Properties of Classes of Random Graphs. - Semantic Scholar
Properties of Classes of Random Graphs. Neal Brand1 and Steve Jackson2 Department of Mathematics University of North Texas Denton, TX 76203
Abstract
In [11] it is shown that the theory of almost all graphs is rst order complete. Furthermore, in [3] a collection of rst order axioms are given from which any rst order property or its negation can be deduced. Here we show that almost all Steinhaus graphs satisfy the axioms of almost all graphs and conclude that a rst order property is true for almost all graphs if and only if it is true for almost all Steinhaus graphs. We also show that certain classes of subgraphs of vertex transitive graphs are rst order complete. Finally, we give a new class of higher order axioms from which it follows that large subgraphs of speci ed type exist in almost all graphs.
1. Introduction.
This paper is motivated by [3],[8] and [11]. In [3] and [11] it is shown that any rst order property of graphs is either satis ed by almost all graphs or else almost all graphs do not satisfy the property. Furthermore, axioms are given, each of which almost all graphs satisfy, such that for any rst order property either the property or its negation can be deduced from a nite number of these axioms. In fact we have an axiom for every natural number k. Axiom k: For any set of 2k distinct vertices fv1; v2 ; v3 ; : : : ; vk ; w1; w2; w3 ; : : : ; wk g, there is a vertex v with vAvi and :vAwi for each 1 i k. Let us de ne a 0-1 matrix (ai;j )ni;j=1 as follows. Start with a 0-1 string (a1;j )nj=2. For 1 < i < j n de ne ai;j inductively by ai;j ai?1;j?1 + ai?1;j (mod 2). We complete (ai;j )1i<jn to an n n matrix by de ning ai;i = 0 and ai;j = aj;i for 1 j < i n. This gives an n n symmetric 0-1 matrix A = (ai;j )ni;j=1 with 0s on the diagonal. The matrix A is the adjacency matrix of some graph. Any graph de ned in this way is called a Steinhaus graph. The string (a1;j )nj=2 is called the generating string for the Steinhaus graph. For convenience we identify the vertex set of a Steinhaus graph with the rst n natural numbers, Vn = f1; 2; 3; : : : ; ng. It is obvious that there are exactly 2n?1 labeled Steinhaus graphs of order n. We use the term Steinhaus graph to mean a labeled Steinhaus graph and assume that the vertex set is Vn. We de ne a probability measure on Steinhaus graphs of order n by requiring that Pr(a1;j = 1) = pn;j where 0 pn;j 1 for each j . We then de ne qn;j = 1 ? pn;j and mn;j = min(pn;j ; qn;j ). Any function f (n) with the properties that for each suciently large n, 0 < f (n) < 1 and mn;j f (n) for each 2 j n is called a probability bound. 1 2
Supported by a University of North Texas Faculty Research Grant. Supported by a University of North Texas Faculty Research Grant and NSF grant number DMS 9007808. 1
Given a probability bound, we will say that almost all Steinhaus graphs have property P if the probability that a Steinhaus graph of order n has the property approaches one as n approaches in nity. Of course, the concept of \almost all" depends on the probability bound. The simplest case is the constant probability bound of 21 . This gives all Steinhaus graphs of order n the same probability. In [8], Brigham and Dutton conjectured that almost all Steinhaus graphs have diameter two. Diameter two is a special case of Axiom 2. The conjecture was proved in [5]. It is natural to ask if all the rst order axioms for almost all graphs are satis ed by Steinhaus graphs. In Section 2 we show that they are satis ed by Steinhaus graphs, which implies that almost all Steinhaus graphs have the same rst order theory as almost all graphs. In Section 3 we look at other classes of graphs whose rst order theory is complete. We consider the set of all spanning subgraphs of a xed vertex transitive graph and de ne a measure on that set. We show that with this measure the rst order theory of almost all spanning subgraphs is complete. Furthermore, we show that with a few technical conditions on the allowed subgraphs, the theory of almost all nite subgraphs is complete and the same as the theory of almost all subgraphs with all vertices. In Section 4 we consider higher order axioms. The rst order axiom scheme described above implies that small structures exist as subgraphs of almost all graphs. The axioms introduced in Section 4 imply the existence of speci ed subgraphs containing all the vertices of the graph for almost all graphs. As a special case, the theorems in Section 4 imply the existence of Hamiltonian cycles in random graphs. Although the probability bounds given in the theorems are not even close to optimal in the case of Hamiltonian cycles, the theorems apply to a large class of subgraphs. Throughout the paper we use log x to denote natural logarithm.
2. Steinhaus graphs.
A Steinhaus triangle is simply the adjacency matrix of a Steinhaus graph above the main diagonal. Steinhaus introduced Steinhaus triangles in [14] where he posed the problem of nding generating strings which give Steinhaus triangles with the same number of zeros as ones. This problem was solved by Harborth in [12], where he constructed such sequences in the cases where there are an even number of entries in the Steinhaus triangle. Since then others have investigated properties of Steinhaus graphs. For example, the maximum clique size is found in [7], regular Steinhaus graphs are constructed in [1], the girth is studied in [10], and conditions under which Steinhaus graphs are bipartite are given in [9]. In [7] it is shown that the maximum diameter of a nontrivial Steinhaus graph is roughly n2 . Furthermore, in [5,6] it is shown that almost all Steinhaus graphs have diameter two. In fact, a slight modi cation of the proof in [5,6] shows that Axiom 1 is satis ed for almost all Steinhaus graphs. It is therefore natural to ask which of the other axioms in the axiom scheme listed above are satis ed by almost all Steinhaus graphs. The main purpose of this section is to prove the following result. Theorem 2.1 Given k 2 N and > 0, almost all Steinhaus graphs with probability bound n? k k + satisfy Axiom k. Before we turn to the proof of Theorem 2.1, we give some consequences of this result. 1 4 (2 +1)
2
Corollary 2.2 Given k 2 N, almost all Steinhaus graphs with constant probability bound satisfy Axiom k. Proof. This is obvious since if k is xed and 0 < < 4k(21k+1) then limn!1 n? 0.
1 4 (2 +1)
k k
+ =
Corollary 2.3 Using a constant probability bound, for any given rst order property, almost all Steinhaus graphs have the property if and only if almost all graphs have the property. Proof. In [3] it is shown that for any rst order property of graphs there is a k such that either the property or its negation can be proved with the axioms of graph theory together with Axioms 1{k. Since the same axioms hold for Steinhaus graphs, the proof of a property for almost all graphs is the same as the proof for almost all Steinhaus graphs. Corollary 2.3 states that random graphs look like random Steinhaus graphs from the point of view of the rst order theory. Since Steinhaus graphs are easily generated, perhaps this fact could be exploited by checking algorithm performance on large random Steinhaus graphs rather than large random graphs. Note that a proof of any true rst order property of graphs requires only a nite number of the listed axioms. Given an upper bound for k, where Axiom k is required in the proof, together with Theorem 2.1 we can obtain a probability bound which implies almost all Steinhaus graphs have the property. As pointed out in [3], there is an algorithm which gives either a proof of a rst order statement or a proof of its negation. Following this algorithm gives an upper bound for k. Many higher order properties follow from rst order properties. For example, diameter two (a rst order property) implies connected (a second order property). Corollary 2.3 implies the following properties of almost all Steinhaus graphs: diameter two, k-connected for any xed k, contain any xed graph as a subgraph, not planar, and have no embedding in a surface of genus g for a xed g. Our proof of Theorem 2.1 is based on 5 lemmas. Lemma 2.4 For any xed 2 k n, let (ai;j ) be the adjacency matrix for a Steinhaus graph with generating string (a1;j ) given by a1;r = k;r (1 if r = k, 0 otherwise). Then for ? i ? 1 any 1 i < j n, ai;j j?k (mod 2): Proof by induction. We have inductively ai;j ai?1;j?1 + ai?1;j ? i?2 . This ? i?is2 easily ? i?veri ed 1 (mod 2). + j ?k?1 j ?k j ?k
?
Given a natural number k, denote by H (k) the number of integers j with pkj odd. For non-negative integers a < b de ne G(b; a) = fkj1 k 2b?1; k = 2al; H (k) 2b?ag. Lemma 2.5 Let a, b be non-negative integers with a + 1 < b. Then jG(b; a)j 2b?a?2: Proof. It is sucient to show the lemma for the special case where a = 0 since the general case follows by dividing by 2a. So we assume that a = 0: 3
?
By Lucas' Theorem [13], a binomial coecient mk is odd if and only if for each w, if there is a zero in the base two representation of k in the 2w place then there is a zero in the 2w place in the base two representation of m. Therefore, k 2 G(b; 0) if and only if the number of ones in the base two representation of k is no more than 2b . Obviously there are at least 21 2b?1 = 2b?2 such values for k: For any pair of vertices (v; w) in a Steinhaus graph, we de ne C (v; w) = jv ? wj + 1. Furthermore, for any subset T Vn and any v 2 Vn we let CT (v) = fC (v; w)jw 2 T g. By Lemma 2.4, given any Steinhaus graph one can change the adjacency of vertices v and w simply by changing the value of the generating string at position C (v; w). Furthermore, this does not change? the adjacency of any pair of vertices (u; z) with CS(u; z) > C (v; w). min f v;w g? 1 We let B(v; w) = fij maxfv;wg?i 1 (mod 2)g. Then we let BT (v) = w2T B(v; w): By Lemma 2.4, BT (v) is the set of all entries in the generating string of a Steinhaus graph which when changed will change the (v; w) entry in the adjacency matrix for some w 2 T . Let T Vn. We say a sequence v1 ; v2 ; : : : ; vr 2 Vn is T -independent if for each 1 i r, CT (vi ) \ BT (vj ) = ; for j < i and jCT (vi )j = jT j for each i: Lemma 2.6 Suppose v1 ; : : : ; vr is a T -independent sequence for some set T Vn: Then for any generating string for a Steinhaus graph with n vertices, by changing only the entries in the generating string indexed by CT (vi ) it is possible to attain any combination of adjacencies between vi and the vertices in T: Furthermore, making these changes does not change the adjacencies between vj and the vertices of T for any j < i. Proof. Let (a1;j ) be an arbitrary generating string for a Steinhaus graph which gives an adjacency matrix (ai;j ). Label the elements in T by w1; : : : ; wt where C (vi; w1) > > C (vi; wt ). It is then clear that by changing the value of a1;C(vi;wr ), the value avi ;wr changes. Furthermore, changing a1;C(vi ;wr ) does not change the value of avi ;wk for k < r, nor does it change the change the value of avj ;ws for any j < i and any s:
Lemma 2.7 Let T = fv1 ; : : : ; vk ; w1; : : : ; wk g Vn and suppose that mn is the proba-
bility bound used for Steinhaus graphs. If there is a T -independent sequence of length r then the probability that Axiom k is true for a Steinhaus graph using the set T is at least 1 ? (1 ? m2nk )r : Proof. Let z1 ; : : : ; zr be a T -independent sequence of length r. Note that the generating strings for Steinhaus graphs for which Axiom k fails for z1; : : : ; zj?1 using T , can be partitioned into subsets each of size 22k by putting two strings in the same subset if the strings agree in every entry except for positions (1; i) where i 2 CT (zj ): By Lemma 2.6, in each subset there is a sequence whose Steinhaus graph satis es Axiom k using T and zj . Therefore, the probability that a Steinhaus graph satis es Axiom k using zj and T given that it does not satisfy Axiom k using T and any zi with i < j is at least m2nk . It is then clear that a lower bound that Axiom k is satis ed using T is 1 ? (1 ? m2nk )r : The goal, based on Lemma 2.7, is to nd a long T -independent sequence. The construction of a long T -independent sequence is sucient (although perhaps not necessary) to prove Theorem 2.1. 4
Lemma 2.8 Given T Vn with jT j = 2k, there is a T -independent sequence of length at least n 32kk ? 1. Proof. Let v1; : : : ; v2k be the elements of T listed in order and write each as vi = n i . Let 0 = 0 and 2k+1 = 1 Pick 0 i 2k so that i+1 ? i 2k1+1 . Certainly there is at least 1 2(2 +1) 2
one such i since the i s break the unit interval into 2k + 1 intervals. Now, let a and b be such that vi < 2a and 2b < vi+1 with b ? a as large as possible. If b a +1 then n k < 8; so the lemma is obviously true. We therefore assume b > a + 1: Then 2b?a 41 n k . Let G = G(b; a): By Lemma 2.5 jGj 2b?a?2: List the elements in G in order and label them h1; h2; : : :. We start a T -independent set by setting x1 = hi where i is as small as possible with fhig T -independent. We then assume inductively that Sr = fx1; : : : ; xr g G is a T -independent set and attempt to add another element xr+1 2 G to S by always choosing xr+1 the smallest element in G so that Sr+1 = Sr [ fxr+1 g is T -independent. There are two required conditions for Sr+1 to be T -independent. First, jCT (xr+1 )j = jT j. This condition fails for at most jT j(jT j ? 1) = 2k(2k ? 1) vertices. The next condition is that CT (xr+1 ) \ BT (xi ) = ; for i r: So, suppose that w1; w2 2 T and C (w1; xr+1 ) 2 B(xi ; w2 ): Every element of T is either less than 2a or else it is larger than 2b. This gives four cases based on the values of w1 and w2: It is easily checked that C (xr+1 ; w1) 62 B(xi ; w2 ) for either w1 or w2 less than 2a: So suppose that w1; w2 > 2b: Note that C (w1; x) = w1 ? x + 1 is a strictly decreasing function of x for x < w1: Furb?a thermore, there are at most r (2 k )2 forbidden values for C (w1; vr+1 ): Therefore as long b?a b ? a ? 2 as 2 ? (2k)r(2k)2 ? 2k(2k ? 1) 1 there is at least one vertex vr+1b?2a?G so that Sr+1 = Sr [ fvr+1 g is T -independent. Solving the inequality gives r < 2 b?a ? 1 2 +1
1 2 +1
2
2
2
2k?b1?a (2k)2 2
1 2(2 +1) 2
? (2k) (21 b?a ) = N: Note that N > n 32kk 2
2
(2k)22
2
? 1; which completes the proof.
Proof of Theorem 2.1. Using Lemmas 2.8 and 2.7, we have that the probability that Axiom k fails is no more than
nn ? k ?
