Independent dominating sets and hamiltonian cycles
Independent dominating sets and hamiltonian cycles Penny Haxell†
Ben Seamone‡
Jacques Verstra¨ete§
†‡§
Department of Combinatorics and Optimization Faculty of Mathematics University of Waterloo 200 University Avenue West Waterloo ON Canada N2L 3G1
Abstract A graph is uniquely hamiltonian if it contains exactly one hamiltonian cycle. In this note we prove that there are no r-regular uniquely hamiltonian graphs when r > 22. This improves upon earlier results of Thomassen. Key words: C-independent set, Lov´asz Local Lemma, Uniquely hamiltonian
1
Introduction
Throughout this paper, we deal only with finite graphs which have no multiple edges or loops. A graph is uniquely hamiltonian if it contains exactly one hamiltonian cycle. The motivation for studying uniquely hamiltonian graphs came from Tait’s approach to the four-colour problem. One of the earliest theorems on hamiltonian graphs is Smith’s Theorem: every three-regular hamiltonian graph contains an even number of hamiltonian cycles through any edge of the graph. The first published proof of this theorem was given by Tutte (6). Smith’s Theorem was later extended to all hamiltonian graphs in which all vertices have odd degree by Thomason (7). A famous conjecture of Sheehan’s (5) states that every four-regular hamiltonian graph has at least two hamiltonian cycles, and this conjecture remains open. In the direction of proving this conjecture, Thomassen (9) showed that there are no uniquely hamiltonian rregular graphs when r > 71. In this note, we extend Thomassen’s result as follows: Preprint submitted to Elsevier Science
20 March 2006
Theorem 1 There are no r-regular uniquely hamiltonian graphs when r > 22, and no r-regular uniquely hamiltonian graphs of girth at least seven when r > 14. Thomassen (9) showed that a graph G with a hamiltonian cycle C has at least two hamiltonian cycles if there is a set I of vertices of G such that no two vertices of I are adjacent on C and every vertex of G is adjacent in G − E(C) to a vertex of I. Such a set will be called a C-independent dominating set of G. To prove Theorem 1, we will show (Section 2) that every r-regular hamiltonian graph G has a C-independent dominating set, where C is a fixed hamiltonian cycle of G. On the other hand one may ask if a smaller value of r is possible in Theorem 1. We will prove that if C is a hamiltonian cycle in a suitable model of random four-regular graphs, then almost surely such a graph contains no C-independent dominating set (Section 3). However the following question is open: does there exist a five-regular graph with a hamiltonian cycle C that does not contain a C-independent dominating set? In another direction, Bondy (see page three of (9)) asked whether every hamiltonian graph of minimum degree at least four has at least two hamiltonian cycles. The only positive result for graphs of large minimum degree was proved in (2): there Bondy and Jackson proved that every n-vertex hamiltonian graph of minimum degree at least c log2 n, for large enough n and c ≈ 2.341, has at least two hamiltonian cycles. The ideas of Theorem 1 can be used to give a proof of this, with a better value of the constant c, namely c ≈ 1.7523 (see Section 2.5).
2
Proof of Theorem 1
Let G be a graph with hamiltonian cycle C. Using the symmetric version of the Lov´asz Local Lemma (4), we will establish the existence of a C-independent dominating set I ∗ ⊂ V (G) when r > 22, and when r > 15 if G has girth at least seven. Since Thomason (7) proved that every r-regular hamiltonian graph has at least two hamiltonian cycles when r is odd, this gives Theorem 1. The symmetric version of the local lemma may be stated as follows: Theorem 2 (Lov´asz Local Lemma) Let A1 , A2 , . . . , An be events in a probability space, and for each Ai let Ji ⊂ {1, 2, . . . , n} be the set of values of j such that Aj depends on Ai . Suppose there exists a real number 0 < γ < 1 such that P(Ai ) < γ(1 − γ)|Ji | . Then P(A1 ∩ A2 ∩ · · · ∩ An ) ≥ (1 − γ)n > 0. 2
2.1 Random independent sets
Let G be a (d + 2)-regular graph with a hamiltonian cycle C. First we describe how to choose the vertices of an independent set I ∗ . For convenience, we assume n is even, and C = (v1 , v2 , . . . , vn , v1 ) is the chosen hamiltonian cycle in G, and let vn+i := vi . The set I of vertices of C is defined recursively, beginning with I0 = V (C). For 1 ≤ k ≤ n/2, define Ik by removing v2k+1 from Ik−1 with probability p and removing both v2k and v2k+2 from Ik−1 with probability 1 − p. The triples {v2k , v2k+1 , v2k+2 } will be called active triples. The set I = In/2 is an independent set in C: in any active triple either the midpoint is not in I or both endpoints are not in I. Finally, we let I ∗ be a maximal independent set in C containing I. We will prove that if d ≥ 22, then with positive probability I ∗ is a dominating set, in the sense that every vertex of C has a neighbour in I ∗ in the graph G − E(C): let A∗h be the event that a vertex vh ∈ V (C) is not dominated by I ∗ and let Ah be the event that vh is not dominated by I. If three neighbours of vh are consecutive on C then vh is dominated by I ∗ , so we consider only those vertices with no three consecutive neighbours on C. If we can prove that with positive probability, all of these vertices are dominated by I, then I ∗ is a C-independent dominating set, and Theorem 1 is verified. By Theorem 2, this holds if there exist p, γ ∈ (0, 1) such that for all vh with no three consecutive neighbours on C, P[Ah ] < γ(1 − γ)δh
(1)
where δh = |Jh | is the number of events Aj which are mutually dependent with Ah . For convenience, we let A = Ah and vh = v.
