 (9 ± 65) and k, m = 2 ± 3, respectively.
For Broom(a, b, c, d, e, f ; 0) we start by showing b = 1. First, we observe that c4 + 1 grows faster than c2 + 2c3 . Next, if b = 2, then we have 5ac + ae + 2a + cd + 2c + de + 2f = 2(acd + ace + acf + ade + aef + cde + cdf + def ) + 1 One can readily check this has no positive integer solutions. Thus, b = 1 is the only possibility. With b = 1 we get instead 3ac + ae + a + cd + c + de + f = acd + ace + acf + ade + aef + cde + cdf + def + 1 In similar fashion we can now get that d = e = f = 1 is the only possibility, and with those we are forced to have a = k and c = m. Thus, Broom(a, b, c, d, e, f ; 0) ∼ Broom(k, 1, m, 1, 1, 1; z).
Finally, using the same idea as with Broom, we have c4 + 1 growing faster than c2 + 2c3 , and we can quickly check that a = b = c = d = e = f = 1 is the only solution to (5.20). With √ these we get k, m = 41 (5 ± i 7). Therefore we conclude that Broom(k, 1, m, 1, 1, 1; z) is DS for all k, m ≥ 1 and z ≥ 0 with the four exceptions Broom(3, 1, 1, 1, 1, 1; z+3), Broom(7, 1, 3, 1, 1, 1; z+1), Broom(1, 1, 1, 1, 1, 1; z+ 1), and Broom(3, 1, 3, 1, 1, 1; z + 5).
128
Broom(1, k, 1, 1, 1, 1; z): d1 = k + z + 5 d2 = 3k + 4 d3 = k d4 = 8k c2 = 3c3 + 4
(5.21)
c4 = 8c3
(5.22)
Starting with Kite we get from (5.22) that d = 8. Using this in (5.21) we get 3abc + 4 = ab + ac + bc + 8a From this we can solve for a = b = 1, c = 5 or a ≥ 1, b = c = 2, which gives Kite(1, 1, 5, 8; z) ∼ Broom(1, 5, 1, 1, 1, 1; z + 5) Kite(a, 2, 2, 8; z + 3a − 7) ∼ Broom(1, 4a, 1, 1, 1, 1; z) For Bull we can check with (5.22) that d, e ≤ 6. With this we can now see from (5.21) that the only possible solutions are a = b = c = 1, d = e = 2; a = b = d = e = 2, b = 1; and a = 1, b ≥ 1, c = d = 2, e = 3. These then give Bull(1, 1, 1, 2, 2; z) ∼ Broom(1, 1, 1, 1, 1, 1; z + 1) Bull(2, 2, 1, 2, 2; z) ∼ Broom(1, 4, 1, 1, 1, 1; z) Bull(1, b, 2, 2, 3; z + b − 3) ∼ Broom(1, 4a, 1, 1, 1, 1; z) Similar to Bull, (5.22) gives d, e ≤ 6 for House. With these we can then see that (5.21) is never satisfied. Just as with Bull and House, (5.22) gives us e, f ≤ 5. We can then use these we get (5.21) is only satisfied when a = c = d = e = f = 1, b ≥ 1, which gives us a graph isomorphic to Broom(1, k, 1, 1, 1, 1; z). P2 × K3 requires a bit more work. From (5.22) we can solve for e. Doing so gives us e=
8bcd − abcd − abdf − acdf − bcdf abc + abf + acd + acf + adf + bcd + bcf + bdf − 8af 129
If we assume a ≥ e ≥ f , then we can show that if a, f ≥ 3 then we must have b + c + d < 8. To see this we can rewrite e ≥ f to get (8a − (ab + ac + ad + bc + bd))f 2 + 8bcd − (abcd + abcf + abdf + 2acdf + 2bcdf ) ≥ 0 Under the assumption that a, f ≥ 3, we have abcd + 2bcdf ≥ 3bcd + 6bcd > 8bcd. This now forces (8a − (ab + ac + ad + bc + bd))f 2 = (a(8 − b − c − d) − bc − bd)f 2 > 0. It is straightforward to see now that b + c + d < 8 must be true. With these we can now check each case where b + c + d < 8, and in doing so we can see that (5.21) and (5.22) are never simultaneously satisfied. Finally, we have to check separately the cases where a = 1 and a = 2. With these we get that (5.21) and (5.22) are simultaneously satisfied only when a = b = 2 and c = d = e = f = 1. This gives P2 × K3 (2, 2, 1, 1, 1, 1; z + 1) ∼ Broom(1, 4, 1, 1, 1, 1; z) Therefore, we may conclude that Broom(1, k, 1, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0 except for when k is a multiple of 4 or the two cases when k = 1, z ≥ 1 and k = 5, z ≥ 5.
130
Broom(1, 1, 1, k, m, 1; z): d1 = k + m + z + 4 d2 = km + k + m + 4 d3 = 1 d4 = 3km + 2k + 2m + 1
c3 = 1
(5.23)
Using c2 = km + k + m + 4 and c4 = 3km + 2k + 2m + 1 we can solve for p 1 2 3c2 − c4 ± (−3c2 + c4 + 11) − 4 (−2c2 + c4 + 7) − 11 2
(5.24)
We saw with Bull(1, 1, 1, k, m) that c3 = 1 implies a = b = c = 1 in Kite, Bull, House, and Broom, and that c3 = 1 is not possible with P2 × K3 . Using (5.24) we can quickly verify that the only possible non-isomorphic solution is Bull(1, 1, 1, 2, 2; z) ∼ Broom(1, 1, 1, 1, 1, 1; z + 1) Therefore, we may conclude that Broom(1, 1, 1, k, m, 1; z) is DS for all k, m ≥ 1 and z ≥ 0 except for when k = m = 1 and z ≥ 1.
131
Broom(1, 1, 1, 1, 1, k; z): d1 = k + z + 5 d2 = k + 6 d3 = 1 d4 = 4k + 4
c3 = 1
(5.25)
c4 = 4c2 − 20
(5.26)
As with our last case we must have a = b = c = 1 in Kite, Bull, House, and Broom, and c3 = 1 never happens with P2 × K3 . We can now very quickly check with (5.26) that the only non-isomorphic solutions are Bull(1, 1, 1, 2, 2; z) ∼ Broom(1, 1, 1, 1, 1, 1; z + 1) Bull(1, 1, 1, 4, 4; z) ∼ Broom(1, 1, 1, 1, 1, 5; z + 1) Therefore, we may conclude that Broom(1, 1, 1, 1, 1, k; z) is DS for all k ≥ 1 and z ≥ 0 except for k = 1, 5 and z ≥ 1.
132
5.9
P2 × K3(a, b, c, d, e, f ; z)
P2 × K3 (k, 1, 1, 1, 1, 1; z): d1 = k + z + 5 d2 = 3k + 6 d3 = k + 1 d4 = 8k + 4
c2 = 3c3 + 3
(5.27)
c4 = 8c3 − 4
(5.28)
Using (5.28) with Kite we can solve for a to get a=
4 bc(d − 8)
We immediately see that a ∈ {1, 2, 4}, d < 8 and bc(d − 8) divides 4. With this we can quickly check that the only solution that also satisfies (5.27) is a = 1, b = c = 2, and d = 7, which gives Kite(1, 2, 2, 7; z) ∼ P2 × K3 (3, 1, 1, 1, 1, 1; z + 4) Similarly, with (5.28) gives a=
4 b(8c − cd − ce − de)
for Bull. This time we can check that the only solution that works with (5.27) is a = b = 1 and c = d = e = 2. This gives Bull(1, 1, 2, 2, 2; z) ∼ P2 × K3 (1, 1, 1, 1, 1, 1; z + 2)
Next, with House we get a=
bcde + 4 b(8c − cd − ce − de)
from (5.28). If we want this to be at least 1, then we get the inequality 8bc − bcd − bce − bde − bcde − 4 = bc(8 − d − e − de) − bde − 4 ≥ 0 133
From this it is clear that d + e + de < 8. d, e ≤ 3 guarantees this. From here we can quickly check that the only solution to (5.27) is a = b = c = d = e = 1. However, this leads to k = 0, which cannot happen. Broom is slightly more complicated. For Broom (5.28) gives f=
8abc − abcd − abce − abde − bcde − 4 b(a + d)(c + e)
For this to be positive we clearly need 8abc − abcd − abce − abde − bcde − 4 > 0. Notice that if d + e ≥ 8 then 8abc − abcd − abcd < 0. Therefore we must have d + e < 8. With this we can then check each such case with (5.27). Doing so shows that no solution exists. Finally with P2 × K3 we have a similar situation as Broom with (5.28). a=
8bcd − bcde − bcdf − bcef − bdef − 4 bcd + bce + bdf + bef + cde + cdf + cef + def − 8ef
The numerator is similar, but we have to be careful with the denominator. This time we can see that if e + f ≥ 8 and b + c + d ≥ 8 then a ≤ 0. Therefore at least one of e + f and b + c + d must be less than 8. This gives a large number of possibilities, but we can check each one with (5.27) and see that the only possible solutions are isomorphic to P2 × K3 (k, 1, 1, 1, 1, 1; z). Therefore, we may conclude that P2 × K3 (k, 1, 1, 1, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0 except for k = 1, z ≥ 2 and k = 3, z ≥ 4.
134
5.10
Coefficients
Similar to λ = −1, we can factor the characteristic polynomials arising from Q4 as xc1 −4 H(x), where H(x) is x4 − c2 x2 + 2c3 x + c4 . P41 (a, b, c, d; z) = a + b + c + d + z P42 (a, b, c, d; z) = ab + bc + cd P43 (a, b, c, d; z) = 0 P44 (a, b, c, d; z) = abcd 2P21 (a, b, c, d; z) = a + b + c + d + z 2P22 (a, b, c, d; z) = ab + cd 2P23 (a, b, c, d; z) = 0 2P24 (a, b, c, d; z) = abcd P51 (a, b, c, d, e; z) = a + b + c + d + e + z P52 (a, b, c, d, e; z) = ab + bc + cd + de P53 (a, b, c, d, e; z) = 0 P54 (a, b, c, d, e; z) = abcd + abde + bcde Kite1 (a, b, c, d; z) = a + b + c + d + z Kite2 (a, b, c, d; z) = ab + ac + ad + bc Kite3 (a, b, c, d; z) = abc Kite4 (a, b, c, d; z) = abcd Bull1 (a, b, c, d, e; z) = a + b + c + d + e + z Bull2 (a, b, c, d, e; z) = ab + ac + ad + bc + be Bull3 (a, b, c, d, e; z) = abc Bull4 (a, b, c, d, e; z) = abcd + abce + abde House1 (a, b, c, d, e; z) = a + b + c + d + e + z House2 (a, b, c, d, e; z) = ab + ac + ae + bc + cd + de House3 (a, b, c, d, e; z) = abc House4 (a, b, c, d, e; z) = abcd + abce + abde + bcde
135
Broom1 (a, b, c, d, e, f ; z) = a + b + c + d + e + f + z Broom2 (a, b, c, d, e, f ; z) = ab + ac + ae + bc + bf + cd + de Broom3 (a, b, c, d, e, f ; z) = abc Broom4 (a, b, c, d, e, f ; z) = abcd + abce + abcf + abde + abef + bcde + bcdf + bdef P2 × K31 (a, b, c, d, e, f ; z) = a + b + c + d + e + f + z P2 × K32 (a, b, c, d, e, f ; z) = ab + ae + af + bc + bd + cd + cf + de + ef P2 × K33 (a, b, c, d, e, f ; z) = aef + bcd P2 × K34 (a, b, c, d, e, f ; z) = abcd + abce + abdf + abef + acde + acdf + acef + adef + bcde + bcdf + bcef + bdef
136
5.11
Bounds on eigenvalues
As with the bounds for P4 , the bounds for Q4 are found by exploiting symmetries in the graphs and applying Theorem 2.1.4.
2P2 (a, b, c, d; z): √ 1,1 √ 1,1 Spec(2P2 (a, b, c, d; z)) = {0a+b+c+d+z−4 , ± ab , ± cd }.
P4 (a, b, c, d; z): Assume a ≥ d. Let M = max{b, c} and m = min{b, c}. Then √
1 ( 4dm 2 √ 1 ( 4dm 2
+ m2 + m) ≤
λ1
≤
+ m2 − m) ≤ λ2 ≤ √ 1 (M − 4aM + M 2 ) ≤ λn−1 ≤ 2 √ 1 4aM + M 2 ) ≤ λn ≤ (−M − 2
√
1 ( 4aM + M 2 + M ) 2 √ 1 ( 4aM + M 2 − M ) 2 √ 1 (m − 4dm + m2 ) 2 √ 1 (−m − 4dm + m2 ) 2
P5 (a, b, c, d, e; z): Assume a ≥ e. Let M = max{b, d} and m = min{b, d}. Then p m(e + 2c) √ em √ − aM p − M (a + 2c)
≤
λ1
≤
≤
λ2
≤
≤ λn−1 ≤
λn
p √
M (a + 2c)
aM √ ≤ − em p ≤ − m(e + 2c)
Kite(a, b, c, d; z): Assume b ≥ c. Then −b ≤ λn−1 ≤ −c, −b ≤ λn ≤ −c, or both.
137
Bull(a, b, c, d, e; z): Assume a ≥ b. Let M = max{d, e} and p 1 ( b(b + 8c + 4m) + b) 2 p 1 ( b(b + 4m) − b) 2 p 1 (− a(a + 4M ) − a) 2 p 1 (− a(a + 8c + 4M ) + a) 2
m = min{d, e}. Then p ≤ λ1 ≤ 21 ( a(a + 8c + 4M ) + a) p ≤ λ2 ≤ 21 ( a(a + 4M ) − a) p ≤ λn−1 ≤ 21 (− b(b + 4m) − b) p ≤ λn ≤ 21 (− b(b + 8c + 4m) + b)
House(a, b, c, d, e; z): Assume a ≥ c. Let M = max{d, e} and m = min{d, e}. Then −(a + M ) ≤ λn−1 ≤ −(c + m), −(a + M ) ≤ λn ≤ −(c + m), or both. Broom(a, b, c, d, e, f ; z): Let M = max{a, b, c, d, e, f } and m = min{a, b, c, d, e, f }. Then √ √ 1 1 (m + 17m) ≤ λ ≤ (M + 17M ) 1 2 2 m ≤ λ2 ≤ M √ √ 1 1 (M − (m − 17M ) ≤ λ ≤ 17m) n−1 2 2 −2M ≤
λn
≤ −2m
Moreover, if a ≥ c, M 0 = max{d, e}, and m0 = min{d, e}, then −(a+M 0 ) ≤ λn−1 ≤ −(c+m0 ), −(a + M 0 ) ≤ λn ≤ −(c + m0 ), or both. P2 × K3 (a, b, c, d, e, f ; z): Let M1 = max{a, d}, M2 = max{b, e}, M3 = max{c, f }, m1 = min{a, d}, m2 = min{b, e}, and m3 = min{c, f }. Then √ √ 1 ( 4m1 + m3 4m2 + m3 + m3 ) 2 √ 1 √ ( m3 4m1 + 4m2 + m3 − m3 ) 2 √ √ 1 (− 4M1 + M3 4M2 + M3 + M3 ) 2 √ √ 1 (− M3 4M1 + 4M2 + M3 − M3 ) 2
≤
λ1
≤
≤
λ2
≤
≤ λn−1 ≤ ≤
λn
≤
√
√
1 ( 4M1 + M3 4M2 + M3 + M3 ) 2 √ √ 1 ( M3 4M1 + 4M2 + M3 − M3 ) 2 √ √ 1 (− 4m1 + m3 4m2 + m3 + m3 ) 2
√ 1 (− m3 2
√ 4m1 + 4m2 + m3 − m3 )
Moreover, if M10 = max{a, b}, M20 = max{d, e}, M30 = max{c, f }, m01 = min{a, b}, m02 = min{d, e}, and m03 = min{c, f }, then m01 + m02 + m03 ≤ λ1 ≤ M10 + M20 + M30 . 138
Chapter 6 Conclusion As we have seen in this thesis it is possible to study graphs with eigenvalues of −1 or 0 of high multiplicity in a simplified manner by first examining a representive class of primitive graphs. We were not able to fully characterize all primitive graphs, but we were able to give an efficient algorithm for calculating the primitive graphs. One area of further work would be to study more closely the structures of the sets Pk and Qk . k
In Chapter 2 it was shown that primitive graphs were bounded in size by O(2 2 ), where k is the multiplicity of −1 or 0. There is a construction that gives graphs that approach this bound. Given a graph G ∈ Pk with n = |V (G)|, we first form the strong product G K2 . Now, to this strong product we add two new vertices. One vertex is attached to all vertices in one copy of G, and the other vertex is attached to all vertices in the other copy of G. Call this new graph Susp2 (G). Susp2 (G) is primitive with 2n + 2 vertices, and it has −1 in its spectrum with multiplicity either n + k − 1 or n + k. If we have the latter, then Susp2 (G) ∈ Pk+2 . By taking the largest graph in Pk , with n vertices, we get that the largest graph in Pk+2 is of size at least 2n + 2. The curious fact is that in all cases thus far this graph is actually the largest in Pk+2 . This leads to the following conjecture. Conjecture 6.0.1. The largest graph in P2k has 2k+1 − 2 vertices. The largest graph in P2k+1 has 2k+1 + 2k−1 − 2 vertices. The Susp2 (G) construction shows that these are lower bounds for P2k and P2k+1 , but it is not guaranteed that they are also upper bounds. A similar construction also works for Qk . 139
Another avenue for further research is in finding more efficient methods of determining whether two graph are isomorphic using information from their spectra. Throughout this thesis we only considered looking purely at when two spectra coincided. By adding more information that is not costly to compute, we may be able to narrow down the search on isomorphic graphs. One natural approach would be to use extra matrices other than only the adjacency matrix. Another approach would be to use the structure of the underlying primitive graph of a blowup or split graph. For example, suppose G = P4 (a, b, c, d). We can quickly compute {a, b, c, d} from AG . Once we know {a, b, c, d}, we can reduce AG to AP4 . Thus, we know the sizes of the blowup partitions as well as AP4 . In this case knowing AP4 does give us P4 since P4 is DS, but in general reducing a graph to its primitive graph only gives us information about the partition of cospectral graphs containing the primitive graph. For example, we would not be able to distinguish K1,4 (a, b, c, d, e) and C4 (a, b, c, d) + K1 (e) from {a, b, c, d, e} and their primitive graphs since K1,4 and C4 + K1 are cospectral. While this approach is still not perfect, it does lead to the notion of a graph being reducibly DS. Definition 6.0.2. Let H be a graph with k = |V (H)|, and let σ be a permutation of [k]. Let H be the set of graphs cospectral with H. If G = H(n1 , . . . , nk ) then define σG = H(nσ(1) , . . . , nσ(k) ). Finally, we say that G is reducibly DS if for every σ such that G ∼ H 0 (nσ(1) , . . . , nσ(k) ) for some H 0 ∈ H, then we have G ∼ = H 0 (nσ(1) , . . . , nσ(k) ). While this definition looks complicated, in practice it greatly reduces the amount of computation needed. For example, since each graph in P4 is DS, we can simply check when G ∼ σG for some fixed blowup partition sizes. In this manner we can easily show that P3 (a, b, c)+K1 (d) is reducibly DS except for P3 (a, b, a)+K1 (b) ∼ P3 (b, a, b)+K1 (a) and P3 (nb(2b − 1), nb, nb(2b − 1)) + K1 (n(2b − 1)) ∼ P3 (nb(2b − 1), n(2b − 1), nb) + K1 (nb(2b − 1)) P3 (2nb(2b + 1), n(2b + 1), 2nb(2b + 1)) + K1 (4nb) ∼ P3 (2nb(2b + 1), 4nb, n(2b + 1)) + K1 (2nb(2b + 1)) Note that P3 (a, b, a) + K1 (b) ∼ P3 (b, a, b) + K1 (a) supersedes the result P3 (k, 1, k) + K1 (1) ∼ P3 (1, k, 1) + K1 (k) that we found in Chapter 4. We can also easily show P4 (a, b, c, d), C4 (a, b, c, d), and K1,3 (a, b, c, d) are reducibly DS for all a, b, c, d ≥ 1. 140
Bibliography [1] Bari, Ruth A., and Frank Harary, eds. Graphs and Combinatorics; Proceedings. Berlin: Springer-Verlag, 1974. [2] Beineke, Lowell W., and Robin J. Wilson, eds. Topics in Algebraic Graph Theory. Cambridge, UK: Cambridge UP, 2004. [3] Bell, F. K., and P. Rowlinson. “On The Multiplicities Of Graph Eigenvalues.” Bulletin of the London Mathematical Society 35.3 (2003): 401-08. [4] Biggs, Norman. Algebraic Graph Theory. London: Cambridge UP, 1974. [5] Cvetkovi´c, Drago˘s M., Peter Rowlinson, and S. Simi´c. An Introduction to the Theory of Graph Spectra. Cambridge, U.K.: Cambridge UP, 2010. [6] Godsil, C. D., and Gordon Royle. Algebraic Graph Theory. New York: Springer, 2001. [7] Kotlov, Andrew, and L´aszl´o Lov´asz. “The Rank and Size of Graphs.” Journal of Graph Theory 23.2 (1996): 185-89. [8] Ma, Haicheng, and Haizhen Ren. “On the Spectral Characterization of the Union of Complete Multipartite Graph and Some Isolated Vertices.” Discrete Mathematics 310.24 (2010): 3648-652.
141
Washington University Open Scholarship Arts & Sciences Electronic Theses and Dissertations
Arts & Sciences
Summer 8-15-2015
Graphs with Eigenvalues of High Multiplicity Casey Boyett Washington University in St. Louis
Follow this and additional works at: http://openscholarship.wustl.edu/art_sci_etds Recommended Citation Boyett, Casey, "Graphs with Eigenvalues of High Multiplicity" (2015). Arts & Sciences Electronic Theses and Dissertations. Paper 636.
This Dissertation is brought to you for free and open access by the Arts & Sciences at Washington University Open Scholarship. It has been accepted for inclusion in Arts & Sciences Electronic Theses and Dissertations by an authorized administrator of Washington University Open Scholarship. For more information, please contact [email protected], [email protected].
WASHINGTON UNIVERSITY IN ST. LOUIS Department of Mathematics
Dissertation Examination Committee: John Shareshian, Chair Renato Feres Robert Pless Ari Stern David Wright
Graphs with Eigenvalues of High Multiplicity by Casey Boyett A dissertation presented to the Graduate School of Arts & Sciences of Washington University in partial fulfillment of the requirements for the degree of Doctor of Philosophy August 2015 St. Louis, Missouri
c 2015, Casey Boyett
TABLE OF CONTENTS List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Abstract of the Dissertation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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1 Introduction . . . . . . . . . 1.1 Background . . . . . . 1.2 Definitions . . . . . . . 1.3 Summary of Chapters
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2 Key Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Classic Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 New Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3 Summary of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4 λ = −1 . . . . . . . . . . . . 4.1 P1 , P2 , and P3 . . . . 4.2 P3 (a, b, c) + K1 (d) . . . 4.3 P4 (a, b, c, d) . . . . . . 4.4 K1,3 (a, b, c, d) . . . . . 4.5 C4 (a, b, c, d) . . . . . . 4.6 P5 (a, b, c, d, e) . . . . . 4.7 C6 (a, b, c, d, e, f ) . . . . 4.8 Coefficients . . . . . . 4.9 Bounds on eigenvalues
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5 λ=0. . . . . . . . . . . . . . 5.1 Q0 , Q2 , and Q3 . . . . . 5.2 2P2 (a, b, c, d; z) . . . . . 5.3 P4 (a, b, c, d; z) . . . . . . 5.4 P5 (a, b, c, d, e; z) . . . . . 5.5 Kite(a, b, c, d; z) . . . . . 5.6 Bull(a, b, c, d, e; z) . . . . 5.7 House(a, b, c, d, e; z) . . . 5.8 Broom(a, b, c, d, e, f ; z) . 5.9 P2 × K3 (a, b, c, d, e, f ; z) 5.10 Coefficients . . . . . . . 5.11 Bounds on eigenvalues .
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6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 ii
LIST OF FIGURES 1.1
C4 + K1 and K1,4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1 2.2 2.3 2.4 2.5
Bow tie and subgraphs . . Blowup and split example P4 . . . . . . . . . . . . . Bipartite Q4 . . . . . . . . Non-bipartite Q4 . . . . .
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ACKNOWLEDGEMENTS
I would first like to thank my advisor, John Shareshian, for allowing me to pursue a topic of my own choosing. While it was more work on me, the satisfaction of working on my own problem as opposed to trudging through one handed to me more than made up for it. Much effort was expended on independent research, but I do not believe I would have enjoyed the alternative. Next, I would like to thank all of the wonderful friends I have made over the last few years as well as those lifelong friends I can never get rid of. Without them I would have easily lost my sanity (what little I still retain). First is my local gaming friends: Sarah Morris, Meredith Sargent, Lauren Katz, and Joseph Colton. I am fortunate to have met each and every one of them, even if occasionally I make a tree fall on them. Next is my “down south” group: Jared Palmer, Lisa Palmer, and Victoria Porter. Even though there are hundreds of miles between us, we still finds ways to keep each other entertained. Many late hours have been spent with them chatting away online. Lastly is Morgan Harrison. Above all others she just gets me. Last, and most important, I would like to thank my family. Without their love and support I would not be where I am today. My mom, my dad, and my niece mean the world to me, and I hope I have made them proud. My only regret is that my dad is not here to see all that I have accomplished as he passed away before I could finish my work. Unfortunately it is not possible to list every single friend I have made along my journey. Even if I am not able to name them all here, they are still very important to me, and I am grateful for all they have done for me.
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Dedicated to my dad.
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ABSTRACT OF THE DISSERTATION Graphs with Eigenvalues of High Multiplicity by Casey Boyett Doctor of Philosophy in Mathematics Washington University in St. Louis, 2015 Professor John Shareshian, Chair Given a graph G we can form a matrix AG indexed by the vertices of G and which encodes the edges of G. AG is called the adjacency matrix of G. From the adjacency matrix we may find the eigenvalues. We would now like to know what information we may garner from the eigenvalues. It turns out quite a bit may be determined from the eigenvalues, collectively called the spectrum.
One big question is to ask whether or not a graph can be uniquely determined by its spectrum. Much research has been done in this area, and it is conjectured that almost all graphs may in fact be determined by their spectra. This is however a difficult task. In this dissertation we look at a subset of all graphs, namely those with either −1 or 0 in their spectrum with a given multiplicity. We first show that any such graph must either be primitive in a sense, or that it is obtained from a primitive graph by an elementary operation of blowing up or splitting vertices. We then show that the set of primitive graphs is finite, for a fixed multiplicity. Lastly, we analyze graphs with −1 or 0 in their spectra with multiplicities up to 4, and show many which are uniquely determined by their spectra.
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Chapter 1 Introduction We assume a basic understanding of graph theory and linear algebra. Beyond these basics we will keep the contents as self-contained as possible. Throughout we will use the notation [n] to mean the set of integers {1, 2, . . . , n}.
1.1
Background
Graph theory is a subject with a myriad of uses. Algebraic, and more specifically spectral, graph theory is no exception. Using the eigenvalues of a graph to study properties of that graph has advantages in that the numbers give quantifiable data, and moreover these eigenvalues can be quickly computed. For example, the spectrum of a graph can give information on structures of the graph such as acyclic digraphs, spanning trees, Hamiltonian cycles, independent sets, cliques, Eulerian orientations, vertex colourings, and so on. Spectral graph theory also has more concrete uses in other areas of study such as information theory, coding theory, chemistry, geographic studies, social sciences, epidemiology, and game theory. Let us look at a couple of these uses in more detail. In information theory one often asks how much information can be transmitted over a channel or a network of channels. The Shannon capacity gives a precise number to how much information can be transmitted, and bounds
1
on the Shannon capacity can be given in terms of the spectrum of the graph representing the channel network. Chemistry is a natural subject in which to use graph theory. The structure of a molecule can easily be modeled by a graph, and information about the spectrum of the graph can quickly give information about the structure of the molecule. One very powerful use of spectral graph theory in chemistry is in determining whether two molecules are the same. For large molecules this can be a difficult task, but by calculating the spectra of the associated graphs and comparing them, it is often possible to determine whether or not the molecules are the same. This is a topic we explore further in this thesis. That is, we look at whether or not certain graphs are uniquely determined by their spectra. Finally, social science is yet another subject in which graph theory has natural uses. With the wealth of social networking sites, a graph representing connections between various individuals is a simple way in which to examine connections through the site. Sites such as Facebook take advantage of these network graphs in order to continually improve their functionality. Spectral graph theory has many theoretical and practical applications. This thesis focuses on the theoretical aspects, and primarily we try to approach the Graph Isomorphism problem through spectral techniques. This approach is not new, but we take a look at a subset of graphs that as of yet have gone untouched.
