JIGSAW PERCOLATION ON ERDÖS-RÉNYI RANDOM GRAPHS

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1 JIGSAW PERCOLATION ON ERDÖS-RÉNYI RANDOM GRAPHS ERIK SLIVKEN Abstract. We extend the jigsaw percolation model to analyze graphs where both underlying people and puzzle graphs are Erdös-Rényi random graphs. Let p ppl and p puz denote the probability that an edge exists in the respective people and puzzle graphs and define p eff = p ppl p puz, the effective probability. We show for constants c 1 > 1 and c 2 > π 2 /6 and c 3 < e 5 if minp ppl, p puz) > c 1 log n/n the critical effective probability p c eff, satisfies c 3 < p c effn log n < c Introduction This paper is an extension of the work done by Brummitt, Chatterjee, Dey, and Sivakoff in [3]. They define a new kind of percolation on finite graphs called jigsaw percolation. It is a model for collaborative networks working together on a problem, where each node represents a person with a unique part to play in solving the problem. It is a process on a graph with two distinct edge sets G = V, E puzzle, E people ). The edges in E people denote acquaintances. The edges in E puzzle denote connections that need to be made in order for the problem to be solved. The process begins with everyone disjoint. Initially if two people share a people and a puzzle edge they merge together to form a cluster. For each subsequent step, clusters merge together if there exists at least one pair of nodes between the clusters connected by a people edge and at least one pair of nodes connected by a puzzle edge. If the process ends with all nodes in a single cluster we say the people graph solves the puzzle graph. This process is an extension of the more widely studied bootstrap percolation from [4], [7], though with a more complicated evolution process. Many of the techniques and even some of the constants π 2 /6) that appear in these papers are used in the study of jigsaw percolation. In this paper we will consider the case where V, E people ) and V, E puzzle ) are Erdös- Rényi random graphs with probabilities p ppl and p puz respectively. The roles of the puzzle graph and people graph are symmetric so without loss of generality we may assume p puz p ppl. For E people to solve E puzzle we need that both V, E people ) and V, E puzzle ) to be connected. Hence for simplicity we will assume for some c 1 > 1, p puz c 1 log n/n, guaranteeing connectivity with high probability for both graphs [5]. Studying random puzzle graphs is akin to asking what type of puzzles can people graphs solve. Social networks are usually modeled with power law graphs [1]. In [3] they show that power law graphs cannot solve a bounded degree puzzle graph with high probability. Networks of people innovate regularly, so it is natural to explore what types of puzzles can be solved by social networks. We hope that this work is a useful step in closing the gap between theory and reality. 1

2 2 ERIK SLIVKEN Definition 1.1. Jigsaw Percolation We repeat the definition of jigsaw percolation in [3]. Jigsaw percolation proceeds on V, E people, E puzzle ) as follows. At every step i 0, we have a partition, C i, of V into disjoint subsets, where if A C i then the nodes of A will have merged by step i. At step i, let E i denote the unordered pairs of clusters in C i that are both people- and puzzle-adjacent. We get C i+1 by merging the connected components of C i, E i ). C 0 = {{v} : v V } C 1 = {A : A is a connected component of V, E puzzle E people )} C i+1 = { A UA : U is a connected component of C i, E i )}. We are particularly interested in the event Solve := {C = {V }}, when the people graph solves the puzzle graph. We will show that in the limit, Solve will only depend on the product p eff = p ppl p puz. We hope to find good bounds for the critical effective probability p c eff, where P p ppl,p puz Solve) = 1/2 if p ppl p puz = p c eff. Theorem 1.1. For any positive constants c 1 > 1, c 2 > π 2 /6, and c 3 < e 5, a) If p eff c 3 n log n, then P p ppl,p puz Solve) 0. b) If p puz c 1 log n/n and p eff c 2 n log n, then P p ppl,p puz Solve) 1. Combined we get the upper and lower bounds c 2 p c eff n log n < c 3. We ll proceed by first proving the lower bound for p c eff. 2. Lower Bound for p c eff Let S i C i denote the largest cluster after i steps. We assume that p eff = e 5 /n log n. For a cluster of vertices, A V, to have merged together at some point in the jigsaw percolation process, the induced graphs A, E people ) and A, E puzzle ) must both be connected. If both subgraphs are connected then both subgraphs must contain a spanning tree. We denote the existence of two such spanning trees on A by the event DualSpan A. Furthermore, we denote with DualSpan k the existence of such a A where A = k. The following few lemmas will show that Solve 2 log n j=log n DualSpan j { S 1 log n} and the probability of the event on the right tends to 0 as n increases. Proof of Theorem 1.1 a), p c eff e 5 n log n. Lemma 2.1. If k s.t. S 1 < k < n/2 and for no j [k, 2k] does DualSpan j occur, then Solve does not occur.