Pn = k
k
1
n 2(2k+1) 1 ? m2nk 32k2 ?1 :
Taking logarithms and substituting mn > n? k k + it is easily shown that Pn ! 0. Note that if we use a constant probability bound mn = m then the above estimate of the probability that Axiom k fails is roughly order can k + where c = 1 ? m2k and a = 321k : In the special case of m = 12 (which means each entry in the generating string n has probability 21 of being a 1), the probability bound can be improved to roughly c k by using di erent arguments which are outlined below. Unfortunately, these arguments fail for any m < 12 : It seems that for m near 21 it should be possible to improve the bound given above to make it closer to the better bound in the case where m = 12 . Theorem 2.9 The ratio of the number of labeled Steinhaus graphs of order n which do not satisfy Axiom k to the total number of labeled Steinhaus graphs of order n is no more 1 4 (2 +1)
1 2(2 +1)
2
8 2
5
2 k ?1 n? ? ? n ? k n than k
k
2
22k
k k?1 k
10 2 + 8 2
:
We do not prove Theorem 2.9, but simply give the essential di erence between the proof of Theorem 2.9 and Theorem 2.1. Instead of looking for a large T -independent set, we look for a large set fv1; : : : ; vr g with the property that the sets CT (vi ) are disjoint and jCT (vi )j = jT j. Obviously it is easier to nd large sets with this property than it is to nd large T -independent sets. In fact, it is not dicult to show there are sets of size at least n?10k +k?1 with the stated property. In either proof the values in CT (vi ) are allowed to 8k change in order to estimate a lower bound for the probability that Axiom k is satis ed by vi and T: In Lemma 2.7 it was necessary to assume that the set v1 ; : : : ; vr is T -independent. The di erence in the case m = 21 is that all the combinations of ones and zeros in CT (vi ) are equally likely. This gives independence of the events that Axiom k is satis ed by T and vi for each of the values of i in the set. 2
2
3. Completeness in Vertex Transitive Graphs.
In this section we show that the rst order theories of some families of graphs are complete in the \almost everywhere" sense, although di erent from the rst order theory of random graphs. To begin we say that M is a master graph if M is a graph with a countable number of vertices v1 ; v2; v3 ; : : :. We consider the collection ? of all spanning subgraphs G of M . Let E (M ) denote the edge set of M . We make ? a probability space by letting the probability of any edge in M be p. More precisely, we have a map : 2E(M ) ! ? given by (f ) = G where (vi ; vj ) is an edge in G if and only if f (vi ; vj ) = 1. The measure on ? is then just the image under of the Bernoulli measure on 2E(M ) with parameter p. Here we let M be a vertex transitive graph with a countably in nite vertex set and with every vertex having nite degree. Given vertices v1 ; v2 : : : ; vn 2 M and N 2 N, the N component CNG(v1 ; : : : ; vn ) of (v1 ; v2; : : : ; vn) in G is de ned to be the set of all vertices v 2 G such that (v; vi) N for some 1 i n, where is the usual edge distance in G. We will think of CNG(v1 ; : : : ; vn) as either a set of vertices or else the subgraph induced by the set of vertices. We take (w; z) = 1 if w and z are in di erent components. Next we let CNG(v1 ; : : : ; vn) = SN1 (v1 ; : : : ; vn) [ [ SNm(v1 ; : : : ; vn) where each SNi (v1 ; : : : ; vn) is a connected component of CnG(v1 ; : : : ; vn). Note that each SNi (v1; : : : ; vn) contains at least one of the vertices v1; : : : ; vn; and the diameter of each SNi (v1 ; : : : ; vn) is not more than 2nN for all i. Let M~ be an isomorphic copy of M . Suppose that T1 ; : : : ; Tm is a collection of nite subgraphs of M~ , each containing a xed vertex v~0. Furthermore, suppose for each 1 i m weShave a subset Bi of fx1; x2 ; : : : ; xng and a function 'i : Bi ! V (Ti ) such that (1) i Bi = fx1 ; x2; : : : ; xn g and the Bi are pairwise disjoint. (2) each Ti is connected, and for each vj 2 Ti, ('(Bi); vj ) N . Also, v~0 2 '(Bi). We call the collection T1 ; : : : ; Tm , together with the corresponding functions '1; : : : ; 'm, an (n; N )-con guration Cn;N . Occasionally we shall refer to the vertex 'j (xi ) as the vertex identi ed with xi , and the set of all the vertices identi ed with some xi as the identi ed vertices. 6
The purpose of de ning (n; N )-con gurations is to translate logical formulae into con gurations. Roughly speaking, for any formula (x1 ; : : : ; xn ) there is an N so that for almost all graphs and all choices of v1; : : : ; vn the truth of (v1 ; : : : ; vn) is determined by the induced subgraph on the vertex set CNG (v1 ; : : : ; vn ). A key point is that any possible nite subgraph of M should normally appear in nitely often in a random graph from ?. Since we assume that M has only nite degree, there are only nitely many Cn;N for each n and N . We say that Cn;N = T1 [ [ Tm and C0n;N = T10 [ [ Tm0 0 are isomorphic if m = m0 and for each i there is a graph isomorphism : Ti ! Ti0 such that 'i(xj ) = '0i(xj ) for each j 2 Bi. Given G 2 ? and vertices v1 ; : : : ; vn in G, we say that (v1 ; : : : ; vn ) satis es Cn;N = T1 [ [ Tm if CN (v1; : : : ; vn) = SN1 (v1 ; : : : ; vn) [ [ SNm(v1 ; : : : ; vn) and for all 1 i m there is a graph isomorphism i : SNi (v1 ; : : : ; vm ) ! Ti such that i (vj ) = 'i(xj ). We say that a graph G 2 ? is good if for any nite connected subgraph S of M containing v0, there are in nitely many v 2 V (G) such that CN (v) is isomorphic with S by an isomorphism sending v to v~0 . Clearly the set of good graphs in ? has probability measure 1. We consider nite sets of (n; N )-con gurations, Dn;N = fC1n;N ; : : : ; Ckn;N g for xed n and N . For any G 2 ? and formula (x1; : : : ; xn ) we say is equivalent to Dn;N if rst for every (v1 ; : : : ; vn ), vertices in G, G j= (v1; : : : ; vn) if and only if some Cjn;N 2 Dn;N satis es (v1; : : : ; vn) in G. We further require that the set Dn;N be maximal in the sense that if G j= (v1; : : : ; vn) and Cn;N satis es (v1 ; : : : ; vn ) then Cn;N 2 Dn;N . Note that Dn;N is closed under isomorphism, that is, if Cn;N ; C0n;N 2 Dn;N are isomorphic and Cn;N 2 Dn;N then C0n;N 2 Dn;N : This follows since Dn;N is required to be maximal. Thus, we may view Dn;N as a set of isomorphism classes of (n; N )-con gurations. If N < N 0 , and C0n;N 0 , Cn;N are con gurations we say that Cn;N is the restriction of C0n;N 0 or that C0n;N 0 is an extension of Cn;N if the con guration obtained from C0n;N 0 by taking the induced subgraphs of C0n;N 0 consisting of all vertices within a distance of N of at least one of the identi ed vertices is isomorphic with Cn;N . For a xed set of con gurations 0 0 which Dn;N and N < N 0 we form the N 0 -extension of Dn;N by simply including all Cn;N extend some Cn;N 2 Dn;N . It is easy to see that if (x1 ; : : : ; xn ) is equivalent to Dn;N then (x1 ; : : : ; xn ) is equivalent to any extension of Dn;N . Theorem 3.1 Let M be a vertex transitive graph of nite degree and ? the corresponding family of graphs. For each formula (x1 ; : : : ; xn) there is an N 2 N and Dn;N , a set of isomorphism classes of (n; N )-con gurations, such that for all good G 2 ?, is equivalent to Dn;N for G. Proof. We rst give a procedure to construct Dn;N which we then show to be equivalent to . We inductively build up the con gurations. As a start assume that is atomic. Suppose (x1 ; : : : ; xn ) = xi Axj . We take N = 1 and let Dn;1 be the set of (n; 1)-con gurations Cn;1 for which xi ; xj are identi ed with vertices v~i; v~j which are connected by an edge in Cn;1. It is clear that is equivalent to Dn;1 . For the other atomic case, xi = xj it is clear that we should take N = 0 and Dn;0 to be the set of (n; 0)-con gurations Cn;0 for which the vertices identi ed with xi and xj are equal. 7
Suppose the theorem holds for formulas and
giving equivalent con gurations
Dn;N and D0n;N 0 respectively. It is easy to nd the appropriate sets of con gurations for the boolean combinations of and . In particular, for : we simply take all Cn;N for which Cn;N 62 Dn;N where is equivalent to Dn;N . In the other two cases we simply take
the intersection or the union of the equivalent con gurations after forming the appropriate extensions so the N parameters are equal. Note that in the Boolean case we may increase the value of N for one of the con gurations but it is only by a simple extension. Assume now without loss of generality that (x1 ; : : : ; xn) = 9xn+1 (x1 ; : : : ; xn ; xn+1), and let Dn+1;N = (C1n+1;N ; : : : ; Ckn+1;N ) be equivalent to . We let D~ n;3N be the set of Cn;3N satisfying one of the following: 1. Cn;3N = T1 [ [ Tm and there is a vertex v~ 2 Ti for some0 1 i m such that v~ is within 2N of v~r for some r and for some 1 j 0 k Cjn+1;N is obtained from 0 Cn;3N by identifying xn+1 with v~ and then \restricting to N ". That is, Cjn+1;N = CNCn; N (~v1; : : : ; v~n; v~) where x1; : : : ; xn are identi ed with v~1 ; : : : ; v~n in Cn;3N : 3
0
2. For some 1 j 0 k, Cjn+1;N is such that v~n+1 is the only identi ed vertex in some 0 component of Cjn+1;N and Cn;3N is a 3N extension of the remaining components of 0 Cnj +1;N . We show that D~ n;3N is equivalent to (x1 ; : : : ; xn): Let G 2 ? be good, and let v1; : : : ; vn 2 G. Assume rst that G j= (v1 ; : : : ; vn ), so let vn+1 2 G be such that G j= (v1 ; : : : ; vn; vn+1). Consider C = CNG(v1 ; : : : ; vn; vn+1). First assume that vn+1 is the only vertex among (v1 ; : : : ; vn; vn+1) in the component of C containing vn+1. Let Cn+1;N be an (n; N )-con guration satisfying v1 ; : : : ; vn+1. Say Cn+1;N = T1 [ [ Tm where Tm is the component containing v~n+1 . By induction Cn+1;N 2 Dn+1;N : If Cn;3N is a 3N -con guration realizing (v1 ; : : : ; vn ) in G, then clearly Cn;3N 2 ~ n;3N from Case 2. D Assume next that each component of C contains an element of fv1; : : : ; vng. It follows that vn+1 2 C2GN (v1 ; : : : ; vn): Let Cn;3N again be the 3N -con guration realizing (v1 ; : : : ; vn ) in G. Let v~n+1 be the vertex corresponding to vn+1 under a xed isomorphism of Cn;3N with C3GN (v1 ; : : : ; vn ). Note that the restriction of Cn;3N to the components of CNCn; N (~v1 ; : : : ; v~n; v~n+1) is isomorphic to Cn+1;N , an (n; N )-con guration realizing v1; : : : ; vn; vn+1. By induction, Cn+1;N 2 Dn+1;N : Hence Cn;3N 2 D~ n;3N by virtue of Case 1. Next suppose v1 ; : : : ; vn 2 G are such that some C~ n;3N 2 D~ n;3N realize v1; : : : ; vn. Suppose rst that C~ n;3N 2 D~ n;3N by virtue of Case 1. Fix an isomorphism between C~ n;3N and C3GN (v1 ; : : : ; vn ). So (~vi ) = vi for 1 i n. Let v~n+1 be the vertex in C~ n;3N as in Case 1, and let vn+1 = (~vn+1 ). Thus, if Cn+1;N is the (n +1; N )-con guration obtained from C~ n;3N by restricting C~n;3N to CNCn; N (~v1 ; : : : ; v~n; v~n+1) then Cn+1;N 2 Dn+1;N and so by induction G j= (v1 ; : : : ; vn ; vn+1), and hence G j= (v1 ; : : : ; vn ). Note that we have not yet used the fact that G is good. 3
3
8
Suppose now that C~ n;3N 2 D~ n;3N by virtue of Case 2. Let Cn+1;N 2 Dn+1;N be as in Case 2. Let Cn+1;N = T1[ [Tm and assume Tm is the component of Cn+1;N with v~n+1 the only identi ed vertex. Since G is good, let vn+1 2 G be such that vn+1 2= C2GN (v1 ; : : : ; vn ) and CNG(vn+1) is isomorphic with Tm . However, then Cn+1;N satis es (v1 ; : : : ; vn; vn+1). So, by induction G j= (v1 ; : : : ; vn ; vn+1) and hence G j= (v1; : : : ; vn). If n = 0 in the above argument then is equivalent to > or ? depending on whether Dn+1;N is non-empty or not. Hence Theorem 3.1 has the following consequences Corollary 3.2 Let M be a vertex transitive graph of nite degree and let ? be the corresponding class of random subgraphs. Then for any sentence in the rst order language of graph theory fG 2 ?jG j= g 2 f0; 1g. A closer examination of the argument above allows us to obtain more. The fact that our reduction worked on rst order formulae (not just sentences) yields the following assertion. Corollary 3.3 With M and ? as above, (x1 ; : : : ; xn ) a rst order formula, and v1; : : : ; vn 2 M , let (v1 ; : : : ; vn) = fG 2 ?jG j= (v1; : : : ; vn)g. Then the set of all possible values for , f (v1 ; : : : ; vn)jv1 ; : : : ; vn 2 M g, is nite. Proof. This follows from the theorem noting that (x1 ; : : : ; xn ) is equivalent to some Dn;N for some N 2 N. Thus, on the measure one set of good G, whether G j= (v1 ; : : : ; vn ) depends only on CNG(v1 ; : : : ; vn). Since M has nite degree, the probability measure can be restricted to a nite probability space. Given the vertex transitive master graph M of nite degree, we say a sequence M1 M2 Mi with Mi M is a reduction of M if the following three conditions are satis ed. 1. Each Mi is a nite spanning subgraph of M . 2. Let r (Mi ) = supv2Mi fsupfn 2 Nj(v; v0 ) n implies v0 2 Mi gg: (So r (Mi ) is the radius of the largest ball in Mi .) Then for every c > 0; r (Mi ) > c log(i) for all i suciently large. This is saying that Mi contains a ball of radius greater than c log(i). 3. For some D, l > 0, Mi contains no more than Dil vertices. As an illustration, let Zrepresent the graph whose vertex set is the set of integers and edges consist of consecutive integers. If we let H = Zn = Z Zwe may take Mi to be the induced subgraph on the vertices with all coordinates between ?i and i. We are then considering random subgraphs of the lattice graph on (?i; i) (?i; i). It is easily seen that this forms a reduction of H . Each Mi determines a nite probability space using the Bernoulli measure in the usual manner with probability of edge connection p. We say that almost all nite spanning subgraphs (relative to the reduction) satisfy if limi!1 P (i) = 1, where P (i) is the probability of a graph in Mi satisfying . Theorem 3.4 Let M be a vertex transitive graph of nite degree and let M1 M2 Mi be a reduction for M . Then for any sentence in the rst order theory 9
of graph theory, either almost all graphs (relative to the reduction) satisfy or almost all satisfy : . Proof. Let G Mi . For xed n, N 2 N we say G is n; N good if for all vertices v1; : : : vn 2 G and T a connected subset of CNM (~v0 ) there exists vn+1 2 G with vn+1 62 C2GN (v1 ; : : : ; vn ) and CNG(vn ) is isomorphic to T by an isomorphism sending vn to v~0 . Investigating the proof of Theorem 3.1 shows that it is sucient to show that almost all G (relative to Mi ) are n; N good, for any xed n, N . We say G is N good relative to v1; : : : ; vn if there is a vn+1 2 G as above for the vertices v1 ; : : : ; vn. Let b(v1 ; : : : ; vn) represent the probability that G is not good relative to v1 ; : : : ; vn. For xed ?i and G Mi , the probabilities that G is not n; N good is at most l Di v ;:::;vn b(v1 ; : : : ; vn ) n b(v1 ; : : : ; vn ) for some v1; : : : ; vn. Fix v1 ; : : : ; vn 2 G. Since r (Mi ) > c log(i) (c suciently large) there are at least c log(i) ?n vertices v 2 G at least 2N +1 apart such that C G (v ) G and v 62 C G (v ; : : : ; v ). n N 2N 1 6N The probabilities that CNG (v) is isomorphic to T , for a xed T , are independent for the c log(i) ? n di erent vertices. The possible number of T s is a function only of N , say w(N ). 6N We divide the vertices v above into w(N ) sets of (roughly) equal size, and within each set attempt to nd at least one v with CNG(v) isomorphic to the corresponding T . The probability we do not succeed is at most 1
?c
log(i) n Nw(N ) ? w(N )
w(N )(1 ? q) for some q > 0. Thus, the probability G is n; N good is at least Dil ?c i ? n 1 ? n w(N )(1 ? q) Nw N w N : Since c can be taken large compared to N and l, the above probability approaches one. It is interesting to note that in the in nite and nite cases one has the same reduction procedure of to > or ?. In particular, the in nite and the nite cases have the same rst order theory as long as the three conditions are met in the nite case. An example of this is the following assertion. Corollary 3.5 For any rst order formula in the language of graph theory, almost all spanning subgraphs of Zn satisfy if and only if almost all spanning subgraphs of (?N; N )n satisfy . Finally we wish to point out that the fact that the master graph has nite degrees is crucial to our argument. It would be interesting to see a proof of Corollary 3.2 which would include the case of in nite degree master graphs. Of course, this proof would then include Corollary 3.2 as well as the fact that the almost all theory of graphs is complete (in this case the master graph is simply the complete graph on the integers). 6
