Thresholds for random distributions on graph sequences with applications to pebbling

Let G = {G1; G2; : : : ; Gn; : : :} be a sequence of graphs with Gn having n vertices and having a random distribution of t(n) pebbles to its vertices. If s¿ 2 is an integer, the event that Gn has s or more vertices with two or more pebbles has threshold t(n) = ( √ n). If t(n) = c √ n, then the limiting distribution for the number of vertices with multiple pebbles is Poisson(c). The threshold for the event that Gn has at least one vertex with s or more pebbles is t(n) = (n(s−1)=s). These results are used to establish new bounds for thresholds for pebbling on sequences of graphs with bounded diameters. If for some d, diameter (Gn)6d for all n, and if for some p ∈ (0; 1], maximum degree (Gn) ⊆ (n), then the threshold th(G) for the solvability of G is in O(n1−0:5p). c © 2002 Elsevier Science B.V. All rights reserved. Let Gn be a connected graph with n vertices and a distribution of t pebbles on the vertices of Gn. A pebbling step consists of removing two pebbles from a vertex v and placing one pebble on an adjacent vertex to v. For a given vertex r, we say the distribution is r-solvable if a pebble may be placed on r by a sequence of pebbling steps. In this case, we say that it is possible to “pebble” to r. The distribution is solvable if it is r-solvable for every vertex r in Gn. A central question is for a given Gn and t, what proportion of all distributions of t pebbles to Gn are solvable? Certain natural families of graphs are best described as sequences of graphs indexed by the number of vertices. The family of complete graphs Kn is a good example. Let G= {G1; G2; : : : ; Gn; : : :} be a sequence of graphs with Gn having n vertices and let E-mail address: [email protected] (J.A. Boyle). 0012-365X/02/$ see front matter c © 2002 Elsevier Science B.V. All rights reserved. PII: S 0012 -365X(02)00445 -4 60 J.A. Boyle /Discrete Mathematics 259 (2002) 59–69 t= t(n) be a natural number. Let PG(n; t) be the proportion of all distributions of t pebbles onto Gn that are solvable. If we think of the distribution as selected at random from all possible distributions of t pebbles on Gn then PG(n; t) is the probability that Gn is solvable. Here, the random selection is made uniformly with all distinguishable distributions having the same probability (the pebbles are indistinguishable, but of course the vertices are distinguishable). For a given graph sequence, we are interested in Fnding a threshold function so that the graphs in the sequence will become solvable with high probability if the number of pebbles is essentially greater than the threshold function. In other words, if the number of pebbles t(n) grows more quickly than the threshold function then the probability Gn is solvable tends to one, and if the number of pebbles grows more slowly than the threshold function the probability Gn is solvable tends to zero. The following deFnitions and notations are taken from [3]. We write f g or g f if lim n→∞ f(n) g(n) = 0: DeFne o(g)= {f |f g} and !(g)= {f | g f}: O(g)= {f | ∃c¿0; & k¿0 such that f(n)=g(n)¡c for all n¿k}; (g)= {f | ∃c¿0; & k¿0 such that g(n)=f(n)¡c for all n¿k}; (g)=O(g) ∩ (g): Then f∈ o(g)⇔ g∈!(f) and o(g)⊂O(g) and !(g)⊂ (g). Let G= {G1; G2; : : : ; Gn; : : :} be a sequence of graphs. We say a function f is a threshold for G and write f∈ th(G) if limn→∞ PG(n; t)= 1 whenever t f and limn→∞ PG(n; t)= 0 whenever t f. More generally, let E be some pebble distribution property and let PG(n; t;E) be the probability that Gn with a random distribution of t pebbles possesses the property E. Then we say the function f is a threshold for property E and write f∈ th(G;E) if limn→∞ PG(n; t;E)=1 whenever t f and limn→∞ PG(n; t;E)=0 whenever t f. For a survey of known results for graph sequences see [3,4]. In [1] it is established that threshold functions always exits for every graph sequence. It is known that the threshold of any sequence belongs to (n1=2) ∩ o(n1+ ). A few graph sequences are known to have the threshold (n1=2). The following theorem and corollary appear in [2]. Theorem 1 and its proof are included here because it is the building block for all of the subsequent results in the paper. Theorem 1. Let G= {G1; G2; : : : ; Gn; : : :} be a graph sequence and let E0 be the event that no vertex of Gn has two or more pebbles. Then limn→∞ PG(n; c √ n;E0)= e−c 2 . The event L E0 that at least one vertex of Gn has two or more pebbles has th(G; L E0)= ( √ n). J.A. Boyle /Discrete Mathematics 259 (2002) 59–69 61 Proof. The total number of distributions of t pebbles to Gn is N = ( n+ t − 1 t ) : The total number of distributions of t pebbles to Gn with no vertex receiving two or more pebbles is N0 = ( n t ). The probability that Gn has no vertices with two or more pebbles is P = PG(n; t;E0)= N0 N = n! t!(n− t)! ÷ (n+ t − 1)! t!(n− 1)! = n · · · (n− t + 1) (n− 1 + t) · · · n = ( n n− 1 + t ) · · · ( n− t + 1 n ) : Notice this last product of fractions is written from largest factor to the smallest. Therefore, ( n− t + 1 n )t ¡P¡ ( n n− 1 + t )t : Letting the number of pebbles be t= c √ n we have ( n− c√n+ 1 n )c√n ¡P¡ ( n n− 1 + c√n )c√n : To simplify these bounds let x= √ n. ( x2 − cx + 1 x2 )cx ¡P¡ ( x2 x2 − 1 + cx )cx or (( 1− c x + 1 x2 )x)c ¡P¡ (( 1− c x +O ( 1 x2 ))x)c : Letting n→∞, we have x→∞ and by an application of l’Hopital’s Rule we get that both of these bounds tend to e−c 2 . Now let L E0 be the event that Gn has at least one vertex with two or more pebbles. Then as n→∞, P( L E0)→ 1− e−c2 . Notice that limc→0 (1 − e−c2 ) = 0 and limc→∞ (1 − e−c2 ) = 1. If the number of pebbles is t √ n then for any c¿0, t¡c √ n for suOciently large n. In this case P( L E0)→ 0. On the other hand, if the number of pebbles is t √ n then for any c¿0; t¿c √ n for suOciently large n. In this case P( L E0)→ 1. This shows that th(G; L E0)= ( √ n). Let K= {K1; K2; : : : ; Kn; : : :} be the sequence of complete graphs. If a vertex v of Kn receives two pebbles then Kn is solvable in one pebbling step since every vertex is adjacent to the vertex v. On the other hand, if t(n)¡n and no vertex of Kn receives two pebbles then Kn is unsolvable. 62 J.A. Boyle /Discrete Mathematics 259 (2002) 59–69 Corollary 2. th(K)= ( √ n). Theorem 3. Let G= {G1; G2; : : : ; Gn; : : :} be a graph sequence and let E be the event that at least s vertices of Gn receive two or more pebbles. Event E has threshold th(G;E)= ( √ n). Proof. Let Ei be the event that exactly i vertices of Gn receive two pebbles and the rest of the vertices have zero or one pebble. Let F be the event that at least one vertex of Gn has three or more pebbles. Then