1.2
Definitions
Definition 1.2.1. A graph G is a pair (V (G), E(G)) where V (G) is a finite set and E(G) is a subset of V (G) × V (G). The elements of V (G) are called the vertices of G, and the elements of E(G) are the edges of G. If e = (v, w) is an edge, then v is called the tail of e and w is the head. An edge of the form (v, v) is called a loop. If both (v, w) and (w, v) are edges, then we call {v, w} an undirected edge. Edges that are not undirected are directed. Similarly, a graph with only undirected edges is called undirected. Unless otherwise noted all graphs are loopless and undirected. Where no confusion will be incurred we will refer to the vertex set V and the edge set E without reference to G. We 2
will commonly label the vertices V (G) = {v1 , . . . , vn }. By abuse of notation an edge {v, w} will often be written vw. A few key properties of graphs will be needed. Definition 1.2.2. The complement of a graph G is the graph G with the same vertex set V (G) and E(G) = {vw : vw 6∈ E(G)}. Definition 1.2.3. A path in a graph is a sequence of distinct vertices v1 v2 · · · vk+1 such that vi vi+1 is an edge for all i ∈ [k]. Such a path is said to have length k. Definition 1.2.4. A cycle in a graph is a sequence of distinct vertices v1 v2 · · · vk such that vi vi+1 is an edge for all i ∈ [k − 1], and v1 vk is also an edge. Such a cycle is said to have length k. Definition 1.2.5. A graph is connected if for every two distinct vertices v and w there is a path containing v and w, otherwise the graph is disconnected. Definition 1.2.6. A forest is an acyclic graph. A tree is a connected acyclic graph. Definition 1.2.7. The neighborhood of v ∈ V (G) is the set N (v) = {w : vw ∈ E(G)}. The closed neighborhood of v is N [v] = N (v) ∪ {v}. Definition 1.2.8. The degree of a vertex v is the number of edges containing v and is denoted by d(v). The degree sequence of G is (d(v1 ), . . . , d(vn )). Definition 1.2.9. The maximum degree of G is ∆(G) = max{d(vi )}. The minimum degree of G is δ(G) = min{d(vi )}. Definition 1.2.10. If S ⊆ V is such that no two vertices in S form an edge, then S is called an independent set. The size of the largest independent set is called the independence number of G and is denoted by α(G). Definition 1.2.11. If S ⊆ V is such that every two distinct vertices in S form an edge, then S is called a clique. The size of the largest clique called the clique number of G and is denoted by ω(G). Definition 1.2.12. Graphs G and H are isomorphic if there exists a bijection φ : V (G) → V (H) such that xy ∈ E(G) iff φ(x)φ(y) ∈ E(H). 3
An easy observation is that α(G) = ω(G) and ω(G) = α(G). There are also a few special classes of graphs and operations on graphs that will be important. Definition 1.2.13. The complete graph Kn has vertex set V (Kn ) = {v1 , . . . , vn } and edge set E(Kn ) = {vi vj : i 6= j}. Definition 1.2.14. A graph is called k-partite if V can be partitioned into k disjoint independent sets. If k = 2 we will call the graph bipartite. Definition 1.2.15. The complete multipartite graph Kn1 ,...,nk has vertex set V (Kn1 ,...,nk ) = V1 ∪· · · · ∪· Vk with |Vi | = ni for all i and edge set E(Kn1 ,...,nk ) = {vi vj : vi ∈ Vi , vj ∈ Vj , i 6= j}. Definition 1.2.16. The cycle graph Cn has vertex set V (Cn ) = {v1 , . . . , vn } and edge set E(Cn ) = {vi vi+1 : 1 ≤ i < n} ∪ {v1 vn }. Definition 1.2.17. The path graph Pn has vertex set V (Pn ) = {v1 , . . . , vn } and edge set E(Pn ) = {vi vi+1 : 1 ≤ i < n}. Definition 1.2.18. Given a subset S ⊆ V , the induced subgraph G[S] has vertex set S and edge set E(G[S]) = {vw : vw ∈ E(G) and v, w ∈ S}. If S = V \T then we will write G − T for G[S] or G − v if T = {v}. Definition 1.2.19. Given two graphs G and H on disjoint vertex sets, we can form the graph G + H with vertex set V (G + H) = V (G) ∪· V (H) and edge set E(G + H) = E(G) ∪· E(H). Since this thesis is primarily focused on the spectrum of a graph we need to define the adjacency matrix of G. Definition 1.2.20. For a graph G we define the adjacency matrix AG as the (0, 1)-matrix indexed by the vertices V (G) = {v1 , . . . vn } with entry aij = 1 when vi vj is an edge of G and aij = 0 otherwise. This is not the only matrix naturally associated with a graph, but it is the only one we will study. Given a matrix it is natural to consider the eigenvalues. This leads to our next definitions. Definition 1.2.21. Given a matrix MG associated with a graph G we define the characteristic polynomial of G with respect to MG as PMG (x) = det (xI − MG ) where I is identity matrix. 4
Definition 1.2.22. Given a matrix MG we call its eigenvalues, treated as a multiset, the M-spectrum of G. We denote this spectrum by M-Spec(G). For our purposes we will eliminate the symbol AG in the above definitions and look solely at the characteristic polynomial PG (x) and its associated spectrum Spec(G). Since we are only considering undirected graphs, the adjacency matrices are all real and symmetric, and therefore all eigenvalues are real. We will often denote Spec(G) = {λ1 , . . . , λn } with λi ≥ λi+1 for all i ∈ [n−1]. Moreover, when explicitly listing the spectrum, superscripts will be used to denote multiplicity, while multiplicity of a single eigenvalue λ will often be denoted mult(λ). We now arrive at two of the most important definitions. Definition 1.2.23. Two graphs G and H are called M-cospectral if M-Spec(G) = M-Spec(H), treated as mutlisets. Equivalently, G and H are M-cospectral if PMG (x) = PMH (x). We denote G and H being M-cospectral by G ∼M H. If a graph G is not M-cospectral to any non-isomorphic graph H, then we say G is determined by its M-spectrum or that it is M-DS for short. As noted above we will drop the A in these definitions and say two graphs are cospectral or that a graph is DS. Since cospectrality is an equivalence relation, this allows us to partition a set of graphs. Ideally we would hope that all graphs are DS, but unfortunately this is not the case. We can see this by looking at K1,4 and C4 + K1 . Both of these graphs have the spectrum {−2, 03 , 2}.
Figure 1.1: C4 + K1 and K1,4
1.3
Summary of Chapters
In Chapter 2 we will explore the key results for this dissertation, starting with the classical theorems and working our way to the new results. In Chapter 3 we give an overview of the classes of graphs that will be proved to be DS or shown to have cospectral mates. Chapters 4 5
and 5 contain the bulk of the calculations that were summarized in Chapter 3. Chapter 4 focuses on graphs with λ = −1 in their spectra with high multiplicity, while Chapter 5 focuses on graphs with λ = 0 with high multiplicity. Finally, Chapter 6 closes out this dissertation with final conclusions, conjectures, and routes for future work.
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Chapter 2 Key Theorems Results without proofs or specific citations can be found in [1, 2, 4, 5, 6].
2.1
Classic Results
The adjacency matrix of a graph has been studied extensively. Many properties of a graph can been obtained from the spectrum while other properties are lost. For example, the number of vertices and the number of edges can be determined from the spectrum while connectedness cannot. The latter can be seen from the K1,4 and C4 + K1 example in Chapter 1. The number of vertices is clearly equal to the number of eigenvalues, while the number of edges can be determined using the following proposition. Proposition 2.1.1. The number of closed k-walks in a graph G is given by λk1 + · · · + λkn . A walk is similar to a path except that the vertices need not be distinct, and a closed walk is one that starts and ends at the same vertex. Therefore, a closed 2-walk is nothing more than an edge, and so the number of edges is given by 12 (λ21 + · · · + λ2n ). Another consequence of this theorem is that λ1 + · · · + λn = 0. This is true because a closed 1-walk would be a loop, and all of our graphs are assumed to have no loops. For a general graph it is difficult to say much about its spectrum, but there are several classes of graphs for which the spectrum is explicitly known. These include the complete 7
graphs, cycle graphs, and path graphs, among many others. A few are summarized in the next proposition. Proposition 2.1.2. • Spec(Kn ) = {−1n−1 , n − 1} • Spec(Kn ) = {0n } √ √ • Spec(Km,n ) = {− mn, 0m+n−2 , mn} • Spec(Cn ) = 2 cos
2kπ n
• Spec(Pn ) = 2 cos
kπ n+1
, k ∈ [n] , k ∈ [n]
One way of calculating the eigenvalues of a graph is by first calculating eigenvectors. The adjacency matrix gives us a way find eigenvectors using the structure of the graph. Proposition 2.1.3. λ ∈ Spec(G) with v as an associated eigenvector iff vi = λ
X
vj .
vj ∈N (vi )
In particular, what Proposition 2.1.3 says is that the entry of an eigenvector associated with vertex vi is dependent upon the entries of the neighbors of vi . Knowledge of the structure of G is often useful in calculating the eigenvalues and eigenvectors of G. Perhaps the most important theorem needed is the Interlacing Theorem. It gives a way of relating eigenvalues of an induced subgraph G[S] with the eigenvalues of G itself. Theorem 2.1.4 (Interlacing Theorem). If λ1 ≥ . . . ≥ λn and µ1 ≥ . . . ≥ µn−1 are the eigenvalues of G and G − v, respectively, then λi ≥ µi ≥ λi+1 for i ∈ [n − 1]. More generally, if S ⊂ V (G) is a subset of m vertices of G and ρi are the eigenvalues of G − S, then λi ≥ ρi ≥ λn−m+i for i ∈ [m]. Corollary 2.1.5. If H is an induced subgraph of G with |V (H)| = |V (G)| − k, and λ ∈ Spec(H) with multiplicity m, then λ ∈ Spec(G) with multiplicity between m − k and m + k. Similarly, if λ ∈ Spec(G) with multiplicity m0 , then λ ∈ Spec(H) with multiplicity between m0 − k and m0 + k. 8
This gives us a way to study a graph by looking at its induced subgraphs. Moreover, since we are primarily interested in graphs with eigenvalues of multiplicity greater than 1, we know that these subgraphs must have these same high multiplicity eigenvalues. Take √ for example the bow tie graph (see Figure 2.1(a)) with eigenvalues { 12 (1 ± 17), −12 , 1}. Since −1 has multiplicity 2, Corollary 2.1.5 guarantees that each subgraph on 4 vertices must have −1 in its spectrum with multiplicity of 1, 2, or 3.
We can quickly check
that each such subgraph has spectrum either {−12 , 12 } (first graph in Figure 2.1(b)) or {−1.4811 . . . , −1, 0.3111 . . . , 2.1700 . . .} (second graph in Figure 2.1(b)).
(a) Bow tie
(b) Vertex-deleted subgraphs
Figure 2.1: Bow tie and subgraphs Knowing that a graph has an eigenvalue with multiplicity greater than 1 is useful for interlacing, and it does indeed guarantee certain eigenvalues in subgraphs, but it would be nice to be able to work up from a given graph and guarantee eigenvalues with increasing multiplicity. That is, we would like to work our way up from the bottom instead of from the top down. We will see in the next section how we can approach this problem. Before investigating further there are a few more properties we can deduce from the Interlacing Theorem. In particular, we can compute elementary bounds on α(G) and ω(G). Proposition 2.1.6. Let n = |V (G)|, n− = |{i : λi < 0}| and n+ = |{i : λi > 0}|. Then α(G) ≤ min{n − n− , n − n+ }. Proof. Let S ⊆ V (G) be a maximal independent set. Then G[S] is a graph with no edges. In particular, each eigenvalue of G[S] is 0. If |S| = s, then Theorem 2.1.4 gives λn−s+i ≤ 0 ≤ λi . From this we can infer that 0 ≤ λs , and that n− ≤ n − s. Similarly, by looking at −AG , we get n+ ≤ n − s. Therefore s = α(G) ≤ min{n − n− , n − n+ }.
9
In a similar fashion, we get the following bounds on ω(G): Proposition 2.1.7. Let n = |V (G)|, N− = |{i : λi ≤ −1}|, and N+ = |{i : λi ≥ −1}|. Then ω(G) ≤ min{N− + 1, N+ + 1, λ1 + 1}. Since the graphs we will be studying will have spectra containing either 0 or −1 with high multiplicity, Propositions 2.1.6 and 2.1.7 tell us that the graphs will contain either large independent sets, in the case of 0, or large cliques, in the case of −1. Another property of a graph we can gather from the spectrum is whether or not the graph is bipartite. This will be useful when examining the λ = 0 case in Chapter 5. Proposition 2.1.8. A graph G is bipartite iff Spec(G) is symmetric about 0. Proof. Let V = S ∪· T be a bipartition of the vertices of G, and let λ ∈ Spec(G). If v is an eigenvector for λ, then v0 defined as having entries vi for each vertex vi ∈ S and −vj for each vertex vj ∈ T is an eigenvector for −λ. For the converse, we need to make use of the Perron-Frobenius theorem for real square matrices, but this is outside of the scope of this thesis. One last key theorem we will utilize relates the characteristic polynomial of a graph to the characteristic polynomial of its vertex-deleted subgraphs. Proposition 2.1.9. If V (G) = {v1 , . . . , vn }, then n
X d PG (x) = PG−vi (x). dx i=1 Proof. The derivative of the characteristic polynomial of an n × n matrix is the sum of the characteristic polynomials of its n principal submatrices obtained by deleting the i-th row and i-th column. The matrix obtained by deleting the i-th row and column of AG is precisely the adjacency matrix of G − vi . Thus, we get n
X d PG (x) = PG−vi (x). dx i=1
10
(a) P4
(b) P4 with blowup
(c) P4 with split
Figure 2.2: Blowup and split example
2.2
New Results
Before we move on to the new results we need some definitions. Definition 2.2.1. If v1 , v2 ∈ V (G) are such that N (v1 )4N (v2 ) = {v1 , v2 }, then we call {v1 , v2 } a blowup or blowup partition. More generally, if S ⊆ V (G) with |S| > 1 is such that {vi , vj } is a blowup for all vi , vj ∈ S, then S is a blowup partition. We call G a blowup graph if it contains a blowup partition. Definition 2.2.2. If v1 , v2 ∈ V (G) are such that N (v1 )4N (v2 ) = ∅, then we call {v1 , v2 } a split or split partition. More generally, if S ⊆ V (G) with |S| > 1 is such that {vi , vj } is a split for all vi , vj ∈ S, then S is a split partition. We call G a split graph 1 if it contains a split partition. The idea of a blowup partition may be thought of as replacing a vertex in a graph with a clique while keeping all adjacencies from the replaced vertex with all vertices in the clique. Similarly, a split partition may be thought of as replacing a vertex with an independent set, again keeping all previous adjacencies. Examples are given in Figure 2.2. Throughout we will use a specific notation for blowup graphs and split graphs. Notation 2.2.3. Let G be a graph with V (G) = {v1 , . . . , vn }, and let ai ∈ N for i ∈ [n]. G(a1 , . . . , an ) is used to denote the blowup graph obtained from G by replacing vertex vi with a clique of size ai while keeping all adjacencies between cliques that replace adjacent vertices. Similarly, G(a1 , . . . , an ; z) is used to denote the split graph with vertices replace by independent sets of size ai along with z isolated vertices. If we know that a graph has a blowup or split partition, then we immediately gain information about the graph’s spectrum. 1
The notion of a split graph sometimes refers to a graph whose vertex set can be partitioned into a clique and an independent set. This distinction should not cause confusion here.
11
Proposition 2.2.4. Suppose G has a blowup partition S of size k ≥ 2. Then −1 ∈ Spec(G) with multiplicity at least k − 1. Similarly, if G has a split partition S of size k ≥ 2, then 0 ∈ Spec(G) with multiplicity at least k − 1. Proof. Suppose S = {v1 , . . . , vk }. Define vectors vi , i ∈ [k − 1] as having entries vik = 1, vii = −1, and vij = 0 otherwise. We need to consider three cases to show that vi is an eigenvector for eigenvalues 0 or −1. First, suppose v ∈ / N (vi ). Then clearly vi ∈ / N (v). Also, since N (vi )4N (vk ) = {vi , vk } or ∅, then vk ∈ / N (v). Thus, using Proposition 2.1.3 we get that the sum over the neighbors of v is 0, as we needed. Next, suppose v ∈ N (vi ) and v 6= vk . This time, the sum over the neighbors of v will include both vi and vk , and this sum is clearly 0. Lastly, suppose v = vi or v = vk . If vi vk is not an edge, then in both cases the sum over the neighbors of v is again 0. This gives λ = 0 as an eigenvalue. On the other hand, if vi vk is an edge, then the only contributing vertices in the sum will be precisely vi or vk . In either case we simply get a change in sign since vii = −vik . This gives λ = −1 as an eigenvalue. Finally, since the set of vi are clearly linearly independent, we get that the multiplicity of λ is at least k − 1. This says that blowup graphs always have −1 in their spectrum, but the converse is not true. Take for example C6 . We have Spec(C6 ) = {−2, −12 , 12 , 2}, but C6 clearly does not have a blowup. Similarly, not all graphs with 0 in their spectrum have a split partition. It would be nice if we could characterize which graphs have this property. Definition 2.2.5. A graph G is called 0-primitive if 0 ∈ / Spec(G), or if 0 ∈ Spec(G) and G does not contain a split partition or a vertex of degree zero. G is called -1-primitive if −1 ∈ / Spec(G) or if −1 ∈ Spec(G) and G does not contain a blowup partition. When 0∈ / Spec(G) or −1 ∈ / Spec(G), then G is called trivially primitive. Primitivity may be with respect to either splits or blowups, and when needed the distinction will be explicitly given. In most cases it will be clear from context. With this new language we are attempting to classify all primitive graphs. A full classification is slightly out of reach, 12
but we are able to give an algorithmic approach to determining all primitive graphs with a fixed number of eigenvalues other than 0 or −1. Before we examine this, we will first show that the algorithm is guaranteed to terminate. Definition 2.2.6. Define Pkn to be the set of -1-primitive graphs with n vertices and k S eigenvalues other than -1, and define Pk = Pkn . Similarly, define Qnk and Qk for 0-primitive graphs. The set P4 is shown in Figure 2.3, and Q4 is shown in Figure 2.4 and Figure 2.5. Q4 is partitioned into bipartite and non-bipartite graphs. This will make the examination in Chapter 5 easier. One key thing to note is that P4 and Q4 are both finite sets. This is not a peculiarity. We will show in Theorem 2.2.13 that Pk and Qk are finite sets for all k ≥ 0. Before we prove Theorem 2.2.13 we first need a theorem on the rank of the adjacency matrix of G and then a way to relate the characteristic polynomial of G with G. Theorem 2.2.7. Let G be a 0-primitive graph on n vertices, and let r be the rank of AG . Then r n = O 22 .
Corollary 2.2.8. Fix k ≥ 0. Then there is an upper bound on the number of vertices of a graph G ∈ Qk . Proof. Let G ∈ Qk with n = |V (G)|. Since 0 ∈ Spec(G) with mult(0) = n − k, we know that the rank of AG is n − mult(0) = n − (n − k) = k. Theorem 2.2.7 then tells us that k n = O 22 .
A proof of Theorem 2.2.7 can be found in [7]. Corollary 2.2.8 tells us that Qk is finite. To show that Pk is also finite we need to relate the characteristic polynomial of G to that of G. Lemma 2.2.9. Suppose A is an invertible matrix and u and v are column vectors. Then det (A + uvT ) = 1 + vT A−1 u det(A). 13
Kd
Kb
Ka
Kc
Kb
Ka
(a) P3 (a, b, c) + K1 (d)
(b) P4 (a, b, c, d)
Kc
Ka
Ka
Kb
Kd
Kc
Kd
Kb (c) K1,3 (a, b, c, d)
Ka
Kc
(d) C4 (a, b, c, d)
Kb
Kc
Kd
(e) P5 (a, b, c, d, e)
Ka
Kb
Kf
Kc
Ke
Kd
(f) C6 (a, b, c, d, e, f )
Figure 2.3: P4
14
Ke
Kd
Ka
Kz
Kc Kz
Kb
Kb
Ka
Kd (a) 2P2 (a, b, c, d; z)
Kc
Kd
(b) P4 (a, b, c, d; z)
Kz
Ka
Kb
Kc
Kd
Ke
(c) P5 (a, b, c, d, e)
Figure 2.4: Bipartite Q4 Let {µ1 , . . . , µm } be the distinct eigenvalues of G. For i ∈ [m], let Pi be the n×n matrix that represents the orthogonal projection of Rn , with respect to the standard basis {e1 , . . . , en }, onto E(µi ), where E(µi ) is the eigenspace of µi . Finally, let βi =
kPi jk √ , n
where j is the all-1
vector. One property we will exploit below is the spectral decomposition of A. We can write A as A = µ 1 P1 + µ 2 P2 + · · · + µ m Pm . Note also that
P
Pi = I, Pi2 = Pi = PiT for all i, and Pi Pj = O for all i 6= j. With this
we can now calculate the characteristic polynomial of G using a mutlilinear determinental expansion. Proposition 2.2.10. For a graph G, the characteristic polynomial of G is given by ! m 2 X β i PG (x) = (−1)n PG (−x − 1) 1 − n . x + 1 + µi i=1 Proof. Let J be the all-1 matrix. Then AG = J − I − AG . Note that J = jjT . From this we
15
Kd
Kc
Ke Ka
Kd
Kb
Kz
Kz
Ka
Kc
Kb (a) Kite(a, b, c, d; z)
(b) Bull(a, b, c, d; z)
Kb Ka Ka
Kc
Ke
Kb
Kf
Kz
Ke Kz
Kc
Kd (c) House(a, b, c, d, e; z)
Kd
(d) Broom(a, b, c, d, e, f ; z)
Ka
Kb
Kf
Kc
Ke
Kd
(e) P2 × K3 (a, b, c, d, e, f ; z)
Figure 2.5: Non-bipartite Q4
16
Kz
get PG (x) = det ((x + 1)I + AG − J) = det ((x + 1)I + AG ) 1 − jT ((x + 1)I + AG )−1 j ! m 2 X β i . = (−1)n PG (−x − 1) 1 − n x + 1 + µ i i=1
Corollary 2.2.11. If λ ∈ Spec(G) with multiplicity m, then −λ − 1 ∈ Spec(G) with multiplicity m − 1, m, or m + 1. We need one last proposition before we can prove Theorem 2.2.13. Proposition 2.2.12. A graph G has a blowup partition iff G has a split partition. Proof. Suppose {v, w} is a blowup partition in G. Then v and w share all neighbors, save for v and w themselves. Therefore, in G, v and w still share all neighbors, and thus {v, w} is a split partition in G. The converse is proved similarly. Theorem 2.2.13. The sets Pk and Qk are finite for all k ≥ 0. Proof. Qk was proved finite in Corollary 2.2.8. To show Pk is finite, we note that for each G ∈ Pk , we have G in either Qk−1 , Qk , or Qk+1 , all of which we know to be finite. Now that we know that Pk and Qk are always finite, it would be useful to know which graphs belong in each set. Unfortunately there is no simple description of each set, but it is possible to build each set algorithmically. Using a proof similar to that of Proposition 2.2.10 we will relate the characteristic polynomial of G to that of G with a vertex added. Let S ⊆ V (G) and define GS = G +S x where x is a new vertex with edges added between x and all vertices in S. Let cS be the characteristic vector of S. Using the same notation as in Proposition 2.2.10, we get the following. Lemma 2.2.14. Let D be a 1 × 1 matrix, and let A, B, and C matrices of appropriate sizes. Then det
A B
! = (D − 1) det(A) + det(A − BC).
C D 17
Proposition 2.2.15. The characteristic polynomial of PGS (x) is given by ! r X kPi cS k2 . PGS (x) = PG (x) x − x − µi i=1 Proof. The characteristic polynomial is given by xI − A −c G S det (xI − AGS ) = −cT x S
= (x − 1) det (xI − AG ) + det (xI − AG − cS cTS ) = det (xI − AG ) x − cTS (xI − AG )−1 cS ! r X kPi cS k2 = PG (x) x − . x − µi i=1
We are now set to demonstrate with the next few propositions how we can build the sets Pk and Qk . Proposition 2.2.16. Let µ ∈ Spec(G) with multiplicity m ≥ 1, and S ⊆ V (G). Then µ ∈ Spec(GS ) with multiplicity at least m iff cS ∈ E(µ)⊥ . Q Proof. Without loss of generality assume µ = µ1 . Write mG (x) = ri=1 (x − µi ) and PG (x) = Qr mi i=1 (x − µi ) . Then Proposition 2.2.15 gives ! r X PG (x) kPi cS k2 mG (x) PGS (x) = xmG (x) − mG (x) x − µi i=1 ! r r r Y X Y mi −1 2 . = (x − µi ) xm (x) − kP c k (x − µ ) G j i S i=1
Define q(x) = xmG (x) −
i=1
r X
2
kPi cS k
i=1
r Y
j=1 j6=i
(x − µj ).
j=1 j6=i
Now, if cS ∈ E(µ1 )⊥ , then kP1 cS k2 = 0. From this we get that x − µ1 is a common factor throughout q(x), and thus q(µ1 ) = 0 which give that µ1 is a root of order at least m1 in PGS (x). 18
On the other hand, suppose cS 6∈ E(µ1 )⊥ . Then kP1 cS k2 6= 0. From this it is now clear that Q q(µ1 ) = kP1 cS k2 rj=2 (µ1 − µj ) 6= 0. Therefore µ1 is a root of order m1 − 1 in PGS (x). Proposition 2.2.17. Let G be a blowup graph and let S ⊆ V (G). Then ei − ej ∈ E(−1) for some i 6= j. If cS ∈ E(−1)⊥ , then GS is a blowup. Proof. The first assertion is trivial. For the second assertion, we have that cS · v = 0 for all v ∈ E(−1). In particular cS · (ei − ej ) = 0. This says that (cS )i = (cS )j . That is, the new vertex x in GS is either adjacent to both vi and vj , or it is adjacent to neither. In both cases we arrive at N (vi )4N (vj ) = {vi , vj }, and thus GS is a blowup. Proposition 2.2.18. If G is primitive with −1 ∈ Spec(G) with multiplicity m and G − v is a blowup for some v ∈ V (G), then −1 ∈ Spec(G − v) with multiplicity m + 1. Proof. Let G0 = G − v and S = N (v) ⊆ V (G). By Proposition 2.2.17, since G0 +S v is not a blowup then cS ∈ / EG0 (−1)⊥ . By Proposition 2.2.16, since cS ∈ / EG0 (−1)⊥ , then the multiplicity of −1 goes down by one from G0 to G. Since the multiplicity of −1 in G is m the multiplicity must be m + 1 in G0 . Proposition 2.2.19. Let G be primitive with −1 ∈ Spec(G) with multiplicity m. Then there exists a v ∈ V (G) such that G − v has −1 ∈ Spec(G − v) with multiplicity m − 1; moreover, G − v is primitive. Proof. From Proposition 2.1.9 we know n
X d PG (x) = PG−vi (x). dx i=1 Since −1 is a root of order m in PG (x) it must be a root of order m − 1 in
d P (x). dx G
By
Theorem 2.1.4 we know that −1 has multiplicity between m−1 and m+1 in each Spec(G−vi ), and therefore we must have that at least one of PG−vi (x) must have −1 as a root of order m − 1. From Proposition 2.2.18 this G − vi must also be primitive. Propositions 2.2.17, 2.2.18, and 2.2.19 apply equally well to λ = 0 with blowup replaced with split. Most importantly they tell us that the primitive graphs with n vertices, say, can 19
be built up from the primitive graphs on n − 1 vertices. We will now give the algorithm in detail. Algorithm 2.2.20. Fix λ ∈ {−1, 0} and k ≥ 0. Set n = k. Let Gi denote the set of all graphs on i vertices. Let Si denote Pki or Qik . Step 1: For each graph G ∈ Gn , compute Spec(G). If λ ∈ / Spec(G), then add G to Sn . Step 2: If Sn is empty, go to Step 3. Else, for each G ∈ Sn and for each S ∈ {0, 1}n ∩EG (λ)⊥ , compute Spec(GS ). If λ ∈ Spec(GS ) and GS is primitive, then add GS to Sn+1 . Increase n by 1. Repeat Step 2. Step 3: If λ = −1, then set Pk =
S
Si . Else, set Qk =
S
Si .