3 JIGSAW PERCOLATION ON ERDÖS-RÉNYI RANDOM GRAPHS 3 Proof. If Solve occurs, then at some point clusters of small size must merge together to form one giant cluster. Although many clusters may merge together in one step, we can break down each step into a series of substeps where exactly two clusters merge. The upper bound for the largest cluster at most doubles in size during each of these substeps. Eventually two clusters of size less than k must merge to form a cluster, A, with A [k, 2k]. Therefore if this does not happen, Solve cannot occur. Lemma 2.2. If p < e 5 /n log n), P S 1 < log n) 1. Proof. Two vertices merge after one step if they are connected by both a people edge and a puzzle edge, so the largest cluster after the first step can be understood by looking at Erdös-Rényi random graphs with edge probability p = p eff. Let T k denote the number of trees of size k in a Erdös-Rényi random graph with such an edge probability. From [2], the expectation of T k is given by ET k = n k ) k k 2 p eff k 1 1 p eff ) kn k)+k 2) k+1 ) n k k 2 p k 1 eff. k Plugging in k = log n in the above expression shows that ET k 0 as n. As T k is the sum of positive random variables, with high probability we have that P S 1 < log n) 1. Lemma 2.3. Let A = k, then and PDualSpan A ) k 2k 4 p eff k 1, PDualSpan k ) ) n k 2k 4 p k 1 eff. k Proof. We use expectation to bound the probabilities in question. Fix a subset of vertices A of size k. Let T ppl and T puz denote the number of spanning trees in A, E people ) and A, E puzzle ) respectively. The event DualSpan A is the intersection of the two independent events {T ppl 1} {T puz 1}. As there are k k 2 possible spanning trees on A we have PT ppl 1) ET ppl = k k 2 p k 1 ppl and PT puz 1) ET puz = k k 2 p k 1 puz. Since the events are independent we may take the product to get PDualSpan A ) = PT ppl 1)PT puz 1) k 2k 4 p eff k 1 as desired. Since there are n k) choices for A the union bound gives ) n PDualSpan k ) k 2k 4 p k 1 eff, k

4 4 ERIK SLIVKEN the expected number of subsets A of size k such that DualSpan A occurs. Lemma 2.4. PSolve) P S 1 log n) + 2 log n k=log n PDualSpan k). Proof. By Lemma 2.1, the event DualSpan k occurs for some k log n, 2 log n). Using the union bound with Lemmas 2.2 and 2.3 gives: PSolve) 2 log n k=log n log n i=0 PDualSpan k ) + P S 1 log n) ne i+log n i + log n) i+log n log ne 5i 5 log n 2πi + log n)log n) i+log n) i + log n) + o1) 4 ne2 log n 2) 2 log n e 2π log n e ) log n 0. 5 log n By Lemma 2.2 the largest cluster in C 1 is bounded by log n with high probability, so PSolve) 0. The constant e 5 is in no way optimal. In Gravner and Sivakoff [6] they show for puzzle graphs with max degree D, if p ppl e 3 /D log n then PSolve) 0. We can show that 1 2 p puz < D < 2np puz with high probability. 3. Upper Bound for p c eff The proof of the upper bound for p c eff consists of two main steps. First we show for some β > 0, there exists an initial vertex which grows to a cluster of size at least n β. Then we show that this cluster of vertices grows large enough that it eventually merges with the remaining vertices We show for any c 1 > 1, and any c 2 > π 2 /6, if and p puz c 1 log n n p puz p ppl c 2 n log n, 2) then Solve occurs a.a.s.. The inequality in 1) guarantees the underlying puzzle- and people-graphs are both connected a.a.s.. We may assume equality in 2) as the event Solve is increasing in both p puz and p ppl. In general we let where 1 < f n < c2 n c 1 log n. p puz = c 1f n log n n and p ppl = c 2 c 1 f n log 2 n 1)