2.2 Estimating P[A] Throughout this section, G is a (d+2)-regular graph and C = (v1 , v2 , . . . , vn , v1 ). For v ∈ V (C), let Γ be the neighbourhood of v in G−E(C) – note that |Γ| = d. A vertex vi is odd if i is odd, and vi is even otherwise. An even or odd neighbour of v is an even or odd vertex of Γ, respectively. Let Q denote the union of all sets of vertices {v2k−3 , v2k−2 , v2k , v2k+2 , v2k+3 } ⊂ Γ. Let S be the set of odd neighbours of v in Γ\Q, and let R be the set of even neighbours of v in Γ\Q which are adjacent on C to an odd neighbour of v.
The events AQ , AR , AS , and AT . Define the events AQ = [Q ∩ I = ∅], AR = [R ∩ I = ∅] and AS = [S ∩ I = ∅]. Let |Q| = 5q and |S| = s. Let ρ be the number of active triples containing a vertex in R and no vertex of S. 3
Then, from the construction of I, P[AQ ∩ AR ∩ AS ] = p2q (1 − p)2q (1 − p)ρ ps . Let T be the set of even neighbours of v in Γ\(R ∪ Q), and let AT be the event [T ∩ I = ∅]. Then, from the construction of I, P[A] = P[AQ ∩ AR ∩ AS ∩ AT ] = P[AQ ∩ AR ∩ AS ]P[AT | AQ ∩ AR ∩ AS ] = p2q (1 − p)2q (1 − p)ρ ps · P[AT | AR ].
(2)
The sets R, S and T are illustrated in the figure below, and Q = ∅ in this case. T T R S
T S
R
Figure 1
Determining P[AT | AR ]. This is the main part of the determination of P[A]. First observe that T admits a partition into maximal non-empty sets T1 , T2 , . . . , Tm of the form {v2k , v2k+2 , . . . , v2k+2j }. If Ai is the event [Ti ∩I = ∅], then m P[AT | AR ] =
Y
i=1
P[Ai | AR ]
since the events Ai are independent. We now determine P[Ai | AR ]. For convenience, suppose Si is the set of odd vertices adjacent on C to a vertex of Ti , but not adjacent on C to any vertex of R and let ti = |Si |. Let Ji ⊂ Si be the random set defined by: Ji = {j : v2j−1 ∈ Si \I}. If Ai occurs, then Ji does not contain two consecutive integers, for if j, j+1 ∈ Ji then v2j ∈ Ti ∩ I. Conversely, if Ji does not contain two consecutive vertices 4
(that is, vertices v2j−1 , v2j+1 ∈ Si for some j) and AR occurs, then Ai occurs. For any set³J ⊂ S´i of size k, the probability that Ji = J is exactly pk (1−p)ti −k . subsets of {1, 2, . . . , ti } of size k which do not contain two There are ti +1−k k consecutive integers – this is a standard combinatorial identity, which can be reduced to a composition problem (3). Therefore, using Lemma 5.1 in the Appendix:
P[Ai | AR ] =
X
2k 1 − p. This allows us to eliminate the variables ρ and q in the above expression: 5
P[A] ≤ p
d−t+m
(1 − p)
t
m Y
φ(ti + 1).
(6)
i=1
In this expression, t = m i=1 ti , and we will assume tm ≥ tm−1 ≥ · · · ≥ t1 . In addition, we observe that 0 ≤ t − m ≤ d. P
2.3 Applying the local lemma To apply the local lemma, we maximize (6) as a function of m and the ti , P subject to the constraints 0 ≤ t − m ≤ d and m i=1 ti = t. It is convenient to replace φ with the function ϕ defined by: β+2 ϕ(x) = 2β + 2
Ã
β+2 2
!x
+
β ¯¯ β + 2 ¯¯x . 1 − 2β + 2 ¯ 2 ¯ ¯
¯
Note that φ(x) ≤ ϕ(x) for all x > 0, and note that concerning ϕ is the following:
β+2 2
(7)
> 1. A useful fact
Fact 1. For any positive integers u and v where v ≥ u, ϕ(u)ϕ(v) < ϕ(u − 1)ϕ(v + 1). Proof. Write ϕ(x) = abx + (1 − a)(b − 1)x where a = since v > u − 1 and 0 < a < 1 and b > 1,
β+2 2β+2
(8)
and b =
β+2 . 2
Then,
ϕ(u − 1)ϕ(v + 1) − ϕ(u)ϕ(v) = (abu−1 + (1 − a)(b − 1)u−1 )(abv+1 + (1 − a)(b − 1)v+1 ) − (abu + (1 − a)(b − 1)u )(abv + (1 − a)(b − 1)v ) = a(1 − a)bv (b − 1)u−1 − a(1 − a)bu−1 (b − 1)v = a(1 − a)[bv (b − 1)u−1 − bu−1 (b − 1)v ] = a(1 − a)bu−1 (b − 1)u−1 [bv−u+1 − (b − 1)v−u+1 ] > 0. 2
Fact 2. If tm ≥ tm−1 ≥ · · · ≥ t1 , and p2 > 1 − p, then with x = t − m, P[A] < pd−x (1 − p)x+1 ϕ(x + 2).