Step 1 computes the trivially primitive graphs. Step 2 iteratively takes the primitive graphs on n vertices, adds a new vertex in such a way to increase the multiplicity of λ, and checks whether or not it is primitive. Finally, Step 3 simply combines the sets produced in Step 2 to create Pk or Qk . The validity of Step 2 is given by Propositions 2.2.16 through 2.2.19, and Theorem 2.2.13 shows that the algorithm will terminate. Using this algorithm we are able to calculate the sets for Pk for 0 ≤ k ≤ 7 and Qk for 0 ≤ k ≤ 6. P0 = ∅ P1 = {K1 } P2 = {2K1 } P3 = {3K1 , P3 } P4 = {4K1 , P3 + K1 , P4 , C4 , P5 , C6 } |P5 | = 30 |P6 | = 422 |P7 | = 8226
20
Q0 = {empty graph} Q1 = ∅ Q2 = {P2 } Q3 = {K3 } Q4 = {K4 , 2P2 , P4 , P5 , Kite, Bull, House, Broom, P2 × K3 } |Q5 | = 25 |Q6 | = 1654 The last part we need in order to study these sets in-depth is a way to calculate the spectra, or equivalently the characteristic polynomials, of the blowup and split graphs. This is easily done using a quotient matrix formed from the adjacency matrix and a partition of the vertices of the graph. Definition 2.2.21. A partition π = {V1 , . . . , Vk } of V is called an equitable partition if for every 1 ≤ i, j ≤ k there exists an integer bij ≥ 0 such that every vertex in Vi is adjacent to exactly bij vertices in Vj . Define the quotient matrix AG /π as having (i, j) entry equal to bij . It is straightforward to see that blowup and split partitions naturally lead to equitable partitions. Suppose S ⊆ V is a split partition. Since all vertices in S have the same neighbors, they will be adjacent to the same number of vertices in any subset of V , including S itself. Within S each vertex is adjacent to 0 other vertices. The same argument holds if S is a blowup partition, except in this each vertex is adjacent to |S| − 1 vertices in S. Take for example the bow tie graph in Figure 2.1. It is a blowup of P3 , where both end-vertices have been blown up with partitions of size 2. Then for the quotient matrix we get 1 1 0 2 0 2 . 0 1 1 The usefulness of having an equitable partition is that we can show that the characteristic polynomial of the quotient matrix induced by the partition divides the characteristic polynomial of G. Proposition 2.2.22. Let π be an equitable partition of V . Then the characteristic polynomial of AG /π divides the characteristic polynomial of AG . 21
A proof of this can be found in [6]. Going back to the bow tie graph example, we can compute both its characteristic polynomial and that of its quotient matrix. Doing so gives PG (x) = −(x − 1)(x + 1)2 x2 − x − 4 PAG /π (x) = −(x − 1) x2 − x − 4 As expected these differ only in the factors containing the eigenvalues of −1 coming from the two blowup partitions. Proposition 2.2.22 now offers us an efficient way to study all blowup graphs coming from Pk and all split graphs coming from Qk . Proposition 2.2.23. Let H = G(a1 , . . . , an ) be a blowup graph, and let A be the quotient P matrix obtained from the n blowup partitions of sizes ai . Let a = ni=1 ai . Then PH (x) = (x + 1)a−n PA (x). Similarly, if K = G(a1 , . . . , an ; z) is a split graph, and A and a are defined similarly, then PK (x) = xa+z−n PA (x). The characteristic polynomials for elements of P3 and P4 are given and analyzed in Chapter 4, while those for Q3 and Q4 are in Chapter 5. Before ending this chapter we need to answer one last question. Up to this point we have only considered graphs with λ = −1 or λ = 0 of high multiplicity. We have shown that graphs can have either of these of arbitrarily high multiplicity. For any λ ∈ / {−1, 0} we can show that mult(λ) forces an upper bound on the size of the graph. Theorem 2.2.24. Let G be a graph with n vertices, and let λ ∈ Spec(G) with λ 6∈ {−1, 0}. Let t = dim E(λ)⊥ . Then we have either (a) n ≤ 12 t(t + 1) or (b) λ = 1 and G = K2 or G = 2K2 . A proof of this can be found in [5]. Since dim E(λ)⊥ = n − mult(λ), we see that the multiplicity of an eigenvalue other than −1 or 0 is bounded by the size of the graph. Some work has been in determining graphs with eigenvalues of high multiplicity other than −1 or 0, such as in [3] in which all graphs with dim E(λ)⊥ ≤ 5 are determined. 22
Chapter 3 Summary of Results For graphs arising from P3 and P4 , we get the following results: • nK1 (a1 , . . . , an ) is DS for all ai ≥ 1 • P3 (a, b, c) is DS for all a, b, c ≥ 1 • P3 (k, m, 1) + K1 (1) is DS for all k, m ≥ 1. • P3 (k, 1, m) + K1 (1) is DS for all k, m ≥ 1 except for when k = m. In that case we have P3 (k, 1, k) + K1 (1) ∼ P3 (1, k, 1) + K1 (k). • P3 (k, 1, 1) + K1 (m) is DS for all k, m ≥ 1. • P3 (1, k, 1) + K1 (m) is DS for all k, m ≥ 1 except for when k = m. In that case we have P3 (k, 1, k) + K1 (1) ∼ P3 (1, k, 1) + K1 (k). • P4 (k, m, 1, 1) is DS for all k, m ≥ 1. • P4 (k, 1, m, 1) is DS for all k, m ≥ 1. • P4 (k, 1, 1, m) is DS for all k, m ≥ 1. • P4 (1, k, m, 1) is DS for all k, m ≥ 1. • K1,3 (k, m, 1, 1) is DS for all k, m ≥ 1 except for C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1) and P3 (a, n + ac − a − c + 1, c) + K1 (1) ∼ K1,3 (k, m, 1, 1) where 2a(a − 1)c(c − 1) = n2 is a 23
perfect square, k = n2 , and m = ac + n2 . • K1,3 (1, k, m, 1) is DS for all k, m ≥ 1. • C4 (k, m, 1, 1) is DS for all k, m ≥ 1 except for C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1). • C4 (k, 1, m, 1) is DS for all k, m ≥ 1 except for C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1) and C4 (12, 1, 6, 1) ∼ P3 (9, 4, 6) + K1 (1). • P5 (k, m, 1, 1, 1) is DS for all k, m ≥ 1. • P5 (k, 1, m, 1, 1) is DS for all k, m ≥ 1 except for P5 (k + 1, 1, k, 1, 1) ∼ P5 (2, k, 1, 1, k) for k ≥ 1. • P5 (k, 1, 1, m, 1) is DS for all k, m ≥ 1. • P5 (1, k, m, 1, 1) is DS for all k, m ≥ 1. • P5 (1, k, 1, m, 1) is DS for all k, m ≥ 1. • C6 (k, m, 1, 1, 1, 1) is DS for all k, m ≥ 1. • C6 (k, 1, m, 1, 1, 1) is DS for all k, m ≥ 1. • C6 (k, 1, 1, m, 1, 1) is DS for all k, m ≥ 1. For graphs arising from Q3 and Q4 , we get the following results: • 2P2 (k, 1, 1, 1; z) is DS iff k = 2 and z ≥ 0. • P4 (k, 1, 1, 1; z) is DS iff k = 2 or k ≥ 1 is odd and z ≥ 0. • P4 (1, k, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0. • P5 (k, 1, 1, 1, 1; z) is DS iff k ≥ 2 is even except for k ∈ {4, 10, 12, 16, 58, 94} and z ≥ 0 or k = 1 and z = 0. • P5 (1, k, 1, 1, 1; z) is DS iff k ∈ {1, 2} and z = 0. • P5 (1, 1, k, 1, 1; z) is DS iff k ≥ 2 is even and z ≥ 0. • Kite(k, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0.
24
• Kite(1, k, 1, 1; z) is DS for all odd k ≥ 1 and z ≥ 0. • Kite(k, 1, 1, m; z) is DS for all k, m ≥ 1 and z ≥ 0. • Bull(2, 1, 1, 1, 1; 0) and Bull(k, 1, 1, 1, 1; z) are DS for all odd k ≥ 1 and z ≥ 0. • Bull(1, 1, k, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0. • Bull(1, 1, 1, k, 1; z) is DS for all k ≥ 1 and z ≥ 0. • Bull(1, 1, 1, k, m; 0) ∼ Broom(1, 1, 1, d, e, f ; z) iff d = d1 d2 , e = e1 e2 , and f = de + d + e ± (d1 e1 + d2 e2 ), in which case k = de + d + e ± d1 e1 , m = de + d + e ± d2 e2 , and z = de. • House(k, 1, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0. • House(1, k, 1, 1, 1; z) is DS for all odd k ≥ 1 and z ≥ 0 or when k = 2 and z = 0. • House(1, 1, 1, k, m; z) is DS for all k, m ≥ 1 and z ≥ 0. • Broom(k, 1, m, 1, 1, 1; z) is DS for all k, m ≥ 1 and z ≥ 0 with four exceptions: Broom(3, 1, 1, 1, 1, 1; z + 3), Broom(7, 1, 3, 1, 1, 1; z + 1), Broom(1, 1, 1, 1, 1, 1; z + 1), and Broom(3, 1, 3, 1, 1, 1; z + 5). • Broom(1, k, 1, 1, 1, 1; z) is DS for all k ≥ 1 not a multiple of 4 and z ≥ 0 except when k = 1 and z ≥ 1 or k = 5 and z ≥ 5. • Broom(1, 1, 1, k, m, 1; z) is DS for all k, m ≥ 1 and z ≥ 0 except for k = m = 1 and z ≥ 1. • Broom(1, 1, 1, 1, 1, k; z) is DS for all k ≥ 1 and z ≥ 0 except for k = 1 or k = 5 and z ≥ 1. • P2 × K3 (k, 1, 1, 1, 1, 1; z) is DS for all k ≥ and z ≥ 0 except for k = 1 and z ≥ 2 or k = 3 and z ≥ 4.
25
Chapter 4 λ = −1 In this chapter we will primarily analyze graphs arising from P4 . First we will show that all graphs coming from P1 , P2 , and P3 are DS. Characteristic polynomials for each can be found at the end of this chapter. Each graph in P4 is given its own section. Within each of these sections we analyze the blowup graphs where all but two partitions are fixed at size 1. For each of these, we first list the relevant coefficients and give relationships found between them. These will be used with the other graphs from P4 to find all cospectral pairs, if any exist. P1 , P2 , and P3 are each analyzed in one section.
4.1
P1, P2, and P3
P1 and P2 are easy to analyze since they each contain one graph. The following proposition will show that each graph arising as a blowup from P1 or P2 is DS. Proposition 4.1.1. If Spec(G) = {−1n1 +...+nk −k , n1 − 1, . . . , nk − 1}, then G is isomorphic to the disjoint union of Kn1 , . . . , Knk . Proof. The spectrum of A+I is {0n1 +...+nk −k , n1 , . . . , nk }. So A+I is positive semi-definite of rank k and can be written as V T V , where V has columns consisting of vectors v1 , . . . , vn ∈ Rk . 26
Since the diagonal entries of A+I are all 1 we get that each vi is a unit vector. Now, since all entries in A + I are either 0 or 1, this means all of the vectors vi are either orthogonal or are equal. This means A + I can be reordered so that it is a block diagonal matrix with all-ones matrices on the diagonal, the sizes of which are n1 , . . . , nk . Therefore A is the adjacency matrix of Kn1 + . . . + Knk . Corollary 4.1.2. All graphs arising as a blowup from P1 or P2 are DS. Not only does this show that all graphs arising from P1 and P2 are DS, but it show that nK1 (a1 , . . . , an ) is DS for all n. Therefore we may exclude nK1 from consideration when determining which graphs are DS or cospectral. Using this logic with P3 , we see that we only need to check when P3 (a, b, c) is cospectral with P3 (d, e, f ). Proposition 4.1.3. P3 (a, b, c) is DS for all a, b, c ≥ 1. Proof. From Section 4.8, we see that the coefficients for P3 (a, b, c) are c1 = a + b + c, c2 = ac, and c3 = abc. Suppose P3 (a, b, c) ∼ P3 (d, e, f ) for some d, e, f ≥ 1. Then ac = df and abc = def imply b = e. We are then left with a + c = d + f and ac = df , which implies {a, c} = {d, f }, and therefore P3 (a, b, c) ∼ = P3 (d, e, f ).
4.2
P3(a, b, c) + K1(d)
P3 (k, m, 1) + K1 (1): d1 = k + m + 2 d2 = 2k + m + 1 d3 = km − k d4 = km
c3 < c 4 c1 + c4 = c2 + c3 + 1
27
(4.1) (4.2)
With c1 = k + m + 2 and c3 = km − k we get q 1 2 c1 − 3 ± c1 − 6c1 − 4c3 + 9 , m = c1 − 2 − k. k= 2 With c2 = 2k + m + 1 and c3 = km − k we get q 1 2 k= c2 − 2 ± c2 − 4c2 − 8c3 + 4 , m = c2 − 1 − 2k. 4 Using c4 = km, these give, respectively, q 1 2 c4 = c1 + 2c3 − 3 ± c1 − 6c1 − 4c3 + 9 2 q 1 2 c4 = c2 + 4c3 − 2 ± c1 − 4c1 − 8c3 + 4 4 After rearranging, squaring, and simplifying, we get c1 c3 + c23 + c24 + 3c4 = c1 c4 + 2c3 c4 + 2c3 c2 c3 + 2c23 + 2c24 + 2c4 = c2 c4 + 4c3 c4 + c3 Using both of these, we can finally simplify to the equation 2c1 c3 + c2 c4 + 4c4 = 2c1 c4 + c2 c3 + 3c3
(4.3)
P3 (a, b, c) + K1 (d) : Using equation (4.3) to solve for a we get a=
−b2 d2 + 3b2 d − 2b2 − bcd2 + 3bcd − 2bc + bd2 − 4bd + 3b − cd2 + 2cd + 2d2 − 3d bcd − bc + bd2 − 3bd + 2b + cd + d2 − 2d
As messy as this is, we can maximize a, with the assumption that c > 1, and show that a ≤ 1 with equality iff d = 1. Therefore P (a, b, c) + K1 (d) ∼ P3 (k, m, 1) + K1 (1) iff a = d = 1 (or similarly c = d = 1). Supposing now that c = d = 1, we have the equations km − k = ab − a and km = ab, which are enough to conclude a = k and b = m. P4 (a, b, c, d) : First suppose d = 1. From equation (4.2) we then get c = ac + bc 28
This implies a + b = 1 which cannot happen with a, b ≥ 1. So we must have a, d > 1. Next we can look at equation (4.3). a2 bc2 d + a2 bcd2 + 2a2 bc + ab2 cd2 + 2ab2 c + 2abc2 + 8abcd + 2b2 cd + 2bc2 d + 2bcd2 = a2 bc2 + 3a2 bcd + 3ab2 cd + 3abc2 d + 3abcd2 + 3abc + b2 cd2 + 3bcd Since bc 6= 0 is a common factor throughout we can reduce this to a2 cd + a2 d2 + 2a2 + abd2 + 2ab + 2ac + 8ad + 2bd + 2cd + 2d2 = a2 c + 3a2 d + 3abd + 3acd + 3ad2 + 3a + bd2 + 3d With the aid of Mathematica we calculate the mimimum of a2 cd + a2 d2 + 2a2 + abd2 + 2ab + 2ac + 8ad + 2bd + 2cd + 2d2 −(a2 c + 3a2 d + 3abd + 3acd + 3ad2 + 3a + bd2 + 3d) as 4. Therefore equation (4.3) is never satisified for a, d ≥ 2 and b, c ≥ 1.
K1,3 (a, b, c, d): Equation (4.2) gives 2abcd + bcd + a + b + c + d = acd + abd + acd + bc + bd + cd + 1 Now, (b − 1)(c − 1)(d − 1) = bcd − bc − bd − cd + b + c + d − 1 ≥ 0 for all b, c, d ≥ 1 implies bcd + b + c + d ≥ bc + bd + cd + 1 with equality when one of b, c, d equals 1. Suppose d = 1. Then a(b − 1)(c − 1) = 2abc − abc − ab − ac + a ≥ 0 for all a, b, c ≥ 1 implies 2abc + a ≥ abc + ab + ac for all a, b, c ≥ 1 with equality when one of b, c equals 1. Suppose c = 1. We now look at (4.3) with c = d = 1. This gives 4a2 b + 2a2 + 8ab2 + 18ab + 4a − 2b2 − 4b = 4a2 b + 8ab2 + 18ab + 4a − 2b2 − 4b which gives 2a2 = 0. This cannot happen with a ≥ 1. 29
C4 (a, b, c, d): With equation (4.2) we get 3abcd + a + b + c + d = abc + abd + acd + bcd + ac + bd + 1 Solving for a we get a=
bcd + bd − b − c − d + 1 3bcd − bc − bd − cd − c + 1
Now we can show that a ≤ 1 unless c = 1 or b = d = 1. To see this we will show 3bcd − bc − bd − cd − c + 1 ≥ bcd + bd − b − c − d + 1 with equality when c = 1 or b = d = 1. This inequality simplifies to 2bcd − 2bd − bc − cd + b + d = (c − 1)(2bd − b − d) ≥ 0 which gives the desired result. Now suppose b = d = 1. Equation (4.3) gives 3a2 c2 + 4a2 c + 2a2 + 4ac2 + 27ac + 4a + 2c2 + 4c = 2a2 c2 + 7a2 c + 7ac2 + 20ac + 4a + 4c which reduces to a2 c2 − 3a2 c + 2a2 − 3ac2 + 7ac + 2c2 = 0 Thinking of this as a quadratic in a we get √ 3c2 − 7c ± c c2 − 18c + 33 a= 2(c2 − 3c + 2) We immediately see that a complex when 3 ≤ c ≤ 15. For c = 1 we get a = −0.5 and with c = 2 we get a = −4. For c ≥ 16 we will show that both roots are in (1, 2). In the above, if we plug in a = 1, 1.55, 2 we get, respectively, 4c + 2 > 0 −0.2475c2 + 3.6425c + 4.805 < 0 2c + 8 > 0 The roots of −0.2475c2 +3.6425c+4.805 are c = −1.2183..., 15.9355..., and it is thus negative for c ≥ 16. A symmetric argument works to for when a = c = 1. Since a is never a positive integer here, we are done. 30
P5 (a, b, c, d, e): Equation (4.2) gives abcd+abce+acde+a+bcde+b+c+d+e = abc−ace+ac+ad+ae+bcd+bd+be+cde+ce+1 Solving for a gives a=
−bcde + bcd + bd + be − b + cde + ce − c − d − e + 1 bcd + bce − bc + cde + ce − c − d − e + 1
Similar to the C4 case we can show that a ≤ 1 with equality iff c = e = 1. Observe bcd + bce − bc + cde + ce − c − d − e + 1 −(−bcde + bcd + bd + be − b + cde + ce − c − d − e + 1) = bcde + bce − bc − bd − be + b = b(cde + ce − c − d − e + 1) And b(cde + ce − c − d − e + 1) ≥ 0 with equality iff c = e = 1. Now with a = c = e = 1, equation (4.3) gives 2b2 d2 +5b2 d+3b2 +5bd2 +26bd+11b+3d2 +11d−6 = b2 d2 +6b2 d+3b2 +6bd2 +23bd+11b+3d2 +11d−6 which simplifies to b2 d2 − b2 d − bd2 + 3bd = bd(bd − b − d + 3) = 0 This cannot happen with b, d ≥ 1.
C6 (a, b, c, d, e, f ): Similar to above, if we solve equation (4.2) for a and show that a < 1. The calculations follow the same pattern, but they are a bit too cumbersome to fit here.
Therefore we may conclude that P3 (k, m, 1) + K1 (1) is DS for all k, m ≥ 1.
31
P3 (k, 1, m) + K1 (1): d1 = k + m + 2 d2 = km + k + m + 1 d3 = 0 d4 = km c3 = 0
(4.4)
c1 + c4 = c2 + 1
(4.5)
Similar to the case of P3 (k, m, 1) + K1 (1), equations (4.4) and (4.5) give that c1 + c4 = c2 + c3 + 1. We can use the information from P3 (k, m, 1) + K1 (1) to aid us.
P3 (a, b, c) + K1 (d): From equation (4.4) we get abc − acd = ac(b − d) = 0 which gives b = d. Equation (4.5) now gives ab2 c + a + 2b + c = ab + ac + b2 + bc + 1 which reduces to ab2 c − ab − ac + a − b2 − bc + 2b + c − 1 = (b − 1)(abc + ac − a − b − c + 1) = 0 We are now left with either b = 1 or a = c = 1. The former gives k+m+2 = a+c+2 km = ac which says a = k, c = m or a = m, c = k. The latter gives k + m + 2 = 2b + 2 km = b2 which gives k = m = b. So we get that P3 (b, 1, b) + K1 (1) ∼ P3 (1, b, 1) + K1 (b) for all b ≥ 1. 32
The stronger result of P3 (a, b, a) + K1 (b) ∼ P3 (b, a, b) + K1 (a) for all a, b ≥ 1 can be shown through direct computations.
P4 (a, b, c, d): c3 = abc + bcd > 0.
K1,3 (a, b, c, d): From the P3 (k, m, 1) + K1 (1) case we know that c = d = 1. Now we have with equation (4.4) 2ab + a − b = 0 But this is not possible with a, b ≥ 1.
C4 (a, b, c, d): c3 = abc + abd + acd + bcd > 0.
P5 (a, b, c, d, e): From the P3 (k, m, 1) + K1 (1) case we have a = c = e = 1. Now we have with equation (4.4) bd + b + d − 1 = 0 But this is not possible with b, d ≥ 1.
C6 (a, b, c, d, e, f ): We know from the P3 (k, m, 1) + K1 (1) case above that c1 + c4 6= c2 + c3 + 1 for any a, b, c, d, e, f ≥ 1.
Therefore we may conclude that P3 (k, 1, m) + K1 (1) is DS for all k 6= m with P3 (k, 1, k) + K1 (1) ∼ P3 (1, k, 1) + K1 (k).
33
P3 (k, 1, 1) + K1 (m): d1 = k + m + 2 d2 = km + k + 2m d3 = k − km d4 = km
c3 ≤ 0 2c1 = c2 + c3 + 4
(4.6) (4.7)
It is straightforward to show that k = c3 + c4 m = c1 − c3 − c4 − 2 Since c4 = km we get c4 = (c3 + c4 )(c1 − c3 − c4 − 2)
(4.8)
P3 (a, b, c) + K1 (d): Using equation (4.8) we get ac(ab2 cd2 + 2ab2 cd + ab2 c − 2abcd2 − 2abcd − abd − ab + acd2 + ad −b2 d − b2 − bcd − bc − bd2 + 3bd + 2b + cd + d2 − 2d) = 0 Since ac ≥ 1 we can focus on the rest. Consider it as a polynomial in b. That is f (a, b, c, d) = b2 (acd2 + 2acd + ac − d − 1) +b(−2acd2 − 2acd − ad − a − cd − c − d2 + 3d + 2) +acd2 + ad + cd + d2 − 2d We can show that when f (a, b, c, d) has real roots, then they must be in (0, 2). This effectively shows that b = 1 is the only possibility. To see this, we first calculate f at b = 0, 1, 2, and 34
we also calculate the inflection points of f which will lie in ( 21 , 2). f (a, 0, c, d) = acd2 + ad + cd + d2 − 2d f (a, 1, c, d) = ac − a − c + 1 f (a, 2, c, d) = acd2 + 4acd + 4ac − ad − 2a − cd − 2c − d2
It is now easy to see that f (a, 0, c, d) > 0, f (a, 1, c, d) ≥ 0, and f (a, 2, c, d) > 0 for a, c, d ≥ 1 with f (a, 1, c, d) = 0 iff a = 1 or c = 1. Now, we also have ∂f (a, b, c, d) = 2abcd2 + 4abcd + 2abc − 2acd2 − 2acd − ad − a − 2bd − 2b − cd − c − d2 + 3d + 2 ∂b This implies that the inflection point is at b∗ = We need to show
1 2
2acd2 + 2acd + ad + a + cd + c + d2 − 3d − 2 2(d + 1)(acd + ac − 1)
< b∗ < 2. For this we look at
2acd2 + 2acd + ad + a + cd + c + d2 − 3d − 2 − 2(2(d + 1)(acd + ac − 1)) = −2acd2 − 6acd − 4ac + ad + a + cd + c + d2 + d + 2 < 0 2(2acd2 + 2acd + ad + a + cd + c + d2 − 3d − 2) − 2(d + 1)(acd + ac − 1) = 2acd2 − 2ac + 2ad + 2a + 2cd + 2c + 2d2 − 4d − 2 > 0 And thus
1 2
< b∗ < 2. Now, if f (a, b∗ , c, d) > 0, then f has no real roots. If f (a, b∗ , c, d) ≤ 0,
then the roots of f are in (0, 1). Since the only integer solution is at b = 1 with a = 1 or c = 1, we are left to consider if P3 (a, 1, 1)+K1 (d) ∼ P3 (k, 1, 1)+K1 (m). Using k +m+2 = a+d+2 and km + k + 2m + ad + a + 2d we conclude that a = k and d = m.
P4 (a, b, c, d): c3 = abc + bcd > 0.
35
K1,3 (a, b, c, d): Similar to P3 (a, b, c) + K1 (d) we need to look at equation (4.8). In this case we get f (a, b, c, d) = 4a2 b2 c2 d2 + 4a2 b2 c2 d + a2 b2 c2 + 4a2 b2 cd2 + 2a2 b2 cd + a2 b2 d2 + 4a2 bc2 d2 +2a2 bc2 d + 2a2 bcd2 − 2a2 bcd − a2 bc − a2 bd + a2 c2 d2 − a2 cd − 4ab2 c2 d2 −2ab2 c2 d − 2ab2 cd2 − 2ab2 cd − ab2 c − ab2 d − 2abc2 d2 − 2abc2 d − abc2 −2abcd2 + 4abcd + 2abc − abd2 + 2abd − ac2 d − acd2 + 2acd + b2 c2 d2 +b2 cd + bc2 d + bcd2 − 2bcd = 0 This time we will consider this as a polynomial in a and will show the real roots are in (0, 1), and therefore has no positive integer roots. Looking at f (1, b, c, d) we have f (1, b, c, d) = b2 c2 d2 + 2b2 c2 d + b2 c2 + 2b2 cd2 + b2 cd − b2 c + b2 d2 − b2 d + 2bc2 d2 +bc2 d − bc2 + bcd2 + bc − bd2 + bd + c2 d2 − c2 d − cd2 + cd From this we can see that f (1, b, c, d) > 0 for all b, c, d ≥ 1. We also need to calculate the inflection point a∗ . As a fraction it is too unwieldy to list here, but from it we can calculate in a similar fashion to P3 (a, b, c) + K1 (d) that 0 < a∗
0.
P5 (a, b, c, d, e): As above we use equation (4.8) to resolve this case. For P5 (a, b, c, d, e) we can factor out a c which leaves us with a linear equation in c left to look at. Solving said linear equation for c leads to a rational function with many terms, but it is straightforward to show from it that c < 1. In particular c can never be a positive integer.
36
C6 (a, b, c, d, e, f ): Using equation (4.8) and solving for a as well as the inflection point a∗ leads to a, a∗ < 0 when a is real. Therefore a is never a positive integer.
Therefore we may conclude that P3 (k, 1, 1) + K1 (m) is DS for all k, m ≥ 1.
