Bootstrap percolation on products of cycles and complete graphs

Bootstrap percolation on a graph iteratively enlarges a set of occupied sites by adjoining points with at least $\theta$ occupied neighbors. The initially occupied set is random, given by a uniform product measure, and we say that spanning occurs if every point eventually becomes occupied. The main question concerns the critical probability, that is, the minimal initial density that makes spanning likely. The graphs we consider are products of cycles of $m$ points and complete graphs of $n$ points. The major part of the paper focuses on the case when two factors are complete graphs and one factor is a cycle. We identify the asymptotic behavior of the critical probability and show that, when $\theta$ is odd, there are two qualitatively distinct phases: the transition from low to high probability of spanning as the initial density increases is sharp or gradual, depending on the size of $m$.


Introduction
Given a graph G = (V, E), bootstrap percolation with threshold θ is a discrete-time growth process that, starting from an initial configuration ω ∈ {0, 1} V , generates an increasing sequence of configurations ω = ω 0 , ω 1 , . . .. Given ω j , j ≥ 0, ω j+1 is given by and ω ∞ is the pointwise limit of ω j as j → ∞. A vertex, v, is called occupied at time j if ω j (v) = 1, and is called empty otherwise; a set of vertices S ⊆ V is called Bootstrap percolation on Z m × K 2 n Figure 1: Cartesian product of Z 5 and K 4 .
graphs under study could be viewed as limiting cases when the Holroyd-Liggett-Romik model [HLR] is combined with anisotropic graphs studied recently in [DE]. These graphs are also of great interest for the study of epidemics due to their community structure, see [LZ] for analysis of an epidemic model on the infinite graphs Z d × K n for d = 2, 3.
For another bootstrap percolation model on graphs that combine local and long-range connections, see [TV].
We are able to prove a general result for V = Z d m × K n , which turns out to be mostly an application of bootstrap percolation results on lattices [Hol1,BBM,BBDM]. Denote by log (k) the kth iterate of log and let λ(d, θ) be the bootstrap percolation scaling constant for the lattice Z d m defined in [BBDM].
Theorem 1.1. Assume bootstrap percolation on V = Z d m × K n for any d ≥ 1 and θ ≥ 2. Then, for θ ≤ d, In all cases, the transition is sharp.
The above theorem, whose proof is relegated to Section 8, demonstrates that there is no transition in behaviors for different γ, therefore this case is not of primary interest and we include it mainly for comparison. Indeed, we will see that the situation is very different when the complete graph K n is replaced by the two-dimensional Hamming torus K 2 n . In this paper we focus on Z m × K 2 n , as the simplest case that exhibits sharp, gradual and hybrid phase transitions, depending on the relative scaling for m and n. We will address the more demanding case Z d1 m × K 2 n in a subsequent paper, although some of the aforementioned phenomena appear to be limited to the case d 1 = 1. Higherdimensional Hamming tori K d2 n , d 2 > 2, are much more complex [GHPS, Sli]. We now state our main results.
Therefore, m ≈ n 1/ marks the boundary between sharp and gradual transition in case of odd-threshold θ, while there is no such boundary when the threshold is even. In fact, the odd threshold case has, when γ > 1/ , another sharp transition. Henceforth, we make the customary identification of the configuration ω ∞ with the occupied set {v : ω ∞ (v) = 1}. Theorem 1.3. Assume bootstrap percolation on Z m × K 2 n , and suppose that θ = 2 + 1 and γ > 1/ . Assume that p ∼ a · (log n) 1/ n 1+1/ . (1.5) In other words, when the scaling constant a of (1.5) is smaller than ( !/ ) 1/ , then a very small proportion of sites becomes occupied during the bootstrap process. When a is larger than ( !/ ) 1/ but smaller than ( 1 2 (γ + 1/ ) !) 1/ , then most, but not all, of the sites become occupied. Finally, when a > ( 1 2 (γ + 1/ ) !) 1/ , then all sites become occupied.
Finally, we give the promised example with mixed phase transition, for which we need to assume that m satisfies a particular boundary scaling with γ = 1/ .