(9) 6
Proof. By (6) and (8), P[A] ≤ pd−t+m (1 − p)t · m i=1 φ(ti + 1) d−t+m t Qm ≤p (1 − p) · i=1 ϕ(ti + 1) P < pd−t+m (1 − p)t · ϕ(2)m−1 ϕ( m−1 i=1 (ti − 1) + tm + 1) d−t+m t m−1 =p (1 − p) · ϕ(2) ϕ(t − m + 2). Q
Finally, use ϕ(2) = φ(2) = 1/(1 − p) to complete the proof. 2 Fact 3. Let p =
33 50
and d ≥ 22. Then P[A] ≤ (1 − p)d+1 ϕ(d + 2).
(10)
Proof. Recall that 0 ≤ t − m ≤ d, and therefore x ∈ [0, d]. The function f (x) = pd−x (1 − p)x+1 ϕ(x + 2) attains a maximum on [0, d] at x = d when 33 and d ≥ 22 – in fact it has a unique local minimum in the range [0, d] p = 50 when d ≥ 22. Therefore, by Fact 2, P[A] ≤ f (d), which completes the proof of Fact 3. 2 2.4 Proof of Theorem 1 Part 1. Regular graphs. By the result of Thomason, every r-regular hamiltonian graph where r is odd has at least two hamiltonian cycles. Now let H be a hamiltonian r-regular graph, where r ≥ 24 is even. If C is a hamiltonian cycle in H, then H − E(C) is (r − 2)-regular. By Petersen’s Theorem, H − E(C) has a 2-factor, so we can remove this 2-factor from H to obtain an (r − 2)-regular graph with hamiltonian cycle C. Continuing in this way, we eventually reach a 24-regular graph, G, with hamiltonian cycle C. Let d = 22. 33 We apply Theorem 2 to G, and show that for p = 50 , and some positive γ < 1, δ P[A] < γ(1−γ) , where δ is the number of events Ah mutually dependent with A. Then, with positive probability, the set I is a C-independent dominating set of G. By definition of I, we have δ ≤ 5d2 − d. By Fact 3, (1) is satisfied if (1 − p)d+1 ϕ(d + 2) < γ(1 − γ)5d for some positive real γ < 1. We take γ = γ(1 − γ)5d
2 −d
1 , 25000
p=
2 −d
33 . 50
(11) Since d = 22, we find
≈ 5 · 10−6 + (1 − p)d+1 ϕ(d + 2). 7
We conclude that every r-regular graph hamiltonian, for r > 22, has at least two hamiltonian cycles. This proves the first statement of Theorem 1. Part 2. Regular graphs of girth at least seven. Let G be an r-regular graph of girth at least seven, where r = d + 2, and let C be a hamiltonian cycle in G. By Petersen’s Theorem, as in Part 1, it suffices to consider the case d = 14. For a vertex vh ∈ V (C), let A∗h denote the event that vh is not dominated by I ∗ (see §2.1), in the sense that neither vh nor any neighbour of vh is in I ∗ . We choose p = 0.657. Then P[w 6∈ I ∗ ] = 2p2 − p3 if w is odd and P[w 6∈ I ∗ ] = 1 − p2 if w is even. The definition of I ∗ gives: P[A∗h ] = (2p2 − p3 )s (1 − p2 )d−s+1
(12)
where s is the number of odd neighbours of vh in G − E(C) (including vh itself if h is odd). Let δh be the number of events A∗j which are mutually dependent with A∗h . Since G has girth at least seven, the events x ∈ I ∗ and y ∈ I ∗ are independent whenever the distance on C from x to y is greater than three. It is then straightforward to check that δh ≤ 5d2 + 8d + 4 + 2(d + 1)s. By differentiating the logarithm of (12), it is easily shown that the ratio of P[A∗h ] to γ(1 − γ)δh is maximized when s = 0 or s = d + 1. In those cases, we 1 respectively have δh = 5d2 + 8d + 4 and δh = 7d2 + 12d + 6. If γ = 20000 and ∗ δh d = 14, then P[Ah ] < γ(1 − γ) , and so Theorem 2 shows that with positive probability, I ∗ is a C-independent dominating set in G. In Figure 2, the two plots for s = 0 and s = d + 1 of P[A∗h ] − γ(1 − γ)δh are shown with d = 14, 1 γ = 20000 , as functions of p. p 0.655 0
0.656
0.657
0.658
-0.00001
-0.00002
-0.00003
-0.00004
-0.00005
Figure 2
8
0.659
0.66
2.5 Graphs of large minimum degree In this section, we prove that an n-vertex hamiltonian graph with minimum degree at least c log2 n contains at least two hamiltonian cycles, if n is large enough and c ≈ 1.7523. Theorem 3 Let G be an n-vertex hamiltonian graph of minimum degree at least c log2 n, where c ≈ 1.7523 and n is large enough. Then G contains at least two hamiltonian cycles. Proof. In fact, we prove a more general statement: if G is a graph and C = (v1 , v2 , . . . , vn , v1 ) is a hamiltonian cycle in G such that vh has degree dh + 2 and there exists p such that p2 > 1 − p and 0 < p < 1 and n X
h=1
(1 − p)dh +1 ϕ(dh + 2) < 1,
(13)
where ϕ is defined by (7), then there is a C-independent dominating set in G. . By Let I be the random independent set in C chosen as in §2.1, with p = 33 50 Fact 3, for any vertex vh ∈ V (C) of degree dh + 2, one has P[Ah ] ≤ (1 − p)dh +1 ϕ(dh + 2). The expected number of vertices which are not dominated by I is therefore at most n X
h=1
(1 − p)dh +1 ϕ(dh + 2)
where ϕ is defined by (7). This sum is less than one whenever (1 − p)dh +1 ϕ(dh + 2)
c log2 n where c = −1/ log2 ((1 − p)(β + 2)/2) = 1.7522..., then the sum is less than 1 for large enough n. So if we take dh = 1.7523 log2 n, then the expected number of vertices which are not dominated by I is less than one, so some independent set I is a dominating set in G − E(C). 9
3
Random regular graphs
Let C denote a cycle of length 4n, say (v1 , v2 , . . . , v4n , v1 ). Define G[C] to be the model of random four-regular multigraphs G = C∪Q1 ∪Q2 ∪· · ·∪Qn , where {Q1 , Q2 , . . . , Qn } is a uniformly chosen set of vertex-disjoint quadrilaterals on V (C). The number of multigraphs of this form is exactly 3n
³ ´³ 4n 4
4n−4 4
n!