37
P3 (1, k, 1) + K1 (m): d1 = k + m + 2 d2 = km + 2m + 1 d3 = k − m d4 = km
c3 < c 4 c1 + c4 = c2 + c3 + 1
(4.9) (4.10)
From c1 = k + m + 2, c3 = k − m, and c4 = km we get 4c4 = (c1 + c3 − 2)(c1 − c3 − 1)
(4.11)
P3 (a, b, c) + K1 (d): Equation (4.10) says abcd − abc + acd − ac − ad + a − bd + b − cd + c + d − 1 = (abc + ac − a − b − c + 1)(d − 1) = 0 This happens iff a = c = 1 or d = 1. If a = c = 1, then b + d + 2 = k + m + 2 and bd + 2d + 1 = km + 2m + 1 imply b = k and d = m. If d = 1, then a + b + c + 1 = k + m + 2, abc − ac = k − m, and abc = km give −a2 b2 c2 + 2a2 bc2 − a2 c2 + a2 − 4abc + 2ab + 2ac − 2a + b2 + 2bc − 2b + c2 − 2c + 1 = 0 Treating this as a polynomial in b we get at b = 0, 1, 2, respectively f (a, 0, c) = −a2 c2 + a2 + 2ac − 2a + c2 − 2c + 1 = −(a − 1)(c − 1)(ac + a + c − 1) f (a, 1, c) = a2 − 2ac + c2 = (a − c)2 f (a, 2, c) = −a2 c2 + a2 − 6ac + 2a + c2 + 2c + 1 From these it is immediately clear that f (a, 0, c) < 0 iff a, c > 1 and f (a, 1, c) ≥ 0 for a, c ≥ 1 with equality iff a = c. A little more work will show that f (a, 2, c) ≤ 0 for a, c ≥ 1 with 38
equality iff a = c = 1. Moreover, we actually have f (1, b, 1) = 0 for all b ≥ 1. We have already checked the case a = c = 1 anyway, so we may ignore it. Having either a > 1 or c > 1, we see from f (a, b, c) that b ∈ (0, 2) with b = 1 iff a = c > 1. From here it is easy to check that a = c = k = m. Thus P3 (1, k, 1) + K1 (k) ∼ P3 (k, 1, k) + K1 (1) for all k ≥ 1, which we have already seen above.
P4 (a, b, c, d): Equation (4.11) gives a2 b2 c2 − a2 + 2ab2 c2 d + 4abcd − abc − 2ab − 2ac − 2ad + 3a +b2 c2 d2 − b2 − bcd − 2bc − 2bd + 3b − c2 − 2cd + 3c − d2 + 3d − 2 = 0 It is easy to show that the left side of the above equation is always greater than 0 except when a = b = c = d = 1. However, we know from equation (4.10) that a, d > 1, so the left hand side is strictly greater than 0 for our purposes.
K1,3 (a, b, c, d): Equation (4.10) gives c = d = 1. Equation (4.11) then gives 2a(2ab2 + 2ab − 2b2 + b − 1) > 0 for a, b ≥ 1.
C4 (a, b, c, d): Equation (4.10) gives b = d = 1. Equation (4.11) then gives 4a2 c2 + 4a2 c + 4ac2 + 10ac − 2a − 2c > 0 for a, c ≥ 1.
P5 (a, b, c, d, e): Equation (4.10) gives a = c = e = 1. Equation (4.11) then gives b2 d2 + 2b2 d + 2bd2 + 5bd − 2b − 2d > 0 39
for b, d ≥ 1.
C6 (a, b, c, d, e, f ): Equation (4.10) is never satisfied for a, b, c, d, e, f ≥ 1.
Therefore we may conclude that P3 (1, k, 1) + K1 (m) is DS whenever k 6= m with P3 (1, k, 1) + K1 (k) ∼ P3 (k, 1, k) + K1 (1) for all k ≥ 1.
40
4.3
P4(a, b, c, d)
P4 (k, m, 1, 1): d1 = k + m + 2 d2 = 2k + m d3 = km + m d4 = km
c3 > c 4 2c1 + c4 = c2 + c3 + 4
(4.12) (4.13)
P3 (a, b, c) + K1 (d): c3 = abc − bcd < abcd = c4 for all a, b, c, d ≥ 1.
P4 (a, b, c, d): For c3 > c4 we need c3 − c4 = abc + bcd − abcd = bc(a + d − ad) > 0 Since bc > 0 for b, c ≥ 1 we must have a + d − ad > 0. This happens when a = 1 or d = 1. Assume d = 1. Then equation (4.13) gives abc + 2a + 2b + 2c + 2 = abc + ac + a + bc + b + 4 This simplifies to a + b + 2c − ac − bc − 2 = (2 − a − b)(c − 1) = 0 This says that either a = b = 1 or c = 1. If a = b = 1, then c + 3 = k + m + 2 and c + 2 = 2k + m imply k = 1 and m = c, which is just P4 (1, c, 1, 1) ∼ = P4 (1, 1, c, 1). On other other hand, if c = 1, then a + b + 2 = k + m + 2 and 2a + b = 2k + m imply a = k and b = m, which is just P4 (a, b, 1, 1) ∼ = P4 (k, m, 1, 1). 41
K1,3 (a, b, c, d): c3 > c4 implies abc + abd + acd > 2abcd + bcd Ignore for a moment the bcd term. Then we can factor abc + abd + acd − 2abcd = a(bc + bd + cd − 2bcd). Now 2bcd ≥ bc + bd + cd when at least two of b, c, d are greater than 1. In that case we would have abc + abd + acd < 2abcd + bcd. Therefore we may assume c = d = 1. With that equation (4.13) now gives 2ab + 2a + 2b + 4 = 2ab + a + b + 5 This simplifies to a + b = 1 which is obviously not possible with a, b ≥ 1.
C4 (a, b, c, d): For c3 > c4 we need abc + abd + acd + bcd > 3abcd Suppose c, d > 1. Then abcd ≥ acd, abcd ≥ bcd and abcd ≥ abc + abd = ab(c + d). This last inequality follows from cd ≥ c + d for c, d ≥ 2. Therefore, by the symmetry of C4 , we may assume that at most one of a, b, c, d is greater than 1. Suppose b = c = d = 1. Then equation (4.13) gives 5a + 6 = 4a + 6 This implies that a = 0, which cannot happen.
P5 (a, b, c, d, e): c3 ≤ c4 for all a, b, c, d, e ≥ 1.
C6 (a, b, c, d, e, f ): c3 ≤ c4 for all a, b, c, d, e, f ≥ 1. Therefore we may conclude that P4 (k, m, 1, 1) is DS for all k, m ≥ 1.
42
P4 (k, 1, m, 1): d1 = k + m + 2 d2 = km + k + 1 d3 = km + m d4 = km
c3 > c 4 c1 + 2c4 = c2 + c3 + 1
(4.14) (4.15)
P3 (a, b, c) + K1 (d): Conveniently (4.12) and (4.14) are the same, so we can reuse the information from the P4 (k, m, 1, 1) case. We know that c3 < c4 for all a, b, c, d ≥ 1.
P4 (a, b, c, d): We may assume d = 1 from (4.14). Equation (4.15) then gives 2abc + a + b + c + 1 = abc + ac + a + bc + b + 1 This simplifies to abc − ac − bc + c = (a − 1)(b − 1)c = 0 Either a = 1 or b = 1. First suppose a = 1. Then b+c+2 = k +m+2, b+c+1 = km+k +1, and 2bc = km + m imply k = 1 and either b = 1 and c = m or b = m and c = 1. In either case we are left with isomorphic graphs. Now suppose b = 1. Then we are once again we get isomorphic graphs from a + c + 2 = k + m + 2, ac + a + 1 = km + k + 1, and ac + c = km + m giving a = k and c = m.
K1,3 (a, b, c, d): Inequality (4.14) gives c = d = 1. Equation (4.15) then gives 4ab + a + b + 2 = 2ab + a + b + 2 43
This simplifies to 2ab = 0, which cannot happen.
C4 (a, b, c, d): Inequality (4.14) implies b = c = d = 1. Equation (4.15) then gives 7a + 3 + 4a + 3 This simplifies to 3a = 0, which cannot happen.
P5 (a, b, c, d, e): Inequality (4.14) is never satisfied with a, b, c, d, e ≥ 1.
C6 (a, b, c, d, e, f ): Inquality (4.14) is never satisfied with a, b, c, d, e, f ≥ 1.
Therefore we may conclude that P4 (k, 1, m, 1) is DS for all k, m ≥ 1.
44
P4 (k, 1, 1, m): d1 = k + m + 2 d2 = km + k + m d3 = k + m d4 = km
c1 = c3 + 2
(4.16)
c2 = c3 + c4
(4.17)
P3 (a, b, c) + K1 (d): Equation (4.16) gives a + b + c + d = abc − bcd + 2 Solving for a we get a=
b+c+d−2 bc − cd − 1
The numerator is always positive and the numerator is positive whenever c > 1 or c = 1 and b > d + 1. In either case we can now use this in equation (4.17). Doing so we get bcd(b + c + d − 2) (c + d − bc + cd)(b + c + d − 2) + (bd + cd)(bc − cd − 1) = bc − cd − 1 bc − cd − 1 Assuming c > 1 we can factor out (bc − cd − 1). Doing so we are left with b2 c + bc2 + 2bcd2 + c2 d2 + cd + 2c + 2d = 2bcd + 3bc + c2 d + c2 + cd2 + d2 It can be shown that the left hand side is always strictly greater than the right side for b, c, d ≥ 1. If c = 1, then we have a=
b+d−1 b−d−1
Equation (4.17) then gives b2 + 2bd2 + 2d + 1 = d2 + 2bd + 2b Once again the left side is always greater than the right.
45
P4 (a, b, c, d): Equation (4.16) gives a + b + c + d = abc + bcd + 2 It is easy to now show that b = c = 1. First observe that abc+1 ≥ a+b for all a, b, c ≥ 1 with equality when either a = c = 1 or b = c = 1. In either case c = 1. Similarly, bcd + 1 ≥ c + d forces b = 1. From here a + d + 2 = k + m + 2 and ad = km gives a = k and d = m or a = m and d = k. In both cases we’re left with isomorphic graphs.
K1,3 (a, b, c, d): Equation (4.17) gives 2abcd + abc + abd + acd = bcd + bc + bd + cd The left side factors as a(2bcd + bc + bd + cd), and thus equality above is clearly never possible.
C4 (a, b, c, d): Equation (4.16) gives a + b + c + d = abc + abd + acd + bcd + 2 Equality is clearly never possible.
P5 (a, b, c, d, e): Equality is never possible in equation (4.16).
C6 (a, b, c, d, e, f ): Equality is never possible in equation (4.16).
Therefore we may conclude that P4 (k, 1, 1, m) is DS for all k, m ≥ 1.
46
P4 (1, k, m, 1): d1 = k + m + 2 d2 = k + m + 1 d3 = 2km d4 = km
c3 > c 4
(4.18)
c1 = c2 + 1
(4.19)
c3 = 2c4
(4.20)
P3 (a, b, c) + K1 (d): Conveniently (4.18) is the same as (4.12) and (4.14). c3 < c4 for all a, b, c, d ≥ 1.
P4 (a, b, c, d): Inequality (4.18) gives d = 1. Equation (4.19) then gives a + b + c + 1 = ac + a + b + 1 This simplifies to ac − c = (a − 1)c = 0. Since c ≥ 1 we must have a = 1. Finally, b + c + 2 = k + m + 2 and bc = km give b = k and c = m or b = m and c = k. In either case we have isomorphic graphs.
K1,3 (a, b, c, d): Inequality (4.18) gives c = d = 1. Equation (4.20) then gives 2ab + a − b = 4ab This simplifies to 2ab − a + b = 0, which is not possible with a, b ≥ 1.
47
C4 (a, b, c, d): Inequality (4.18) gives b = c = d = 1. Equation (4.19) then gives a+3=a+2 This implies a = 0, which is not possible.
P5 (a, b, c, d, e): Inequality (4.18) is never satisfied with a, b, c, d, e ≥ 1.
C6 (a, b, c, d, e, f ): Inquality (4.18) is never satisfied with a, b, c, d, e, f ≥ 1.
Therefore we may conclude that P4 (1, k, m, 1) is DS for all k, m ≥ 1.
48
4.4
K1,3(a, b, c, d)
K1,3 (k, m, 1, 1): d1 = k + m + 2 d2 = 2m + 1 d3 = 2km + k − m d4 = 2km c1 + c4 = c2 + c3 + 1 2c4 = (2c1 − c2 − 3)(c2 − 1)
(4.21) (4.22)
P3 (a, b, c) + K1 (d): Equation (4.21) gives abcd + a + b + c + d = abc − acd + ac + ad + bd + cd + 1 This simplifies to abcd − abc + acd − ac − ad + a − bd + b − cd + c + d − 1 = (abc + ac − a − b − c + 1)(d − 1) = 0 This implies either a = c = 1 or d = 1. If a = c = 1, then equation (4.22) gives b2 d2 − 2b2 d + 2bd2 − 2bd = bd(bd − 2b + 2d − 2) = 0 Since bd ≥ 1 we must have bd − 2b + 2d − 2 = 0. If we solve this for b we get b= From this we quickly get that b ≥ 1 iff
4 3
2 − 2d d−2
≤ d < 2, and therefore d is not an integer. On the
other hand, if d = 1 then equation (4.22) gives a2 c2 − a2 + 2abc − 2ab − 2ac + 2a − b2 − 2bc + 2b − c2 + 2c − 1 = 0 Unfortunately from here we run into a problem. If we solve this for b and take the positive answer we get b=
p 2a(a − 1)c(c − 1) + ac − a − c + 1 49
This is an integer when 2a(a − 1)c(c − 1) is a perfect square. This is solved in a lemma below, but the result shows that for almost all a ∈ N there are infinitely many c ∈ N such that 2a(a − 1)c(c − 1) is a perfect square.
P4 (a, b, c, d): We have k = 21 (2c1 − c2 − 3). For this to be positive we obviously need 2c1 > c2 + 3. From this we get that −bd + 2b + 2c + 2d − 3 c+d−2 For b, c ≥ 1 and d ≥ 2 we get that the right side is always at most 3, and as such we get 1≤a
c2 + 3 gives 2a + 2b + 2c + 2d + 2e + 2f > ac + ad + ae + bd + be + bf + ce + cf + df + 3 This is easily shown to never be true with a, b, c, d, e, f ≥ 1.
Therefore we may conclude that K1,3 (k, m, 1, 1) is DS for all k, m ≥ 1 except for C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1) and P3 (a, n + ac − a − c + 1, c) + K1 (1) ∼ K1,3 (k, m, 1, 1) where 2a(a−1)c(c− 1) = n2 is a perfect square, k = n2 , and m = ac + n2 . When is 2a(a − 1)c(c − 1) a perfect square with a, c ≥ 2? If 2a(a − 1) = m2 n, where n is squarefree, then we have the equation m2 nc(c−1) = k 2 . Clearly for this to have a solution we need nc(c−1) = k 02 . Therefore we may first consider when nc(c−1) is a perfect square. Lemma 4.4.1. If (x, y) ∈ Z2 solves the equation x2 − ny 2 = k with k|x, then x2 x2 nxy 2 n −1 = k k k 2 2 is an integral solution to nc(c − 1) = k 02 . If xk < 0 then we may take xk and
x2 ( − 1) k for
positive solutions. Proof. Since x2 − ny 2 = k we get x2 − k y = n Multiplying both sides by n2 x2 gives nx2 (x2 − k) = n2 x2 y 2 . Now, since k|x we can divide 2
both sides by k 2 to get x2 n k
x2 nxy 2 −1 = k k 52
This is clearly an integral solution to nc(c − 1) = k 02 . Lemma 4.4.2. If nc(c − 1) is a perfect square, then c = n1 c21 and c − 1 = n2 c22 where n = n1 n2 . Corollary 4.4.3. Using the notation of the previous lemma, if nc(c − 1) is a perfect square, then (n1 c1 )2 − nc22 = n1 . With these lemmas we have that solving nc(c − 1) = k 02 is equivalent to looking for solutions of x2 − ny 2 = k with the conditions that k|x and k|n. Luckily this is a problem of antiquity. Firstly we can show that knowing two solution (x1 , y1 , k1 ) and (x2 , y2 , k2 ) to x2 − ny 2 = k we can get a third solution, and thus infinitely many. Lemma 4.4.4. If (x1 , y1 , k1 ) and (x2 , y2 , k2 ) both solve x2 −ny 2 = k, then (x1 x2 +ny1 y2 , x1 y2 + x2 y1 , k1 k2 ) is another solution. Proof. This follows immediately from Brahmagupta’s identity which states (x21 − ny12 )(x22 − ny22 ) = (x1 x2 + ny1 y2 )2 − n(x1 y2 + x2 y1 )2 .
What is even better is that we really only need one solution (x1 , y1 , k1 ) of x2 − ny 2 = k to generate the rest through recursion with xi+1 = x1 xi + ny1 yi yi+1 = x1 yi + y1 xi ki+1 = k1 ki The only concern with this now is whether or not ki |xi and ki |n. If k1 = 1, then ki = 1 for all i, and Lagrange showed that x2 − ny 2 = 1 has infinitely many solutions as long as n is not a perfect square. So we are guaranteed solutions, but we need not always have ki = 1. For example we have 22 − 6 · 12 = −2, and −2|2 and −2|6 which gives 6 · 3 · (3 − 1) = 62 . If we use the Lemma 4.4.4 with (2, 1, −2) then we get (10, 4, 4) at the next iteration, and we can check that 102 − 6 · 42 = 4, but 46 | 6. To remedy this we can divide x and y by −2 and 53
k by 4. This gives the triple (−5, −2, 1). We still have (−5)2 − 6 · (−2)2 = 1 and 1|6. To account for this we can modify the lemma above. Lemma 4.4.5. If (x1 , y1 , k1 ) and (x2 , y2 , k1 ) are integral solutions to x2 − ny 2 = k with k1 |n and k1 |xi , then
x1 x2 + ny1 y2 x1 y2 + x2 y1 , ,1 k1 k1
is another integral solution. Proof. This again follows immediately from Brahmagupta’s identity, but this time dividing both sides by k12 . This gives k2 (x2 − ny12 )(x22 − ny22 ) 1 = 12 = 1 = k1 k12
x1 x2 + ny1 y2 k1
2
−n
x1 y2 + x2 y1 k1
2 .
This also gives us a modified recursion that guarantees ki |n and ki |xi : x1 xi + ny1 yi ki x1 y i + y 1 xi = ki k1 = ki
xi+1 = yi+1 ki+1
It can be shown that solutions to x2 − ny 2 = k show up in the study of continued fractions of √ n, but that is not necessary for our purposes. It can also be shown that all solutions can be found using a fundamental solution (x1 , y1 , k1 ) to x2 − ny 2 = k and the recursion above, but again this is not necessary for our purposes. Now that we have solutions to nc(c − 1) = k 02 , we need to consider when 2a(a − 1) = m2 n. In particular we want to know when n = 1. This, however, is just a particular case of our examination above. It is straightforward to get that a fundamental solution for x2 − 2y 2 = k is (1, 1, −1). This leads to the set of solutions for {a, a − 1} as {2Pn2 , 2Pn2 + (−1)n } where Pn is the n-th Pell number which can be defined by Pn =
(1 +
√ n √ 2) − (1 − 2)n √ 2 2
54
These are important because they are the values for a for which 2a(a − 1)c(c − 1) is never a perfect square. The first few values of a in this sequence are {2, 9, 50, 289, . . .}. All other values of a lead back to the equation nc(c − 1) = k 02 .
55
K1,3 (1, k, m, 1): d1 = k + m + 2 d2 = km + k + m d3 = k + m d4 = 2km
c1 = c3 + 2 2c3 + c4 = 2c2
(4.23) (4.24)
Equation (4.23) is the same as (4.16) from P4 (k, 1, 1, m), so we get to reuse that data.
P3 (a, b, c) + K1 (d): We will start by showing that the only possibility is k = m = d + 1. For this we need to look at PK1,3 (1,k,m,1) (d). Since d ∈ Spec(P3 (a, b, c) + K1 (d + 1)) we need PK1,3 (1,k,m,1) (d) = 0. To that effect we can plug d into PK1,3 (1,k,m,1) to get (d + 1)k+m−2 (d4 − d3 k − d3 m + 2d3 + d2 km − 2d2 k − 2d2 m + 2dkm − 2d − km + k + m − 1) For this to vanish we clearly need d4 − d3 k − d3 m + 2d3 + d2 km − 2d2 k − 2d2 m + 2dkm − 2d − km + k + m − 1 = 0 since d ≥ 1. Solving this for k we get k=
d4 − d3 m + 2d3 − 2d2 m − 2d + m − 1 d3 − d2 m + 2d2 − 2dm + m − 1
We will now show that k ∈ (d − 1, d) for m ∈ (1, d − 1); k ≤ −1 for m = d; k = d + 1 for m = d + 1; and k ∈ (d, d + 1) for m ≥ d + 2. The easiest to check are when m = d, d + 1. For m = d it is straightforward to calculate k=
−d − 1 d−1
Similarly for m = d + 1 we get k=
−d(d + 1)2 =d+1 −d(d + 1)
56
These give the desired results for m = d, d + 1. For the other two claims we need to look at d4 − d3 m + 2d3 − 2d2 m − 2d + m − 1 − (d − 1)(d3 − d2 m + 2d2 − 2dm + m − 1) = d3 − d2 m + 2d2 − 3dm − d + 2m − 2 d4 − d3 m + 2d3 − 2d2 m − 2d + m − 1 − (d + 1)(d3 − d2 m + 2d2 − 2dm + m − 1) = −d(d + 1)(d − m + 1) We also need
From
∂k ∂m
∂k d2 (d − 1)2 =− 3 ∂m (d − d2 m + 2d2 − 2dm + m − 1)2
we see that k is decreasing for all m ≥ 1 except when d3 −d2 m+2d2 −2dm+m−1 =
0, which is at m = m
0 when
≤ d − 1 and −d(d + 1)(d − m + 1) > 0 when d ≥ 1 and m ≥ d + 1. This
shows k > d − 1 for 1 ≤ m ≤ d − 1 and k < d + 1 for m > d + 1. To get the bound with d we can look at k when m = 1 and when m = d + 2. We get respectively at m = 1 and m=d+2 k=
(d + 1)(d2 − 2) d 2d2 + 3d − 1 Since k is decreasing we have the desired result. Therefore we must have k = m = d + 1. k=
Now we can use c1 = 2d + 4 and c2 = (d + 1)(d + 3) to get 2c2 − 2cd − 8c + 3 c c2 − cd − 4c + 3 b = c
a = −
Finally, c4 = 2(d + 1)2 reduces to 2c4 d − 4c3 d2 − 16c3 d + 2c2 d3 + 16c2 d2 + 41c2 d − 7cd2 − 32cd + 2c + 9d =0 c But 2c4 d − 4c3 d2 − 16c3 d + 2c2 d3 + 16c2 d2 + 41c2 d − 7cd2 − 32cd + 2c + 9d > 0 for all c, d ≥ 1.
P4 (a, b, c, d): Equation (4.23) implies b = c = 1, and we have already seen that P4 (a, 1, 1, d) is DS. 57
K1,3 (a, b, c, d): Equation (4.24) gives 2abcd + 2(abc + abd + acd − bcd) = 2(bc + bd + cd) This reduces to 2(a − 1)(bcd + bc + bd + cd) = 0, which implies a = 1. Now equation (4.23) gives b + c + d + 1 = −bcd + bc + bd + cd + 2 This reduces to (b − 1)(c − 1)(d − 1) = 0, which implies one of b, c, d is 1. Suppose d = 1. Then k + m + 2 = b + c + 2 and km + k + m = bc + b + c imply b = k and c = m or b = m and c = k. In either case we have isomorphic graphs.
C4 (a, b, c, d): Equality is never possible in equation (4.23).
P5 (a, b, c, d, e): Equality is never possible in equation (4.23).
C6 (a, b, c, d, e, f ): Equality is never possible in equation (4.23).
Therefore we may conclude that K1,3 (1, k, m, 1) is DS for all k, m ≥ 1.
58
4.5
C4(a, b, c, d)
C4 (k, m, 1, 1): d1 = k + m + 2 d2 = k + m d3 = 2km + k + m d4 = 3km
c1 = c2 + 2 3c2 + 2c4 = 3c3
P3 (a, b, c) + K1 (d): Equality is never possible in equation (4.25).
P4 (a, b, c, d): Equality is never possible in equation (4.25).
K1,3 (a, b, c, d): Equation (4.25) gives a = bc + bd + cd − b − c − d + 2 Now, k + m = c2 and 2km = c3 − c2 gives q 1 k, m = c2 ± c22 + 2c2 − 2c3 2 For these to be integers we need c22 + 2c2 − 2c3 ≥ 0. This gives −b2 c2 − 2b2 cd + 2b2 c − b2 d2 + 2b2 d − 2bc2 d + 2bc2 −2bcd2 + 8bcd − 2bc + 2bd2 − 2bd − c2 d2 + 2c2 d + 2cd2 − 2cd ≥ 0
59
(4.25) (4.26)
The only positive integer solutions, with b ≥ c ≥ d ≥ 1, to this are b ≥ 1, c = d = 1; b = c = 2, d = 1; and b = 3, c = 2, d = 1. We can quickly check that these give 4b + 1, 8, and 1, respectively. We can immediately eliminate b = c = 2, d = 1 since 8 is not a perfect square. For b = 3, c = 2, d = 1, we get a = 1, k = 4, m = 1. Finally, for b ≥ 1, c = d = 1 we have 4b + 1 is a perfect square exact when b = r(r + 1) for some positive integer r. This gives a = r(r + 1) + 1, k = (r + 1)2 , and m = r2 . Now for c4 = 3km we have 2(r(r + 1) + 1)r(r + 1) = 3r2 (r + 1)2 which reduces to (r − 1)r(r + 1)(r + 2) = 0. Since r ≥ 1 we must have r = 1, but this once again gives a = 3, b = 2, k = 4, and m = 1.
C4 (a, b, c, d): Equation (4.25) gives c = d = 1. We then get from k + m = a + b and 3km = 3ab that a = k and b = m or a = m and b = k. In either case we have isomorphic graphs.
P5 (a, b, c, d, e): Equality is never possible in equation (4.25).
C6 (a, b, c, d, e, f ): Equality is never possible in equation (4.25).
Therefore we may conclude that C4 (k, m, 1, 1) is DS for all k, m ≥ 1 except for C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1).
60
C4 (k, 1, m, 1): d1 = k + m + 2 d2 = km + 1 d3 = 2km + k + m d4 = 3km c1 + 2c2 = c3 + 4
(4.27)
3c2 = c4 + 3
(4.28)
3c1 + 2c4 = 3c3 + 6
(4.29)
Note that 0 ∈ Spec(C4 (k, 1, m, 1)). Also note that from c1 = k + m + 2 and c2 = km + 1 we get m2 + (2 − c1 )m + c2 − 1 = 0
(4.30)
P3 (a, b, c) + K1 (d): 0 ∈ Spec(P3 (a, b, c) + K1 (d)) iff a = c = 1 or d = 1. If a = c = 1 then equation (4.27) reduces to 2(b + 3)d = 0, which cannot happen. Suppose then that d = 1. Then we can solve equation (4.27) for b=
3(ac + a + c − 1) ac − 3
We can now use this with (4.30) to get (ac − 3)m2 − ac(a + c + 2)m + ac(ac + a + c − 1) =0 ac − 3 Let f (a, c, m) = (ac − 3)m2 − ac(a + c + 2)m + ac(ac + a + c − 1). Then we can show that the only positive integer roots for f are at m = 6, 12. To do this we need to assume a ≥ c and then look at f (a, c, c − 2) = 3a2 c − 3ac2 + 7ac − 3c2 + 12c − 12 f (a, c, c − 1) = 2a2 c − 2ac2 + 2ac − 3c2 + 6c − 3 f (a, c, c) = c(a2 − ac − a − 3c) f (a, c, c + 1) = −2ac − 3c2 − 6c − 3 61
From these we can show that f (a, c, c−2) > 0 for a, c ≥ 1; f (a, c, c−1) > 0 for 1 ≤ c ≤ a−1; and f (a, c, c + 1) > 0 for a, c ≥ 1. This gives m ∈ (c − 1, c + 1) for 1 ≤ c < a and m ∈ (c − 2, c + 1) for a = c. We can do better and show that m ∈ (c − 2, c) for a = c. For the latter case this would imply m = c−1 is the only possible integer root if a = c, but when a = c √ we get f (c, c, c−1) = −c2 +6c−3 which has roots c = 3± 6. Therefore we may eliminate this possibility. For c < a, on the other hand, we must look at f (a, c, c) = c(a2 − ac − a − 3c) = 0. Since c 6= 0 we must have a2 − ac − a − 3c = 0, which gives c=
a(a − 1) a+3
From this we can show that a − 4 < c < a − 3 if a > 9. For a ≤ 9 the only positive integer solutions are a = 3, c = 1 and a = 9, c = 6. For a = 3, c = 1 we would have b=
3(ac+a+c−1) ac−3
undefined. For a = 9, c = 3 we get b = 4, k = 12, and m = 6. That is,
P3 (9, 4, 6) + K1 (1) ∼ C4 (12, 1, 6, 1).