6
4. Higher order axioms.
log( ) ( )
(
)
We consider now some higher order properties of graphs which generalize the rst order axioms given in Section 1. The axioms of Section 1 imply the existence of \local" 10
structures. For example, for any xed graph almost all graphs have a subgraph isomorphic to the xed graph. The axioms of this section imply the existence of \global" structures, such as hamiltonian cycles in almost all graphs. The global structures are constructed according to a \pattern", which is made precise in the de nition of a constructing family. We prove two theorems in this section. Theorem 4.1 asserts the existence of large subgraphs and furthermore allows negative edge requirements. In Theorem 4.4 we only assert the existence of large subgraphs but with improved estimates. A constructing family of functions = fFn(k;l)g of type (k; l) is a collection of funtions F) n(k;l) for each n 2 N such that ( k;l 1) Fn has domain tuples (S0; <S ; : : : ; Sj ; <Sj ; (x1 ; : : : ; xk ); (y1 : : : ; yl )) where Si f1; : : : ; ng with jSij < n for each i, <Si is a linear ordering of a subset of Si for each i, x1 ; : : : ; xk 2 Sj are distinct, and y1 ; : : : ; yl 2 f1; 2; : : : ; ng ? Sj are distinct, 2) Fn(k;l)(S0; <S ; : : : ; Sj ; <Sj ; (x1 ; : : : ; xk ); (y1 ; : : : ; yl )) = (Sj+1; <Sj ), where Sj+1 = Sj [ fy1; : : : ; yl g, and <Sj is a linear ordering of a subset of Sj+1. The intended meaning of a constructing family is to give a procedure for constructing a large subgraph according to a speci c pattern. For example, to construct a hamiltonian cycle we start with a small cycle and add one vertex at each step. The vertex is chosen to be adjacent with consecutive vertices around the cycle, increasing the length of the cycle by one. This corresponds to a construction family of type (2; 1). In particular, Si is the set of vertices in the cycle in step i while <Si is the order around the cycle starting at any xed vertex. The set Si+1 is the new cycle with the added vertex and <Si is again the order around the newly formed cycle. Note in this case Fn(k;l)(S0; <S ; : : : ; Sj ; <Sj ; (x1 ; x2 ); (y1 )) depends only on (Sj ; <Sj ; (x1 ; x2 ); (y1 )). We say that the functions fn : N ! R are bounding functions for fFn(k;l)g if, with the above notation, j eld(<Sj )j fn (jSj j). Here eld refers to the set of elements x where either x <Sj y or else y <Sj x for some y. The graph in Figure 2a can also be constructed with a (2; 1) family of functions. In this graph we have approximately n1=16 paths of length n1=16 which we will refer to as the cycles. We require that the end vertices of each path are not connected (the dotted line in the gure), they are connected as shown, and have attached paths of lengths approximately n7=8 as shown. In the rst n3=16 steps we build up a subgraph consisting of all the cycles and attached paths of length n1=16. For j in this range we take the eld of <Sj to consist of only two vertices. For j n3=16 the constructing function adds at each step a single vertex to one of the paths. At each step the path chosen is the next one clockwise from the previous one. The eld of Sj+1 is simply the next path to be added to. The order <Sj is the order along this path. For j > n3=16 the size of the eld of Sj is at least jSj j1=3 . We consider the following axiom scheme for a graph G with n vertices. The axioms say that the construction procedure corresponding to Fnk;l can be carried out in G. Axiom A(k;l): Let d be a graph with vertex set fa1 : : : ; ak ; b1 ; : : : blg and let din be a graph with vertex set fc1;: : : ; cr g where k r k + l. There exists a sequence (S0 ; <S ), : : : , (Sm ; <Sm ) with m = n?l r , where Si is a subset of the vertices of G and 1) the subgraph of G induced from the vertices S0 is isomorphic with din. Also, the ordering <S on S0 corresponds to an ordering c1 < c2 < < cr under the isomor0
0
+1
+1
+1
0
+1
0
0
11
phism; 2) for every 0 j m?1, Fn(k;l)(S0; <S ; : : : ; Sj ; <Sj ; (x1 ; : : : ; xk ); (y1 : : : ; yl )) = (Sj+1; <Sj ) for some elements x1 ; : : : ; xk 2 Sj , y1; : : : ; yl 2= Sj , and where x1 ; : : : ; xk are consecutive elements of Sj with respect to <Sj ; also, xi Ays in G if and only if ai Abs in d and yi Ays ?in G if andonly if bi Abs in d; 3) if e = n ? r + l n?l r > 0, then there are y1 ; : : : ; ye , vertices in G which are not in Sm, and consecutive elements (using order <Sm ) x1; : : : ; xk such that xi Ays in G if and only if ai Abs in d and yi Ays in G if and only if bi Abs in d. If the vertices x1 , : : : , xk , y1, : : : , yw (for w = l or w = e) satisfy Property 2 or Property 3 above then we say that x1 ; : : : ; xk ; y1 ; : : : ; yw t d. Note that Axiom (k; l) has Fn(k;l) as a parameter in its statement. Given a constructing family fFn(k;l)g, it makes sense to ask if a graph G satis es Axiom (k; l). We will show below that for fFn(k;l)g having suitable bounding functions f (n), for each k; l and d; din Axiom (k; l) holds for \almost all graphs". We interpret almost all as in Bollobas [4], where for each n we let p = pn be a number between 0 and 1. We then make the set of all labeled graphs with n vertices into a probability space by requiring the probability of each possible edge to be p, independent of the other edges. Then we say almost all graphs have any given property P if and only if Pr(G 2 P ) ! 1 as n ! 1. Before verifying the axioms, we remark how they generalize the axioms in Section 1. The axioms of Section 1 implied the existence of \small" structures in G. The Axioms (k; l) imply G possesses \large" (i.e. of size n) structures. As indicated above, Axiom (2; 1) easily implies hamiltonicity of G. Figure 1 illustrates some other structures G must contain given G satis es certain of these axioms (in each case the structure contains all the vertices of G). In Figure 1a Fn(2;1)(S0; <S ; : : : ; Si; <Si ; (x1 ; x2 ); y1 ) adds y1 to the top branch of the graph if i 0 (mod 6), to the middle branch if i 1; 2 (mod 6) and to the bottom branch if i 3; 4; 5 (mod 6), each time connecting y to consecutive elements x1; x2 2 Si . (Here we take d to be a path of length two with end vertices a1 and a2 and middle vertex b1 ). The reader can easily discern the relevant constructing functions for the other structures. (Note in Figure 1c and 1d k = 2 and l = 3.) Before stating the theorem, we give a few technical conditions which are required in the theorem. Let 0 < < 1 be xed numbers and de ne w = minf 101 ( ? l l ); (4k) ( ? ); 10 (3 + ); 4(k) (3 + ); 2(1r) g > 0 if l > 1 and w = minf 101 ( ? ); 10 (3 + ); 2(1r) g > 0 if l = 1: Next we let 0 < pn < 1 be a sequence of numbers and de ne qn = 1 ? pn. Finally, let mn = minfpn; qn g. Theorem 4.1 Let fFn(k;l)g be a constructing family with bounding functions fn satisfying fn (m) > m for all m n , and let pn be the probability of an edge in a labeled graph with n vertices. Suppose that there is an > 0 such that if n is suciently large then ? l(wl ) + if l > 1 and mn > n?w+ if l = 1. Then almost all graphs satisfy mn > n Axiom (k; l). 0
0
2
2
2
2
2
Before giving the proof we point out the following two lemmas which will be used in 12
+1
the proof of Theorem 4.1. Let Zn represent the integers modulo n. Lemma 4.2 Let V = Zl Zn. For each 0 i nl and 0 j < n, let Ti;j = f(r; j + ri)j0 r < lg V , where the second coordinate is interpreted modulo n. Then for any (i; j ) 6= (k; t) with 0 i; k nl and 0 j; t < n, jTi;j \ Tk;t j < 2. The proof of Lemma 4.2 is straightforward. We call fTi;j ji = i0g the i0 sweep. Note that in each sweep each pair in V occurs exactly once. Next we need a technical lemma which allows us to partition the vertex set of almost all graphs into the the basic building blocks needed in Theorem 4.1. Given a \probabilistic" construction based on a random graph, we say that an edge fv; wg is used if the estimate of the probability that the construction succeeds depends on whether or not fv; wg is an edge in the graph. Otherwise the edge is said to be unused. So, if an edge is unused, the success of the construction is independent of whether or not the graph p has the given edge. p In the proofs below we will ignore the di erence between terms such as x and [ x] which do not change the calculations in any signi cant manner. Lemma 4.3 Suppose p?n, wthe+probability of any given edge in a labeled graph with n l vertices, satis es mn > n l( ) if l > 1 and mn > n?w+ if l = 1 for some > 0 and all n suciently large. Let d be a graph with l vertices and din be a graph with r k + l vertices. Then almost all graphs can be partitioned into one set of size r which induces a n ? subgraph isomorphic with din, l r of size l each inducing a subgraph isomorphic with d, and one with the residual vertices inducing a subgraph isomorphic with the induced graph on the rst vertices of d. Furthermore, with this construction given any vertex v, there are w + at most O(n ) used edges involving v. Proof. Let n 2 N and consider the probability space of all labeled graphs on n vertices. pn We identify the vertex set of the graphs with V = f 1 ; 2 ; : : : ; n g . Let V be the last 0 p vertices of V . We partition the set V0 into rn sets of consecutive integers each of size r (with perhaps a few residual vertices). The probability that the induced subgraphr onpat n least one of these subsets is isomorphic with din is bounded below by 1 ? (1 ? m(n ) ) r . By taking logarithms and using the fact that w 2(1r) , it follows that this probability approaches one as n approaches in nity. Note that in this construction each vertex has at most only r used edges and if fi; j g is used then both i and j are in V0. Assuming there is such a subgraph we call it Din. The next step is to partition the rest of the vertices into subsets of size l so the induced graph on each subset is isomorphic with d. If l = 1 then this step is trivial since d consists of just a vertex. So we assume here that l > 1. To this end, we partition the vertices of V1 = V ? V0 into two subsets, one consisting of the rst bjV1j vertices of V , which we call p V2, and the other of size (1 ? b)(n ? n), which we call V3 , where b = n?w . We further partition V3 into two subsets, V4 with b2 n = lu vertices and V5 with jV3j ? jV4j vertices. w 1+ We follow the following algorithm until V5 = ; or for n steps, whichever comes rst. We index the vertices of V4 by Zl Zu. We list the sets of Lemma 4.2 in the order T0;0; T0;1; : : : ; T0;u?1; : : : ; T ul ;0; : : : ; T ul ;u?1. Starting with the rst set S = T0;0 we check if the subgraph induced by the vertices with indices in S is isomorphic with d (A success is 2
5 2
2
2
2
13
when the induced subgraph is isomorphic with d and a failure is when it is not isomorphic with d). In the case of a success, we add the vertices indexed by S to the set W (which will contain sets of size l each inducing a subgraph isomorphic to d), remove the next l vertices from V5 and index them by the indices in S . We then proceed to the next step which is to set S to the next set in the list of Ti;j and continue as before. Note that the number of Ti;j s is O(n2(1?2w)) > n1+ w since w < 101 ( ? ) < 92 . l w l At each step the probability of a success is bounded below by m(n ) n? +( ). Since at each step only unused edges are considered, the probability of exhausting V5 before w w 1+ 1+ n steps is bounded below byl the probability that in n independent Bernoulli trials with probability of success m(n ) one attains at least O(n) successes. It is clear that this probability approaches 1 as n approaches in nity using, for example Chebychev's inequality by the ordering of the sets Ti;j it is clear that for Pr(jX ? j > a) < a1 . Furthermore, w w n each vertex v at most u = O(n ) edges involving v are used, and each of the used edges has both vertices in V3. At this point in the algorithm have one subgraph, Din, which is isomorphic to din, pn)?we (1 ? b )( n ? b n and W which contains subgraphs each isomorphic with d. The rest of the l vertices are in a subset, V6 V4 [ (V0 ? Din) of size approximately b2 n and V2 of size approximately bn. First note that no edge between V6 and V2 is used. We next nd jV6j subsets of size l from V2 [ V6 by partitioning the set V2 into approximately lbn ?1 subsets of size l ? 1 each and attempting to match up each vertex in V6 with a subset of size l ? 1 from V2 such that the induced subgraph from these vertices is isomorphic to d. For each vertex in V6 we examine (l?11)b subsets of size l ? 1 from V2 to check if one of the induced subgraphs is isomorphic with d.l The probability that we succeed for each vertex in V2 is bounded below by (1 ? (1 ? m(n ) ) b l? )b n. By taking logarithms it is easy to check this approaches 1 as n approaches in nity. Next, the remaining vertices in V2 need to be partitioned into subsets of size l so each induced subgraph is isomorphic to d. We do this by rst partitioning the remaining vertices of V2 into subsets of size l (with perhaps one of the subsets having size less than l). We then index the subsets in the partition of W with Zl?1 Zn0 where n0 = O(n). We de ne Ti;j as in Lemma 4.2 and order the Ti;j s as before. The subsets in the partition of V2 are ordered arbitrarily. Note that Ti;j consists of l ? 1 subgraphs each isomorphic to d: We call an index set, Ti;j , bad if there is a used edge between two of the subgraphs indexed by Ti;j , otherwise the index set is good. We call sweep i0 acceptable if no more that half of the index sets Ti ;j in the sweep are bad. Note that there are at most O(n w ) sweeps which are not acceptable since the total number of used edges is O(n1+ w ). Therefore there are at least O(n) > jV2 j acceptable sweeps. To each subset Y in the partition of V2 we associate an acceptable sweep. For each of the rst nw good sets S in that sweep we attempt to repartition the vertices of S and Y into l subsets of size l so that each subset induces a subgraph isomorphic with d: Speci cally we arbitrarily number the vertices in each set of S from 1 to l and number the vertices in Y from 1 to l and then check if the vertices having the same number form a graph isomorphic 2