´
···
³ ´ 4 4
=
(4n)! 8n n!
(14)
and each is an equally likely outcome in G[C]. If a graph G ∈ G[C] contains a C-independent dominating set, I, then each Qi contains two vertices of I, so |I| = 2n, and there are only two possible sets I ⊂ V (C). Fixing one of these sets I, the number of ways of choosing {Q1 , Q2 , . . . , Qn } so that I ∩ V (Qi ) = 2 for 1 ≤ i ≤ n is exactly 3n
³ ´2 ³ 2n 2
´ 2n−2 2 2
n!
···
³ ´2 2 2
=
3n (2n)!2 . 4n n!
(15)
The expected number of independent dominating sets in G ∈ G[C] is therefore 2 · 6n
1
³ ´ = O(n 2 ( 3 )n ). 8 4n 2n
Therefore G almost surely has no C-independent dominating sets. Furthermore, the expected number of multiple edges in G is easily shown to be a constant c < 1, so many simple graphs G ∈ G[C] do not have a C-independent dominating set. It is not hard to construct explicit examples of four regular graphs without independent dominating sets: for example consider the square of a cycle C of length n ≥ 5 where n 6≡ 0 modulo three. If I is a C-independent dominating set, then i ∈ I implies both vertices at distance three from i on C must also be in I, otherwise one of the vertices adjacent to i is not dominated in G − E(C). It would be interesting to see whether every 4-regular graph of sufficiently large girth has a C-independent dominating set.
4
Concluding Remarks
With substantially more computations, the first statement of Theorem 1 can be show to hold for r > 20 instead of r > 22. However, the local lemma no 10
longer works when applied to the independent set I ∗ in §2.1 if we allow r ≤ 20. We leave the following two open problems:
Problem 1. Does there exist a five-regular hamiltonian graph G with a hamiltonian cycle C containing no C-independent dominating set? Problem 2. Does every four-regular graph hamiltonian graph with a hamiltonian cycle C of sufficiently large girth have a C-independent dominating set? It would also be interesting to tie the existence of C-independent dominating sets to enumeration of hamiltonian cycles. We ask the following question:
Problem 3. If d is large enough, does every d-regular hamiltonian graph on n vertices contain cn hamiltonian cycles for some c > 1? The graph obtained by taking a cycle labelled with the integers modulo 2n consisting of the edges {j, 2n − j − 1} for all j : 2 ≤ j ≤ n − 3, and the edges {0, 2n − 2}, {1, 2n − 1}, {n − 2, n}, {n − 1, n + 1} has exactly four hamiltonian cycles, so d ≥ 4 is required in Problem 3. 5
Appendix
Lemma 5.1 For α > 0 and β =
X
2k≤n
Ã
!
β+2 n−k k α = k 2β + 2
Ã
√ 1 + 4α − 1, we have β+2 2
!n
Ã
β β+2 + 1− 2β + 2 2
!n
.
(16)
Proof. Let φ(n) denote the expression on the right. The generating function for the expression on the left is
X
n
Pn (α)x =
n≥0
X X
n≥0 n≥k≥0
=
XX
j≥0 k≥0
=
X
Ã
!
n−k (αx)k xn−k k
à !
j (αx)k xj k
(1 + αx)j xj
j≥0
11
=
1 1 − x − αx2 Ã
β+2 (β + 2)x = 1− 2β + 2 2 =
X
!−1
Ã
"
# !−1
β β+2 + 1− 1− x 2β + 2 2
φ(n)xn .
n≥0
6
Acknowledgements
We would like to thank Ian Goulden and Bruce Richmond for indicating Lemma 5.1.