P4 (a, b, c, d): Equation (4.27) gives a + b + c + d + 2(ac + ad + bd) = abc + bcd + 4 We can solve this for a to get a=
−bcd + 2bd + b + c + d − 4 bc − 2c − 2d − 1
We can now use this with equation (4.28) to solve for b as q (c+6)d−6 (c + d − 1) +2 cd b= (c − 2)d + 2 If d = 1 then b = 3 for all c ≥ 1. This gives a = −2. If c ≥ 3 then 1 ≤ b ≤ 3 for all d ≥ 1. Using this we can show that the only integer solutions with d ≥ 4 is b = 1, c = 6, d = 25. This gives a =
24 . 19
Finally we must consider the cases c = 2. With c = 2 we have ! r 3 1 4− +2 b = (d + 1) 2 d
√ From this it is clear that d must equal 3, but even with that we would have b = 4 + 2 3. We have thus exhausted all possibilities. 62
K1,3 (a, b, c, d): Equation (4.27) gives a + b + c + d + 2(bc + bd + cd) = abc + abd + acd − bcd + 4 We can solve this for bc + bd + cd =
a + b + c + d + bcd − 4 a−2
This with equation (4.28) gives a + b + c + d + bcd − 4 3 − 3 = 2abcd a−2 This simplifies to 2a2 bcd − 4abcd − 3bcd − 3b − 3c − 3d + 6 = 0 If we look at the left side as a polynomial f (a, b, c, d), then we can show that the roots in a are in [−1, 0] ∪ [2, 3]. To do this we look at f (−1) = f (3) = 3(bcd − b − c − d + 2) ≥ 0 f (0) = f (2) = −3(bcd + b + c + d − 2) ≤ 0 Equality above happens when at least two of b, c, d are 1. Suppose c = d = 1. Equation (4.27) then reduces to 2b(a − 3) = 0, which implies a = 3. Finally, from here we can solve for b = 2, k = 4, and m = 1. That is, C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1), which we saw above.
C4 (a, b, c, d): Equation (4.28) gives 3ac + 3bd = 3abcd + 3 This simplifies to (ac − 1)(bd − 1) = 0. Thus we have a = c = 1 or b = d = 1, both of which are equivalent. Suppose b = d = 1. We can then use k +m+2 = a+c+2 and km+1 = ac+1 to get a = k and c = m or a = m or c = k. In either case we have isomorphic graphs.
P5 (a, b, c, d, e): Equation (4.29) gives 3(a + b + c + d + e) + 2(abcd + abce + acde + bcde) = 3(abc + bcd + cde − acd) + 6 63
Since 3(a+b+c+d+e) > 6 for all a, b, c, d, e ≥ 1, we must have 2(abcd+abce+acde+bcde) < 3(abc + bcd + cde − acd) for equality above. But now this reduces to 2 (abd + abe + ade + bde) + ae < ab + bd + de 3 This can easily be shown to always be false with a, b, c, d, e ≥ 1.
C6 (a, b, c, d, e, f ): Similar to the P5 (a, b, c, d, e) case above we can show that equation (4.29) is never satisfied.
Therefore, we may conclude that C4 (k, 1, m, 1) is DS for all k, m ≥ except for C4 (4, 1, 1, 1) ∼ K1,3 (3, 2, 1, 1) and C4 (12, 1, 6, 1) ∼ P3 (9, 4, 6) + K1 (1).
64
4.6
P5(a, b, c, d, e)
P5 (k, m, 1, 1, 1): d1 = k + m + 3 d2 = 3k + 2m + 1 d3 = km − k + m + 1 d4 = 2km + k + m 9c1 + c4 = 4c2 + 2c3 + 21
(4.31)
Using c1 = k + m + 1 and c2 = 3k + 2m + 1 we get k = −2c1 + c2 + 5 and m = 3c1 − c2 − 8. So we also get c3 = −6c21 + 5c1 c2 + 36c1 − c22 − 15c2 − 52
(4.32)
c4 = −12c21 + 10c1 c2 + 63c1 − 2c22 − 26c2 − 83
(4.33)
Since c1 , c2 > 0, we can also show that c3 , c4 > 0 with (4.32) and (4.33).
P3 (a, b, c) + K1 (d): Equation (4.31) gives 9(a + b + c + d) + abcd = 4(ac + ad + bd + cd) + 2(abc − acd) + 21 Solving for a gives 4bd − 9b + 4cd − 9c − 9d + 21 bcd − 2bc + 2cd − 4c − 4d + 9 We can now use (4.32) to show that 1 ≤ b ≤ 3. Let f (b, c, d) be the numerator of (4.32) a=
(with c3 moved to the right side). (f (b, c, d) is a very ugly polynomial that is omitted here.) Then the first thing we can show is that f (3, c, d) ≥ 0 for c, d ≥ 1 with equality iff c = d = 1. Moreover, f (b, c, d) > 0 for b ≥ 4 and c, d ≥ 1.
Now, using c3 > 0, we also get that b > d. Since 1 ≤ b ≤ 3 and d ≥ 1, we can eliminate b = 1. This leaves us with only b = 2 and d = 1, or b = 3 and d = 1 (since b = 3 iff c = d = 1). Now, if b = 2 and d = 1, then we have f (2, c, 1) = −(c2 − 2c + 2)(3c2 − 2c + 1) = 0 65
√ This gives c = 1 ± i, 13 (1 ± i 2). Finally, if b = 3 and c = d = 1, then a is undefined as the denominator above is 0.
P4 (a, b, c, d): Equation (4.31) gives 9(a + b + c + d) + abcd = 4(ac + ad + bd) + 2(abc + bcd) + 21 Solving this for a gives a=
2bcd + 4bd − 9b − 9c − 9d + 21 bcd − 2bc − 4c − 4d + 9
Now we can use this in equation (4.32), just as above, and show that b ≤ 1. Moreover we have b = 1 iff c = 1. This would then give P4 (a, 1, 1, d) which we know is DS.
K1,3 (a, b, c, d): Equation (4.31) gives 9(a + b + c + d) + 2abcd = 2(abc + abd + acd − bcd) + 4(bc + bd + cd) + 21 Solving for a gives a=
−2bcd + 4bc + 4bd − 9b + 4cd − 9c − 9d + 21 2bcd − 2bc − 2bd − 2cd + 9
From this we get that a < 1 if b, c, d ≥ 2. To show this we look at −2bcd + 4bc + 4bd − 9b + 4cd − 9c − 9d + 21 − (2bcd − 2bc − 2bd − 2cd + 9) = −4bcd + 6bc + 6bd − 9b + 6cd − 9c − 9d + 12 This is always less than 0 when b, c, d ≥ 2. Thus we are forced to have at least one of b, c, d equal 1. Suppose d = 1. Now a=−
2bc − 5b − 5c + 12 2b + 2c − 9
With the assumption that b ≥ c, this gives that a ≥ 1 iff b ≥ 4 and c = 1, or b = 3 and c = 2. If c = 1 then we have K1,3 (a, b, 1, 1) which we know is not cospectral with P5 (a, b, c, d, e). If b = 3 and c = 2, then a = 1. This gives K1,3 (1, 3, 2, 1), which we know is DS. 66
C4 (a, b, c, d): Equation (4.31) gives 9(a + b + c + d) + 3abcd = 4(ac + bd) + 2(abc + abd + acd + bcd) + 21 Solving for a gives 2bcd + 4bd − 9b − 9c − 9d + 21 3bcd − 2bc − 2bd − 2cd − 4c + 9 Now, if b, c, d ≥ 2 then a < 2. To see this we look at a=
2bcd+4bd−9b−9c−9d+21−2(3bcd−2bc−2bd−2cd−4c+9) = −4bcd+4bc+8bd−9b+4cd−c−9d+3 This is less than 0 for all b, c, d ≥ 2. With this we may assume that one of, and exactly one of, a, b, c, d equals 1. We may assume any due to the symmetry of C4 and exactly one because C4 (k, m, 1, 1) and C4 (k, 1, m, 1) are never cospectral with P5 (a, b, c, d, e). Suppose d = 1. Then equation (4.32) gives 2b2 − (2ac − 8a − 8c + 14)b + a2 c2 − 5a2 c + 6a2 − 5ac2 + 23ac − 24a + 6c2 − 24c + 22 = 0 If we let b = 0, then the left side becomes a2 (c2 − 5c + 6) + a(−5c2 + 23c − 24) + 6c2 − 24c + 22 This is always greater than 0 for a, c ≥ 2. This implies that b < 0, which cannot happen.
P5 (a, b, c, d, e): Equation (4.31) gives 9(a + b + c + d + e) + abcd + abce + acde + bcde = 4(ac + ad + ae + bd + be + ce) + 2(abc − ace + bcd + cde) + 21 Solving this for c gives c=
4ad + 4ae − 9a + 4bd + 4be − 9b − 9d − 9e + 21 abd + abe − 2ab + ade + 2ae − 4a + bde − 2bd − 2de − 4e + 9
We can narrow down the cases by considering when a, e are 1 or not. If a = e = 1, then c ≤ 0 for b, d ≥ 2. If b or d is also 1 (and the other is not 2), then c = 1. If b = 2 and d = 1, 67
then c can be anything. Now, if a, e ≥ 2 then c < 2, which implies c must be 1. The case for a ≥ 2 and e = 1 is a little more subtle. If a = 2 and d ≥ 3, then c < 7. If a = d = 2, then c = 3b. Finally, if a ≥ 3 and b, d ≥ 1, then c ≤ 3 .
If a = d = e = 1 and b = 2, then equation (4.32) reduces to 2(c − 1) = 0 This implies c = 1. Now, using k + m + 3 = c1 and 3k + 2m + 1 = c2 we get k = 1 and m = 2. These are isomorphic.
If a, e ≥ 2 and c = 1, then we can use equations (4.31) and (4.32) to show that e ≤ 1 if b ≥ 1 and d ≥ 2, which is obviously a contradiction. If d = 1, then e = 1. We are then left with k + m + 3 = a + b + 3 and 3k + 2m + 1 = 3a + 2a + 1 which gives a = k and b = m, which leaves us with isomorphic graphs.
If a = 2 and e = 1, then equation (4.32) reduces to b2 (−d2 + 3d − 2) + b(−2cd + d2 − 5d + 6) + 2d − 2 = 0 If d ≥ 3 and c ≤ 6, then this is never satisfied. With d = 2 and c = 3b, this then becomes 2(6b2 − 1) = 0 This would imply b =
q
1 , 6
which is absurd.
Finally, we can look at a ≥ 3, b, d ≥ 1 and 1 ≤ c ≤ 3. First suppose c = 1. Then we can show that d = 1 is the only integer solution to equation (4.32). We are left with the same situation as above and get a = k and b = m. If c = 2, then we can show again that d < 2. Now, for d = 1 we get b = 2a − a2 , which implies b < 0 for a ≥ 3. This is cannot happen. Lastly, if c = 3 we once again have d < 2 with d = 1 implying 2b = −3a2 + 7a − 2. This again implies b < 0 for a ≥ 3. We have thus exhausted all possibilities.
68
C6 (a, b, c, d, e, f ): We can use equation (4.31) to solve for a and show that a < 1 for all b, c, d, e, f ≥ 1. The fraction for a is omitted for space.
Therefore we may conclude that P5 (k, m, 1, 1, 1) is DS for all k, m ≥ 1.
69
P5 (k, 1, m, 1, 1): d1 = k + m + 3 d2 = km + 2k + m + 2 d3 = 2m d4 = 3km + m
6c1 + c4 = 3c2 + 2c3 + 12 Also, we can easily get that m =
c3 2
c4
and k = c1 − c23 − 3. With these we get 3 3 = c3 c1 − c3 − 4 2 4
(4.34)
(4.35)
P3 (a, b, c) + K1 (d): Equation (4.34) gives 6(a + b + c + d) + abcd = 3(ac + ad + bd + cd) + 2(abc − acd) + 12 Solving for d gives d=
2abc + 3ac − 6a − 6b − 6c + 12 abc + 2ac − 3a − 3b − 3c + 6
Now, if a, c ≥ 2 then d < 2. That would imply d = 1. If d = 1, then 0 ∈ Spec(P3 (a, b, c) + K1 (d)), but we can easily check that 0 ∈ Spec(P5 (k, 1, m, 1, 1)) iff k = m = 1. If a = c = 1 and b ≥ 2, then 2 < d < 3. We will check a = b = c = 1 separately. Finally, if a = 2, c = 1, and b ≥ 2, then 2 > d ≥ 4, with d = 4 iff b = 2 and d = 3 iff b = 3. Similarly we will check a = 2 and b = c = 1 separately.
For (a, b, c, d) = (2, 2, 1, 3) we get k = 8 and m = −2. For (a, b, c, d) = (2, 3, 1, 3) we get k = 6 and m = 0. For (a, b, c, d) = (1, 1, 1, d) equation (4.35) reduces to (3d − 1)2 = 0, which gives d =
1 . 3
For (a, b, c, d) = (2, 1, 1, d) equation (4.35) reduces to 6d2 − 3d − 1 = 0, which gives d = √ 1 (3 ± 33). 12 70
P4 (a, b, c, d): Equation (4.34) gives 6(a + b + c + d) + abcd = 3(ac + ad + bd) + 2(abc + bcd) + 12 Solving for a gives 2bcd + 3bd − 6b − 6c − 6d + 12 bcd − 2bc − 3c − 3d + 6 We can now use this with equation (4.35). That then reduces to a=
(b3 d4 − 8b2 d3 − 8b2 d2 + 21bd2 + 36bd + 12b − 18d − 36)c3 +(4b3 d3 − 8b3 d2 − 8b2 d4 + 8b2 d3 − 4b2 d2 + 24b2 d +24b2 + 42bd3 − 12bd2 − 80bd − 104b − 54d2 + 168)c2 +(3b3 d2 − 12b3 d + 12b3 − 14b2 d3 + 36b2 d2 + 18b2 d −68b2 + 21bd4 − 48bd3 − 2bd2 − 64bd + 184b − 54d3 +108d2 + 120d − 240)c + 6bd3 − 36bd2 + 72bd − 48b −18d4 + 72d3 − 48d2 − 96d + 96 = 0 Call this polynomial f (b, c, d). We can use this to show that if b, d ≥ 3 then c < 2. For this we can calculate f (b, 2, d) > 0 for all b, d ≥ 3 and f (b, c, d) is strictly increasing for b, d ≥ 3 and c > 2. Suppose c = 1. Then we have f (b, 1, d) is (d−1)(b3 d3 +5b3 d2 −12b3 −8b2 d3 −22b2 d2 +2b2 d+44b2 +21bd3 +21bd2 −8bd−44b−18d3 +6d+12) We can now use this to show that d < 2 for all b ≥ 3. But then we would have P4 (a, b, 1, 1), which we know is DS. Therefore, we must have b = 1, 2 and d ≥ 1 or b ≥ 1 and d = 1, 2. Suppose first that b = 1. Then d cannot be 1. If d = 2 then c = 2. This then gives a = 32 . If d ≥ 13 then we can show that 2 < c < 4. If b = 1 and c = 3, then a < 0 for all d ≥ 13. For 3 ≤ d ≤ 12 we can directly show that there are no integer solutions for c.
Now suppose b = 2. If d = 1 then c = 1, which gives P4 (a, 2, 1, 1), which is DS. If d = 2, then c = 4. This gives a = − 23 . For 3 ≤ d ≤ 37 we can directly show there are no integer solutions for c. Finally, if d ≥ 38, then we can show 1 < c < 2.
Next suppose d = 1. Then b ≥ 2. For 2 ≤ c ≤ 27 we can show that b = 1 is the only integer 71
solution. For c ≥ 28 we can show that b = 1 is again the only integer solution with the other solutions being in (5, 6) ∪ (9, 10). Finally, suppose d = 2. If b = 1 then c = 2. This gives a = 23 . If b = 2 then c = 4. This gives a = − 23 . If b = 3, then c = 0. Finally, if b ≥ 4 then c ∈ [0, 1). We have thus exhausted all possibilities.
K1,3 (a, b, c, d): Equation (4.34) gives 6(a + b + c + d) + 2abcd = 2(abc + abd + acd − bcd) + 3(bc + bd + cd) + 12 Solving for a gives a=
−2bcd + 3bc + 3bd − 6b + 3cd − 6c − 6d + 12 2(bcd − bc − bd − cd + 3)
If b ≥ 3 and c, d ≥ 2, then a < 0 except for b = 3 and c = d = 2 in which case a is undefined. If c = d = 1, then 1 < a < 2 for b ≥ 3. a = 0.5 when b = 1 and a is undefined when b = 2. For c = 2 and d = 1 we have a ≤ 1 for b ≥ 2 with equality iff b = 2. This would give K1,3 (1, 2, 2, 1), which we know is DS. Finally, when c = d = 2 we have a = b. With this last case we get from k + m + 3 = 2b + 4 and 2m = 4b2 that k = −2b2 + 2b + 1 and m = 2b2 . This means that k ≤ 0 for b ≥ 2.
C4 (a, b, c, d): Equation (4.34) gives 6(a + b + c + d) + 3abcd = 3(ac + bd) + 2(abc + abd + acd + bcd) + 12 Solving for a gives a=
2bcd + 3bd − 6b − 6c − 6d + 12 3bcd − 2bc − 2bd − 2cd − 3c + 6
72
This with equation (4.35) reduces to 0 = (9b4 d4 − 36b3 d3 − 24b3 d2 − 24b2 d3 + 45b2 d2 +60b2 d + 12b2 + 60bd2 + 6bd − 36b + 12d2 − 36d)c4 +(−24b4 d2 + 24b3 d3 + 24b3 d2 + 72b3 d + 24b3 −24b2 d4 + 24b2 d3 − 48b2 d2 − 88b2 d − 104b2 + 72bd3 −88bd2 − 88bd + 168b + 24d3 − 104d2 + 168d)c3 +(18b4 d4 − 24b4 d3 − 9b4 d2 + 12b4 d + 12b4 − 24b3 d4 −12b3 d3 + 24b3 d2 + 10b3 d − 68b3 − 9b2 d4 + 24b2 d3 +148b2 d2 − 116b2 d + 184b2 + 12bd4 + 10bd3 −116bd2 + 128bd − 240b + 12d4 − 68d3 + 184d2 − 240d)c2 +(−12b4 d2 + 24b4 d + 32b3 d2 − 64b3 d − 12b2 d4 + 32b2 d3 −8b2 d2 + 8b2 d − 48b2 + 24bd4 − 64bd3 + 8bd2 + 96b − 48d2 + 96d)c +9b4 d4 − 24b4 d3 + 12b4 d2 − 24b3 d4 + 46b3 d3 +4b3 d2 + 12b2 d4 + 4b2 d3 − 32b2 d2 − 48b2 d − 48bd2 + 96bd As daunting as this polynomial is, we can use it to quickly narrow our search. Call the polynomial f (b, c, d). First, if c = 0 then f (b, 0, c) = (b − 2)b(d − 2)d(9b2 d2 − 6b2 d − 6bd2 − 14bd + 24) This is easily seen to be strictly positive for b, d ≥ 3. Moreover, we can also show that f (b, c, d) is strictly increasing for b, d ≥ 3 and c ≥ 0. Thus we must have one of b, d equal to 1 or 2. By the symmetry of C4 we may assume that d = 1, 2.
First suppose d = 1. Then f (b, c, 1) = (9b4 − 60b3 + 93b2 + 30b − 24)c4 + (−24b4 + 144b3 − 240b2 + 64b + 88)c3 +(9b4 − 70b3 + 231b2 − 206b − 112)c2 + (12b4 − 32b3 − 28b2 + 64b + 48)c +48b − 3b4 + 26b3 − 64b2 For 1 ≤ b ≤ 8 we can directly calculate that there are no positive integer solutions for c. For b ≥ 9 we can show that c ∈ (−1, 0) ∪ (0, 1) ∪ (1, 2). 73
Now suppose d = 2. Then f (b, c, 2) = 8c2 (18b4 c2 − 12b4 c + 12b4 − 48b3 c2 + 57b3 c − 54b3 +15b2 c2 − 83b2 c + 74b2 + 27bc2 + 27bc − 22b − 3c2 + 14c − 12) Ignoring the 8c2 factor we can show that the rest of f (b, c, 2) is strictly greater than 0 for b ≥ 3 and is also strictly increasing. For b = 2 we get c = 0.
P5 (a, b, c, d, e): Equation (4.34) gives 6(a + b + c + d + e) + abcd + abce + acde + bcde = 3(ac + ad + ae + bd + be + ce) + 2(abc − ace + bcd + cde) + 12 Solving for a gives a=
−bcde + 2bcd + 3bd + 3be − 6b + 2cde + 3ce − 6c − 6d − 6e + 12 bcd + bce − 2bc + cde + 2ce − 3c − 3d − 3e + 6
Using this with equation (4.35) we can show that if c, d ≥ 2 then all real roots b of the resulting polynomial are strictly less than 1 for all e ≥ 1. Thus, we must have either c = 1 or d = 1.
First suppose c = 1. If d ≥ 3 and e ≥ 2, then b < 2 with b = 1 iff e = 2 and d ≥ 3 or d = e = 3. We are left to check when d = 1, 2 or e = 1.
If e = 1 then b ∈ (−1, 2) ∪ (2, 3] with b = 3 iff d = 1. b = 1 is a root for all d ≥ 1.
If d = 1, then the above polynomial becomes (b − 1)e(b − e)(3b2 e − 2b2 − 3be2 − 3be + 2b + e2 + 2e) = 0 From this we can easily show that b = 1 and b = e are the only integer solutions for e ≥ 2 (the other two roots are in (0, 1) ∪ (e + 1, e + 2)), and b = 1 and b = 3 are solutions when e=1
74
Finally, if d = 2, then we can show that for e ≥ 13 the only positive root for b is in ( e+2 , e+3 ). 3 3 For 1 ≤ e ≤ 12 we can directly calculate that b = 1 is the only positive integer root, and this happens iff e = 1 or e = 2.
Now we must check with d = 1. If c = 3 then b ≤ 6. If c = 4 then b ≤ 3. If c = 5 then b ≤ 2 and if c ≥ 6 then b ≤ 1. Finally, if c = 2 then we have the same argument as with d = 2 and c = 1 above. That is, b = 1 is the only solution, and this happens iff e = 1 or e = 2.
Finally, of those possibilities listed above we can check to see that all either give isomorphic graphs or no solutions for k, m except for when a = 2, b = e, c = 1, d = 1, e ≥ 1, in which case k = e + 1 and m = e. This gives P5 (k + 1, 1, k, 1, 1) ∼ P5 (2, k, 1, 1, k) for all k ≥ 1. We can actually do better and show P5 (a(k + 1), a, ak, a, a) ∼ P5 (2a, ak, a, a, ak) for all a, k ≥ 1.
C6 (a, b, c, d, e, f ): Solving equation (4.34) for a we can show that a < 1 for all b, c, d, e, f ≥ 1.
Therefore we may conclude that P5 (k, 1, m, 1, 1) is DS for all k, m ≥ 1 except for P5 (k + 1, 1, k, 1, 1) ∼ P5 (2, k, 1, 1, k) for k ≥ 1.
75
P5 (k, 1, 1, m, 1): d1 = k + m + 3 d2 = km + 2k + m + 2 d3 = 2m d4 = 2km + k + m 3c1 + c4 = 2c2 + c3 + 5 Similar to P5 (k, 1, 1, m, 1) we have m = c4 = −
c3 2
and k = c1 −
c3 2
(4.36) − 3. With these we get
c23 + c1 c3 − 3c3 + c1 − 3 2
(4.37)
P3 (a, b, c) + K1 (d): Equation (4.36) gives 3(a + b + c + d) + abcd = 2(ac + ad + bd + cd) + abc − acd + 5 Solving for d gives d=
abc + 2ac − 3a − 3b − 3c + 5 abc + ac − 2a − 2b − 2c + 3
Now equation (4.37) reduces to (a4 c4 − 6a3 c3 + 12a2 c2 − 8ac)b4 +(−6a4 c3 − 6a3 c4 + 16a3 c3 + 24a3 c2 + 24a2 c3 − 64a2 c2 − 24a2 c − 24ac2 + 68ac − 8)b3 +(−4a4 c4 + 4a4 c3 + 12a4 c2 + 4a3 c4 + 34a3 c3 − 78a3 c2 − 24a3 c + 12a2 c4 − 78a2 c3 +28a2 c2 + 148a2 c − 24ac3 + 148ac2 − 164ac − 24a − 24c + 36)b2 +(12a4 c3 − 14a4 c2 − 8a4 c + 12a3 c4 − 52a3 c3 − 2a3 c2 + 92a3 c − 14a2 c4 − 2a2 c3 +176a2 c2 − 212a2 c − 24a2 − 8ac4 + 92ac3 − 212ac2 + 88ac + 72a − 24c2 + 72c − 52)b +a4 c4 − 8a4 c3 − 5a4 c2 + 12a4 c − 8a3 c4 + 6a3 c3 + 44a3 c2 − 48a3 c − 8a3 −5a2 c4 + 44a2 c3 − 95a2 c2 + 38a2 c + 36a2 + 12ac4 − 48ac3 +38ac2 + 46ac − 52a − 8c3 + 36c2 − 52c + 24 = 0 With this we can show that b < 4 if a = 1 and c ≥ 2 or c = 1 and a ≥ 2; and b < 3 if a, c ≥ 2. From this we immediately get that if b = 3 then one of a, c is 1. Assuming b = 3 76
and c = 1 we can then use this with our equation for d above to show that d < 1 for all a ≥ 3. For a = 1 we get d = 35 , and for a = 2 we get d = 3. Next, if b = 1, then we can show that d < 2 for all a, c ≥ 1. Finally, if b = 2 we can show d ≤ 2 with d = 1 iff a = c = 2 and d = 2 iff a = 2 and c = 1.
With only these three possibilities we can directly calculate that no positive integer solutions for k, m exist.
P4 (a, b, c, d): Equation (4.36) gives 3(a + b + c + d) + 2abcd = 2(ac + ad + bd) + abc + bcd + 5 Solving for a gives a=
bcd + 2bd − 3b − 3c − 3d + 5 bcd − bc − 2c − 2d + 3
Equation (4.37) then reduces to (c4 d4 + 2c3 d3 − 4c3 d2 − 2c2 d + 3c2 )b4 +(−6c4 d3 − 4c4 d2 − 6c3 d4 + 8c3 d3 − 2c3 d2 +12c3 d + 6c3 − 4c2 d3 + 14c2 d2 − 2c2 d − 16c2 )b3 +(12c4 d2 + 14c4 d + 3c4 + 24c3 d3 − 22c3 d2 − 26c3 d − 28c3 + 12c2 d4 −36c2 d3 + 20c2 d2 − 20c2 d + 49c2 − 8cd2 + 24cd − 18c)b2 +(−8c4 d − 12c4 − 24c3 d2 + 20c3 d + 48c3 − 24c2 d3 +76c2 d2 − 20c2 d − 62c2 − 8cd4 + 44cd3 − 68cd2 + 12cd + 26c)b −8c3 − 24c2 d + 36c2 − 24cd2 + 72cd − 52c − 8d3 + 36d2 − 52d + 24 = 0 With this we can now show that b < 3 if d ≥ 2 and c ≥ 1. If d = 1 then b ∈ (−1, 1] ∪ (3, 4) ∪ (5, 6) for all c ≥ 5. For 1 ≤ c ≤ 4 we can directly show that b = 1 is the only integer solution. In fact b = 1 is always an integer solution with d = 1. Regardless, this would give P4 (a, 1, c, 1), which we know is DS. Now using our equation for a above we can show that with b = 1 that a ≤ 7 for all c ≥ 1 and d ≥ 2. With b = 2 we get a ≤ 8 for all c ≥ 1 and d ≥ 2. We can then calculate 77
the only possible combinations for (a, b, c, d) are (2, 1, 4, 8), (2, 1, 8, 4), (3, 1, 3, 11), (4, 1, 6, 4), (5, 1, 4, 5), (7, 1, 3, 7), (2, 2, 3, 8), (2, 2, 8, 3), (3, 2, 2, 8), (8, 2, 2, 3). Finally, with these we can directly compute that no solutions for k, m exist.
K1,3 (a, b, c, d): Equation (4.36) gives 3(a + b + c + d) + 2abcd = abc + abd + acd − bcd + 2(bc + bd + cd) + 5 Solving for a gives a=
−bcd + 2bc + 2bd − 3b + 2cd − 3c − 3d + 5 2bcd − bc − bd − cd + 3
From this we can show that a < 1 for all b, c, d ≥ 1.
C4 (a, b, c, d): Equation (4.36) gives 3(a + b + c + d) + 3abcd = 2(ac + bd) + abc + abd + acd + bcd + 5 Solving for a gives bcd + 2bd − 3b − 3c − 3d + 5 3bcd − bc − bd − cd − 2c + 3 With this we can show a < 2 for b ≥ 2 and c, d ≥ 1. If a = 1, then c = 1. Finally, if b = 1, a=
then a < 1 if d ≥ 2. These give C4 (1, b, 1, d) and C4 (a, 1, c, 1), respectively, which we know are not cospectral with P5 (k, 1, 1, m, 1).