Theorem 1.4. Assume bootstrap percolation on Z m × K 2 n , and suppose that θ = 2 + 1, (1.6) and that p satisfies (1.5). Then (1.7) The proofs of Theorems 1.2-1.4 are completed in Section 7, after the auxiliary results are established in Sections 2-7. In particular, we use a result on the birthday problem in a "slightly supercritical" regime, which is proved in Section 2.

Two simple lemmas
Lemma 2.1. In a sequence of k independent Bernoulli random variables, which are 1 with a small probability r, P (no two consecutive 1s) = exp(−kr 2 + O(kr 3 + r 2 )).
Proof. Let a k be the probability of no two consecutive 1s in a sequence of length k. Then a k satisfies the recursion a k = r(1 − r)a k−2 + (1 − r)a k−1 for k ≥ 2, and a 0 = a 1 = 1. Let S = √ 1 + 2r − 3r 2 . Then the solution of the above recursion is , and the result follows by Taylor expansion.
We will also use the following well-known Chernoff bounds.

Birthday problem
In this self-contained section, we use m and n as is customary in the classic birthday problem, therefore these variables do not have the same meaning as in the rest of the paper.
The k-coincidence birthday problem asks, "What is the probability that, among m people with birthdays chosen independently and uniformly at random from [n] = {1, . . . , n}, there exists a set of k people that have the same birthday?" Let A = A n,m,k be the event that such a k-coincidence exists. An equivalent formulation of the birthday problem, relevant to our setting, is as follows. Given an m-by-n binary array, chosen uniformly at random from arrays that have a single 1 on every row, what is the probability that there exists a column with at least k 1's? This formulation appears in the proof of Lemma 3.1, where the following lemma is applied. Lemma 2.3. Assume k is fixed, n is large, and m depends on n in such a way that m k+1 n k . Then It is not difficult to show, using Poisson approximation, that (2.1) holds when m k /n k−1 approaches a constant [AGG]. However, we need the formula when m is larger by a multiplicative power of log n, in which case the standard upper bound for the error in Poisson approximation [BHJ] is too large. Instead, we use the following asymptotic expansion result. For a function f analytic in a neighborhood of 0, we denote by f [m] the coefficient of z m in its power series expansion.
Theorem 2.4 (Theorem 4 in [Gar]). Let f (z) = ∞ i=0 a i z i , and suppose that a 0 > 0, a 1 > 0, a i ≥ 0 for i ≥ 2, and the series has positive radius of convergence. Suppose m, n → ∞ such that m = o(n). Define ρ > 0 to be the unique positive solution of (1 + o(1)).
Proof of Lemma 2.3. We will first give a formula for P (A) using embedding into a Poisson process.
Suppose {ξ i s } n i=1 is a collection of n i.i.d. Poisson Processes with rate 1/n, so ξ i s is the number of people with birthday i at time s, where people arrive at rate 1. Therefore, letting e k (z) = (2.2) as m, n → ∞.
In order to apply Theorem 2.4, we need to estimate ρ and f (ρ) when f = e k . We observe that From this, after factoring out e ρ from both f (ρ) and f (ρ) and a short computation Next we observe and therefore

Single copies of Hamming planes
In the sequel, we will refer to a copy of K 2 n , by which we mean a (deterministic) subgraph {i} × K 2 n , for some i ∈ Z m . Its four subsquares are obtained by division of , and a vertical line is a subset of the form {(i, x, y) : y ∈ [n]} for a fixed x ∈ [n]. We use the following terminology to describe the properties of the initial configuration ω 0 within a copy of K 2 n or, in one case, within a copy and its neighboring copies. For a fixed k ≥ 1, call a copy of K 2 n : • k-viable if it contains a horizontal or a vertical line with at least k initially occupied sites; • k-internally spanned (k-IS ) if the bootstrap dynamics with threshold k (rather than θ) restricted to it spans it; • k-internally inert (k-II ) if the bootstrap dynamics with threshold k restricted to it does not change the initial configuration; • k-inert if no site in it becomes occupied at time 1 in the bootstrap dynamics with threshold k (perhaps with help of neighboring planes); and • k-proper if within each of the four subsquares there are at least θ horizontal and θ vertical lines, such that each line contains at least k initially occupied sites.