2
2
2
2
2
1+ 2
5 2
2
2
(
1
1)
2
5 2
0
5 2
14
2
2
with d. that the probability that we succeed for every Y is bounded below It is easy to lverify n n l where n1 = O(nw ) and n2 = O(n1?w ). It is easily checked by 1 ? 1 ? mn( ) that this probability bound approaches 1. Clearly for each vertex v at most O(nw ) edges containing v become used in this process. Now we are ready to prove Theorem 4.1. The proof is much like the proof of Lemma 4.3, in that the same ideas of partitioning, looking for edges, and estimating probabilities of success are used. Proof. For a xed constructing family of functions fFn(k;l)g with bounding functions fn, and and as above, we x d and din and consider the probability that a graph G on n vertices satis es Axiom A(k;l). To begin, let G be a graph with n vertices. We assume that n jdinj (mod jdj) in order to avoid inessential detail relating to how to deal with the special case at the end where fewer than jdj vertices are left over. We rst use Lemma 4.3 to partition the vertex set of G into one subset S0, inducing a subgraph of G isomorphic with din, and the rest into subsets of size l, each inducing a subgraph of G isomorphic with the subgraph of d induced by the vertices b1 ; : : : ; bl . Let s = n?jldinj and h1; : : : ; hs be the sets in the partition de ned above. We partition the set fh1; : : : ; hsg into two subsets, A and B with jAj = s and jBj = s ? s . We further partition A into j = s + sets, each of size s( ? ), which we call T1 ; T2; : : : ; Tj : We now attempt to form S1; S2; : : : ; Sj as in the statement of A(k;l). Given the sets S1; S2 : : : ; Si, we look for h 2 Ti+1 such that if (~c1; : : : ; c~k ) are the <Si rst k vertices in Si (with order <Si given inductively from Fn(k;l)), then the subgraph of G induced from w fc~1; : : : ; c~k g[ h ts d. Since the construction of Lemma 4.3 used at most O(n ) edges for each vertex, there are at least s ? ?O(n w ) choices of sets in Ti+1 which involve unused edges. It follows that the probability that the rst s + steps can be accomplished is at least 2
1
2
+ 2
3 4
1 4
1 4
+ 2
1 4
5 2
1 4
1 4
5 2
3 4
? P1 = 1 ? 1 ? mkln s
1 4
1 4
+ ? 41 ?O(n 52w ) s 4 4 3
1
We let S denote the vertices and <S denote the order produced to this point. Next we let the sets in B be denoted U1; U2; : : : ; Us?s and succesively adjoin the l vertices Ui+1 = f~b1 ; : : : ; ~blg to k consecutive vertices c1; : : : ; ck in S [ (U1 [ [ Ui ), (with order given inductively by Fnl;k ) so that the subgraph of G induced by fc1; : : : ; ck g [ Ui+1 ts d. An easy computation gives the following lower bound for the probability of success for every Ui as + 2
0 ls B ? kl P2 = B @1 ? 1 ? mn
(3 4
+1 4
k
15
)
1s?s ?lO(n w ) C C : A + 2
5 2
Last, we denote the subsets of size l in A ? S by V1 , V2, : : : , and at step i + 1 attempt to join Vi+1 to k consecutive vertices in S [ B [ (V1 [ [ Vi) as required. Now at each the domain of the linear ordering step there are at least (l(s ? s )) ? ls vertices in ( k;l ) given by Fn which lie in B. Since there are only ls vertices in A, at least + 2
+ 2
+ 2
(l(s ? s )) ? ls 1 s ? l ?1 2 ls (for large enough n) vertices in B must be consecutive in that ordering. This easily gives the following lower bound for the probability of success: + 2
+ 2
2
+ 2
P3 = 1 ?
!
s l ?1 s ?2 ?lO(n 52w ) 2k kl 1?m
?
n
+ 2
Since only unused edges were considered in the estimates of probabilities P1, P2 , and P3, we get P1P2 P3 as a lower bound for the probability that the graph satis es A(k;l) given that Lemma 4.3 is satis ed for xed d and din. Let P4 be the minimum probability that Lemma 4.3 is satis ed for a graph with n vertices over all d and din with parameters r; l; k. r) kl ( l ) ( As we range over all possible choices for d and din we get (1 ? 2 2 2 (1 ? P1P2 P3P4)) as a lower bound for the probability that G satis es axiom A(k;l) for the given Fn(k;l). It is enough to show Pi ! 1 as n ! 1 for i = 1; 2; 3. This follows easily in all ?three cases by l ? ? taking logarithms. We point out that the bounds w < 10 and w < 4k 2 are needed l for estimating P1, w < 10 (3 + ) and w < 4(k) (3 + ) are needed for estimating P2 l and w < ?5 (which follows from w < 101 ( ? )) and w < (4k) ( ? ) are needed for estimating P3. We consider now a variation A~(k;l) of the axiom scheme A(k;l). Here we do not allow negative edge requirements. Axiom A~ (k;l): Let d be a graph with vertex set fa1; : : : ; ak ; b1 ; : : : ; bl g and let din be a graph with vertex set fc1; : : : ; crg where k r k + l. Then there is a sequence (S0; <S ); : : : ; (Sm ; <Sm ) with m = n?l r , where Si is a subset of the vertices of G and 1) The subgraph of G induced from the vertices S0 is isomorphic with d0in where d0in has the same vertex set as din and contains all the edges of din, but may contain more edges. Also, the ordering <S on S0 corresponds to the ordering c1 < c2 < < cr under the isomorphism. 2) Fn(k;l)(S0; <S ; : : : ; Sj ; <Sj ; (x1 ; : : : ; xk ); (y1 ; : : : ; yl )) = (Sj+1 ; <Sj ) for some elements x1; : : : ; xk 2 Sj , y1; : : : ; yl 2= Sj , and where x1 ; : : : ; xk are consecutive elements of S with respect? to 0, then there are y1 ; : : : ; ye , vertices in G which are not in Sm, and consecutive elements (using order <Sm ) x1; : : : ; xk such that xi Ays in G if ai Abs and yi Ayj in G if bi Abj in d. We have the analog of Theorem 4.1. 2
2
2
2
0
0
0
+1
16
Theorem 4.4 Let fFn(k;l)g be a constructing family with bounding function fn for Fn(k;l). Suppose that for all c > 0, for all suciently large n, fn (m) > c log m for all n < m n
for some < 1. If pn, the probability of any given edge on a labeled graph with n vertices is bounded below by (logLn) kl for some constant L, then the probability that a graph on n vertices satis es A~(k;l) tends to 1 as n ! 1. Proof. We x a suitable large c > 0 and only consider n such that fn(m) > c log m for 1+
0 n < m n. Let = 2 . We exploit a device of [4]. Instead of considering graphs with probability of an edge pn, we consider three graphs with the same vertex set and color the edges in one graph red, one graph blue, and the other graph green. We denote the probability of a red edge by pr , the probability of a blue edge by pb and the probability of a green edge pg : We then de ne the graph G to have the same vertex set as the three graphs and let fv; wg be an edge of G if and only if fv; wg is either a blue, red or green edge. Note that the probability that fv; wg is an edge in G is 1 ? (1 ? pr )(1 ? pb)(1 ? pg ) p and the edges are independent. By choosing pr = pb = pg = 1 ? 1 ? pn we get pn for the probability of each edge in G. (Note that for small pn, the common value of pr = pb = pg is approximately 13 pn.) We rst apply Lemma 4.3 using the green graph. (Note that the probability bound in Lemma 4.3 is satis ed.) Let P0 denote the probability that the green graph satis es Lemma 4.3. We will call each of the0 subgraphs isomorphic to d a block. We next partition the rst n blocks of G into n disjoint sets, each of size n 0? , say T0; T1 ; : : :. We start with din and successively nd a block in Ti+1 that can be added to S0 [ S1 [ [ Si according to the function F (k;l) using only blue edges. A computation as before yields the following lower bound for the probability of success: 1
3
? n 0? n kl P1 = 1 ? 1 ? p b :
So far approximately n blocks have been included. Next, we attempt to succesively add the remaining blocks to consecutive vertices already included using only red edges. This gives the following lower bound for the probability of success:
? c P2 = 1 ? 1 ? pklr
n )
log(
k
nl
:
It is easy to check that P0 , P1 and P2 all approach one as n approaches in nity. We leave it to the reader to check that if we replace c log n in the statement of Theorem 4.4 with the bound for fn (m) in Theorem 4.1 then one can improve the estimate for pn in Theorem 4.4 to the requirement for pn in Theorem 4.1. Suppose we x 0 < p < 1 and let pn = p: It is interesting to note that Theorem 4.1 implies that almost all graphs contain the structure a) in Figure 2 (using = 31 , = 163 and a suitable fFn(k;l)g as described earlier), while Theorem 4.4 implies almost all graphs contain the structure b) (which has only positive requirements). Neither theorem implies almost all graphs contain the graph of c). We do not know if this is the case. 17
The probability estimates are by no means best possible for any given spanning subgraph. As an example, Hamiltonian cycles exist with probability approaching one in n+!(n) where ! (n) ! 1 as n ! 1 (see for example random graphs with pn log n+log log n [4]). The bound from Theorem 4.1 is n =1 for any > 0, which is much larger than log n+log log n+!(n) . In fact, Theorem 4.1 only asserts the existence of spanning subgraphs n when the probability bound is nw for some ?1 < w < 0. However, it is easy to check for pn < n =1 the probability approaches zero that one can partition a random graph on n vertices into n3 3-cycles. Of course the existence of such a partition follows from Theorem 4.1 with pn > n 1= . Although there is a big gap between 32 and 901 , this example does show that a general theorem asserting the existence of both a Hamiltonian cycle and a partition of the vertices into triangles cannot do better than nw for some ?1 < w < 0. We wish to thank the referee for useful comments and suggestions for improvements to some of our results. 1 10+
2 3+
1 90
References [1] C. K. Bailey and W. M. Dymacek, Regular Steinhaus graphs, Congressus Numerantium 66(1988),45{47. [2] P. Billingsley, Probability and Measure, John Wiley and Sons, New York, 1979. [3] A. Blass and F. Harary, Properties of almost all graphs and complexes, J. Graph Theory 3(1976) 225{240. [4] B. Bollobas, Random Graphs, Academic Press, London, 1985. [5] N. Brand, Almost all Steinhaus graphs have diameter two, Journal of Graph Theory 16(1992) 213{219. [6] N. Brand, S. Curran, S. Das, T. Jacob, Probability of diameter two for Steinhaus graphs, Discrete Applied Math 41(1993) 165{171. [7] R. C. Brigham, N. Deo and R. D. Dutton, Some properties of Steinhaus graphs, preprint. [8] R. C. Brigham and R. D. Dutton, Distances and diameters in Steinhaus graphs, Congressus Numerantium 76(1990) 7{14. [9] W. M. Dymacek, Bipartite Steinhaus graphs, Discrete Math. 59(1986) 9{20. [10] W. M. Dymacek, Small cycles in Steinhaus graphs, Congressus Numerantium 70(1990) 41{52. [11] R. Fagin, Probabilities on nite models, J. Symbolic Logic 41(1976) 50{58. [12] H. Harborth, Solution to Steinhaus' problem with plus and minus signs, J. Combinatorial Theory 12(A)(1972) 253{259. [13] D. Knuth, The Art of Computer Programming. Volume 1: Fundamentals of Algorithms, Addison-Wesley, Reading, Mass. (1973). [14] H. Steinhaus, One Hundred Problems in Elementary Mathematics, Elinsford, New York, 1963. 18
~ n/3
~ n 7/8
~ n/6 ~ n/2
(a) (b) ~ n 1/8
n/3 triangles
n/3 triangles
(c)
(d)
Figure 1.