P5 (a, b, c, d, e): Solving equation (4.36) for c gives c=
2ad + 2ae − 3a + 2bd + 2be − 3b − 3d − 3e + 5 abd + abe − ab + ade + ae − 2a + bde − bd − de − 2e + 3
With this we can show that c < 2 for all a, b, d, e ≥ 1 with c = 1 iff a, d ≥ 1 and b = e = 1 or a = d = 1 and b, e ≥ 1. These are isomorphic, so we may assume b = e = 1. Now k + m + 3 = a + d + 3 and 2m = 2d imply that a = k and d = m. These are isomorphic graphs. 78
C6 (a, b, c, d, e, f ): Using equation (4.36), we can show that a < 1 for all b, c, d, e, f ≥ 1.
Therefore we can conclude that P5 (k, 1, 1, m, 1) is DS for all k, m ≥ 1.
79
P5 (k, 1, 1, 1, m): d1 = k + m + 3 d2 = km + 2k + 2m + 1 d3 = −km + k + m + 1 d4 = 2km + k + m
3c1 = c2 + c3 + 7 c2 = c3 + c4
(4.38) (4.39)
P3 (a, b, c) + K1 (d): Equations (4.38) and (4.39) together reduce to (−a2 c2 + 5ac − 3)b2 + (−a2 c2 + 5a2 c + 5ac2 − 13ac − 6a − 6c + 7)b +2a2 c2 − 3a2 c − 3a2 − 3ac2 + 4ac + 7a − 3c2 + 7c = 0 With this we can show that b ∈ (−2, 1) ∪ (1, 2) if a ≥ 3 and c ≥ 2. If we assume a ≥ c (and we can due to symmetry), then we must have either c = 1 and a ≥ 1 or c = 2 and a = 2. If c = 1 then b ∈ (2, 3) for a ≥ 9. For a ≤ 8 we can calculate that the only integer solutions for b are b = 1, 8 at a = 1 and b = 8 at a = 5. These give d = 0, d = 38 , and d = − 35 , respectively. Now, for a = c = 2 we get b = 1 or b = 4, but then we get d = 0 or d = −3.
P4 (a, b, c, d): Equation (4.38) gives 3(a + b + c + d) = abc + ac + ad + bcd + bd + 7 Solving for d gives d=
−abc − ac + 3a + 3b + 3c − 7 a + bc + b − 3
From this we can show that d ≤ 1 with equality iff exactly two of a, b, c are 1. Regardless, we reduce to cases of P4 (a, b, c, d) that we have already shown to be DS.
80
K1,3 (a, b, c, d): In the P4 (k, 1, 1, m) case we already showed that equation (4.39) is never satisfied with a, b, c, d ≥ 1.
C4 (a, b, c, d): Similar to P4 (a, b, c, d) we can solve equation (4.38) for d to get d=
−abc − ac + 3a + 3b + 3c − 7 ab + ac + bc + b − 3
This time we get that d < 1 for all a, b, c ≥ 1.
P5 (a, b, c, d, e): Solving (4.38) for c gives c=
−ad − ae + 3a − bd − be + 3b + 3d + 3e − 7 ab − ae + a + bd + de + e − 3
We can then show that c ≤ 1 with equality iff b = d = 1. Finally, with k + m + 3 = a + e + 3 and km + 2k + 2m + 1 = ae + 2a + 2e + 1 we get k = a and m = e or k = e and m = a.
C6 (a, b, c, d, e, f ): Solving (4.38) for f gives f=
−abc + ace − ac − ad − ae + 3a − bcd − bd − be + 3b − cde − ce + 3c + 3d + 3e − 7 ab + ae − bd + b + c + de + d − 3
With this we have f < 1 for all a, b, c, d, e ≥ 1.
Therefore we may conclude that P5 (k, 1, 1, 1, m) is DS for all k, m ≥ 1.
81
P5 (1, k, m, 1, 1): d1 = k + m + 3 d2 = 2k + 2m + 2 d3 = 2km d4 = 3km + m
2c1 = c2 + 4 9c23 + 6c1 c3 + 4c24 + 12c4 = 4c1 c4 + 12c3 c4 + 16c3
(4.40) (4.41)
P3 (a, b, c) + K1 (d): Equation (4.40) gives 2(a + b + c + d) = ac + ad + bd + cd + 4 Solving for d gives d=
−ac + 2a + 2b + 2c − 4 a+b+c−2
From this we can show that d < 2 for all a, b, c ≥ 1. This would imply that d = 1. More importantly that means 0 ∈ Spec(P3 (a, b, c) + K1 (d)), but we can show that 0 ∈ Spec(P5 (1, k, m, 1, 1)) iff m = 1. For that we observe PP5 (1,k,m,1,1) (0) = −k(m − 1) We can now conclude that P3 (a, b, c) + K1 (d) is not cospectral with P5 (1, k, 1, 1, 1) from our previous results with P5 (k, m, 1, 1, 1).
P4 (a, b, c, d): Equation (4.40) gives 2(a + b + c + d) = ac + ad + bd + 4 Solving for d gives d=
−ac + 2a + 2b + 2c − 4 a+b−2 82
If a ≥ 3 and b, c ≥ 1, then d < 2 which means d = 1 is the only possibility. If a = 2 and b, c ≥ 1, then d = 2. The a = 1 case is symmetric with the d = 1 case from before.
For a = d = 2, equation (4.41) reduces to 2bc + b + c + 2 = 0 This is clearly not possible with b, c ≥ 1. On the other hand, if d = 1, then a=
b + 2c − 2 c−1
With this (4.41) reduces to −12b4 c2 + c(−60c2 + 56c + 13)b3 + c(−72c3 + 128c2 − 14c − 36)b2 +(−12c4 + 37c3 − 14c2 + 19c − 30)b + 18c3 + 42c2 − 138c + 78 = 0 With this we can now show that b < 1 for all c ≥ 1.
K1,3 (a, b, c, d): Equation (4.40) gives 2(a + b + c + d) = bc + bd + cd + 4 Solving for d gives d=
2a − bc + 2b + 2c − 4 b+c−2
Using this with equation (4.41) we can then show that b < 1 for all c ≥ 1.
C4 (a, b, c, d): Equation (4.40) gives 2(a + b + c + d) = ac + bd + 4 Solving for d gives d=
−ac + 2a + 2b + 2c − 4 b−2
Using this with equation (4.41) we can show that b ≤ 2 for a, c ≥ 1, with b = 2 iff a = c = 4 or a = 6 and c = 3. We can then show directly that no positive integer solutions exist for
83
k, m. For b = 1, we can show that the only positive solutions for c are in (0, 1) ∪ (1, 2) for a ≥ 3 and that no positive integer solutions exist for a = 1, 2.
P5 (a, b, c, d, e): Equation (4.40) gives 2(a + b + c + d + e) = ac + ad + ae + bd + be + ce + 4 Solving for a gives a=
−bd − be + 2b − ce + 2c + 2d + 2e − 4 c+d+e−2
From this we can show that a ≤ 1 for all b, c, d, e ≥ 1 with equality iff b = e = 1 or d = e = 1. These are isomorphic, so we may assume that a = d = e = 1. It is then straightforward to show that k = b and m = c.
C6 (a, b, c, d, e, f ): Equation (4.40) gives 2(a + b + c + d + e + f ) = ac + ad + ae + bd + be + bf + ce + cf + df + 4 Solving for a gives a=
−bd − be − bf + 2b − ce − cf + 2c − df + 2d + 2e + 2f − 4 c+d+e−2
From this we can show that a < 1 for all b, c, d, e, f ≥ 1.
Therefore we may conclude that P5 (1, k, m, 1, 1) is DS for all k, m ≥ 1.
84
P5 (1, k, 1, m, 1): d1 = k + m + 3 d2 = km + k + m + 3 d3 = km + k + m − 1 d4 = 2km + k + m c2 = c3 + 4 c1 + c4 + 3 = 2c2 c1 + c4 = 2c3 + 5
(4.42) (4.43) (4.44)
Note that 0 ∈ Spec(P5 (1, k, 1, m, 1)).
P3 (a, b, c) + K1 (d): For 0 ∈ Spec(P3 (a, b, c) + K1 (d)) we need either a = c = 1 or d = 1. If a = c = 1, then we have P3 (1, b, 1) + K1 (d), which we know is not cospectral with any P5 (a, b, c, d, e). Suppose then that d = 1. Equation (4.42) reduces to a + b + c + ac = abc − ac + 4 Solving for a gives b+c−4 bc − 2c − 1 From this we can show that a ≤ 1 for all b, c ≥ 1 unless b = 3 and c = 1. However, a=
in all cases we reduce to P3 (1, b, c) + K1 (1), which we know is not cospectral with any P5 (a, b, c, d, e).
P4 (a, b, c, d): Equations (4.42) and (4.44) together reduce to (b2 d2 − 2bd + b − 1)c2 + (b2 d2 + 2b2 d + b2 − 2bd2 − 4bd − 4b − 2d − 5)c − d2 − 5d − 4 = 0 With this we can show that c ∈ (1, 2) for b = 1 and d ≥ 22 and that c ∈ (0, 1) if b = 2 and d ≥ 5, b = 3 and d ≥ 2, and b ≥ 4 and d ≥ 1. For all other cases missed we can directly check that there are no positive integer roots for c. 85
K1,3 (a, b, c, d): Equations (4.42) and (4.43) reduce to (2c2 d2 + 2c2 d − 2c2 + 2cd2 − 4cd + c − 2d2 + d)b2 +(2c2 d2 − 4c2 d + c2 − 4cd2 − 4cd + 4c + d2 + 4d)b −2c2 d2 + c2 d + cd2 + 4cd − 4 = 0 With this we can show that b ≤ 1 for all c, d ≥ 1 with equality iff c = 1 or d = 1. In either case we have K1,3 (a, 1, 1, d), which we know is not cospectral with any P5 (a, b, c, d, e).
C4 (a, b, c, d): Equation (4.42) gives ac + bd = abc + abd + acd + bcd + 4 Solving for a gives a=
−bcd + bd − 4 bc + bd + cd − c
With this we can show that a < 1 for all b, c, d ≥ 1.
P5 (a, b, c, d, e): Solving equation (4.43) for c gives c=
2ad + 2ae − a + 2bd + 2be − b − d − e − 3 abd + abe + ade − 2a + bde − 2e + 1
With this we can show that c < 3 for all a, b, d, e ≥ 1. Moreover, we can show that c1 + c4 > 2c3 + 5 for all a, e ≥ 2 and b, c, d ≥ 1. Therefore we may assume e = 1. With this we can now use equations (4.42) and (4.43) to reduce to d(3a2 c2 + a2 cd − 2acd − 6ac − c2 d + 3cd − d + 3) = 0 When c = 1 this factors as (a − 1)2 d(d + 3) = 0 which has a = 1 as the only positive root. From here it is easily shown that either k = b and m = d or k = d and m = b. When c = 2 it is d(2a2 d + 12a2 − 4ad − 12a + d + 3) = 0, which only has (a, d) = (1, 3) as a positive integer solution. This then implies b = 0.
86
C6 (a, b, c, d, e, f ): c1 + c4 > 2c3 + 5 for all a, b, c, d, e, f ≥ 1. Therefore we may conclude that P5 (1, k, 1, m, 1) is DS for all k, m ≥ 1
87
4.7
C6(a, b, c, d, e, f )
C6 (k, m, 1, 1, 1, 1): d1 = k + m + 4 d2 = 3k + 3m + 3 d3 = 2km + 2 d4 = 5km + 3k + 3m + 1 3c1 = c2 + 9
(4.45)
5c3 + 2c2 = 2c4 + 14
(4.46)
5c3 + 6c1 = 2c4 + 16
(4.47)
0 is never in the spectrum of C6 (k, m, 1, 1, 1, 1), and 1 is in the spectrum iff k = 1 or m = 1. Finally, using the bounds on the eigenvalues, we can show that 1 ≤ λ3 < 1.5.
P3 (a, b, c) + K1 (d): Solving equation (4.45) for d gives d=
3a + 3b + 3c − ac − 9 a+b+c−3
From this we can show that d < 3 for all a, b, c ≥ 1. We know that 0 is never in the spectrum of C6 (k, m, 1, 1, 1, 1), so we can assume that d = 2 and m = 1. With m = 1, our system of equations simplifies greatly. We can deduce that 2c1 = c3 + 8. With this added relation we get the equation a2 c2 − a2 c − ac2 − ac − 2 = 0 This equation has no positive integer solutions.
P4 (a, b, c, d): Equations (4.45) and (4.46) reduce to a2 (2c2 d − 5c2 + 2cd2 − 11cd + 15c) +a(−11c2 d + 15c2 − 11cd2 + 48cd − 51c) +15c2 d + 15cd2 − 39cd + 6d2 − 32d + 42 = 0 88
From this we can show that if d = 1 then a ∈ (2, 3) for all c ≥ 2 and a = 1 if c = 1; if d = 2 then a ∈ (2, 3) for all c ≥ 1; if d = 3 then a ∈ {3} ∪ (15, 16) for all c ≥ 7, a = 21 if c = 1, a = 18 if c = 2, a = 17 if c = 3, and a = 16 if c = 6; if d = 4 then a ∈ (3, 4) ∪ (6, 7] for all c ≥ 1 with a = 7 iff c = 1; and if d ≥ 5 then a ∈ (3, 5) for all c ≥ 1, and a 6= 4 for any c. (This includes only positive real roots.)
Finally, having narrowed down the possibilities we can directly computer that no positive integer solutions exist.
K1,3 (a, b, c, d): Solving equations (4.45) and (4.47) for a and then setting those equal to one another gives b2 (4c2 d − 5c2 + 4cd2 − 22cd + 15c − 5d2 + 15d) +b(4c2 d2 − 22c2 d + 15c2 − 22cd2 + 96cd − 51c + 15d2 − 51d) −5c2 d2 + 15c2 d + 15cd2 − 51cd − 6 = 0 With this we can show that b ∈ (0, 7) (whenever b is positive and real) for all c ≥ d ≥ 1 with b = 4 iff c = d = 1 and b = 1 iff c = 4 and d = 1. In both cases we get that a = 0.
C4 (a, b, c, d): Assume a ≥ b, c, d and b ≥ d. Then using C4 (c, d, c, d) ≤ C4 (a, b, c, d) ≤ C4 (a, b, a, b) we get d − 1 ≤ λ3 ≤ b − 1. We also know that 1 ≤ λ3 < 1.5. Therefore d − 1 < 1.5 which implies d = 1, 2. With d = 1 equations (4.45) and (4.47) give 2b2 c − 10b2 − 2bc2 − 22bc + 36b − 21c2 + 40c + 6 =0 c−3 On the other hand, using (4.46) and (4.47) we get 2 (2b2 c − 10b2 − 2bc2 − 14bc − 4b − 21c2 − 42) =0 bc − 5b − 7c These together then simplify to bc − 5b − 5c − 6 = 0 which only has {b, c} = {6, 36} as positive integer solutions. With these we can directly compute that there are no solutions
89
for a, k, m. A similar argument for d = 2 leads to 7bc − 10b − 10c − 6 = 0 which has no positive integer solutions.
P5 (a, b, c, d, e): Assuming a ≥ e, similar to C4 (a, b, c, d) we can show that min{c, e}−1 ≤ λ3 ≤ max{a, c}−1. Since λ3 < 1.5 we must one of c, e in {1, 2}. Using these with equations (4.45), (4.46), and (4.47) we get, from c = 1, 2 and e = 1, 2 respectively, the equations 2bd + 2be − 5b + 2de + 5e − 6 = 0 2bd + 2be − 5b + 2de + 5e − 3 = 0 2bcd − 3bc + 2cd + 5c − 6 = 0 2bcd − bc + 4cd + 10c − 6 = 0 Finally, with these we get the possible combinations for (b, c, d, e) the following: (1, 1, 1, 1), (4, 1, 1, 1), (4, 2, 1, 1), (5, 3, 1, 1), (6, 6, 1, 1). From these we can directly compute that no integer solutions then exist for a, k, m.
C6 (a, b, c, d, e, f ): Equation (4.45) gives a=
−bd − be − bf + 3b − ce − cf + 3c − df + 3d + 3e + 3f − 9 c+d+e−3
From this we can show that a ≤ 1 with equality iff only at most two consecutive variables are greater than 1. That is C6 (a, b, c, d, e, f ) ∼ = C6 (k 0 , m0 , 1, 1, 1, 1). From here it is easy to show that k 0 = k and m0 = m.
Therefore we may conclude that C6 (k, m, 1, 1, 1, 1) is DS for all k, m ≥ 1.
90
C6 (k, 1, m, 1, 1, 1): d1 = k + m + 4 d2 = km + 2k + 2m + 4 d3 = 2k + 2m d4 = 5km + 3k + 3m + 1 2c1 = c3 + 8 10c2 = 7c3 + 2c4 + 38 7c1 + c4 = 5c2 + 9
(4.48) (4.49) (4.50)
Note that k ∈ Spec(C6 (k, 1, m, 1, 1, 1)) iff k = m; k + 1 ∈ Spec(C6 (k, 1, m, 1, 1, 1)) iff k = m = 1. We also have the bounds k ≤ λ1 ≤ k + 2, m ≤ λ2 ≤ m + 1, 1 ≤ λ3 < 1.25, and −4 < λn ≤ −2.
P3 (a, b, c) + K1 (d): Since d − 1 ∈ Spec(P3 (a, b, c) + K1 (d)) we must have k = m = d − 1 by the note above. With this we can already compare coefficients to see that there are no integer solutions for a, b, c, d.
P4 (a, b, c, d): From equation (4.48) we get a=
−bcd + 2b + 2c + 2d − 8 bc − 2
This gives that a ≥ 1 iff b = c = 1 and 1 ≤ d ≤ 3. Regardless we get P4 (a, 1, 1, d) which we know is DS.
K1,3 (a, b, c, d): If b ≥ c ≥ d, then we have d − 1 ≤ λ3 ≤ b − 1. Since 1 ≤ λ3 < 1.25, we can conclude that d = 1 or d = 2. Moreover, we can show that if d = 1, then 0 ≤ λ3 < 1. Therefore d = 2 is the only possibility. With this, equation (4.48) gives c=−
2(ab − a − b + 2) ab + 2a − 2b − 2 91
This gives that c 6≥ 2.
C4 (a, b, c, d): We may assume d = 1 or d = 2. With these we get, respectively, from equation (4.48) a=
−bc + 2b + 2c − 6 bc + b + c − 2
a=−
2(bc − b − c + 2) bc + 2b + 2c − 2
In both cases we have a < 1.
P5 (a, b, c, d, e): Assume a ≥ e. Since 1 ≤ λ3 < 1.25, we must have min{c, e} ≤ 2. Using equations (4.48) and (4.49) we can then show that the only possibilities for (b, c, d, e) are (1, 1, d, 1), (1, 1, 2, 3), (2, 1, 4, 10), and (1, 1, 4, 13). With (1, 1, d, 1) we get a = 1, but we know P5 (1, 1, 1, d, 1) is DS. With the other possibilities we have a 6∈ Z. C6 (a, b, c, d, e, f ): Since 1 ≤ λ3 < 1.25 we must have m1 + m2 − 1 < 1.25, where m1 = min{a, c, e} and m2 = min{b, d, f }. In particular this means m1 = m2 = 1. Suppose m2 = f = 1. Then equation (4.50) gives a=
−bcde − bcd − bde + 5bd + 5be − 2b − cde + 5ce − 2c − 2d − 7e + 2 bcd + bce + bc + bd + be + cde + ce − 5c + de − 5d − 5e + 7
With this we can then show that a < 2 for all b, c, d, e ≥ 1, and a = 1 iff b = d = 1 or c = e = 1. That is, we have C6 (1, 1, c, 1, e, 1) or C6 (1, b, 1, d, 1, 1), and from here is it straightforward to show {c, e} = {k, m} or {b, d} = {k, m}.
Therefore we may conclude that C6 (k, 1, m, 1, 1, 1) is DS for all k, m ≥ 1.
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C6 (k, 1, 1, m, 1, 1): d1 = k + m + 4 d2 = km + 2k + 2m + 4 d3 = 2k + 2m d4 = 4km + 4k + 4m
2c1 = c3 + 8 4c1 + c4 = 4c2 4c2 = 2c3 + c4 + 16
(4.51) (4.52) (4.53)
Assume k ≥ m. We can bound the eigenvalues as k ≤ λ1 ≤ k + 1, m < λ2 < m + 1, λ3 = 1, and −3 < λn ≤ −2.
P3 (a, b, c) + K1 (d): Assume a ≥ c. −1 ∈ Spec(P3 (a, b, c) + K1 (d)) iff d = 2; a = c = 2; a = 5, b = 6, c = 1; a = 6, b = 4, c = 1; or a = 8, b = 3, c = 1. In all cases, for λ3 = 1 we must have d ≥ 2.
If d = 2, then a=
2(b + c − 2) bc − 2c − 2
We can then show that k, m ≥ 2 iff b = 3 and c ≥ 3. Now, if b = 3 then 2 < a ≤ 8 for all c ≥ 3 with a = 8 and c = 3 or a = 5 and c = 4 the only integral possibilities. From these it is easily shown that k, m 6∈ Z. If a = c = 2, then equation (4.51) gives b = 3d. Since b > d, this implies that λ2 = d − 1, but m < λ2 < m + 1 means that λ2 6∈ Z. Thus, this cannot happen. If a = 6, b = 4, and c = 1, then d = 45 . If a = 8, b = 3, and c = 1, then d = 54 . Finally, if a = 5, b = 6, and c = 1, then d = 2. This again gives that λ2 = 1 ∈ Z, which cannot happen.
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P4 (a, b, c, d): From equation (4.51) we get a=
−bcd + 2b + 2c + 2d − 8 bc − 2
This gives that a ≥ 1 iff b = c = 1 and 1 ≤ d ≤ 3. Regardless we get P4 (a, 1, 1, d) which we know is DS.
K1,3 (a, b, c, d): If b ≥ c ≥ d, then we have d − 1 ≤ λ3 ≤ b − 1. Since λ3 = 1, we can conclude that d = 1 or d = 2. Moreover, we can show that if d = 1, then 0 ≤ λ3 < 1. Therefore d = 2 is the only possibility. With this, equation (4.51) gives c=−
2(ab − a − b + 2) ab + 2a − 2b − 2
This gives that c 6≥ 2.
C4 (a, b, c, d): We may assume d = 1 or d = 2. With these we get, respectively, from equation (4.51) a=
−bc + 2b + 2c − 6 bc + b + c − 2
a=−
2(bc − b − c + 2) bc + 2b + 2c − 2
In both cases we have a < 1.
P5 (a, b, c, d, e): Assume a ≥ e. For λ3 = 1 we must have min{c, e} = 1, 2. If either c or e is 2, then we must have a = c = e = 2. If c = 1 then a ≥ 4 and e ≥ 3. If e = 1 then a or c is at least 3. We can immediately check that with a = c = e = 2 there are no solutions for k, m.
Suppose c = 1. We can use equations (4.51) and (4.52) to show that a ≥ 4 and e ≥ 3 iff one of the following is true: b = 1, d ≥ 14 and e = 3; b = 1, 8 ≤ d ≤ 11, and e = 4; b = 6, d = 1, 94
and e = 3; or b = 7, d = 1, and e = 4. With only these possibilities we can immediately check that equations (4.51) and (4.52) are never simultaneously satisfied.
Now suppose e = 1. Requiring that no three of a, b, c, d, e are 1, we can use equations (4.51) and (4.52) to show that a ≥ 1 iff b = 1 and one of the following is true: c = 2 and d ≥ 2; c = 3 and 2 ≤ d ≤ 4; or d = 2 and 4 ≤ c ≤ 7. With only these possibilities we can immediately check that equations (4.51) and (4.52) are never simultaneously satisfied.
C6 (a, b, c, d, e, f ): For λ3 = 1 we need m1 = m2 = 1. Suppose f = 1. Then equation (4.52) gives a=
−bcde − bcd − bde + 4bd + 4be − cde + 4ce − 4e − 4 bcd + bce + bc + bd + be + cde + ce − 4c + de − 4d − 4e + 4
With this we can show that C6 (a, b, c, d, e, f ) ∼ = C6 (k 0 , 1, 1, m0 , 1, 1) with k 0 ≥ m0 , and then it is immediate that k = k 0 and m = m0 .
Therefore we may conclude that C6 (k, 1, 1, m, 1, 1) is DS for all k, m ≥ 1.
95
4.8
Coefficients
The coefficients presented here are for the characteristic polynomials arising from P3 and P4 , excluding 3K1 and 4K1 . They are factored as (x + 1)c1 −k H(x), where k = 3 or k = 4, depending on whether we are looking at P3 or P4 , and H(x) is x3 + (3 − c1 )x2 + (3 − 2c1 + c2 )x + (1 − c1 + c2 + c3 ) for P3 and x4 + (4 − c1 )x3 + (6 − 3c1 + c2 )x2 + (4 − 3c1 + 2c2 + c3 )x + (1 − c1 + c2 + c3 − c4 ) for P4 . The characteristic polynomials have been factored in this way to facilitate ease in comparing different graphs with one another. P31 (a, b, c) = a + b + c P32 (a, b, c) = ac P33 (a, b, c) = abc P3 + K11 (a, b, c, d) = a + b + c + d P3 + K12 (a, b, c, d) = ac + ad + bd + cd P3 + K13 (a, b, c, d) = abc − acd P3 + K14 (a, b, c, d) = abcd P41 (a, b, c, d) = a + b + c + d P42 (a, b, c, d) = ac + ad + bd P43 (a, b, c, d) = abc + bcd P44 (a, b, c, d) = abcd K1,31 (a, b, c, d) = a + b + c + d K1,32 (a, b, c, d) = bc + bd + cd K1,33 (a, b, c, d) = abc + abd + acd − bcd K1,34 (a, b, c, d) = 2abcd C41 (a, b, c, d) = a + b + c + d C42 (a, b, c, d) = ac + bd C43 (a, b, c, d) = abc + abd + acd + bcd C44 (a, b, c, d) = 3abcd 96
P51 (a, b, c, d, e) = a + b + c + d + e P52 (a, b, c, d, e) = ac + ad + ae + bd + be + ce P53 (a, b, c, d, e) = abc + bcd + cde − ace P54 (a, b, c, d, e) = abcd + abce + acde + bcde C61 (a, b, c, d, e, f ) = a + b + c + d + e + f C62 (a, b, c, d, e, f ) = ac + ad + ae + bd + be + bf + ce + cf + df C63 (a, b, c, d, e, f ) = abc + bcd + cde + def + aef + abf − ace − bdf C64 (a, b, c, d, e, f ) = abcd + abce + abcf + abdf + abef + acde + acef + adef + bcde + bcdf + bdef + cdef
97
4.9
Bounds on eigenvalues
The bounds in this section were found by exploiting symmetries in the graphs and using Theorem 2.1.4.
P3 (a, b, c) + K1 (d): d − 1 is always in Spec(P3 (a, b, c) + K1 (d)). With the assumption that a ≥ c, the other three eigenvalues can be bounded by √ c2 + 6cb + b2 + c + b − 2 ≤ λ01 ≤
1 2
√
1 2
a2 + 6ab + b2 + a + b − 2
c − 1 ≤ λ02 ≤ a − 1 √ √ − a2 + 6ab + b2 + a + b − 2 ≤ λn ≤ 21 − c2 + 6cb + b2 + c + b − 2
1 2
λ01 and λ02 have been used here as we do not necessarily know where d − 1 falls with respect to these.