References [1] Bollob´as, B. Random graphs. Second edition. Cambridge Studies in Advanced Mathematics, 73. Cambridge University Press, Cambridge, 2001. [2] Bondy, J. A.; Jackson, B. Vertices of small degree in uniquely Hamiltonian graphs. J. Combin. Theory Ser. B 74 (1998), no. 2, 265–275. [3] Goulden, I. P; Jackson, D. M. Combinatorial Enumeration. WileyInterscience Series in Discrete Mathematics. John Wiley & Sons, Inc., New York, 1983. [4] Erd˝os, P; L. Lov´asz, L. Problems and results on 3-chromatic hypergraphs and some related questions. In A. Hajnal et al. (eds.), Infinite and Finite Sets, North-Holland, Amsterdam (1975) pp. 609–628. [5] J. Sheehan. The multiplicity of Hamiltonian circuits in a graph, in: M. Fiedler (Ed.), Recent Advances in Graph Theory, Academia, Prague, 1975, pp. 447–480. [6] Tutte, W. T. On Hamiltonian circuits. J. London Math. Soc. 21, (1946). 98–101. [7] Thomason, A. G. Hamiltonian cycles and uniquely edge colourable graphs. Advances in graph theory (Cambridge Combinatorial Conf., Trinity College, Cambridge, 1977). Ann. Discrete Math. 3 (1978), Exp. No. 13, 3 pp. [8] Thomassen, C. On the number of Hamiltonian cycles in bipartite graphs. Combin. Probab. Comput. 5 (1996), no. 4, 437–442. [9] Thomassen, C. Independent dominating sets and a second Hamiltonian cycle in regular graphs. J. Combin. Theory Ser. B 72 (1998), no. 1, 104– 109.
12
Ben Seamone‡
Jacques Verstra¨ete§
†‡§
Department of Combinatorics and Optimization Faculty of Mathematics University of Waterloo 200 University Avenue West Waterloo ON Canada N2L 3G1
Abstract A graph is uniquely hamiltonian if it contains exactly one hamiltonian cycle. In this note we prove that there are no r-regular uniquely hamiltonian graphs when r > 22. This improves upon earlier results of Thomassen. Key words: C-independent set, Lov´asz Local Lemma, Uniquely hamiltonian
1
Introduction
Throughout this paper, we deal only with finite graphs which have no multiple edges or loops. A graph is uniquely hamiltonian if it contains exactly one hamiltonian cycle. The motivation for studying uniquely hamiltonian graphs came from Tait’s approach to the four-colour problem. One of the earliest theorems on hamiltonian graphs is Smith’s Theorem: every three-regular hamiltonian graph contains an even number of hamiltonian cycles through any edge of the graph. The first published proof of this theorem was given by Tutte (6). Smith’s Theorem was later extended to all hamiltonian graphs in which all vertices have odd degree by Thomason (7). A famous conjecture of Sheehan’s (5) states that every four-regular hamiltonian graph has at least two hamiltonian cycles, and this conjecture remains open. In the direction of proving this conjecture, Thomassen (9) showed that there are no uniquely hamiltonian rregular graphs when r > 71. In this note, we extend Thomassen’s result as follows: Preprint submitted to Elsevier Science
20 March 2006
Theorem 1 There are no r-regular uniquely hamiltonian graphs when r > 22, and no r-regular uniquely hamiltonian graphs of girth at least seven when r > 14. Thomassen (9) showed that a graph G with a hamiltonian cycle C has at least two hamiltonian cycles if there is a set I of vertices of G such that no two vertices of I are adjacent on C and every vertex of G is adjacent in G − E(C) to a vertex of I. Such a set will be called a C-independent dominating set of G. To prove Theorem 1, we will show (Section 2) that every r-regular hamiltonian graph G has a C-independent dominating set, where C is a fixed hamiltonian cycle of G. On the other hand one may ask if a smaller value of r is possible in Theorem 1. We will prove that if C is a hamiltonian cycle in a suitable model of random four-regular graphs, then almost surely such a graph contains no C-independent dominating set (Section 3). However the following question is open: does there exist a five-regular graph with a hamiltonian cycle C that does not contain a C-independent dominating set? In another direction, Bondy (see page three of (9)) asked whether every hamiltonian graph of minimum degree at least four has at least two hamiltonian cycles. The only positive result for graphs of large minimum degree was proved in (2): there Bondy and Jackson proved that every n-vertex hamiltonian graph of minimum degree at least c log2 n, for large enough n and c ≈ 2.341, has at least two hamiltonian cycles. The ideas of Theorem 1 can be used to give a proof of this, with a better value of the constant c, namely c ≈ 1.7523 (see Section 2.5).
2
Proof of Theorem 1
Let G be a graph with hamiltonian cycle C. Using the symmetric version of the Lov´asz Local Lemma (4), we will establish the existence of a C-independent dominating set I ∗ ⊂ V (G) when r > 22, and when r > 15 if G has girth at least seven. Since Thomason (7) proved that every r-regular hamiltonian graph has at least two hamiltonian cycles when r is odd, this gives Theorem 1. The symmetric version of the local lemma may be stated as follows: Theorem 2 (Lov´asz Local Lemma) Let A1 , A2 , . . . , An be events in a probability space, and for each Ai let Ji ⊂ {1, 2, . . . , n} be the set of values of j such that Aj depends on Ai . Suppose there exists a real number 0 < γ < 1 such that P(Ai ) < γ(1 − γ)|Ji | . Then P(A1 ∩ A2 ∩ · · · ∩ An ) ≥ (1 − γ)n > 0. 2
2.1 Random independent sets
Let G be a (d + 2)-regular graph with a hamiltonian cycle C. First we describe how to choose the vertices of an independent set I ∗ . For convenience, we assume n is even, and C = (v1 , v2 , . . . , vn , v1 ) is the chosen hamiltonian cycle in G, and let vn+i := vi . The set I of vertices of C is defined recursively, beginning with I0 = V (C). For 1 ≤ k ≤ n/2, define Ik by removing v2k+1 from Ik−1 with probability p and removing both v2k and v2k+2 from Ik−1 with probability 1 − p. The triples {v2k , v2k+1 , v2k+2 } will be called active triples. The set I = In/2 is an independent set in C: in any active triple either the midpoint is not in I or both endpoints are not in I. Finally, we let I ∗ be a maximal independent set in C containing I. We will prove that if d ≥ 22, then with positive probability I ∗ is a dominating set, in the sense that every vertex of C has a neighbour in I ∗ in the graph G − E(C): let A∗h be the event that a vertex vh ∈ V (C) is not dominated by I ∗ and let Ah be the event that vh is not dominated by I. If three neighbours of vh are consecutive on C then vh is dominated by I ∗ , so we consider only those vertices with no three consecutive neighbours on C. If we can prove that with positive probability, all of these vertices are dominated by I, then I ∗ is a C-independent dominating set, and Theorem 1 is verified. By Theorem 2, this holds if there exist p, γ ∈ (0, 1) such that for all vh with no three consecutive neighbours on C, P[Ah ] < γ(1 − γ)δh
(1)
where δh = |Jh | is the number of events Aj which are mutually dependent with Ah . For convenience, we let A = Ah and vh = v.