Some structures of size n almost all graphs of size n contain according to Theorem 4.1. A solid line indicates an edge, a broken line indicates no edge, and neither indicates there is no requirement either for an edge or for no edge. 19
1/16
~n
7/8
~n
1/16
~n
(a)
1/8
~n
3/4
~n 1/8
~n
(b)
1/8
~n
3/4
~n ~n 1/8
(c)
Figure 2. As in Figure 1, a solid line indicates an edge, a broken line indicates there is no edge, and no line indicates there is no requirement concerning an edge. The graph in a) follows from Theorem 4.1, the graph in b) follows from Theorem 4.4, while the graph in c) does not follow from either theorem. 20
Abstract
In [11] it is shown that the theory of almost all graphs is rst order complete. Furthermore, in [3] a collection of rst order axioms are given from which any rst order property or its negation can be deduced. Here we show that almost all Steinhaus graphs satisfy the axioms of almost all graphs and conclude that a rst order property is true for almost all graphs if and only if it is true for almost all Steinhaus graphs. We also show that certain classes of subgraphs of vertex transitive graphs are rst order complete. Finally, we give a new class of higher order axioms from which it follows that large subgraphs of speci ed type exist in almost all graphs.
1. Introduction.
This paper is motivated by [3],[8] and [11]. In [3] and [11] it is shown that any rst order property of graphs is either satis ed by almost all graphs or else almost all graphs do not satisfy the property. Furthermore, axioms are given, each of which almost all graphs satisfy, such that for any rst order property either the property or its negation can be deduced from a nite number of these axioms. In fact we have an axiom for every natural number k. Axiom k: For any set of 2k distinct vertices fv1; v2 ; v3 ; : : : ; vk ; w1; w2; w3 ; : : : ; wk g, there is a vertex v with vAvi and :vAwi for each 1 i k. Let us de ne a 0-1 matrix (ai;j )ni;j=1 as follows. Start with a 0-1 string (a1;j )nj=2. For 1 < i < j n de ne ai;j inductively by ai;j ai?1;j?1 + ai?1;j (mod 2). We complete (ai;j )1i<jn to an n n matrix by de ning ai;i = 0 and ai;j = aj;i for 1 j < i n. This gives an n n symmetric 0-1 matrix A = (ai;j )ni;j=1 with 0s on the diagonal. The matrix A is the adjacency matrix of some graph. Any graph de ned in this way is called a Steinhaus graph. The string (a1;j )nj=2 is called the generating string for the Steinhaus graph. For convenience we identify the vertex set of a Steinhaus graph with the rst n natural numbers, Vn = f1; 2; 3; : : : ; ng. It is obvious that there are exactly 2n?1 labeled Steinhaus graphs of order n. We use the term Steinhaus graph to mean a labeled Steinhaus graph and assume that the vertex set is Vn. We de ne a probability measure on Steinhaus graphs of order n by requiring that Pr(a1;j = 1) = pn;j where 0 pn;j 1 for each j . We then de ne qn;j = 1 ? pn;j and mn;j = min(pn;j ; qn;j ). Any function f (n) with the properties that for each suciently large n, 0 < f (n) < 1 and mn;j f (n) for each 2 j n is called a probability bound. 1 2
Supported by a University of North Texas Faculty Research Grant. Supported by a University of North Texas Faculty Research Grant and NSF grant number DMS 9007808. 1
Given a probability bound, we will say that almost all Steinhaus graphs have property P if the probability that a Steinhaus graph of order n has the property approaches one as n approaches in nity. Of course, the concept of \almost all" depends on the probability bound. The simplest case is the constant probability bound of 21 . This gives all Steinhaus graphs of order n the same probability. In [8], Brigham and Dutton conjectured that almost all Steinhaus graphs have diameter two. Diameter two is a special case of Axiom 2. The conjecture was proved in [5]. It is natural to ask if all the rst order axioms for almost all graphs are satis ed by Steinhaus graphs. In Section 2 we show that they are satis ed by Steinhaus graphs, which implies that almost all Steinhaus graphs have the same rst order theory as almost all graphs. In Section 3 we look at other classes of graphs whose rst order theory is complete. We consider the set of all spanning subgraphs of a xed vertex transitive graph and de ne a measure on that set. We show that with this measure the rst order theory of almost all spanning subgraphs is complete. Furthermore, we show that with a few technical conditions on the allowed subgraphs, the theory of almost all nite subgraphs is complete and the same as the theory of almost all subgraphs with all vertices. In Section 4 we consider higher order axioms. The rst order axiom scheme described above implies that small structures exist as subgraphs of almost all graphs. The axioms introduced in Section 4 imply the existence of speci ed subgraphs containing all the vertices of the graph for almost all graphs. As a special case, the theorems in Section 4 imply the existence of Hamiltonian cycles in random graphs. Although the probability bounds given in the theorems are not even close to optimal in the case of Hamiltonian cycles, the theorems apply to a large class of subgraphs. Throughout the paper we use log x to denote natural logarithm.
2. Steinhaus graphs.
A Steinhaus triangle is simply the adjacency matrix of a Steinhaus graph above the main diagonal. Steinhaus introduced Steinhaus triangles in [14] where he posed the problem of nding generating strings which give Steinhaus triangles with the same number of zeros as ones. This problem was solved by Harborth in [12], where he constructed such sequences in the cases where there are an even number of entries in the Steinhaus triangle. Since then others have investigated properties of Steinhaus graphs. For example, the maximum clique size is found in [7], regular Steinhaus graphs are constructed in [1], the girth is studied in [10], and conditions under which Steinhaus graphs are bipartite are given in [9]. In [7] it is shown that the maximum diameter of a nontrivial Steinhaus graph is roughly n2 . Furthermore, in [5,6] it is shown that almost all Steinhaus graphs have diameter two. In fact, a slight modi cation of the proof in [5,6] shows that Axiom 1 is satis ed for almost all Steinhaus graphs. It is therefore natural to ask which of the other axioms in the axiom scheme listed above are satis ed by almost all Steinhaus graphs. The main purpose of this section is to prove the following result. Theorem 2.1 Given k 2 N and > 0, almost all Steinhaus graphs with probability bound n? k k + satisfy Axiom k. Before we turn to the proof of Theorem 2.1, we give some consequences of this result. 1 4 (2 +1)
2
Corollary 2.2 Given k 2 N, almost all Steinhaus graphs with constant probability bound satisfy Axiom k. Proof. This is obvious since if k is xed and 0 < < 4k(21k+1) then limn!1 n? 0.
1 4 (2 +1)
k k
+ =
Corollary 2.3 Using a constant probability bound, for any given rst order property, almost all Steinhaus graphs have the property if and only if almost all graphs have the property. Proof. In [3] it is shown that for any rst order property of graphs there is a k such that either the property or its negation can be proved with the axioms of graph theory together with Axioms 1{k. Since the same axioms hold for Steinhaus graphs, the proof of a property for almost all graphs is the same as the proof for almost all Steinhaus graphs. Corollary 2.3 states that random graphs look like random Steinhaus graphs from the point of view of the rst order theory. Since Steinhaus graphs are easily generated, perhaps this fact could be exploited by checking algorithm performance on large random Steinhaus graphs rather than large random graphs. Note that a proof of any true rst order property of graphs requires only a nite number of the listed axioms. Given an upper bound for k, where Axiom k is required in the proof, together with Theorem 2.1 we can obtain a probability bound which implies almost all Steinhaus graphs have the property. As pointed out in [3], there is an algorithm which gives either a proof of a rst order statement or a proof of its negation. Following this algorithm gives an upper bound for k. Many higher order properties follow from rst order properties. For example, diameter two (a rst order property) implies connected (a second order property). Corollary 2.3 implies the following properties of almost all Steinhaus graphs: diameter two, k-connected for any xed k, contain any xed graph as a subgraph, not planar, and have no embedding in a surface of genus g for a xed g. Our proof of Theorem 2.1 is based on 5 lemmas. Lemma 2.4 For any xed 2 k n, let (ai;j ) be the adjacency matrix for a Steinhaus graph with generating string (a1;j ) given by a1;r = k;r (1 if r = k, 0 otherwise). Then for ? i ? 1 any 1 i < j n, ai;j j?k (mod 2): Proof by induction. We have inductively ai;j ai?1;j?1 + ai?1;j ? i?2 . This ? i?is2 easily ? i?veri ed 1 (mod 2). + j ?k?1 j ?k j ?k
?
Given a natural number k, denote by H (k) the number of integers j with pkj odd. For non-negative integers a < b de ne G(b; a) = fkj1 k 2b?1; k = 2al; H (k) 2b?ag. Lemma 2.5 Let a, b be non-negative integers with a + 1 < b. Then jG(b; a)j 2b?a?2: Proof. It is sucient to show the lemma for the special case where a = 0 since the general case follows by dividing by 2a. So we assume that a = 0: 3
?
By Lucas' Theorem [13], a binomial coecient mk is odd if and only if for each w, if there is a zero in the base two representation of k in the 2w place then there is a zero in the 2w place in the base two representation of m. Therefore, k 2 G(b; 0) if and only if the number of ones in the base two representation of k is no more than 2b . Obviously there are at least 21 2b?1 = 2b?2 such values for k: For any pair of vertices (v; w) in a Steinhaus graph, we de ne C (v; w) = jv ? wj + 1. Furthermore, for any subset T Vn and any v 2 Vn we let CT (v) = fC (v; w)jw 2 T g. By Lemma 2.4, given any Steinhaus graph one can change the adjacency of vertices v and w simply by changing the value of the generating string at position C (v; w). Furthermore, this does not change? the adjacency of any pair of vertices (u; z) with CS(u; z) > C (v; w). min f v;w g? 1 We let B(v; w) = fij maxfv;wg?i 1 (mod 2)g. Then we let BT (v) = w2T B(v; w): By Lemma 2.4, BT (v) is the set of all entries in the generating string of a Steinhaus graph which when changed will change the (v; w) entry in the adjacency matrix for some w 2 T . Let T Vn. We say a sequence v1 ; v2 ; : : : ; vr 2 Vn is T -independent if for each 1 i r, CT (vi ) \ BT (vj ) = ; for j < i and jCT (vi )j = jT j for each i: Lemma 2.6 Suppose v1 ; : : : ; vr is a T -independent sequence for some set T Vn: Then for any generating string for a Steinhaus graph with n vertices, by changing only the entries in the generating string indexed by CT (vi ) it is possible to attain any combination of adjacencies between vi and the vertices in T: Furthermore, making these changes does not change the adjacencies between vj and the vertices of T for any j < i. Proof. Let (a1;j ) be an arbitrary generating string for a Steinhaus graph which gives an adjacency matrix (ai;j ). Label the elements in T by w1; : : : ; wt where C (vi; w1) > > C (vi; wt ). It is then clear that by changing the value of a1;C(vi;wr ), the value avi ;wr changes. Furthermore, changing a1;C(vi ;wr ) does not change the value of avi ;wk for k < r, nor does it change the change the value of avj ;ws for any j < i and any s:
Lemma 2.7 Let T = fv1 ; : : : ; vk ; w1; : : : ; wk g Vn and suppose that mn is the proba-
bility bound used for Steinhaus graphs. If there is a T -independent sequence of length r then the probability that Axiom k is true for a Steinhaus graph using the set T is at least 1 ? (1 ? m2nk )r : Proof. Let z1 ; : : : ; zr be a T -independent sequence of length r. Note that the generating strings for Steinhaus graphs for which Axiom k fails for z1; : : : ; zj?1 using T , can be partitioned into subsets each of size 22k by putting two strings in the same subset if the strings agree in every entry except for positions (1; i) where i 2 CT (zj ): By Lemma 2.6, in each subset there is a sequence whose Steinhaus graph satis es Axiom k using T and zj . Therefore, the probability that a Steinhaus graph satis es Axiom k using zj and T given that it does not satisfy Axiom k using T and any zi with i < j is at least m2nk . It is then clear that a lower bound that Axiom k is satis ed using T is 1 ? (1 ? m2nk )r : The goal, based on Lemma 2.7, is to nd a long T -independent sequence. The construction of a long T -independent sequence is sucient (although perhaps not necessary) to prove Theorem 2.1. 4
Lemma 2.8 Given T Vn with jT j = 2k, there is a T -independent sequence of length at least n 32kk ? 1. Proof. Let v1; : : : ; v2k be the elements of T listed in order and write each as vi = n i . Let 0 = 0 and 2k+1 = 1 Pick 0 i 2k so that i+1 ? i 2k1+1 . Certainly there is at least 1 2(2 +1) 2
one such i since the i s break the unit interval into 2k + 1 intervals. Now, let a and b be such that vi < 2a and 2b < vi+1 with b ? a as large as possible. If b a +1 then n k < 8; so the lemma is obviously true. We therefore assume b > a + 1: Then 2b?a 41 n k . Let G = G(b; a): By Lemma 2.5 jGj 2b?a?2: List the elements in G in order and label them h1; h2; : : :. We start a T -independent set by setting x1 = hi where i is as small as possible with fhig T -independent. We then assume inductively that Sr = fx1; : : : ; xr g G is a T -independent set and attempt to add another element xr+1 2 G to S by always choosing xr+1 the smallest element in G so that Sr+1 = Sr [ fxr+1 g is T -independent. There are two required conditions for Sr+1 to be T -independent. First, jCT (xr+1 )j = jT j. This condition fails for at most jT j(jT j ? 1) = 2k(2k ? 1) vertices. The next condition is that CT (xr+1 ) \ BT (xi ) = ; for i r: So, suppose that w1; w2 2 T and C (w1; xr+1 ) 2 B(xi ; w2 ): Every element of T is either less than 2a or else it is larger than 2b. This gives four cases based on the values of w1 and w2: It is easily checked that C (xr+1 ; w1) 62 B(xi ; w2 ) for either w1 or w2 less than 2a: So suppose that w1; w2 > 2b: Note that C (w1; x) = w1 ? x + 1 is a strictly decreasing function of x for x < w1: Furb?a thermore, there are at most r (2 k )2 forbidden values for C (w1; vr+1 ): Therefore as long b?a b ? a ? 2 as 2 ? (2k)r(2k)2 ? 2k(2k ? 1) 1 there is at least one vertex vr+1b?2a?G so that Sr+1 = Sr [ fvr+1 g is T -independent. Solving the inequality gives r < 2 b?a ? 1 2 +1
1 2 +1
2
2
2
2k?b1?a (2k)2 2
1 2(2 +1) 2
? (2k) (21 b?a ) = N: Note that N > n 32kk 2
2
(2k)22
2
? 1; which completes the proof.
Proof of Theorem 2.1. Using Lemmas 2.8 and 2.7, we have that the probability that Axiom k fails is no more than
nn ? k ?