P4 (a, b, c, d): Assume a ≥ d. Let M = max{b, c} and m = min{b, c}. Then √
1 ( 2
1 (− 2
√
d2 + 4m2 + d + 2m − 2) ≤ λ1 ≤ √ √ 1 d d + 4m + d − 2) ≤ λ2 ≤ ( 2
d2 + 4m2 + d + 2m − 2) ≤ λ3 ≤ √ √ 1 (− a a + 4M + a − 2) ≤ λn ≤ 2
√
1 ( a2 + 4M 2 + a + 2M − 2) 2 √ 1 √ a a + 4M + a − 2) ( 2 √ 1 (− a2 + 4M 2 + a + 2M − 2) 2 √ √ 1 (− d d + 4m + d − 2) 2
K1,3 (a, b, c, d): Assume b ≥ c ≥ d. Then √
1 ( 2
a2 + 10ad + d2 + a + d − 2) ≤ λ1 ≤
√
1 ( 2
a2 + 10ab + b2 + a + b − 2)
d − 1 ≤ λ2 ≤ b − 1 √
1 (− 2
d − 1 ≤ λ3 ≤ b − 1 √ a2 + 10ab + b2 + a + b − 2) ≤ λn ≤ 21 (− a2 + 10ad + d2 + a + d − 2)
98
C4 (a, b, c, d): Assume a ≥ b, c, d and b ≥ d. Then 1 ( 2
√ c2 + 14cd + d2 + c + d − 2) ≤ λ1 ≤
1 ( 2
√ a2 + 14ab + b2 + a + b − 2)
c − 1 ≤ λ2 ≤ a − 1 d − 1 ≤ λ3 ≤ b − 1 √ √ 1 (− a2 + 14ab + b2 + a + b − 2) ≤ λn ≤ 12 (− c2 + 14cd + d2 + c + d − 2) 2
P5 (a, b, c, d, e): Let M1 = max{a, c, e}, M2 = max{b, d}, m1 = min{a, c, e}, and m2 = min{b, d}. Then p f (m1 , m2 ) + m1 + m2 − 2) ≤ λ1 ≤
1 ( 2
p f (M1 , M2 ) + M1 + M2 − 2)
1 ( 2
m1 + m2 − 1 ≤ λ2 ≤ M1 + M2 − 1 m1 − 1 ≤ λ3 ≤ M1 − 1 p p 1 (− f (M1 , M2 ) + M1 + M2 − 2) ≤ λn ≤ 12 (− f (m1 , m2 ) + m1 + m2 − 2) 2 where f (x, y) = x2 + 10xy + y 2 .
C6 (a, b, c, d, e, f ): Let M1 = max{a, c, e}, M2 = max{b, d, f }, m1 = min{a, c, e}, and m2 = min{b, d, f }. Then p f (m1 , m2 ) + m1 + m2 − 2) ≤ λ1 ≤
1 ( 2
p f (M1 , M2 ) + M1 + M2 − 2)
1 ( 2
m1 + m2 − 1 ≤ λ2 ≤ M1 + M2 − 1 1 (− 2
m1 + m2 − 1 ≤ λ3 ≤ M1 + M2 − 1 p p f (M1 , M2 ) + M1 + M2 − 2) ≤ λn ≤ 12 (− f (m1 , m2 ) + m1 + m2 − 2)
where f (x, y) = x2 + 14xy + y 2 .
99
Chapter 5 λ=0 This chapter is broken down similarly to Chapter 4. We will begin with a section dedicated to Q0 , Q2 , and Q3 . The sections following it will focus on Q4 . For these graphs, we fix all but one split partition at size one, with a few exceptions. We also partition and analyze Q4 through bipartite and non-bipartite graphs. We do this because the spectrum of a graph can determine whether or not it is bipartite, and a split graph is bipartite if and on if it comes from a primitive bipartite graph.
5.1
Q0, Q2, and Q3
For Q0 , we get always get nK1 . Since the spectrum determines the number of vertices and edges (see Proposition 2.1.1), it is straightforward to see that all split graphs coming from Q0 are DS as there is exactly one graph with n vertices and 0 edges. Q2 and Q3 are not quite as simple to analyze as Q0 or even P1 , P2 , or P3 . Recall that Q2 = {K2 } and Q3 = {K3 }. Notice that Kn (a1 , . . . , an ; z) is precisely that complete multipartite graph Ka1 ,...,an with z isolated vertices. Unlike the complete graphs and union of complete graphs, we do not get that complete multipartite graphs are DS. The simplest example of this is K1,4 and C4 + K1 , as we saw in Chapter 2. Notice that C4 is K2,2 . In fact, we saw √ √ in Proposition 2.1.2 that Spec(Ka,b ) = {− ab, 0a+b−2 , ab}, which quickly leads us to the following proposition. 100
Proposition 5.1.1. Let a, b ≥ 1 and let n = ab. Then K2 (a, b; z) is DS iff n is prime or a and b are such that they minimize {x + y : x, y ≥ 1 and xy = n}. In the first case we can allow z ≥ 0. In the second case we need z to be less than a + b minus the second smallest element in {x + y : x, y ≥ 1 and xy = n}. Proof. Without loss of generality we may assume a ≥ b. Now, suppose that K2 (a, b; z) is cospectral with K2 (k, m; x) for some k, m ≥ 1 and x ≥ 0. We know that ab = km and a + b + z = k + m + x by equating their spectra. Now, if ab is prime, then clearly we get {a, b} = {k, m}, and thus the two graphs are isomorphic. Thus, K2 (a, b; z) is always DS when ab is prime. On the other hand, if ab is not prime, then from a + b + z = k + m + x we get that z = k + m + x − a − b ≥ 0. Now, if a + b = k + m, then {a, b} = {k, m}. Suppose then that a + b < k + m. Then we see that z is never less than x + y − a − b. If instead a + b > k + m, then z is always 0 or greater. Therefore, in this case K2 (a, b; z) is DS iff a + b < k + m and 0 ≤ z < k + m − a − b. The case for K3 (a, b, c; z) is not quite as simple. Some work has been done in the case of general complete multipartite graphs, but it is difficult to characterize DS graphs as definitively as in Proposition 5.1.1. However, we can show similarly that K3 (a, b, c; z) is DS when abc is prime or when a + b + c is minimal. Perhaps the most useful theorem about complete multipartite graphs is the characterization of their spectra as having only one positive eigenvalue. Theorem 5.1.2. G has exactly one positive eigenvalue iff it is the union of a complete multipartite graph and k isolated vertices with k ≥ 0. This result and more can be found in [8]. Theorem 5.1.2 will allow us to exclude examining K4 (a, b, c, d; z) in the rest of this chapter.
101
5.2
2P2(a, b, c, d; z)
2P2 (k, 1, 1, 1; z): d1 = k + z + 3 d2 = k + 1 d3 = 0 d4 = k
c2 = c4 + 1
(5.1)
2P2 (a, b, c, d; 0) : c2 = c4 + 1 gives ab + cd = abcd + 1. This can only happen if ab = 1 or cd = 1. Suppose c = d = 1. From here we can quickly solve for a, b ≥ 1, ab = k, and z = a + b − ab − 1 = −(a − 1)(b − 1). Since z ≤ 0, we can move the isolated points to 2P2 (a, b, c, d; z). Thus, we have 2P2 (a, b, 1, 1; z + (a − 1)(b − 1)) ∼ 2P2 (ab, 1, 1, 1; z) for all a, b ≥ 1 and z ≥ 0. In particular 2P2 (k, 1, 1, 1; z) has a cospectral mate for all even k ≥ 4.
P4 (a, b, c, d; 0) : We may assume a ≥ d. c2 = c4 + 1 gives ab + bc + cd = abcd + 1. From this we solve for a. a=
bc + cd − 1 b(cd − 1)
With a ≥ d ≥ 1 we can get that either d = 1 and b, c ≥ 1 or d = 2 and b = c = 1. Finally, from these we get P4 (3, 1, 2, 1; z+2) ∼ 2P2 (6, 1, 1, 1; z), P4 (2, 2, 3, 1; z+7) ∼ 2P2 (12, 1, 1, 1; z), and P4 (2, 1, 1, 2; z + 1) ∼ 2P2 (4, 1, 1, 1; z) for z ≥ 0.
P5 (a, b, c, d, e; 0) : Using the bounds on eigenvalues we know that λ2 = 1 and λn−1 = −1. These imply √
em ≤
1
≤
√
aM √ √ − aM ≤ −1 ≤ − em 102
Therefore aM = em = 1, and thus a = b = d = e = 1. Finally we can solve for k = 2c + 1, and thus P5 (1, 1, c, 1, 1; z + c) ∼ 2P2 (2c + 1, 1, 1, 1; z) for c ≥ 1 and z ≥ 0. This says that 2P2 (k, 1, 1, 1; z) has a cospectral mate for all odd k ≥ 3. Therefore we conclude that 2P2 (k, 1, 1, 1; z) is DS iff k = 2 and z ≥ 0.
103
5.3
P4(a, b, c, d; z)
P4 (k, 1, 1, 1; z): d1 = k + z + 3 d2 = k + 2 d3 = 0 d4 = k c2 = c4 + 2
(5.2)
2P2 (a, b, c, d; 0) : c2 = c4 + 2 gives ab + cd = abcd + 2, but for all a, b, c, d ≥ 1 we have ab + cd < abcd + 2.
P4 (a, b, c, d; 0) : We may assume a ≥ d. c2 = c4 + 2 gives ab + bc + cd = abcd + 2. From this we solve for a. bc + cd − 2 b(cd − 1) With a ≥ d ≥ 1 we can only have d = 1. With d = 1 we now have a=
a=
bc + c − 2 b(c − 1)
If we solve for b instead and let c = 1, then b = 1, which just gives P4 (a, 1, 1, 1; z) ∼ P4 (k, 1, 1, 1; z). So we may assume c ≥ 2. If we also assume b ≥ 2, then we 1 < a ≤ 2, with a = 2 iff c = 2. On the other hand, if b = 1, then a = 2 for all c ≥ 2. From these we can then solve for P4 (2, b, 2, 1; z +3b−2) ∼ P4 (4b, 1, 1, 1; z) and P4 (2, 1, c, 1; z +c−1) ∼ P4 (2c, 1, 1, 1; z) for all b, c ≥ 1 and z ≥ 0. In particular P4 (k, 1, 1, 1; z) has a cospectral mate for all even k ≥ 4.
P5 (a, b, c, d, e; 0) : √ Using the bounds on eigenvalues with the fact that λ2 = 2em ≤ k + 2 −
√ k+2− k2 +4 √ 2
√ k 2 + 4 ≤ 2aM
104
we get
√ k 2 + 4 ≤ 2aM , but for k ≥ 2 we have 1 < k + 2 − k 2 + 4 < 2. √ For k = 1 we have 2em < 2 − 3 < 1. Since 2em ≥ 2 we have a contradiction. Therefore 2em ≤ k + 2 −
√
Therefore we conclude that P4 (k, 1, 1, 1; z) is DS iff k = 2 or k ≥ 1 is odd and z ≥ 0.
105
P4 (1, k, 1, 1; z): d1 = k + z + 3 d2 = 2k + 1 d3 = 0 d4 = k
c2 = 2c4 + 1
(5.3)
2P2 (a, b, c, d; 0) : c2 = 2c4 +1 gives ab+cd = 2abcd+1, but for all a, b, c, d ≥ 1 we have ab+cd < 2abcd+1.
P4 (a, b, c, d; 0) : We may assume a ≥ d. Solving (5.3) for a, we get a=
bc + cd − 1 b(2cd − 1)
This implies that a ≤ 1, with a = 1 iff b ≥ 1 and c = d = 1 or c ≥ 1 and b = d = 1. In either case we have P4 (a, b, c, d; z) ∼ = P4 (1, k, 1, 1; z).
P5 (a, b, c, d, e; 0) : √
√ 2k+1− 4k2 +1
√ Similar to P4 (k, 1, 1, 1; z), with the bounds on the eigenvalues and using λ2 = 2 √ √ 2 2 we get that 2em ≤ 2k + 1 − 4k + 1 ≤ 2aM . However, 2k < 4k + 1 < 2k + 1, and thus √ 2em ≤ 2k + 1 − 4k 2 + 1 < 1 for all k ≥ 1, a contradiction.
Therefore we conclude that P4 (1, k, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0.
106
5.4
P5(a, b, c, d, e; z)
P5 (k, 1, 1, 1, 1; z): d1 = k + z + 4 d2 = k + 3 d3 = 0 d4 = 2k + 1
2c2 = c4 + 5
(5.4)
2P2 (a, b, c, d; 0) : Equation (5.4) gives 2ab + 2cd = abcd + 5. We get equality iff ab = 3 and cd = 1 or ab = 1 and cd = 3. Assume a = 3 and b = c = d = 1. This gives k = 1, and thus 2P2 (3, 1, 1, 1; z) ∼ P5 (1, 1, 1, 1, 1; z + 1). This agrees with our analysis of 2P2 (k, 1, 1, 1; z).
P4 (a, b, c, d; 0) : From equation (5.4) we can solve for a=
2bc + 2cd − 5 b(cd − 2)
If we assume a ≥ d, then we can use this to show that d ≤ 3. If d = 3, then we must have a = 3 and b = c = 1. This leads to P4 (3, 1, 1, 3; z) ∼ P5 (4, 1, 1, 1, 1; z). If d = 2, then we must have c = 1, 2 and b ≥ 1, or c ≥ 3 and 1 ≤ b ≤ 3. If c ≥ 3, then 3 < a < 4 if b = 1 and 2 < a < 3 if b = 2. We can only have b = 3 if c = 3 in which case a=
13 . 4
If c = 2, then a =
7 2
if b = 1 and 1 < a < 2 if b ≥ 1. If c = 1, then using c2 = k + 3
and c4 = 2k + 1 we get that b = 12 . Finally we consider d = 1. If b = c = 1, then a = 1 and k = 0. If c = 2, then similar to above we can solve for b = 14 . With c ≥ 3 it can be shown that a ∈ (2, 7] with a = 3 when b = 3 and c = 13 or b = 9 and c = 7, a = 5 when b = 1 and c = 5, and a = 7 when 107
b = 1 and c = 3. These lead to P4 (3, 3, 13, 1; z + 42) ∼ P5 (58, 1, 1, 1, 1; z), P4 (3, 9, 7, 1; z + 78) ∼ P5 (94, 1, 1, 1, 1; z), P4 (5, 1, 5, 1; z + 4) ∼ P5 (12, 1, 1, 1, 1; z), and P4 (7, 1, 3, 1; z + 2) ∼ P5 (10, 1, 1, 1, 1; z).
P5 (a, b, c, d, e; 0) : √
√ k+3− k2 −2k+5
√ we get that 2em ≤ With the bounds on the eigenvalues and using λ2 = 2 √ k + 3 − k 2 − 2k + 5 < 4. This implies that 2em = 2, and thus e = m = 1. If b = 1 or
d = 1 we can show that 1 < a ≤ 3. With b = 1 we have a = 3 if d = 1 and c ≥ 1. With d = 1 we have a = 2 with b = c = 3 and a = 3 with b = 1 and c ≥ 1. With these we get P5 (3, 1, c, 1, 1; z+c−1) ∼ P5 (2c+1, 1, 1, 1, 1; z) and P5 (2, 3, 3, 1, 1; z+10) ∼ P5 (16, 1, 1, 1, 1; z). In particular P5 (k, 1, 1, 1, 1; z) has a cospectral mate for all odd k ≥ 3 and z ≥ 0. Therefore we may conclude that P5 (k, 1, 1, 1, 1; z) is DS iff k ≥ 2 is even except for k ∈ {4, 10, 12, 16, 58, 94} and z ≥ 0 or k = 1 and z = 0.
108
P5 (1, k, 1, 1, 1; z): d1 = k + z + 4 d2 = 2k + 2 d3 = 0 d4 = 3k
3c2 = 2c4 + 6
(5.5)
2P2 (a, b, c, d; 0): Equation (5.5) gives 3ab + 3cd = 2abcd + 6. We get equality iff ab = 3 and cd = 1 or ab = 1 and cd = 3. Assume a = 3 and b = c = d = 1. This gives k = 1, and thus 2P2 (3, 1, 1, 1; z) ∼ P5 (1, 1, 1, 1, 1; z + 1). This agrees with our analysis of 2P2 (k, 1, 1, 1; z).
P4 (a, b, c, d; 0): From equation (5.5) we can solve for a=
3(bc + cd − 2) b(2cd − 3)
If we assume a ≥ d, then we can use this to show that d ≤ 2. If d = 2, then we get a = 3, c = 1, b ≥ 1 as the only solution. This leads to P4 (3, b, 1, 2; z + b − 2) ∼ P5 (1, 2b, 1, 1, 1; z) for b ≥ 1 and z ≥ 0, with the one exception of b = 1 and z = 0.
If d = 1, then we get many solutions. The solutions for (a, b, c) are (6, b, 2), (4, 1, 3), (3, 2, 4),
109
(2, 2, 12), (2, 4, 18), (2, 6, 10), (2, 7, 9), (2, 9, 8), and (2, 15, 7). These give P4 (6, b, 2, 1; z + 3b − 5) ∼ P5 (1, 4b, 1, 1, 1; z) P4 (4, 1, 3, 1; z) ∼ P5 (1, 4, 1, 1, 1; z + 1) P4 (3, 2, 4, 1; z + 2) ∼ P5 (1, 8, 1, 1, 1; z) P4 (2, 4, 18, 1; z + 27) ∼ P5 (1, 48, 1, 1, 1; z) P4 (2, 2, 12, 1; z + 3) ∼ P5 (1, 16, 1, 1, 1; z) P4 (2, 6, 10, 1; z + 25) ∼ P5 (1, 40, 1, 1, 1; z) P4 (2, 7, 9, 1; z + 27) ∼ P5 (1, 42, 1, 1, 1; z) P4 (2, 9, 8, 1; z + 32) ∼ P5 (1, 48, 1, 1, 1; z) P4 (2, 15, 7, 1; z + 49) ∼ P5 (1, 70, 1, 1, 1; z) With these we see that P5 (1, k, 1, 1, 1; z) has a cospectral mate for all even k ≥ 2 and z ≥ 0, except when k = 2 and z = 0, and that there is much overlap when k is a multiple of 4.
P5 (a, b, c, d, e; 0): p √ With the bounds on the eigenvalues and using λ2 = k + 1 − k 2 − k + 1 we get that √ em ≤ k + 1 − k 2 − 2 + 1 < 2. This implies that em = 1, and thus e = m = 1. If b = 1 or d = 1 we can show that 1 ≤ a < 2. In either case we arrive at P5 (1, 3, c, 1, 1; z + c − 1) ∼ P5 (1, 2c + 1, 1, 1, 1; z). This is very similar to the P5 (k, 1, 1, 1, 1; z) case. In particular we see that P5 (1, k, 1, 1, 1; z) has a cospectral mate for all odd k ≥ 3 and z ≥ 0.
Therefore we conclude that P5 (1, k, 1, 1, 1; z) is DS iff k = 1, 2 and z = 0.
110
P5 (1, 1, k, 1, 1; z): d1 = k + z + 4 d2 = 2k + 2 d3 = 0 d4 = 2k + 1 c2 = c4 + 1
(5.6)
√ 1,1 Spec(P5 (1, 1, k, 1, 1; z)) = {0k+z , ±11,1 , ± 2k + 1 }.
2P2 (a, b, c, d; 0): Since we know Spec(2P2 (a, b, c, d; z)), we can directly compute c = d = 1 and ab = 2k + 1. Since ab is odd we need both a and b to be odd. Therefore, we have 2P2 (2a + 1, 2b + 1, 1, 1; z + 2ab − a − b) ∼ P5 (1, 1, 2ab + a + b, 1, 1; z) for all a ≥ 1, b ≥ 0, and z ≥ 0. In particular with b = 0, we get that P5 (1, 1, k, 1, 1; z) has a cospectral mate for all odd k ≥ 3 and z ≥ 0.
P4 (a, b, c, d; 0): From the bounds on eigenvalues, and using λ2 = 1, we get 1 √ 4dm + m2 − m ≤ 1 2 This reduces to dm − m − 1 ≤ 0 This is only true if d = 1 and m ≥ 1 or d = 2 and m = 1. We can then use PP4 (a,b,c,d;z) (1) = 0 to get abcd − ab − bc − cd + 1 = 0 This along with d = 1, 2 gives us the only possible solutions as (3, 1, 2, 1), (2, 2, 3, 1), and (2, 1, 1, 2). These give k = 52 , 11 , 3 , respectively, of which none can happen. 2 2 P5 (a, b, c, d, e; 0): 111
With the bounds on the eigenvalues and using λ2 = 1 we get that em ≤ 1, and thus e = m = 1. If d = 1 then PP5 (a,b,c,1,1;z) (1) = 0 gives c(ab − 1) = 0, and thus a = b = 1. On the other hand, if b = 1, then PP5 (a,1,c,d,1;z) (1) = 0 gives acd + ad − a − c − d + 1 = 0. This, too, can be shown to only be true when a = d = 1. In both cases we are led to P5 (1, 1, c, 1, 1; z) ∼ = P5 (1, 1, k, 1, 1; z). Therefore we conclude that P5 (1, 1, k, 1, 1; z) is DS iff k ≥ 2 is even and z ≥ 0.
112
5.5
Kite(a, b, c, d; z)
Kite(k, m, n, 1; z): d1 = k + m + n + z + 4 d2 = km + kn + mn + k d3 = kmn d4 = kmn
c3 = c4
(5.7)
It is easy to see that (5.7) is not satisfied for Bull, House, Broom, or P2 × K3 . Kite(a, b, c, d; z) to satisfy (5.7) then d = 1. Now, with d = 1 we can show that
1 4
For
< λ2 < 21 .
For this we look at 1 PKite(a,b,c,1;z) = 2−2a−2b−2c−2z−2 (128abc − 16ab − 16ac − 16a − 16bc + 1) 4 1 PKite(a,b,c,1;z) = −2−a−b−c−z−1 (4ab + 4ac + 4a + 4bc − 1) 2 From these it is easy to see that PKite(a,b,c,1;z) 41 > 0 and PKite(a,b,c,1;z) 12 < 0 for all a, b, c ≥ 1. Moreover, we can also show that λ2 < 31 iff b = c = 1. For this we look at 1 PKite(a,b,c,1;z) = 3−a−b−c−z−1 (27abc − 9ab − 9ac − 9a − 9bc + 1) 3 This is negative iff b = c = 1. These are enough to show that Kite(k, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0.
On the other hand, if we assume k = 1, then we can solve for p a + ab + ac + bc − abc − 1 ± (a + ab + ac + bc − abc − 1)2 − 4abc m, n = 2 With the assumption that b ≥ c, for these to be real and positive we need a = 1 and 1 ≤ c ≤ b; a ≥ 2, b ≥ 1, and c = 1; or a ≥ 9 and b = c = 2.
With the a = 1 case we get m = b and n = c, and thus Kite(1, b, c, 1; z) ∼ Kite(1, m, n, 1; z). 113
With the a ≥ 2, b ≥ 1, and c = 1 case we get √ 1 2 2 m, n = 2a + b − 1 ± 4a + b − 4a − 2b + 1 2 b 2
for b > 2. More precisely, n ≤ a − 1 b−1 2 for a ≤ ( 2 ) . for b ≤ a(a − 1), with n = a − 1 iff b = a(a − 1). Similarly n ≤ b−1 2 b−1 It is immediately clear that for given m, n, z, there are at most min{a − 1, 2 } possible From this we can quickly see that n < a and n
3c3 .
For Kite(a, b, c, d; 0) we quickly see from (5.11) that d = 3. Now assume m = 1. We can now solve ab + ac + bc + 3a = 3k + 2 abc = k to get k = 2a, b = 2, and c = 1. This shows that Kite(a, 2, 1, 3; z+a−2) ∼ Bull(2a, 1, 1, 1, 1; z) and Kite(1, 2, 1, 3; z) ∼ Bull(2, 1, 1, 1, 1; z + 1) for all a ≥ 2 and z ≥ 0. We can also show the stronger result that for all a, m ≥ 1 and z ≥ 0 we have Kite(a, 2m, 1, 3; z + a − m − 2) ∼ Bull(2a, m, 1, 1, 1; z) or Kite(a, 2m, 1, 3; z) ∼ Bull(2a, m, 1, 1, 1; z − a + m + 2), depending on whether a − m − 2 is positive or negative. Unfortunately this does not encompass all solutions as can be seen from Kite(6, 2, 2, 3; z + 1) ∼ Bull(8, 3, 1, 1, 1; z).
For Bull(a, b, c, d, e; 0) equation (5.11) gives c = d = e = 1 from which we quickly get Bull(a, b, c, d, e; z) ∼ = Bull(k, m, 1, 1, 1; z).
Finally, for House(a, b, c, d, e; 0) we get from equation (5.11) that a = c = 2 and d = e = 1. Solving for k, m, and b from here gives b = 21 (2z + 3), which is never an integer.
118
Therefore we conclude that Bull(2, 1, 1, 1, 1; 0) and Bull(k, 1, 1, 1, 1; z) are DS for all odd k ≥ 1 and z ≥ 0.
119
Bull(1, 1, k, 1, 1; z): d1 = k + z + 4 d2 = 2k + 3 d3 = k d4 = 2k + 1
c2 = 2c3 + 3
(5.12)
c4 = 2c3 + 1
(5.13)
c2 = c4 + 2
(5.14)
We saw previously that with Broom and P2 × K3 that c4 > 3c3 ≥ 2c3 + 1. We can also show that with House that c4 > 2c3 + 1.
For Kite(a, b, c, d; 0) we get abcd = 2abc + 1 from (5.13), which gives a = b = c = 1 and d = 3. However, with these we get c2 = 6 6= 5 = 2c3 + 3.
Finally, for Bull(a, b, c, d, e; 0) (5.13) gives abcd + abce + abde = 2abc + 1 This factors as ab(cd + ce + de − 2c) = 1 which implies ab = 1 and cd + ce + de − 2c = 1. The former implies a = b = 1 while the latter can be seen to give d = e = 1. Therefore c = k, and thus Bull(a, b, c, d, e; z) ∼ = Bull(1, 1, k, 1, 1; z). Therefore we conclude that Bull(1, 1, k, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0.
120
Bull(1, 1, 1, k, m; z): d1 = k + m + z + 3 d2 = k + m + 3 d3 = 1 d4 = km + k + m
c3 = 1
(5.15)
We can immediately rule out P2 × K3 from (5.15). For each of Kite, Bull, House, and Broom (5.15) forces a = b = c = 1.
From c2 = k + m + 3 and c4 = km + k + m we can solve for q 1 2 k, m = c2 − 3 ± c2 − 2c2 − 4c3 − 3 2 From this we get k, m = 0, d with Kite; k, m = d, e with Bull; and k, m = 0, 2de + d + e with House. Broom is explored in full detail below. In particular we will show that if m = 1, then no Broom is cospectral.
Therefore we conclude that Bull(1, 1, 1, k, 1; z) is DS for all k ≥ 1 and z ≥ 0.
From (5.15) we get for Broom(1, 1, 1, d, e, f ; 0) k, m =
p 1 de + d + e + f ± (de + d + e + f )2 − 4(def + de + df + ef ) 2
For this to be an integer it is clear that we need (de + d + e + f )2 − 4(def + de + df + ef ) = n2 for some n ∈ N0 . If we solve this for f we get f = de + d + e ±
√ n2 + 4de
We now need to solve for which n makes n2 + 4de a perfect square. First suppose n = 2r and n2 + 4de = (2s)2 . Then we get de = (r + s)(r − s). On the other hand, if n = 2r + 1 and n2 + 4de = (2s + 1)2 , then de = (r + s + 1)(r − s). We now see that r and s depend on how we can factor de. Suppose we factor de = pq. For n even, if we set r + s = p and r − s = q, 121
then we get r = then we get r =
p+q , and thus n 2 p+q−1 , and once 2
= p + q. For n odd, if we set r + s + 1 = p and r − s = q, again n = p + q. Finally, we arrive at f = de + d + e ± (p + q)
where de = pq. This gives us the following proposition:
Proposition 5.6.1. Bull(1, 1, 1, k, m; 0) is cospectral with Broom(1, 1, 1, d, e, f ; z) iff d = d1 d2 , e = e1 e2 , and f = de + d + e ± (d1 e1 + d2 e2 ) in which case k = de + d + e ± d1 e1 , m = de + d + e ± d2 e2 and z = de. From this proposition we quickly see that k, m ≥ 2 since di ei ≤ de. As an example of this proposition in action consider d = 6 and e = 2. One possible way to factor de = 12 is d1 = 2, d2 = 3, e1 = 2, and e2 = 1. Taking the negatives above we are led to f = 12 + 6 + 2 − (2 · 3 + 2 · 1) = 12, k = 12 + 6 + 2 − 2 · 3 = 14, m = 12 + 6 + 2 − 2 · 1 = 18, and z = 12. The full list for d = 6 and e = 2 is: Bull(1, 1, 1, 19, 8; 0) ∼ Broom(1, 1, 1, 6, 2, 7; 12) Bull(1, 1, 1, 18, 14; 0) ∼ Broom(1, 1, 1, 6, 2, 12; 12) Bull(1, 1, 1, 17, 16; 0) ∼ Broom(1, 1, 1, 6, 2, 13; 12) Bull(1, 1, 1, 24, 23; 0) ∼ Broom(1, 1, 1, 6, 2, 27; 12) Bull(1, 1, 1, 26, 22; 0) ∼ Broom(1, 1, 1, 6, 2, 28; 12) Bull(1, 1, 1, 32, 21; 0) ∼ Broom(1, 1, 1, 6, 2, 33; 12)
122
5.7
House(a, b, c, d, e; z)
House(k, 1, 1, 1, 1; z): d1 = k + z + 4 d2 = 3k + 3 d3 = k d4 = 3k + 1
c4 = 3c3 + 1
(5.16)
We will use (5.16) to quickly show that House(k, 1, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0. Starting with Kite, (5.16) gives abcd = 3abc + 1 This gives abc(d − 3) = 1, and thus a = b = c = 1, and d = 4. We have already shown above that Kite(1, 1, 1, 4; z) is DS for all z ≥ 0. Continuing with Bull, we get abcd + abce + abde = 3abc + 1 This gives ab(cd + ce + de − 3c) = 1, and thus a = b = 1 and cd + ce + de − 3c = 1. If d > 1 or e > 1, then cd + ce ≥ 3c and de > 1, so we require d = e = 1. This then forces c = 0. Next, with House we get abcd + abce + abde + bcde = 3abc + 1 This gives b(acd + ace + ade + cde − 3ac) = 1. Similar to Bull above we are forced to have b = d = e = 1 which leaves a + c − ac = 1. This is true iff a or c is 1. Since we may assume a ≥ c, we can set c = 1, and we are left with a = k, and thus House(a, b, c, d, e; z) ∼ = House(k, 1, 1, 1, 1; z). Lastly we will look at Broom and P2 × K3 . With Broom we have abcd + abce + abcf ≥ 3c for all d, e, f ≥ 1, and thus c4 > 3c3 + 1. We can show this true with P2 × K3 as well with a similar argument. Therefore we conclude that House(k, 1, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0.