In this section, we will assume that p is of the form for some a > 0, where = (θ − 1)/2 . We will now briefly explain why this is the critical scaling for spanning, by sketching a simplified argument that does not establish a critical constant a but illustrates some of our arguments. When θ − 1 is odd, the probability that a copy of K 2 n is not -viable and the probability that it is not (θ − 1)-IS are both about exp(−2n(np) / !). When θ − 1 is even, the second probability goes up to about exp(−n(np) / !). In either case, if a is large enough, every copy of K 2 n is (θ − 1)-IS, and all we need for spanning is a single θ-IS copy of K 2 n , which will appear if γ is large. On the other hand, take any pair of consecutive Hamming planes {i, i + 1} × K 2 n , and form a new configuration on a Hamming plane K 2 n in which x ∈ K 2 n is initially occupied if either (i, x) or (i + 1, x) is initially occupied. If such configuration does not span K 2 n with threshold θ − 1, which certainly happens if such configuration is not -viable, then the original configuration on Z m × K 2 n never adds an occupied point on {i, i + 1} × K 2 n even if all other points eventually become occupied. Existence of such a "blocking pair" is guaranteed if a is small enough.
The precise sufficient and necessary conditions for spanning, which yield the correct critical constant in (3.1), are given in Lemmas 4.1 and 5.1. To estimate the probabilities that these conditions are met, we need to carefully estimate the probabilities that a single Hamming plane contains various internal initial configurations.
Throughout the rest of the paper, we make use of the standard notation O(·) and o(·) to describe asymptotic properties of quantities that depend on n → ∞, with implied constants that may depend on θ and γ.
Lemma 3.1. If p is given by (3.1) and ≥ 2, then Proof. The binomial probability that a fixed line contains at least initially occupied sites is so the probability of the event H that at least one of n horizontal lines contains at least initially occupied sites satisfies as n +2 p +1 is n −1/ times a power of log n. Conditioned on H c , the configurations on horizontal lines are independent and the conditional probability that any one fixed horizontal line contains exactly one initially occupied site is bounded below by To connect to the birthday problem, focus on these horizontal lines with a single initially occupied site. The location of the initially occupied site is uniform on each of these lines, a "birthday." We are looking for a vertical line with initially occupied sites, which will happen if of these "birthdays" coincide. Thus, by Lemma 2.3 P p (no vertical line contains at least initially occupied sites | H c ) (3.4) By the FKG inequality, we also have which, together with (3.3) and (3.4), finishes the proof.
Lemma 3.2. Assume that p is given by (3.1) and ≥ 1. Then Proof. Let G be the event that K 2 n is ( − 1)-proper, and G 1 the event that the top left (say) square contains θ horizontal lines containing at least − 1 initially occupied points. As P p (G c ) ≤ 8P (G c 1 ), we in fact need to get the upper bound in (3.5) for P p (G c 1 ). Arguing as for (3.2), the expected number of horizontal lines in the top left square containing at Bootstrap percolation on Z m × K 2 n least − 1 initially occupied points is Ω(n(np) −1 ). Taking = 1/2 in Lemma 2.2, we get a constant c > 0 so that which is clearly enough for (3.5). To prove (3.6), we use analogous definitions of G and G 1 , and then (3.7) is replaced with which establishes (3.6).
Lemma 3.3. If p is given by (3.1) and ≥ 2, then n is -viable}, and then Lemma 3.1 and (3.5) finish the proof.