Pn = k
k
1
n 2(2k+1) 1 ? m2nk 32k2 ?1 :
Taking logarithms and substituting mn > n? k k + it is easily shown that Pn ! 0. Note that if we use a constant probability bound mn = m then the above estimate of the probability that Axiom k fails is roughly order can k + where c = 1 ? m2k and a = 321k : In the special case of m = 12 (which means each entry in the generating string n has probability 21 of being a 1), the probability bound can be improved to roughly c k by using di erent arguments which are outlined below. Unfortunately, these arguments fail for any m < 12 : It seems that for m near 21 it should be possible to improve the bound given above to make it closer to the better bound in the case where m = 12 . Theorem 2.9 The ratio of the number of labeled Steinhaus graphs of order n which do not satisfy Axiom k to the total number of labeled Steinhaus graphs of order n is no more 1 4 (2 +1)
1 2(2 +1)
2
8 2
5
2 k ?1 n? ? ? n ? k n than k
k
2
22k
k k?1 k
10 2 + 8 2
:
We do not prove Theorem 2.9, but simply give the essential di erence between the proof of Theorem 2.9 and Theorem 2.1. Instead of looking for a large T -independent set, we look for a large set fv1; : : : ; vr g with the property that the sets CT (vi ) are disjoint and jCT (vi )j = jT j. Obviously it is easier to nd large sets with this property than it is to nd large T -independent sets. In fact, it is not dicult to show there are sets of size at least n?10k +k?1 with the stated property. In either proof the values in CT (vi ) are allowed to 8k change in order to estimate a lower bound for the probability that Axiom k is satis ed by vi and T: In Lemma 2.7 it was necessary to assume that the set v1 ; : : : ; vr is T -independent. The di erence in the case m = 21 is that all the combinations of ones and zeros in CT (vi ) are equally likely. This gives independence of the events that Axiom k is satis ed by T and vi for each of the values of i in the set. 2
2
3. Completeness in Vertex Transitive Graphs.
In this section we show that the rst order theories of some families of graphs are complete in the \almost everywhere" sense, although di erent from the rst order theory of random graphs. To begin we say that M is a master graph if M is a graph with a countable number of vertices v1 ; v2; v3 ; : : :. We consider the collection ? of all spanning subgraphs G of M . Let E (M ) denote the edge set of M . We make ? a probability space by letting the probability of any edge in M be p. More precisely, we have a map : 2E(M ) ! ? given by (f ) = G where (vi ; vj ) is an edge in G if and only if f (vi ; vj ) = 1. The measure on ? is then just the image under of the Bernoulli measure on 2E(M ) with parameter p. Here we let M be a vertex transitive graph with a countably in nite vertex set and with every vertex having nite degree. Given vertices v1 ; v2 : : : ; vn 2 M and N 2 N, the N component CNG(v1 ; : : : ; vn ) of (v1 ; v2; : : : ; vn) in G is de ned to be the set of all vertices v 2 G such that (v; vi) N for some 1 i n, where is the usual edge distance in G. We will think of CNG(v1 ; : : : ; vn) as either a set of vertices or else the subgraph induced by the set of vertices. We take (w; z) = 1 if w and z are in di erent components. Next we let CNG(v1 ; : : : ; vn) = SN1 (v1 ; : : : ; vn) [ [ SNm(v1 ; : : : ; vn) where each SNi (v1 ; : : : ; vn) is a connected component of CnG(v1 ; : : : ; vn). Note that each SNi (v1; : : : ; vn) contains at least one of the vertices v1; : : : ; vn; and the diameter of each SNi (v1 ; : : : ; vn) is not more than 2nN for all i. Let M~ be an isomorphic copy of M . Suppose that T1 ; : : : ; Tm is a collection of nite subgraphs of M~ , each containing a xed vertex v~0. Furthermore, suppose for each 1 i m weShave a subset Bi of fx1; x2 ; : : : ; xng and a function 'i : Bi ! V (Ti ) such that (1) i Bi = fx1 ; x2; : : : ; xn g and the Bi are pairwise disjoint. (2) each Ti is connected, and for each vj 2 Ti, ('(Bi); vj ) N . Also, v~0 2 '(Bi). We call the collection T1 ; : : : ; Tm , together with the corresponding functions '1; : : : ; 'm, an (n; N )-con guration Cn;N . Occasionally we shall refer to the vertex 'j (xi ) as the vertex identi ed with xi , and the set of all the vertices identi ed with some xi as the identi ed vertices. 6
The purpose of de ning (n; N )-con gurations is to translate logical formulae into con gurations. Roughly speaking, for any formula (x1 ; : : : ; xn ) there is an N so that for almost all graphs and all choices of v1; : : : ; vn the truth of (v1 ; : : : ; vn) is determined by the induced subgraph on the vertex set CNG (v1 ; : : : ; vn ). A key point is that any possible nite subgraph of M should normally appear in nitely often in a random graph from ?. Since we assume that M has only nite degree, there are only nitely many Cn;N for each n and N . We say that Cn;N = T1 [ [ Tm and C0n;N = T10 [ [ Tm0 0 are isomorphic if m = m0 and for each i there is a graph isomorphism : Ti ! Ti0 such that 'i(xj ) = '0i(xj ) for each j 2 Bi. Given G 2 ? and vertices v1 ; : : : ; vn in G, we say that (v1 ; : : : ; vn ) satis es Cn;N = T1 [ [ Tm if CN (v1; : : : ; vn) = SN1 (v1 ; : : : ; vn) [ [ SNm(v1 ; : : : ; vn) and for all 1 i m there is a graph isomorphism i : SNi (v1 ; : : : ; vm ) ! Ti such that i (vj ) = 'i(xj ). We say that a graph G 2 ? is good if for any nite connected subgraph S of M containing v0, there are in nitely many v 2 V (G) such that CN (v) is isomorphic with S by an isomorphism sending v to v~0 . Clearly the set of good graphs in ? has probability measure 1. We consider nite sets of (n; N )-con gurations, Dn;N = fC1n;N ; : : : ; Ckn;N g for xed n and N . For any G 2 ? and formula (x1; : : : ; xn ) we say is equivalent to Dn;N if rst for every (v1 ; : : : ; vn ), vertices in G, G j= (v1; : : : ; vn) if and only if some Cjn;N 2 Dn;N satis es (v1; : : : ; vn) in G. We further require that the set Dn;N be maximal in the sense that if G j= (v1; : : : ; vn) and Cn;N satis es (v1 ; : : : ; vn ) then Cn;N 2 Dn;N . Note that Dn;N is closed under isomorphism, that is, if Cn;N ; C0n;N 2 Dn;N are isomorphic and Cn;N 2 Dn;N then C0n;N 2 Dn;N : This follows since Dn;N is required to be maximal. Thus, we may view Dn;N as a set of isomorphism classes of (n; N )-con gurations. If N < N 0 , and C0n;N 0 , Cn;N are con gurations we say that Cn;N is the restriction of C0n;N 0 or that C0n;N 0 is an extension of Cn;N if the con guration obtained from C0n;N 0 by taking the induced subgraphs of C0n;N 0 consisting of all vertices within a distance of N of at least one of the identi ed vertices is isomorphic with Cn;N . For a xed set of con gurations 0 0 which Dn;N and N < N 0 we form the N 0 -extension of Dn;N by simply including all Cn;N extend some Cn;N 2 Dn;N . It is easy to see that if (x1 ; : : : ; xn ) is equivalent to Dn;N then (x1 ; : : : ; xn ) is equivalent to any extension of Dn;N . Theorem 3.1 Let M be a vertex transitive graph of nite degree and ? the corresponding family of graphs. For each formula (x1 ; : : : ; xn) there is an N 2 N and Dn;N , a set of isomorphism classes of (n; N )-con gurations, such that for all good G 2 ?, is equivalent to Dn;N for G. Proof. We rst give a procedure to construct Dn;N which we then show to be equivalent to . We inductively build up the con gurations. As a start assume that is atomic. Suppose (x1 ; : : : ; xn ) = xi Axj . We take N = 1 and let Dn;1 be the set of (n; 1)-con gurations Cn;1 for which xi ; xj are identi ed with vertices v~i; v~j which are connected by an edge in Cn;1. It is clear that is equivalent to Dn;1 . For the other atomic case, xi = xj it is clear that we should take N = 0 and Dn;0 to be the set of (n; 0)-con gurations Cn;0 for which the vertices identi ed with xi and xj are equal. 7
Suppose the theorem holds for formulas and
giving equivalent con gurations
Dn;N and D0n;N 0 respectively. It is easy to nd the appropriate sets of con gurations for the boolean combinations of and . In particular, for : we simply take all Cn;N for which Cn;N 62 Dn;N where is equivalent to Dn;N . In the other two cases we simply take
the intersection or the union of the equivalent con gurations after forming the appropriate extensions so the N parameters are equal. Note that in the Boolean case we may increase the value of N for one of the con gurations but it is only by a simple extension. Assume now without loss of generality that (x1 ; : : : ; xn) = 9xn+1 (x1 ; : : : ; xn ; xn+1), and let Dn+1;N = (C1n+1;N ; : : : ; Ckn+1;N ) be equivalent to . We let D~ n;3N be the set of Cn;3N satisfying one of the following: 1. Cn;3N = T1 [ [ Tm and there is a vertex v~ 2 Ti for some0 1 i m such that v~ is within 2N of v~r for some r and for some 1 j 0 k Cjn+1;N is obtained from 0 Cn;3N by identifying xn+1 with v~ and then \restricting to N ". That is, Cjn+1;N = CNCn; N (~v1; : : : ; v~n; v~) where x1; : : : ; xn are identi ed with v~1 ; : : : ; v~n in Cn;3N : 3
0
2. For some 1 j 0 k, Cjn+1;N is such that v~n+1 is the only identi ed vertex in some 0 component of Cjn+1;N and Cn;3N is a 3N extension of the remaining components of 0 Cnj +1;N . We show that D~ n;3N is equivalent to (x1 ; : : : ; xn): Let G 2 ? be good, and let v1; : : : ; vn 2 G. Assume rst that G j= (v1 ; : : : ; vn ), so let vn+1 2 G be such that G j= (v1 ; : : : ; vn; vn+1). Consider C = CNG(v1 ; : : : ; vn; vn+1). First assume that vn+1 is the only vertex among (v1 ; : : : ; vn; vn+1) in the component of C containing vn+1. Let Cn+1;N be an (n; N )-con guration satisfying v1 ; : : : ; vn+1. Say Cn+1;N = T1 [ [ Tm where Tm is the component containing v~n+1 . By induction Cn+1;N 2 Dn+1;N : If Cn;3N is a 3N -con guration realizing (v1 ; : : : ; vn ) in G, then clearly Cn;3N 2 ~ n;3N from Case 2. D Assume next that each component of C contains an element of fv1; : : : ; vng. It follows that vn+1 2 C2GN (v1 ; : : : ; vn): Let Cn;3N again be the 3N -con guration realizing (v1 ; : : : ; vn ) in G. Let v~n+1 be the vertex corresponding to vn+1 under a xed isomorphism of Cn;3N with C3GN (v1 ; : : : ; vn ). Note that the restriction of Cn;3N to the components of CNCn; N (~v1 ; : : : ; v~n; v~n+1) is isomorphic to Cn+1;N , an (n; N )-con guration realizing v1; : : : ; vn; vn+1. By induction, Cn+1;N 2 Dn+1;N : Hence Cn;3N 2 D~ n;3N by virtue of Case 1. Next suppose v1 ; : : : ; vn 2 G are such that some C~ n;3N 2 D~ n;3N realize v1; : : : ; vn. Suppose rst that C~ n;3N 2 D~ n;3N by virtue of Case 1. Fix an isomorphism between C~ n;3N and C3GN (v1 ; : : : ; vn ). So (~vi ) = vi for 1 i n. Let v~n+1 be the vertex in C~ n;3N as in Case 1, and let vn+1 = (~vn+1 ). Thus, if Cn+1;N is the (n +1; N )-con guration obtained from C~ n;3N by restricting C~n;3N to CNCn; N (~v1 ; : : : ; v~n; v~n+1) then Cn+1;N 2 Dn+1;N and so by induction G j= (v1 ; : : : ; vn ; vn+1), and hence G j= (v1 ; : : : ; vn ). Note that we have not yet used the fact that G is good. 3
3
8
Suppose now that C~ n;3N 2 D~ n;3N by virtue of Case 2. Let Cn+1;N 2 Dn+1;N be as in Case 2. Let Cn+1;N = T1[ [Tm and assume Tm is the component of Cn+1;N with v~n+1 the only identi ed vertex. Since G is good, let vn+1 2 G be such that vn+1 2= C2GN (v1 ; : : : ; vn ) and CNG(vn+1) is isomorphic with Tm . However, then Cn+1;N satis es (v1 ; : : : ; vn; vn+1). So, by induction G j= (v1 ; : : : ; vn ; vn+1) and hence G j= (v1; : : : ; vn). If n = 0 in the above argument then is equivalent to > or ? depending on whether Dn+1;N is non-empty or not. Hence Theorem 3.1 has the following consequences Corollary 3.2 Let M be a vertex transitive graph of nite degree and let ? be the corresponding class of random subgraphs. Then for any sentence in the rst order language of graph theory fG 2 ?jG j= g 2 f0; 1g. A closer examination of the argument above allows us to obtain more. The fact that our reduction worked on rst order formulae (not just sentences) yields the following assertion. Corollary 3.3 With M and ? as above, (x1 ; : : : ; xn ) a rst order formula, and v1; : : : ; vn 2 M , let (v1 ; : : : ; vn) = fG 2 ?jG j= (v1; : : : ; vn)g. Then the set of all possible values for , f (v1 ; : : : ; vn)jv1 ; : : : ; vn 2 M g, is nite. Proof. This follows from the theorem noting that (x1 ; : : : ; xn ) is equivalent to some Dn;N for some N 2 N. Thus, on the measure one set of good G, whether G j= (v1 ; : : : ; vn ) depends only on CNG(v1 ; : : : ; vn). Since M has nite degree, the probability measure can be restricted to a nite probability space. Given the vertex transitive master graph M of nite degree, we say a sequence M1 M2 Mi with Mi M is a reduction of M if the following three conditions are satis ed. 1. Each Mi is a nite spanning subgraph of M . 2. Let r (Mi ) = supv2Mi fsupfn 2 Nj(v; v0 ) n implies v0 2 Mi gg: (So r (Mi ) is the radius of the largest ball in Mi .) Then for every c > 0; r (Mi ) > c log(i) for all i suciently large. This is saying that Mi contains a ball of radius greater than c log(i). 3. For some D, l > 0, Mi contains no more than Dil vertices. As an illustration, let Zrepresent the graph whose vertex set is the set of integers and edges consist of consecutive integers. If we let H = Zn = Z Zwe may take Mi to be the induced subgraph on the vertices with all coordinates between ?i and i. We are then considering random subgraphs of the lattice graph on (?i; i) (?i; i). It is easily seen that this forms a reduction of H . Each Mi determines a nite probability space using the Bernoulli measure in the usual manner with probability of edge connection p. We say that almost all nite spanning subgraphs (relative to the reduction) satisfy if limi!1 P (i) = 1, where P (i) is the probability of a graph in Mi satisfying . Theorem 3.4 Let M be a vertex transitive graph of nite degree and let M1 M2 Mi be a reduction for M . Then for any sentence in the rst order theory 9
of graph theory, either almost all graphs (relative to the reduction) satisfy or almost all satisfy : . Proof. Let G Mi . For xed n, N 2 N we say G is n; N good if for all vertices v1; : : : vn 2 G and T a connected subset of CNM (~v0 ) there exists vn+1 2 G with vn+1 62 C2GN (v1 ; : : : ; vn ) and CNG(vn ) is isomorphic to T by an isomorphism sending vn to v~0 . Investigating the proof of Theorem 3.1 shows that it is sucient to show that almost all G (relative to Mi ) are n; N good, for any xed n, N . We say G is N good relative to v1; : : : ; vn if there is a vn+1 2 G as above for the vertices v1 ; : : : ; vn. Let b(v1 ; : : : ; vn) represent the probability that G is not good relative to v1 ; : : : ; vn. For xed ?i and G Mi , the probabilities that G is not n; N good is at most l Di v ;:::;vn b(v1 ; : : : ; vn ) n b(v1 ; : : : ; vn ) for some v1; : : : ; vn. Fix v1 ; : : : ; vn 2 G. Since r (Mi ) > c log(i) (c suciently large) there are at least c log(i) ?n vertices v 2 G at least 2N +1 apart such that C G (v ) G and v 62 C G (v ; : : : ; v ). n N 2N 1 6N The probabilities that CNG (v) is isomorphic to T , for a xed T , are independent for the c log(i) ? n di erent vertices. The possible number of T s is a function only of N , say w(N ). 6N We divide the vertices v above into w(N ) sets of (roughly) equal size, and within each set attempt to nd at least one v with CNG(v) isomorphic to the corresponding T . The probability we do not succeed is at most 1
?c
log(i) n Nw(N ) ? w(N )
w(N )(1 ? q) for some q > 0. Thus, the probability G is n; N good is at least Dil ?c i ? n 1 ? n w(N )(1 ? q) Nw N w N : Since c can be taken large compared to N and l, the above probability approaches one. It is interesting to note that in the in nite and nite cases one has the same reduction procedure of to > or ?. In particular, the in nite and the nite cases have the same rst order theory as long as the three conditions are met in the nite case. An example of this is the following assertion. Corollary 3.5 For any rst order formula in the language of graph theory, almost all spanning subgraphs of Zn satisfy if and only if almost all spanning subgraphs of (?N; N )n satisfy . Finally we wish to point out that the fact that the master graph has nite degrees is crucial to our argument. It would be interesting to see a proof of Corollary 3.2 which would include the case of in nite degree master graphs. Of course, this proof would then include Corollary 3.2 as well as the fact that the almost all theory of graphs is complete (in this case the master graph is simply the complete graph on the integers). 6