5 JIGSAW PERCOLATION ON ERDÖS-RÉNYI RANDOM GRAPHS 5 Let W denote a subset of V. A sequence of vertices {v i } k i=0 W will grow into a cluster of size k if the following hold for 0 i < k: v i, v i+1 ) E puzzle, v i, v i+1 ) E people for some 0 i i. We say such a sequence is good and let BigSeed β W ) denote the event that a good sequence of size k = n β exists in W. Lemma 3.1. For any c 1 > 1 and any c 2 > π 2 /6, there exists positive β and ɛ both sufficiently small such that if p eff = c 2 /n log n, p puz c 1 log n/n and W 1 ɛ)n, BigSeed β W ) occurs a.a.s.. Proof. We define a process that creates a sequence of sequences {S j } j=0 n1 ν where ν is a small positive constant to be determined later. For each j, S j will consist of a sequence of vertices {s j i }nβ i=0. For j n1 ν we recursively define the following sets for 0 < i < n β and 0 j < n 1 ν. U j i = U j i 1 N j i 1. N j i = {w W \U j i sj i, w) E puzzle}, the puzzle-neighborhood of s j i in W \U j i. N j i N j i is a subset of size at most np puz = c 1 f n log n chosen uniformly from the vertices of N j i. S j i = {sj k }i k=0, the subsequence of the eventual Sj M j i = {v N j i such that for s S j i, s, v ) E people }. E j i = Ej i 1 {s, w) E people, s S j k, w W } {sj i, w) E puzzle, w W }. For i > 0 the set U j i denotes the used vertices in the bootstrapping process i.e. vertices with an incident people-edge exposed). The set E j i denotes the used vertices. During the creation of a sequence we must set aside each exposed neighborhood, N j i as each has a people edge with a previous s j l exposed for some l < i. However at the start of a new sequence we can reset U j 0 to only add the sequence Sj 1 to U j 1 0 since S j is disjoint from S j 1. We have U0 0 = and U j 0 = U j 1 0 Sj 1 for j > 0. We also let E0 0 denote all the edges in both E people and E puzzle that are incident to s 0 0 and w W. For j > 0 we let E j 0 = Ej 1 S j 1 1 {sj 0, w) E people, s j 0, w) E puzzle}. The introduction of the set N j i allows us to control the size of U j i. At most n 1 2 new vertices are added to U j i at each increase of i. A sequence is at most n β in length so at most n β vertices are added by an increase in j. There are at most n 1 ν sequences created so U j 0 < n1 ν+β, and U j i < n1 ν+β + n 1 2 +β. The constant β will be chosen once ν is determined so that satisfy β < ν/2. We grow the sequence in the following manner. First choose a vertex, s j 0, uniformly from W \U j 0. For 0 < i < nβ if M j i 1 is non-empty we choose a sj i from it uniformly and create the above sets and proceed. If M j i 1 is empty, we let sj i = sj i 1 for all remaining i < n β and choose s j+1 0 uniformly from W \U j+1 0.

6 6 ERIK SLIVKEN Whether or not M j i By the Chernoff bounds for 0 < α < 1, If α = ) N j i < 1 α)µj i E j i < j is empty will depend on the size of N i. Let [ µ j i = E j N i ] E j i > 1 2ɛ)c 1 f n log n. e α 1 α) 1 α) ) µ j i < exp α1 2ɛ)c1 log n). 3) ɛ 1 2ɛ, we may restate 3) in the following useful form N j i < 1 3ɛ)c 1f n log n ) E j i < n ɛc 1. 4) Given a choice of ɛ > 0 we may then choose β > 0 so that n ɛc 1 < n 3β. By construction S j i has i + 1 vertices. Each of those vertices has exactly N j i possible j people connections to the vertices of N i. Each of those people connections exist with probability p ppl independent of everything that has been revealed thus far E j i ). Hence the probability that M j i is empty is at most ) M j i = Ej i 1 p ppl ) i+1) N j i. 5) Let q j i denote the probability both N j i 1 3ɛ)c 1f n log n and M j i is nonempty. By 4) and 5) we have or ) q j i 1 c 2 n 3β ) 1 1 c 1 f n log 2 n )i+1)1 3ɛ)c 1f n log n q j i 1 n 3β) 1 exp i + 1)1 3ɛ)c )) 2. log n The conditional probability S j has n β unique vertices is given by n β 1 i=0 nβ q j i 1 1 n 3β ) nβ i=0 1 exp i + 1)1 3ɛ)c 2 / log n)) 6) n β 1 o1)) exp log 1 exp i + 1)1 3ɛ)c 2 / log n)) 7) i=0 log n 1 o1)) exp 1 3ɛ)c 2 0 log 1 e x) ) dx 1 o1))n π ɛ)c 2 9) Since c 2 > π 2 /6 we may choose both ɛ and ν small enough so that 9) is at least n 1+2ν. Let β < minν/2, ɛc 1 /6). 8)