2.2 Estimating P[A] Throughout this section, G is a (d+2)-regular graph and C = (v1 , v2 , . . . , vn , v1 ). For v ∈ V (C), let Γ be the neighbourhood of v in G−E(C) – note that |Γ| = d. A vertex vi is odd if i is odd, and vi is even otherwise. An even or odd neighbour of v is an even or odd vertex of Γ, respectively. Let Q denote the union of all sets of vertices {v2k−3 , v2k−2 , v2k , v2k+2 , v2k+3 } ⊂ Γ. Let S be the set of odd neighbours of v in Γ\Q, and let R be the set of even neighbours of v in Γ\Q which are adjacent on C to an odd neighbour of v.
The events AQ , AR , AS , and AT . Define the events AQ = [Q ∩ I = ∅], AR = [R ∩ I = ∅] and AS = [S ∩ I = ∅]. Let |Q| = 5q and |S| = s. Let ρ be the number of active triples containing a vertex in R and no vertex of S. 3
Then, from the construction of I, P[AQ ∩ AR ∩ AS ] = p2q (1 − p)2q (1 − p)ρ ps . Let T be the set of even neighbours of v in Γ\(R ∪ Q), and let AT be the event [T ∩ I = ∅]. Then, from the construction of I, P[A] = P[AQ ∩ AR ∩ AS ∩ AT ] = P[AQ ∩ AR ∩ AS ]P[AT | AQ ∩ AR ∩ AS ] = p2q (1 − p)2q (1 − p)ρ ps · P[AT | AR ].
(2)
The sets R, S and T are illustrated in the figure below, and Q = ∅ in this case. T T R S
T S
R
Figure 1
Determining P[AT | AR ]. This is the main part of the determination of P[A]. First observe that T admits a partition into maximal non-empty sets T1 , T2 , . . . , Tm of the form {v2k , v2k+2 , . . . , v2k+2j }. If Ai is the event [Ti ∩I = ∅], then m P[AT | AR ] =
Y
i=1
P[Ai | AR ]
since the events Ai are independent. We now determine P[Ai | AR ]. For convenience, suppose Si is the set of odd vertices adjacent on C to a vertex of Ti , but not adjacent on C to any vertex of R and let ti = |Si |. Let Ji ⊂ Si be the random set defined by: Ji = {j : v2j−1 ∈ Si \I}. If Ai occurs, then Ji does not contain two consecutive integers, for if j, j+1 ∈ Ji then v2j ∈ Ti ∩ I. Conversely, if Ji does not contain two consecutive vertices 4
(that is, vertices v2j−1 , v2j+1 ∈ Si for some j) and AR occurs, then Ai occurs. For any set³J ⊂ S´i of size k, the probability that Ji = J is exactly pk (1−p)ti −k . subsets of {1, 2, . . . , ti } of size k which do not contain two There are ti +1−k k consecutive integers – this is a standard combinatorial identity, which can be reduced to a composition problem (3). Therefore, using Lemma 5.1 in the Appendix:
P[Ai | AR ] =
X
2k 1 − p. This allows us to eliminate the variables ρ and q in the above expression: 5
P[A] ≤ p
d−t+m
(1 − p)
t
m Y
φ(ti + 1).
(6)
i=1
In this expression, t = m i=1 ti , and we will assume tm ≥ tm−1 ≥ · · · ≥ t1 . In addition, we observe that 0 ≤ t − m ≤ d. P
2.3 Applying the local lemma To apply the local lemma, we maximize (6) as a function of m and the ti , P subject to the constraints 0 ≤ t − m ≤ d and m i=1 ti = t. It is convenient to replace φ with the function ϕ defined by: β+2 ϕ(x) = 2β + 2
Ã
β+2 2
!x
+
β ¯¯ β + 2 ¯¯x . 1 − 2β + 2 ¯ 2 ¯ ¯
¯
Note that φ(x) ≤ ϕ(x) for all x > 0, and note that concerning ϕ is the following:
β+2 2
(7)
> 1. A useful fact
Fact 1. For any positive integers u and v where v ≥ u, ϕ(u)ϕ(v) < ϕ(u − 1)ϕ(v + 1). Proof. Write ϕ(x) = abx + (1 − a)(b − 1)x where a = since v > u − 1 and 0 < a < 1 and b > 1,
β+2 2β+2
(8)
and b =
β+2 . 2
Then,
ϕ(u − 1)ϕ(v + 1) − ϕ(u)ϕ(v) = (abu−1 + (1 − a)(b − 1)u−1 )(abv+1 + (1 − a)(b − 1)v+1 ) − (abu + (1 − a)(b − 1)u )(abv + (1 − a)(b − 1)v ) = a(1 − a)bv (b − 1)u−1 − a(1 − a)bu−1 (b − 1)v = a(1 − a)[bv (b − 1)u−1 − bu−1 (b − 1)v ] = a(1 − a)bu−1 (b − 1)u−1 [bv−u+1 − (b − 1)v−u+1 ] > 0. 2
Fact 2. If tm ≥ tm−1 ≥ · · · ≥ t1 , and p2 > 1 − p, then with x = t − m, P[A] < pd−x (1 − p)x+1 ϕ(x + 2).