6
4. Higher order axioms.
log( ) ( )
(
)
We consider now some higher order properties of graphs which generalize the rst order axioms given in Section 1. The axioms of Section 1 imply the existence of \local" 10
structures. For example, for any xed graph almost all graphs have a subgraph isomorphic to the xed graph. The axioms of this section imply the existence of \global" structures, such as hamiltonian cycles in almost all graphs. The global structures are constructed according to a \pattern", which is made precise in the de nition of a constructing family. We prove two theorems in this section. Theorem 4.1 asserts the existence of large subgraphs and furthermore allows negative edge requirements. In Theorem 4.4 we only assert the existence of large subgraphs but with improved estimates. A constructing family of functions = fFn(k;l)g of type (k; l) is a collection of funtions F) n(k;l) for each n 2 N such that ( k;l 1) Fn has domain tuples (S0; <S ; : : : ; Sj ; <Sj ; (x1 ; : : : ; xk ); (y1 : : : ; yl )) where Si f1; : : : ; ng with jSij < n for each i, <Si is a linear ordering of a subset of Si for each i, x1 ; : : : ; xk 2 Sj are distinct, and y1 ; : : : ; yl 2 f1; 2; : : : ; ng ? Sj are distinct, 2) Fn(k;l)(S0; <S ; : : : ; Sj ; <Sj ; (x1 ; : : : ; xk ); (y1 ; : : : ; yl )) = (Sj+1; <Sj ), where Sj+1 = Sj [ fy1; : : : ; yl g, and <Sj is a linear ordering of a subset of Sj+1. The intended meaning of a constructing family is to give a procedure for constructing a large subgraph according to a speci c pattern. For example, to construct a hamiltonian cycle we start with a small cycle and add one vertex at each step. The vertex is chosen to be adjacent with consecutive vertices around the cycle, increasing the length of the cycle by one. This corresponds to a construction family of type (2; 1). In particular, Si is the set of vertices in the cycle in step i while <Si is the order around the cycle starting at any xed vertex. The set Si+1 is the new cycle with the added vertex and <Si is again the order around the newly formed cycle. Note in this case Fn(k;l)(S0; <S ; : : : ; Sj ; <Sj ; (x1 ; x2 ); (y1 )) depends only on (Sj ; <Sj ; (x1 ; x2 ); (y1 )). We say that the functions fn : N ! R are bounding functions for fFn(k;l)g if, with the above notation, j eld(<Sj )j fn (jSj j). Here eld refers to the set of elements x where either x <Sj y or else y <Sj x for some y. The graph in Figure 2a can also be constructed with a (2; 1) family of functions. In this graph we have approximately n1=16 paths of length n1=16 which we will refer to as the cycles. We require that the end vertices of each path are not connected (the dotted line in the gure), they are connected as shown, and have attached paths of lengths approximately n7=8 as shown. In the rst n3=16 steps we build up a subgraph consisting of all the cycles and attached paths of length n1=16. For j in this range we take the eld of <Sj to consist of only two vertices. For j n3=16 the constructing function adds at each step a single vertex to one of the paths. At each step the path chosen is the next one clockwise from the previous one. The eld of Sj+1 is simply the next path to be added to. The order <Sj is the order along this path. For j > n3=16 the size of the eld of Sj is at least jSj j1=3 . We consider the following axiom scheme for a graph G with n vertices. The axioms say that the construction procedure corresponding to Fnk;l can be carried out in G. Axiom A(k;l): Let d be a graph with vertex set fa1 : : : ; ak ; b1 ; : : : blg and let din be a graph with vertex set fc1;: : : ; cr g where k r k + l. There exists a sequence (S0 ; <S ), : : : , (Sm ; <Sm ) with m = n?l r , where Si is a subset of the vertices of G and 1) the subgraph of G induced from the vertices S0 is isomorphic with din. Also, the ordering <S on S0 corresponds to an ordering c1 < c2 < < cr under the isomor0
0
+1
+1
+1
0
+1
0
0
11
phism; 2) for every 0 j m?1, Fn(k;l)(S0; <S ; : : : ; Sj ; <Sj ; (x1 ; : : : ; xk ); (y1 : : : ; yl )) = (Sj+1; <Sj ) for some elements x1 ; : : : ; xk 2 Sj , y1; : : : ; yl 2= Sj , and where x1 ; : : : ; xk are consecutive elements of Sj with respect to <Sj ; also, xi Ays in G if and only if ai Abs in d and yi Ays ?in G if andonly if bi Abs in d; 3) if e = n ? r + l n?l r > 0, then there are y1 ; : : : ; ye , vertices in G which are not in Sm, and consecutive elements (using order <Sm ) x1; : : : ; xk such that xi Ays in G if and only if ai Abs in d and yi Ays in G if and only if bi Abs in d. If the vertices x1 , : : : , xk , y1, : : : , yw (for w = l or w = e) satisfy Property 2 or Property 3 above then we say that x1 ; : : : ; xk ; y1 ; : : : ; yw t d. Note that Axiom (k; l) has Fn(k;l) as a parameter in its statement. Given a constructing family fFn(k;l)g, it makes sense to ask if a graph G satis es Axiom (k; l). We will show below that for fFn(k;l)g having suitable bounding functions f (n), for each k; l and d; din Axiom (k; l) holds for \almost all graphs". We interpret almost all as in Bollobas [4], where for each n we let p = pn be a number between 0 and 1. We then make the set of all labeled graphs with n vertices into a probability space by requiring the probability of each possible edge to be p, independent of the other edges. Then we say almost all graphs have any given property P if and only if Pr(G 2 P ) ! 1 as n ! 1. Before verifying the axioms, we remark how they generalize the axioms in Section 1. The axioms of Section 1 implied the existence of \small" structures in G. The Axioms (k; l) imply G possesses \large" (i.e. of size n) structures. As indicated above, Axiom (2; 1) easily implies hamiltonicity of G. Figure 1 illustrates some other structures G must contain given G satis es certain of these axioms (in each case the structure contains all the vertices of G). In Figure 1a Fn(2;1)(S0; <S ; : : : ; Si; <Si ; (x1 ; x2 ); y1 ) adds y1 to the top branch of the graph if i 0 (mod 6), to the middle branch if i 1; 2 (mod 6) and to the bottom branch if i 3; 4; 5 (mod 6), each time connecting y to consecutive elements x1; x2 2 Si . (Here we take d to be a path of length two with end vertices a1 and a2 and middle vertex b1 ). The reader can easily discern the relevant constructing functions for the other structures. (Note in Figure 1c and 1d k = 2 and l = 3.) Before stating the theorem, we give a few technical conditions which are required in the theorem. Let 0 < < 1 be xed numbers and de ne w = minf 101 ( ? l l ); (4k) ( ? ); 10 (3 + ); 4(k) (3 + ); 2(1r) g > 0 if l > 1 and w = minf 101 ( ? ); 10 (3 + ); 2(1r) g > 0 if l = 1: Next we let 0 < pn < 1 be a sequence of numbers and de ne qn = 1 ? pn. Finally, let mn = minfpn; qn g. Theorem 4.1 Let fFn(k;l)g be a constructing family with bounding functions fn satisfying fn (m) > m for all m n , and let pn be the probability of an edge in a labeled graph with n vertices. Suppose that there is an > 0 such that if n is suciently large then ? l(wl ) + if l > 1 and mn > n?w+ if l = 1. Then almost all graphs satisfy mn > n Axiom (k; l). 0
0
2
2
2
2
2
Before giving the proof we point out the following two lemmas which will be used in 12
+1
the proof of Theorem 4.1. Let Zn represent the integers modulo n. Lemma 4.2 Let V = Zl Zn. For each 0 i nl and 0 j < n, let Ti;j = f(r; j + ri)j0 r < lg V , where the second coordinate is interpreted modulo n. Then for any (i; j ) 6= (k; t) with 0 i; k nl and 0 j; t < n, jTi;j \ Tk;t j < 2. The proof of Lemma 4.2 is straightforward. We call fTi;j ji = i0g the i0 sweep. Note that in each sweep each pair in V occurs exactly once. Next we need a technical lemma which allows us to partition the vertex set of almost all graphs into the the basic building blocks needed in Theorem 4.1. Given a \probabilistic" construction based on a random graph, we say that an edge fv; wg is used if the estimate of the probability that the construction succeeds depends on whether or not fv; wg is an edge in the graph. Otherwise the edge is said to be unused. So, if an edge is unused, the success of the construction is independent of whether or not the graph p has the given edge. p In the proofs below we will ignore the di erence between terms such as x and [ x] which do not change the calculations in any signi cant manner. Lemma 4.3 Suppose p?n, wthe+probability of any given edge in a labeled graph with n l vertices, satis es mn > n l( ) if l > 1 and mn > n?w+ if l = 1 for some > 0 and all n suciently large. Let d be a graph with l vertices and din be a graph with r k + l vertices. Then almost all graphs can be partitioned into one set of size r which induces a n ? subgraph isomorphic with din, l r of size l each inducing a subgraph isomorphic with d, and one with the residual vertices inducing a subgraph isomorphic with the induced graph on the rst vertices of d. Furthermore, with this construction given any vertex v, there are w + at most O(n ) used edges involving v. Proof. Let n 2 N and consider the probability space of all labeled graphs on n vertices. pn We identify the vertex set of the graphs with V = f 1 ; 2 ; : : : ; n g . Let V be the last 0 p vertices of V . We partition the set V0 into rn sets of consecutive integers each of size r (with perhaps a few residual vertices). The probability that the induced subgraphr onpat n least one of these subsets is isomorphic with din is bounded below by 1 ? (1 ? m(n ) ) r . By taking logarithms and using the fact that w 2(1r) , it follows that this probability approaches one as n approaches in nity. Note that in this construction each vertex has at most only r used edges and if fi; j g is used then both i and j are in V0. Assuming there is such a subgraph we call it Din. The next step is to partition the rest of the vertices into subsets of size l so the induced graph on each subset is isomorphic with d. If l = 1 then this step is trivial since d consists of just a vertex. So we assume here that l > 1. To this end, we partition the vertices of V1 = V ? V0 into two subsets, one consisting of the rst bjV1j vertices of V , which we call p V2, and the other of size (1 ? b)(n ? n), which we call V3 , where b = n?w . We further partition V3 into two subsets, V4 with b2 n = lu vertices and V5 with jV3j ? jV4j vertices. w 1+ We follow the following algorithm until V5 = ; or for n steps, whichever comes rst. We index the vertices of V4 by Zl Zu. We list the sets of Lemma 4.2 in the order T0;0; T0;1; : : : ; T0;u?1; : : : ; T ul ;0; : : : ; T ul ;u?1. Starting with the rst set S = T0;0 we check if the subgraph induced by the vertices with indices in S is isomorphic with d (A success is 2
5 2
2
2
2
13
when the induced subgraph is isomorphic with d and a failure is when it is not isomorphic with d). In the case of a success, we add the vertices indexed by S to the set W (which will contain sets of size l each inducing a subgraph isomorphic to d), remove the next l vertices from V5 and index them by the indices in S . We then proceed to the next step which is to set S to the next set in the list of Ti;j and continue as before. Note that the number of Ti;j s is O(n2(1?2w)) > n1+ w since w < 101 ( ? ) < 92 . l w l At each step the probability of a success is bounded below by m(n ) n? +( ). Since at each step only unused edges are considered, the probability of exhausting V5 before w w 1+ 1+ n steps is bounded below byl the probability that in n independent Bernoulli trials with probability of success m(n ) one attains at least O(n) successes. It is clear that this probability approaches 1 as n approaches in nity using, for example Chebychev's inequality by the ordering of the sets Ti;j it is clear that for Pr(jX ? j > a) < a1 . Furthermore, w w n each vertex v at most u = O(n ) edges involving v are used, and each of the used edges has both vertices in V3. At this point in the algorithm have one subgraph, Din, which is isomorphic to din, pn)?we (1 ? b )( n ? b n and W which contains subgraphs each isomorphic with d. The rest of the l vertices are in a subset, V6 V4 [ (V0 ? Din) of size approximately b2 n and V2 of size approximately bn. First note that no edge between V6 and V2 is used. We next nd jV6j subsets of size l from V2 [ V6 by partitioning the set V2 into approximately lbn ?1 subsets of size l ? 1 each and attempting to match up each vertex in V6 with a subset of size l ? 1 from V2 such that the induced subgraph from these vertices is isomorphic to d. For each vertex in V6 we examine (l?11)b subsets of size l ? 1 from V2 to check if one of the induced subgraphs is isomorphic with d.l The probability that we succeed for each vertex in V2 is bounded below by (1 ? (1 ? m(n ) ) b l? )b n. By taking logarithms it is easy to check this approaches 1 as n approaches in nity. Next, the remaining vertices in V2 need to be partitioned into subsets of size l so each induced subgraph is isomorphic to d. We do this by rst partitioning the remaining vertices of V2 into subsets of size l (with perhaps one of the subsets having size less than l). We then index the subsets in the partition of W with Zl?1 Zn0 where n0 = O(n). We de ne Ti;j as in Lemma 4.2 and order the Ti;j s as before. The subsets in the partition of V2 are ordered arbitrarily. Note that Ti;j consists of l ? 1 subgraphs each isomorphic to d: We call an index set, Ti;j , bad if there is a used edge between two of the subgraphs indexed by Ti;j , otherwise the index set is good. We call sweep i0 acceptable if no more that half of the index sets Ti ;j in the sweep are bad. Note that there are at most O(n w ) sweeps which are not acceptable since the total number of used edges is O(n1+ w ). Therefore there are at least O(n) > jV2 j acceptable sweeps. To each subset Y in the partition of V2 we associate an acceptable sweep. For each of the rst nw good sets S in that sweep we attempt to repartition the vertices of S and Y into l subsets of size l so that each subset induces a subgraph isomorphic with d: Speci cally we arbitrarily number the vertices in each set of S from 1 to l and number the vertices in Y from 1 to l and then check if the vertices having the same number form a graph isomorphic 2