7 JIGSAW PERCOLATION ON ERDÖS-RÉNYI RANDOM GRAPHS 7 The probability that all n 1 ν sequences each do not have n β unique elements is bounded above by 1 n 1+2ν ) n1 ν 1 2 exp nν ). 10) Therefore we may conclude BigSeed β W ) occurs a.a.s.. Now that we have a seed of size n β we will show this creates a large cluster in the remaining ɛn vertices of V \W. Then there is a simple argument for to show this larger cluster will eventually merge with all remaining vertices. Lemma 3.2. Let p puz c 1 log n/n and p eff > c 2 /n log n for c 1 > 1 and c 2 > π 2 /6, and ɛ > 0, depending on c 1 and c 2 and β > 0 depending on c 1, c 2 and ɛ. Let V = W {V \W } be a disjoint partition of vertices with W = 1 ɛ)n. If there exists a cluster of merged vertices in W of size n β, then they eventually merge with everything in V. Proof. Split V into two disjoint sets W and V \ W where W = 1 ɛ)n. Let S denote a good sequence in W of size n β. Let S 0 = S 0 = S, and R 0 = V \ W. For i > 0 define recursively the following sets: S i+1 := {v R i there exists u, w S i with u, v) E people, w, v) E puzzle }, the vertices that have both a people and puzzle edge connecting them to S i. S i+1 := a subset of S i+1 of size at most n 1.5iβ. R i+1 := {v V \ W ) \ i+1 S j=1 j }, the vertices in V \ W not yet merged with the big cluster. We let Epeople i and Ei puzzle denote the subsets of E people and E puzzle with one endpoint in S i and one in R i. The important thing to note is that Epeople i and Ei puzzle are disjoint from j<i E j people and j<ie j puzzle respectively. Denote the events T i := i j=1 { S j n 1.5jβ }, so we may prove Lemma 3.2 by showing: PSolve T 1 ) = 1 o1). 11) For each v R i, the vertices of S i will come from R i and will be determined by edges from E i puzzle and Ei people. Let u, u S i and v R i. We have u, v) Epuzzle T i ) = ppuz, and similarly, u, v) E people T i ) = pppl. The probability that v connects to S i hence v S i+1 ) is given by ) ) Pv S i+1 T i ) 1 1 p puz ) S i 1 1 p ppl ) S i. 12) This probability is the same for all vertices of R i. Moreover if S i = op 1 ppl ) then the Inequality 12) can be simplified to P ) v S i+1 1 T i 4 S i 2 p puz p ppl n21.5i)β c 2 4n log n 13)

8 8 ERIK SLIVKEN The size of S i+1 conditioned on T i stochastically dominates a binomial random variable X i+1 Binɛn on), n iβ 1 ). Therefore if T i occurs, the expected value of S i+1 is at least the expected value EX i+1 n i )β. Chernoff bounds then give ) Ti+1 T i PXi+1 n 1.5i+1β ) 1 exp 12 ) nβ 14) which for finite i gives ) Ti+1 T0 1 exp 1 )) i+1 2 nβ = 1 o1) 15) As long as n 1.5iβ p ppl = o1), we can grown the size of S i with high probability. Let τ = inf{i : n 1.5iβ p ppl = o1)}. We have a little wiggle room with β, and can choose it so that τ satisfies n 1.5τ+1β p ppl = n γ for some small γ > 0 Since p ppl > n log n) 1/2 then τ < log1/2)/ log1.5β) and is finite. We denote the large cluster C τ = {S k } τ+1 k=0. The size of C τ will be within a constant multiple of n 1.5τ+1β = on). By the definition of τ the probability a vertex v V \ C τ is not people-connected to C τ is given by 1 p ppl ) Cτ < 1 p ppl ) n1.5τ+1β < exp n γ ). The probability that every vertex in V \ C τ is people-connected to C τ is bounded below by 1 n exp n γ ). Since the puzzle graph is connected we can find a path from a vertex in C τ to a vertex in V \ C τ. Given that every vertex is either in C τ or connected to C τ by a people edge, we can merge each vertex along the path until every vertex in V \ C τ has been added to the large cluster. Combining Lemma 3.1 with Lemma 3.2 proves the second half of Theorem 1.1. Acknowledgments The author would like to thank Christopher Hoffman, Janko Gravner, and David Sivakoff for all their useful suggestions along the way. References [1] A.-L. Barabâsi and R. Albert, Emergence of scaling in random networks, Science, ), pp [2] B. Bollobás, Random Graphs 2nd Ed., Cambridge University Press, [3] C. D. Brummitt, S. Chatterjee, P. S. Dey, and D. Sivakoff, Jigsaw percolation: What social networks can collaboratively solve a puzzle, arxiv, 2012). [4] J. Chalupa, P. L. Leath, and G. R. Reich, Bootstrap percolation on a Bethe lattice, J. Phys. C, ), pp. L31 L35. [5] P. Erdős and A. Rényi, On the strength of connectedness of a random graph, Acta Mathematica Hungarica, ), p [6] J. Gravner and D. Sivakoff, Nucleation scaling in jigsaw percolation, arxiv, 2013). [7] A. E. Holroyd, Sharp metastability threshold for two-dimensional bootstrap percolation, Probability Theory Related Fields, ), pp