(9) 6
Proof. By (6) and (8), P[A] ≤ pd−t+m (1 − p)t · m i=1 φ(ti + 1) d−t+m t Qm ≤p (1 − p) · i=1 ϕ(ti + 1) P < pd−t+m (1 − p)t · ϕ(2)m−1 ϕ( m−1 i=1 (ti − 1) + tm + 1) d−t+m t m−1 =p (1 − p) · ϕ(2) ϕ(t − m + 2). Q
Finally, use ϕ(2) = φ(2) = 1/(1 − p) to complete the proof. 2 Fact 3. Let p =
33 50
and d ≥ 22. Then P[A] ≤ (1 − p)d+1 ϕ(d + 2).
(10)
Proof. Recall that 0 ≤ t − m ≤ d, and therefore x ∈ [0, d]. The function f (x) = pd−x (1 − p)x+1 ϕ(x + 2) attains a maximum on [0, d] at x = d when 33 and d ≥ 22 – in fact it has a unique local minimum in the range [0, d] p = 50 when d ≥ 22. Therefore, by Fact 2, P[A] ≤ f (d), which completes the proof of Fact 3. 2 2.4 Proof of Theorem 1 Part 1. Regular graphs. By the result of Thomason, every r-regular hamiltonian graph where r is odd has at least two hamiltonian cycles. Now let H be a hamiltonian r-regular graph, where r ≥ 24 is even. If C is a hamiltonian cycle in H, then H − E(C) is (r − 2)-regular. By Petersen’s Theorem, H − E(C) has a 2-factor, so we can remove this 2-factor from H to obtain an (r − 2)-regular graph with hamiltonian cycle C. Continuing in this way, we eventually reach a 24-regular graph, G, with hamiltonian cycle C. Let d = 22. 33 We apply Theorem 2 to G, and show that for p = 50 , and some positive γ < 1, δ P[A] < γ(1−γ) , where δ is the number of events Ah mutually dependent with A. Then, with positive probability, the set I is a C-independent dominating set of G. By definition of I, we have δ ≤ 5d2 − d. By Fact 3, (1) is satisfied if (1 − p)d+1 ϕ(d + 2) < γ(1 − γ)5d for some positive real γ < 1. We take γ = γ(1 − γ)5d
2 −d
1 , 25000
p=
2 −d
33 . 50
(11) Since d = 22, we find
≈ 5 · 10−6 + (1 − p)d+1 ϕ(d + 2). 7
We conclude that every r-regular graph hamiltonian, for r > 22, has at least two hamiltonian cycles. This proves the first statement of Theorem 1. Part 2. Regular graphs of girth at least seven. Let G be an r-regular graph of girth at least seven, where r = d + 2, and let C be a hamiltonian cycle in G. By Petersen’s Theorem, as in Part 1, it suffices to consider the case d = 14. For a vertex vh ∈ V (C), let A∗h denote the event that vh is not dominated by I ∗ (see §2.1), in the sense that neither vh nor any neighbour of vh is in I ∗ . We choose p = 0.657. Then P[w 6∈ I ∗ ] = 2p2 − p3 if w is odd and P[w 6∈ I ∗ ] = 1 − p2 if w is even. The definition of I ∗ gives: P[A∗h ] = (2p2 − p3 )s (1 − p2 )d−s+1
(12)
where s is the number of odd neighbours of vh in G − E(C) (including vh itself if h is odd). Let δh be the number of events A∗j which are mutually dependent with A∗h . Since G has girth at least seven, the events x ∈ I ∗ and y ∈ I ∗ are independent whenever the distance on C from x to y is greater than three. It is then straightforward to check that δh ≤ 5d2 + 8d + 4 + 2(d + 1)s. By differentiating the logarithm of (12), it is easily shown that the ratio of P[A∗h ] to γ(1 − γ)δh is maximized when s = 0 or s = d + 1. In those cases, we 1 respectively have δh = 5d2 + 8d + 4 and δh = 7d2 + 12d + 6. If γ = 20000 and ∗ δh d = 14, then P[Ah ] < γ(1 − γ) , and so Theorem 2 shows that with positive probability, I ∗ is a C-independent dominating set in G. In Figure 2, the two plots for s = 0 and s = d + 1 of P[A∗h ] − γ(1 − γ)δh are shown with d = 14, 1 γ = 20000 , as functions of p. p 0.655 0
0.656
0.657
0.658
-0.00001
-0.00002
-0.00003
-0.00004
-0.00005
Figure 2
8
0.659
0.66
2.5 Graphs of large minimum degree In this section, we prove that an n-vertex hamiltonian graph with minimum degree at least c log2 n contains at least two hamiltonian cycles, if n is large enough and c ≈ 1.7523. Theorem 3 Let G be an n-vertex hamiltonian graph of minimum degree at least c log2 n, where c ≈ 1.7523 and n is large enough. Then G contains at least two hamiltonian cycles. Proof. In fact, we prove a more general statement: if G is a graph and C = (v1 , v2 , . . . , vn , v1 ) is a hamiltonian cycle in G such that vh has degree dh + 2 and there exists p such that p2 > 1 − p and 0 < p < 1 and n X
h=1
(1 − p)dh +1 ϕ(dh + 2) < 1,
(13)