2
2
2
2
2
1+ 2
5 2
2
2
(
1
1)
2
5 2
0
5 2
14
2
2
with d. that the probability that we succeed for every Y is bounded below It is easy to lverify n n l where n1 = O(nw ) and n2 = O(n1?w ). It is easily checked by 1 ? 1 ? mn( ) that this probability bound approaches 1. Clearly for each vertex v at most O(nw ) edges containing v become used in this process. Now we are ready to prove Theorem 4.1. The proof is much like the proof of Lemma 4.3, in that the same ideas of partitioning, looking for edges, and estimating probabilities of success are used. Proof. For a xed constructing family of functions fFn(k;l)g with bounding functions fn, and and as above, we x d and din and consider the probability that a graph G on n vertices satis es Axiom A(k;l). To begin, let G be a graph with n vertices. We assume that n jdinj (mod jdj) in order to avoid inessential detail relating to how to deal with the special case at the end where fewer than jdj vertices are left over. We rst use Lemma 4.3 to partition the vertex set of G into one subset S0, inducing a subgraph of G isomorphic with din, and the rest into subsets of size l, each inducing a subgraph of G isomorphic with the subgraph of d induced by the vertices b1 ; : : : ; bl . Let s = n?jldinj and h1; : : : ; hs be the sets in the partition de ned above. We partition the set fh1; : : : ; hsg into two subsets, A and B with jAj = s and jBj = s ? s . We further partition A into j = s + sets, each of size s( ? ), which we call T1 ; T2; : : : ; Tj : We now attempt to form S1; S2; : : : ; Sj as in the statement of A(k;l). Given the sets S1; S2 : : : ; Si, we look for h 2 Ti+1 such that if (~c1; : : : ; c~k ) are the <Si rst k vertices in Si (with order <Si given inductively from Fn(k;l)), then the subgraph of G induced from w fc~1; : : : ; c~k g[ h ts d. Since the construction of Lemma 4.3 used at most O(n ) edges for each vertex, there are at least s ? ?O(n w ) choices of sets in Ti+1 which involve unused edges. It follows that the probability that the rst s + steps can be accomplished is at least 2
1
2
+ 2
3 4
1 4
1 4
+ 2
1 4
5 2
1 4
1 4
5 2
3 4
? P1 = 1 ? 1 ? mkln s
1 4
1 4
+ ? 41 ?O(n 52w ) s 4 4 3
1
We let S denote the vertices and <S denote the order produced to this point. Next we let the sets in B be denoted U1; U2; : : : ; Us?s and succesively adjoin the l vertices Ui+1 = f~b1 ; : : : ; ~blg to k consecutive vertices c1; : : : ; ck in S [ (U1 [ [ Ui ), (with order given inductively by Fnl;k ) so that the subgraph of G induced by fc1; : : : ; ck g [ Ui+1 ts d. An easy computation gives the following lower bound for the probability of success for every Ui as + 2
0 ls B ? kl P2 = B @1 ? 1 ? mn
(3 4
+1 4
k
15
)
1s?s ?lO(n w ) C C : A + 2
5 2
Last, we denote the subsets of size l in A ? S by V1 , V2, : : : , and at step i + 1 attempt to join Vi+1 to k consecutive vertices in S [ B [ (V1 [ [ Vi) as required. Now at each the domain of the linear ordering step there are at least (l(s ? s )) ? ls vertices in ( k;l ) given by Fn which lie in B. Since there are only ls vertices in A, at least + 2
+ 2
+ 2
(l(s ? s )) ? ls 1 s ? l ?1 2 ls (for large enough n) vertices in B must be consecutive in that ordering. This easily gives the following lower bound for the probability of success: + 2
+ 2
2
+ 2
P3 = 1 ?
!
s l ?1 s ?2 ?lO(n 52w ) 2k kl 1?m
?
n
+ 2
Since only unused edges were considered in the estimates of probabilities P1, P2 , and P3, we get P1P2 P3 as a lower bound for the probability that the graph satis es A(k;l) given that Lemma 4.3 is satis ed for xed d and din. Let P4 be the minimum probability that Lemma 4.3 is satis ed for a graph with n vertices over all d and din with parameters r; l; k. r) kl ( l ) ( As we range over all possible choices for d and din we get (1 ? 2 2 2 (1 ? P1P2 P3P4)) as a lower bound for the probability that G satis es axiom A(k;l) for the given Fn(k;l). It is enough to show Pi ! 1 as n ! 1 for i = 1; 2; 3. This follows easily in all ?three cases by l ? ? taking logarithms. We point out that the bounds w < 10 and w < 4k 2 are needed l for estimating P1, w < 10 (3 + ) and w < 4(k) (3 + ) are needed for estimating P2 l and w < ?5 (which follows from w < 101 ( ? )) and w < (4k) ( ? ) are needed for estimating P3. We consider now a variation A~(k;l) of the axiom scheme A(k;l). Here we do not allow negative edge requirements. Axiom A~ (k;l): Let d be a graph with vertex set fa1; : : : ; ak ; b1 ; : : : ; bl g and let din be a graph with vertex set fc1; : : : ; crg where k r k + l. Then there is a sequence (S0; <S ); : : : ; (Sm ; <Sm ) with m = n?l r , where Si is a subset of the vertices of G and 1) The subgraph of G induced from the vertices S0 is isomorphic with d0in where d0in has the same vertex set as din and contains all the edges of din, but may contain more edges. Also, the ordering <S on S0 corresponds to the ordering c1 < c2 < < cr under the isomorphism. 2) Fn(k;l)(S0; <S ; : : : ; Sj ; <Sj ; (x1 ; : : : ; xk ); (y1 ; : : : ; yl )) = (Sj+1 ; <Sj ) for some elements x1; : : : ; xk 2 Sj , y1; : : : ; yl 2= Sj , and where x1 ; : : : ; xk are consecutive elements of S with respect? to 0, then there are y1 ; : : : ; ye , vertices in G which are not in Sm, and consecutive elements (using order <Sm ) x1; : : : ; xk such that xi Ays in G if ai Abs and yi Ayj in G if bi Abj in d. We have the analog of Theorem 4.1. 2
2
2
2
0
0
0
+1
16
Theorem 4.4 Let fFn(k;l)g be a constructing family with bounding function fn for Fn(k;l). Suppose that for all c > 0, for all suciently large n, fn (m) > c log m for all n < m n
for some < 1. If pn, the probability of any given edge on a labeled graph with n vertices is bounded below by (logLn) kl for some constant L, then the probability that a graph on n vertices satis es A~(k;l) tends to 1 as n ! 1. Proof. We x a suitable large c > 0 and only consider n such that fn(m) > c log m for 1+
0 n < m n. Let = 2 . We exploit a device of [4]. Instead of considering graphs with probability of an edge pn, we consider three graphs with the same vertex set and color the edges in one graph red, one graph blue, and the other graph green. We denote the probability of a red edge by pr , the probability of a blue edge by pb and the probability of a green edge pg : We then de ne the graph G to have the same vertex set as the three graphs and let fv; wg be an edge of G if and only if fv; wg is either a blue, red or green edge. Note that the probability that fv; wg is an edge in G is 1 ? (1 ? pr )(1 ? pb)(1 ? pg ) p and the edges are independent. By choosing pr = pb = pg = 1 ? 1 ? pn we get pn for the probability of each edge in G. (Note that for small pn, the common value of pr = pb = pg is approximately 13 pn.) We rst apply Lemma 4.3 using the green graph. (Note that the probability bound in Lemma 4.3 is satis ed.) Let P0 denote the probability that the green graph satis es Lemma 4.3. We will call each of the0 subgraphs isomorphic to d a block. We next partition the rst n blocks of G into n disjoint sets, each of size n 0? , say T0; T1 ; : : :. We start with din and successively nd a block in Ti+1 that can be added to S0 [ S1 [ [ Si according to the function F (k;l) using only blue edges. A computation as before yields the following lower bound for the probability of success: 1
3
? n 0? n kl P1 = 1 ? 1 ? p b :
So far approximately n blocks have been included. Next, we attempt to succesively add the remaining blocks to consecutive vertices already included using only red edges. This gives the following lower bound for the probability of success:
? c P2 = 1 ? 1 ? pklr
n )
log(
k
nl
:
It is easy to check that P0 , P1 and P2 all approach one as n approaches in nity. We leave it to the reader to check that if we replace c log n in the statement of Theorem 4.4 with the bound for fn (m) in Theorem 4.1 then one can improve the estimate for pn in Theorem 4.4 to the requirement for pn in Theorem 4.1. Suppose we x 0 < p < 1 and let pn = p: It is interesting to note that Theorem 4.1 implies that almost all graphs contain the structure a) in Figure 2 (using = 31 , = 163 and a suitable fFn(k;l)g as described earlier), while Theorem 4.4 implies almost all graphs contain the structure b) (which has only positive requirements). Neither theorem implies almost all graphs contain the graph of c). We do not know if this is the case. 17
The probability estimates are by no means best possible for any given spanning subgraph. As an example, Hamiltonian cycles exist with probability approaching one in n+!(n) where ! (n) ! 1 as n ! 1 (see for example random graphs with pn log n+log log n [4]). The bound from Theorem 4.1 is n =1 for any > 0, which is much larger than log n+log log n+!(n) . In fact, Theorem 4.1 only asserts the existence of spanning subgraphs n when the probability bound is nw for some ?1 < w < 0. However, it is easy to check for pn < n =1 the probability approaches zero that one can partition a random graph on n vertices into n3 3-cycles. Of course the existence of such a partition follows from Theorem 4.1 with pn > n 1= . Although there is a big gap between 32 and 901 , this example does show that a general theorem asserting the existence of both a Hamiltonian cycle and a partition of the vertices into triangles cannot do better than nw for some ?1 < w < 0. We wish to thank the referee for useful comments and suggestions for improvements to some of our results. 1 10+
2 3+
1 90
References [1] C. K. Bailey and W. M. Dymacek, Regular Steinhaus graphs, Congressus Numerantium 66(1988),45{47. [2] P. Billingsley, Probability and Measure, John Wiley and Sons, New York, 1979. [3] A. Blass and F. Harary, Properties of almost all graphs and complexes, J. Graph Theory 3(1976) 225{240. [4] B. Bollobas, Random Graphs, Academic Press, London, 1985. [5] N. Brand, Almost all Steinhaus graphs have diameter two, Journal of Graph Theory 16(1992) 213{219. [6] N. Brand, S. Curran, S. Das, T. Jacob, Probability of diameter two for Steinhaus graphs, Discrete Applied Math 41(1993) 165{171. [7] R. C. Brigham, N. Deo and R. D. Dutton, Some properties of Steinhaus graphs, preprint. [8] R. C. Brigham and R. D. Dutton, Distances and diameters in Steinhaus graphs, Congressus Numerantium 76(1990) 7{14. [9] W. M. Dymacek, Bipartite Steinhaus graphs, Discrete Math. 59(1986) 9{20. [10] W. M. Dymacek, Small cycles in Steinhaus graphs, Congressus Numerantium 70(1990) 41{52. [11] R. Fagin, Probabilities on nite models, J. Symbolic Logic 41(1976) 50{58. [12] H. Harborth, Solution to Steinhaus' problem with plus and minus signs, J. Combinatorial Theory 12(A)(1972) 253{259. [13] D. Knuth, The Art of Computer Programming. Volume 1: Fundamentals of Algorithms, Addison-Wesley, Reading, Mass. (1973). [14] H. Steinhaus, One Hundred Problems in Elementary Mathematics, Elinsford, New York, 1963. 18
~ n/3
~ n 7/8
~ n/6 ~ n/2
(a) (b) ~ n 1/8
n/3 triangles
n/3 triangles
(c)
(d)
Figure 1.
Some structures of size n almost all graphs of size n contain according to Theorem 4.1. A solid line indicates an edge, a broken line indicates no edge, and neither indicates there is no requirement either for an edge or for no edge. 19
1/16
~n
7/8
~n
1/16
~n
(a)
1/8
~n
3/4
~n 1/8
~n
(b)
1/8
~n
3/4
~n ~n 1/8
(c)
Figure 2. As in Figure 1, a solid line indicates an edge, a broken line indicates there is no edge, and no line indicates there is no requirement concerning an edge. The graph in a) follows from Theorem 4.1, the graph in b) follows from Theorem 4.4, while the graph in c) does not follow from either theorem. 20