123
House(1, k, 1, 1, 1; z): d1 = k + z + 4 d2 = 2k + 4 d3 = k d4 = 4k
c2 = 2c3 + 4
(5.17)
c4 = 4c3
(5.18)
Starting with Kite(a, b, c, d; 0), (5.18) immediately gives d = 4. Using this with (5.17) gives ab + ac + bc + 4a = 2abc + 4 The solutions to this are a = 2, b = 6, and c = 1 or a ≥ 1 and b = c = 2. Thus, Kite(2, 6, 1, 4; z + 3) ∼ House(1, 12, 1, 1, 1; z), Kite(1, 2, 2, 4; z) ∼ House(1, 4, 1, 1, 1; z + 1), and Kite(a, 2, 2, 4; z + 3a − 4) ∼ House(1, 4a, 1, 1, 1; z) for all a ≥ 2 and z ≥ 0.
Next with Bull(a, b, c, d, e; 0), (5.18) gives a, b ≥ 1, c = d = 2, and e = 1. With these (5.17) becomes 3ab − 4a − 3b + 4 = (a − 1)(3b − 4) = 0 Clearly a = 1, and we get Bull(1, 1, 2, 2, 1; z) ∼ House(1, 2, 1, 1, 1; z+1) and Bull(1, b, 2, 2, 1; z+ b − 2) ∼ House(1, 2b, 1, 1, 1; z) for b ≥ 2 and z ≥ 0.
Continuing with House(a, b, c, d, e; 0), (5.18) gives d + e < 4. If we assume d ≥ e, then we have either d = 2 and e = 1 or d = e = 1. With the former (5.18) becomes ac − 2a − 2c = 0 which has solutions a = 3, c = 6; a = c = 4; and a = 6, c = 3. Solving for b in (5.17) gives 31 13 , 27 12
and
28 , 27
respectively. On the other hand, with d = e = 1, (5.18) becomes 2ac − a − c = 0
124
which only has a = c = 1 as a solution. We can then solve for b = k, and thus we get House(a, b, c, d, e; z) ∼ = House(1, k, 1, 1, 1; z).
In a similar fashion as above we get for Broom(a, b, c, d, e, f ; 0) that a = 7, c = 3, d = e = f = 1, and b =
28 . 31
For P2 × K3 we have no simultaneous solutions to (5.17) and (5.18).
Therefore we conclude that House(1, k, 1, 1, 1; z) is DS for all odd k ≥ 1 and z ≥ 0 or when k = 2 and z = 0.
125
House(1, 1, 1, k, m; z): d1 = k + m + z + 3 d2 = km + k + m + 3 d3 = 1 d4 = 2km + k + m
c3 = 1
(5.19)
We saw above that c3 = 1 forces a = b = c = 1 in Kite, Bull, House, and Broom, and that c3 = 1 is not possible with P2 × K3 . Solving for k, m we get p 1 2 2c2 − c4 − 6 ± (2c2 − c4 − 6) + 4(c2 − c4 − 3) k, m = 2 From this we can quickly calculate k = d, m = 0 with Kite(1, 1, 1, d; z) and k = d, m = e with House(1, 1, 1, d, e; 0). With Bull and Broom we get that m < 0, when it is real, for all d, e, f ≥ 1.
Therefore we conclude that House(1, 1, 1, k, m; z) is DS for all k, m ≥ 1 and z ≥ 0.
126
5.8
Broom(a, b, c, d, e, f ; z)
Broom(k, 1, m, 1, 1, 1; z): d1 = k + m + z + 4 d2 = km + 2k + 2m + 2 d3 = km d4 = 3km + 2k + 2m + 1
c2 + 2c3 = c4 + 1
k, m =
(5.20)
p 1 c2 − c3 − 2 ± (c2 − c3 − 2)2 − 16c3 4
Starting with Kite(a, b, c, d; 0), we will first show c = 1. From (5.20) we solve for a a=
bc − 1 bcd − 2bc − b − c − d
For a ≥ 1 and k, m to be real we get that one of c = 1, if we assume b ≥ c. Moreover, for k, m to be real we also get b ≥ 6 and d = 4; b = 4, 5 and d = 5; b = 3 and d = 6; or b = 2 and d = 8. Checking these possibilities, we get the only positive integer solutions for a, k, m are with Kite(1, 3, 1, 6; z) ∼ Broom(3, 1, 1, 1, 1, 1; z + 3) and Kite(3, 7, 1, 4; z) ∼ Broom(7, 1, 3, 1, 1, 1; z + 1).
Next, for Bull(a, b, c, d, e; 0) to satisfy (5.20) we get the possibilities a = b = c = e = 1 and d = 4; a = 3, b = d = 2 and c = e = 1; a = c = 2, b = e = 1, and d = 3; or a = b = 1, d = e = 2, and c ≥ 1. Of these we can only have the last for k, m to be real and positive. Moreover, we have k, m =
√ 1 c + 3 ± c2 − 10c + 9 4
in this case. From here we can show that c2 − 10c + 9 is a perfect square iff c ∈ {1, 9, 10}, and of these c = 10 gives a rational value for m. Thus, we have Bull(1, 1, 1, 2, 2; z) ∼ Broom(1, 1, 1, 1, 1, 1; z + 1) and Bull(1, 1, 9, 2, 2; z) ∼ Broom(3, 1, 3, 1, 1, 1; z + 5).
127
House(a, b, c, d, e; 0) is simpler. (5.20) is only satisfied with a = b = c = e = 1 and d = 4 √ √ or a = b = c = 1 and d = e = 2. These give k, m = 14 (9 ± 65) and k, m = 2 ± 3, respectively.
For Broom(a, b, c, d, e, f ; 0) we start by showing b = 1. First, we observe that c4 + 1 grows faster than c2 + 2c3 . Next, if b = 2, then we have 5ac + ae + 2a + cd + 2c + de + 2f = 2(acd + ace + acf + ade + aef + cde + cdf + def ) + 1 One can readily check this has no positive integer solutions. Thus, b = 1 is the only possibility. With b = 1 we get instead 3ac + ae + a + cd + c + de + f = acd + ace + acf + ade + aef + cde + cdf + def + 1 In similar fashion we can now get that d = e = f = 1 is the only possibility, and with those we are forced to have a = k and c = m. Thus, Broom(a, b, c, d, e, f ; 0) ∼ Broom(k, 1, m, 1, 1, 1; z).
Finally, using the same idea as with Broom, we have c4 + 1 growing faster than c2 + 2c3 , and we can quickly check that a = b = c = d = e = f = 1 is the only solution to (5.20). With √ these we get k, m = 41 (5 ± i 7). Therefore we conclude that Broom(k, 1, m, 1, 1, 1; z) is DS for all k, m ≥ 1 and z ≥ 0 with the four exceptions Broom(3, 1, 1, 1, 1, 1; z+3), Broom(7, 1, 3, 1, 1, 1; z+1), Broom(1, 1, 1, 1, 1, 1; z+ 1), and Broom(3, 1, 3, 1, 1, 1; z + 5).
128
Broom(1, k, 1, 1, 1, 1; z): d1 = k + z + 5 d2 = 3k + 4 d3 = k d4 = 8k c2 = 3c3 + 4
(5.21)
c4 = 8c3
(5.22)
Starting with Kite we get from (5.22) that d = 8. Using this in (5.21) we get 3abc + 4 = ab + ac + bc + 8a From this we can solve for a = b = 1, c = 5 or a ≥ 1, b = c = 2, which gives Kite(1, 1, 5, 8; z) ∼ Broom(1, 5, 1, 1, 1, 1; z + 5) Kite(a, 2, 2, 8; z + 3a − 7) ∼ Broom(1, 4a, 1, 1, 1, 1; z) For Bull we can check with (5.22) that d, e ≤ 6. With this we can now see from (5.21) that the only possible solutions are a = b = c = 1, d = e = 2; a = b = d = e = 2, b = 1; and a = 1, b ≥ 1, c = d = 2, e = 3. These then give Bull(1, 1, 1, 2, 2; z) ∼ Broom(1, 1, 1, 1, 1, 1; z + 1) Bull(2, 2, 1, 2, 2; z) ∼ Broom(1, 4, 1, 1, 1, 1; z) Bull(1, b, 2, 2, 3; z + b − 3) ∼ Broom(1, 4a, 1, 1, 1, 1; z) Similar to Bull, (5.22) gives d, e ≤ 6 for House. With these we can then see that (5.21) is never satisfied. Just as with Bull and House, (5.22) gives us e, f ≤ 5. We can then use these we get (5.21) is only satisfied when a = c = d = e = f = 1, b ≥ 1, which gives us a graph isomorphic to Broom(1, k, 1, 1, 1, 1; z). P2 × K3 requires a bit more work. From (5.22) we can solve for e. Doing so gives us e=
8bcd − abcd − abdf − acdf − bcdf abc + abf + acd + acf + adf + bcd + bcf + bdf − 8af 129
If we assume a ≥ e ≥ f , then we can show that if a, f ≥ 3 then we must have b + c + d < 8. To see this we can rewrite e ≥ f to get (8a − (ab + ac + ad + bc + bd))f 2 + 8bcd − (abcd + abcf + abdf + 2acdf + 2bcdf ) ≥ 0 Under the assumption that a, f ≥ 3, we have abcd + 2bcdf ≥ 3bcd + 6bcd > 8bcd. This now forces (8a − (ab + ac + ad + bc + bd))f 2 = (a(8 − b − c − d) − bc − bd)f 2 > 0. It is straightforward to see now that b + c + d < 8 must be true. With these we can now check each case where b + c + d < 8, and in doing so we can see that (5.21) and (5.22) are never simultaneously satisfied. Finally, we have to check separately the cases where a = 1 and a = 2. With these we get that (5.21) and (5.22) are simultaneously satisfied only when a = b = 2 and c = d = e = f = 1. This gives P2 × K3 (2, 2, 1, 1, 1, 1; z + 1) ∼ Broom(1, 4, 1, 1, 1, 1; z) Therefore, we may conclude that Broom(1, k, 1, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0 except for when k is a multiple of 4 or the two cases when k = 1, z ≥ 1 and k = 5, z ≥ 5.
130
Broom(1, 1, 1, k, m, 1; z): d1 = k + m + z + 4 d2 = km + k + m + 4 d3 = 1 d4 = 3km + 2k + 2m + 1
c3 = 1
(5.23)
Using c2 = km + k + m + 4 and c4 = 3km + 2k + 2m + 1 we can solve for p 1 2 3c2 − c4 ± (−3c2 + c4 + 11) − 4 (−2c2 + c4 + 7) − 11 2
(5.24)
We saw with Bull(1, 1, 1, k, m) that c3 = 1 implies a = b = c = 1 in Kite, Bull, House, and Broom, and that c3 = 1 is not possible with P2 × K3 . Using (5.24) we can quickly verify that the only possible non-isomorphic solution is Bull(1, 1, 1, 2, 2; z) ∼ Broom(1, 1, 1, 1, 1, 1; z + 1) Therefore, we may conclude that Broom(1, 1, 1, k, m, 1; z) is DS for all k, m ≥ 1 and z ≥ 0 except for when k = m = 1 and z ≥ 1.
131
Broom(1, 1, 1, 1, 1, k; z): d1 = k + z + 5 d2 = k + 6 d3 = 1 d4 = 4k + 4
c3 = 1
(5.25)
c4 = 4c2 − 20
(5.26)
As with our last case we must have a = b = c = 1 in Kite, Bull, House, and Broom, and c3 = 1 never happens with P2 × K3 . We can now very quickly check with (5.26) that the only non-isomorphic solutions are Bull(1, 1, 1, 2, 2; z) ∼ Broom(1, 1, 1, 1, 1, 1; z + 1) Bull(1, 1, 1, 4, 4; z) ∼ Broom(1, 1, 1, 1, 1, 5; z + 1) Therefore, we may conclude that Broom(1, 1, 1, 1, 1, k; z) is DS for all k ≥ 1 and z ≥ 0 except for k = 1, 5 and z ≥ 1.
132
5.9
P2 × K3(a, b, c, d, e, f ; z)
P2 × K3 (k, 1, 1, 1, 1, 1; z): d1 = k + z + 5 d2 = 3k + 6 d3 = k + 1 d4 = 8k + 4
c2 = 3c3 + 3
(5.27)
c4 = 8c3 − 4
(5.28)
Using (5.28) with Kite we can solve for a to get a=
4 bc(d − 8)
We immediately see that a ∈ {1, 2, 4}, d < 8 and bc(d − 8) divides 4. With this we can quickly check that the only solution that also satisfies (5.27) is a = 1, b = c = 2, and d = 7, which gives Kite(1, 2, 2, 7; z) ∼ P2 × K3 (3, 1, 1, 1, 1, 1; z + 4) Similarly, with (5.28) gives a=
4 b(8c − cd − ce − de)
for Bull. This time we can check that the only solution that works with (5.27) is a = b = 1 and c = d = e = 2. This gives Bull(1, 1, 2, 2, 2; z) ∼ P2 × K3 (1, 1, 1, 1, 1, 1; z + 2)
Next, with House we get a=
bcde + 4 b(8c − cd − ce − de)
from (5.28). If we want this to be at least 1, then we get the inequality 8bc − bcd − bce − bde − bcde − 4 = bc(8 − d − e − de) − bde − 4 ≥ 0 133
From this it is clear that d + e + de < 8. d, e ≤ 3 guarantees this. From here we can quickly check that the only solution to (5.27) is a = b = c = d = e = 1. However, this leads to k = 0, which cannot happen. Broom is slightly more complicated. For Broom (5.28) gives f=
8abc − abcd − abce − abde − bcde − 4 b(a + d)(c + e)
For this to be positive we clearly need 8abc − abcd − abce − abde − bcde − 4 > 0. Notice that if d + e ≥ 8 then 8abc − abcd − abcd < 0. Therefore we must have d + e < 8. With this we can then check each such case with (5.27). Doing so shows that no solution exists. Finally with P2 × K3 we have a similar situation as Broom with (5.28). a=
8bcd − bcde − bcdf − bcef − bdef − 4 bcd + bce + bdf + bef + cde + cdf + cef + def − 8ef
The numerator is similar, but we have to be careful with the denominator. This time we can see that if e + f ≥ 8 and b + c + d ≥ 8 then a ≤ 0. Therefore at least one of e + f and b + c + d must be less than 8. This gives a large number of possibilities, but we can check each one with (5.27) and see that the only possible solutions are isomorphic to P2 × K3 (k, 1, 1, 1, 1, 1; z). Therefore, we may conclude that P2 × K3 (k, 1, 1, 1, 1, 1, 1; z) is DS for all k ≥ 1 and z ≥ 0 except for k = 1, z ≥ 2 and k = 3, z ≥ 4.
134
5.10
Coefficients
Similar to λ = −1, we can factor the characteristic polynomials arising from Q4 as xc1 −4 H(x), where H(x) is x4 − c2 x2 + 2c3 x + c4 . P41 (a, b, c, d; z) = a + b + c + d + z P42 (a, b, c, d; z) = ab + bc + cd P43 (a, b, c, d; z) = 0 P44 (a, b, c, d; z) = abcd 2P21 (a, b, c, d; z) = a + b + c + d + z 2P22 (a, b, c, d; z) = ab + cd 2P23 (a, b, c, d; z) = 0 2P24 (a, b, c, d; z) = abcd P51 (a, b, c, d, e; z) = a + b + c + d + e + z P52 (a, b, c, d, e; z) = ab + bc + cd + de P53 (a, b, c, d, e; z) = 0 P54 (a, b, c, d, e; z) = abcd + abde + bcde Kite1 (a, b, c, d; z) = a + b + c + d + z Kite2 (a, b, c, d; z) = ab + ac + ad + bc Kite3 (a, b, c, d; z) = abc Kite4 (a, b, c, d; z) = abcd Bull1 (a, b, c, d, e; z) = a + b + c + d + e + z Bull2 (a, b, c, d, e; z) = ab + ac + ad + bc + be Bull3 (a, b, c, d, e; z) = abc Bull4 (a, b, c, d, e; z) = abcd + abce + abde House1 (a, b, c, d, e; z) = a + b + c + d + e + z House2 (a, b, c, d, e; z) = ab + ac + ae + bc + cd + de House3 (a, b, c, d, e; z) = abc House4 (a, b, c, d, e; z) = abcd + abce + abde + bcde
135
Broom1 (a, b, c, d, e, f ; z) = a + b + c + d + e + f + z Broom2 (a, b, c, d, e, f ; z) = ab + ac + ae + bc + bf + cd + de Broom3 (a, b, c, d, e, f ; z) = abc Broom4 (a, b, c, d, e, f ; z) = abcd + abce + abcf + abde + abef + bcde + bcdf + bdef P2 × K31 (a, b, c, d, e, f ; z) = a + b + c + d + e + f + z P2 × K32 (a, b, c, d, e, f ; z) = ab + ae + af + bc + bd + cd + cf + de + ef P2 × K33 (a, b, c, d, e, f ; z) = aef + bcd P2 × K34 (a, b, c, d, e, f ; z) = abcd + abce + abdf + abef + acde + acdf + acef + adef + bcde + bcdf + bcef + bdef
136
5.11
Bounds on eigenvalues
As with the bounds for P4 , the bounds for Q4 are found by exploiting symmetries in the graphs and applying Theorem 2.1.4.
2P2 (a, b, c, d; z): √ 1,1 √ 1,1 Spec(2P2 (a, b, c, d; z)) = {0a+b+c+d+z−4 , ± ab , ± cd }.
P4 (a, b, c, d; z): Assume a ≥ d. Let M = max{b, c} and m = min{b, c}. Then √
1 ( 4dm 2 √ 1 ( 4dm 2
+ m2 + m) ≤
λ1
≤
+ m2 − m) ≤ λ2 ≤ √ 1 (M − 4aM + M 2 ) ≤ λn−1 ≤ 2 √ 1 4aM + M 2 ) ≤ λn ≤ (−M − 2
√
1 ( 4aM + M 2 + M ) 2 √ 1 ( 4aM + M 2 − M ) 2 √ 1 (m − 4dm + m2 ) 2 √ 1 (−m − 4dm + m2 ) 2
P5 (a, b, c, d, e; z): Assume a ≥ e. Let M = max{b, d} and m = min{b, d}. Then p m(e + 2c) √ em √ − aM p − M (a + 2c)
≤
λ1
≤
≤
λ2
≤
≤ λn−1 ≤
λn
p √
M (a + 2c)
aM √ ≤ − em p ≤ − m(e + 2c)
Kite(a, b, c, d; z): Assume b ≥ c. Then −b ≤ λn−1 ≤ −c, −b ≤ λn ≤ −c, or both.
137
Bull(a, b, c, d, e; z): Assume a ≥ b. Let M = max{d, e} and p 1 ( b(b + 8c + 4m) + b) 2 p 1 ( b(b + 4m) − b) 2 p 1 (− a(a + 4M ) − a) 2 p 1 (− a(a + 8c + 4M ) + a) 2
m = min{d, e}. Then p ≤ λ1 ≤ 21 ( a(a + 8c + 4M ) + a) p ≤ λ2 ≤ 21 ( a(a + 4M ) − a) p ≤ λn−1 ≤ 21 (− b(b + 4m) − b) p ≤ λn ≤ 21 (− b(b + 8c + 4m) + b)
House(a, b, c, d, e; z): Assume a ≥ c. Let M = max{d, e} and m = min{d, e}. Then −(a + M ) ≤ λn−1 ≤ −(c + m), −(a + M ) ≤ λn ≤ −(c + m), or both. Broom(a, b, c, d, e, f ; z): Let M = max{a, b, c, d, e, f } and m = min{a, b, c, d, e, f }. Then √ √ 1 1 (m + 17m) ≤ λ ≤ (M + 17M ) 1 2 2 m ≤ λ2 ≤ M √ √ 1 1 (M − (m − 17M ) ≤ λ ≤ 17m) n−1 2 2 −2M ≤
λn
≤ −2m
Moreover, if a ≥ c, M 0 = max{d, e}, and m0 = min{d, e}, then −(a+M 0 ) ≤ λn−1 ≤ −(c+m0 ), −(a + M 0 ) ≤ λn ≤ −(c + m0 ), or both. P2 × K3 (a, b, c, d, e, f ; z): Let M1 = max{a, d}, M2 = max{b, e}, M3 = max{c, f }, m1 = min{a, d}, m2 = min{b, e}, and m3 = min{c, f }. Then √ √ 1 ( 4m1 + m3 4m2 + m3 + m3 ) 2 √ 1 √ ( m3 4m1 + 4m2 + m3 − m3 ) 2 √ √ 1 (− 4M1 + M3 4M2 + M3 + M3 ) 2 √ √ 1 (− M3 4M1 + 4M2 + M3 − M3 ) 2
≤
λ1
≤
≤
λ2
≤
≤ λn−1 ≤ ≤
λn
≤
√
√
1 ( 4M1 + M3 4M2 + M3 + M3 ) 2 √ √ 1 ( M3 4M1 + 4M2 + M3 − M3 ) 2 √ √ 1 (− 4m1 + m3 4m2 + m3 + m3 ) 2
√ 1 (− m3 2
√ 4m1 + 4m2 + m3 − m3 )
Moreover, if M10 = max{a, b}, M20 = max{d, e}, M30 = max{c, f }, m01 = min{a, b}, m02 = min{d, e}, and m03 = min{c, f }, then m01 + m02 + m03 ≤ λ1 ≤ M10 + M20 + M30 . 138
Chapter 6 Conclusion As we have seen in this thesis it is possible to study graphs with eigenvalues of −1 or 0 of high multiplicity in a simplified manner by first examining a representive class of primitive graphs. We were not able to fully characterize all primitive graphs, but we were able to give an efficient algorithm for calculating the primitive graphs. One area of further work would be to study more closely the structures of the sets Pk and Qk . k
In Chapter 2 it was shown that primitive graphs were bounded in size by O(2 2 ), where k is the multiplicity of −1 or 0. There is a construction that gives graphs that approach this bound. Given a graph G ∈ Pk with n = |V (G)|, we first form the strong product G K2 . Now, to this strong product we add two new vertices. One vertex is attached to all vertices in one copy of G, and the other vertex is attached to all vertices in the other copy of G. Call this new graph Susp2 (G). Susp2 (G) is primitive with 2n + 2 vertices, and it has −1 in its spectrum with multiplicity either n + k − 1 or n + k. If we have the latter, then Susp2 (G) ∈ Pk+2 . By taking the largest graph in Pk , with n vertices, we get that the largest graph in Pk+2 is of size at least 2n + 2. The curious fact is that in all cases thus far this graph is actually the largest in Pk+2 . This leads to the following conjecture. Conjecture 6.0.1. The largest graph in P2k has 2k+1 − 2 vertices. The largest graph in P2k+1 has 2k+1 + 2k−1 − 2 vertices. The Susp2 (G) construction shows that these are lower bounds for P2k and P2k+1 , but it is not guaranteed that they are also upper bounds. A similar construction also works for Qk . 139
Another avenue for further research is in finding more efficient methods of determining whether two graph are isomorphic using information from their spectra. Throughout this thesis we only considered looking purely at when two spectra coincided. By adding more information that is not costly to compute, we may be able to narrow down the search on isomorphic graphs. One natural approach would be to use extra matrices other than only the adjacency matrix. Another approach would be to use the structure of the underlying primitive graph of a blowup or split graph. For example, suppose G = P4 (a, b, c, d). We can quickly compute {a, b, c, d} from AG . Once we know {a, b, c, d}, we can reduce AG to AP4 . Thus, we know the sizes of the blowup partitions as well as AP4 . In this case knowing AP4 does give us P4 since P4 is DS, but in general reducing a graph to its primitive graph only gives us information about the partition of cospectral graphs containing the primitive graph. For example, we would not be able to distinguish K1,4 (a, b, c, d, e) and C4 (a, b, c, d) + K1 (e) from {a, b, c, d, e} and their primitive graphs since K1,4 and C4 + K1 are cospectral. While this approach is still not perfect, it does lead to the notion of a graph being reducibly DS. Definition 6.0.2. Let H be a graph with k = |V (H)|, and let σ be a permutation of [k]. Let H be the set of graphs cospectral with H. If G = H(n1 , . . . , nk ) then define σG = H(nσ(1) , . . . , nσ(k) ). Finally, we say that G is reducibly DS if for every σ such that G ∼ H 0 (nσ(1) , . . . , nσ(k) ) for some H 0 ∈ H, then we have G ∼ = H 0 (nσ(1) , . . . , nσ(k) ). While this definition looks complicated, in practice it greatly reduces the amount of computation needed. For example, since each graph in P4 is DS, we can simply check when G ∼ σG for some fixed blowup partition sizes. In this manner we can easily show that P3 (a, b, c)+K1 (d) is reducibly DS except for P3 (a, b, a)+K1 (b) ∼ P3 (b, a, b)+K1 (a) and P3 (nb(2b − 1), nb, nb(2b − 1)) + K1 (n(2b − 1)) ∼ P3 (nb(2b − 1), n(2b − 1), nb) + K1 (nb(2b − 1)) P3 (2nb(2b + 1), n(2b + 1), 2nb(2b + 1)) + K1 (4nb) ∼ P3 (2nb(2b + 1), 4nb, n(2b + 1)) + K1 (2nb(2b + 1)) Note that P3 (a, b, a) + K1 (b) ∼ P3 (b, a, b) + K1 (a) supersedes the result P3 (k, 1, k) + K1 (1) ∼ P3 (1, k, 1) + K1 (k) that we found in Chapter 4. We can also easily show P4 (a, b, c, d), C4 (a, b, c, d), and K1,3 (a, b, c, d) are reducibly DS for all a, b, c, d ≥ 1. 140
Bibliography [1] Bari, Ruth A., and Frank Harary, eds. Graphs and Combinatorics; Proceedings. Berlin: Springer-Verlag, 1974. [2] Beineke, Lowell W., and Robin J. Wilson, eds. Topics in Algebraic Graph Theory. Cambridge, UK: Cambridge UP, 2004. [3] Bell, F. K., and P. Rowlinson. “On The Multiplicities Of Graph Eigenvalues.” Bulletin of the London Mathematical Society 35.3 (2003): 401-08. [4] Biggs, Norman. Algebraic Graph Theory. London: Cambridge UP, 1974. [5] Cvetkovi´c, Drago˘s M., Peter Rowlinson, and S. Simi´c. An Introduction to the Theory of Graph Spectra. Cambridge, U.K.: Cambridge UP, 2010. [6] Godsil, C. D., and Gordon Royle. Algebraic Graph Theory. New York: Springer, 2001. [7] Kotlov, Andrew, and L´aszl´o Lov´asz. “The Rank and Size of Graphs.” Journal of Graph Theory 23.2 (1996): 185-89. [8] Ma, Haicheng, and Haizhen Ren. “On the Spectral Characterization of the Union of Complete Multipartite Graph and Some Isolated Vertices.” Discrete Mathematics 310.24 (2010): 3648-652.
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