where ϕ is defined by (7), then there is a C-independent dominating set in G. . By Let I be the random independent set in C chosen as in §2.1, with p = 33 50 Fact 3, for any vertex vh ∈ V (C) of degree dh + 2, one has P[Ah ] ≤ (1 − p)dh +1 ϕ(dh + 2). The expected number of vertices which are not dominated by I is therefore at most n X
h=1
(1 − p)dh +1 ϕ(dh + 2)
where ϕ is defined by (7). This sum is less than one whenever (1 − p)dh +1 ϕ(dh + 2)
c log2 n where c = −1/ log2 ((1 − p)(β + 2)/2) = 1.7522..., then the sum is less than 1 for large enough n. So if we take dh = 1.7523 log2 n, then the expected number of vertices which are not dominated by I is less than one, so some independent set I is a dominating set in G − E(C). 9
3
Random regular graphs
Let C denote a cycle of length 4n, say (v1 , v2 , . . . , v4n , v1 ). Define G[C] to be the model of random four-regular multigraphs G = C∪Q1 ∪Q2 ∪· · ·∪Qn , where {Q1 , Q2 , . . . , Qn } is a uniformly chosen set of vertex-disjoint quadrilaterals on V (C). The number of multigraphs of this form is exactly 3n
³ ´³ 4n 4
4n−4 4
n!
´
···
³ ´ 4 4
=
(4n)! 8n n!
(14)
and each is an equally likely outcome in G[C]. If a graph G ∈ G[C] contains a C-independent dominating set, I, then each Qi contains two vertices of I, so |I| = 2n, and there are only two possible sets I ⊂ V (C). Fixing one of these sets I, the number of ways of choosing {Q1 , Q2 , . . . , Qn } so that I ∩ V (Qi ) = 2 for 1 ≤ i ≤ n is exactly 3n
³ ´2 ³ 2n 2
´ 2n−2 2 2
n!
···
³ ´2 2 2
=
3n (2n)!2 . 4n n!
(15)
The expected number of independent dominating sets in G ∈ G[C] is therefore 2 · 6n
1
³ ´ = O(n 2 ( 3 )n ). 8 4n 2n
Therefore G almost surely has no C-independent dominating sets. Furthermore, the expected number of multiple edges in G is easily shown to be a constant c < 1, so many simple graphs G ∈ G[C] do not have a C-independent dominating set. It is not hard to construct explicit examples of four regular graphs without independent dominating sets: for example consider the square of a cycle C of length n ≥ 5 where n 6≡ 0 modulo three. If I is a C-independent dominating set, then i ∈ I implies both vertices at distance three from i on C must also be in I, otherwise one of the vertices adjacent to i is not dominated in G − E(C). It would be interesting to see whether every 4-regular graph of sufficiently large girth has a C-independent dominating set.
4
Concluding Remarks
With substantially more computations, the first statement of Theorem 1 can be show to hold for r > 20 instead of r > 22. However, the local lemma no 10
longer works when applied to the independent set I ∗ in §2.1 if we allow r ≤ 20. We leave the following two open problems:
Problem 1. Does there exist a five-regular hamiltonian graph G with a hamiltonian cycle C containing no C-independent dominating set? Problem 2. Does every four-regular graph hamiltonian graph with a hamiltonian cycle C of sufficiently large girth have a C-independent dominating set? It would also be interesting to tie the existence of C-independent dominating sets to enumeration of hamiltonian cycles. We ask the following question:
Problem 3. If d is large enough, does every d-regular hamiltonian graph on n vertices contain cn hamiltonian cycles for some c > 1? The graph obtained by taking a cycle labelled with the integers modulo 2n consisting of the edges {j, 2n − j − 1} for all j : 2 ≤ j ≤ n − 3, and the edges {0, 2n − 2}, {1, 2n − 1}, {n − 2, n}, {n − 1, n + 1} has exactly four hamiltonian cycles, so d ≥ 4 is required in Problem 3. 5
Appendix
Lemma 5.1 For α > 0 and β =
X
2k≤n
Ã
!
β+2 n−k k α = k 2β + 2
Ã
√ 1 + 4α − 1, we have β+2 2
!n
Ã
β β+2 + 1− 2β + 2 2
!n
.
(16)
Proof. Let φ(n) denote the expression on the right. The generating function for the expression on the left is
X
n
Pn (α)x =
n≥0
X X
n≥0 n≥k≥0
=
XX
j≥0 k≥0
=
X
Ã
!
n−k (αx)k xn−k k
à !
j (αx)k xj k
(1 + αx)j xj
j≥0
11
=
1 1 − x − αx2 Ã
β+2 (β + 2)x = 1− 2β + 2 2 =
X
!−1
Ã
"
# !−1
β β+2 + 1− 1− x 2β + 2 2
φ(n)xn .
n≥0
6
Acknowledgements
We would like to thank Ian Goulden and Bruce Richmond for indicating Lemma 5.1.
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