Lemma 3.4. If p is given by (3.1) and ≥ 2, then Proof. Let H (resp., V ) be the event that at least one horizontal (resp., vertical) line contains at least initially occupied sites. Yet again, let G be the event that K 2 n is ( − 1)-proper. Further, let F be the event that there exists a line with at least + 1 initially occupied sites. By the same reasoning as for (3.2), the probability that a fixed line contains at least + 1 initially occupied sites is O((np) +1 ), so and, by FKG inequality and (3.5), (3.10) n for any constant L > 0. To find an appropriate lower bound for P p (H • V ), write H as a disjoint union H = ∪ n k=1 H k , where H k is the event that, counted from the top, the kth horizontal line is the first to contain at least initially occupied sites. Let V k be the event that there exists a vertical line that contains at least initially occupied sites outside of the kth horizontal line. Then Given H k , the horizontal lines are independent, the first k − 1 have configurations conditioned on containing at most − 1 initially occupied sites, and those after the kth horizontal line have the independent Bernoulli configuration. An identical argument as in equation (3.4), with the same values of q n and , shows that where the o(1) is uniform over all k. Then, from (3.11), which, together with (3.10), provides the matching bound to (3.9) and ends the proof.
Then, adapting the proof of Lemma 3.3 in [GHPS], the occurrence of the event (H ∩ V ) \ (H • V ) implies that there exists an initially occupied site with 2 additional initially occupied sites in its Hamming neighborhood, so for all ≥ 1. It then follows from FKG and BK inequalities that Therefore, . On the other hand, let G be the event that K 2 n is -proper. By (3.6), P (G) → 1. Then which establishes the desired asymptotics.
Lemma 3.7. If p is given by (3.1) and ≥ 1, then the probability that any fixed copy of K 2 n contains a site that has an initially occupied Z-neighbor, and at least θ − 1 initially occupied K-neighbors, is Proof. The probability in question is O(n 2 p(np) θ−1 ), which implies the O bounds. Then Lemmas 3.4 and 3.5 finish the proof.
Lemma 3.8. If p is given by (3.1) and ≥ 1, then then the probability that any fixed copy of K 2 n has a site that that has an initially occupied Z-neighbor, and at least θ − 2 initially occupied K-neighbors, is Proof. The probability in question is O(1/(np)) times the one in previous lemma.

Spanning: sufficient condition
Lemma 4.1. Assume the following three conditions are satisfied: (1) Every copy of K 2 n is (θ − 2)-IS.
(2) Between any pair (if any) of copies of K 2 n that are not (θ − 1)-IS there is a copy of K 2 n that is θ-IS. (3) There is at least one θ-IS copy of K 2 n . Then Z m × K 2 n is spanned.
Proof. Assume that the graph is not spanned. Then, by (3), there is a contiguous interval, {i, i + 1, . . . , j} × K 2 n , such that no Hamming plane in the interval is fully occupied in the final configuration, ω ∞ . By (1), every such interval contains at least two Hamming planes. The boundary planes of this interval, {i} × K 2 n and {j} × K 2 n , are not (θ − 1)-IS. By (2), a plane in the interval must be θ-IS, and thus is fully occupied in the final configuration, a contradiction.
n Call a Hamming plane exceptional if it is either θ-IS or not (θ − 1)-IS. Let K be the number of exceptional planes. Each of these planes is not (θ − 1)-IS with probability p bad /(p good + p bad ) ∼ p bad /p good . Moreover, by Lemma 2.2, P p 1 2 p good m ≤ K ≤ 2p good m → 1.
Proof. Observe that condition (3) now holds a. a. s. as soon as m → ∞, due to Lemma 3.4.
The rest of the proof is similar to the one for the previous lemma.
Proof. Let γ ∈ (1/ , γ) be such that a / ! > (γ + 1/ )/2, and let m = n γ . Divide the cycle, Z m , into m/m intervals of length m (leaving out any leftover interval of smaller length), and identify each interval with a subgraph of Z m × (K n ) 2 in the obvious way. Denote these subgraphs by R 1 , . . . , R m/m . As in the proof of Lemma 4.2, call a Hamming plane exceptional if it is either θ-IS or not (θ − 1)-IS.
Observe that a subgraph, R i , is internally spanned if it satisfies conditions (1), (2) and (3) of Lemma 4.1, in addition to (4) The first and last exceptional Hamming planes in R i are θ-IS.
If G 4 is the event that condition (4) is satisfied, then Lemmas 3.4 and 3.5 imply that P p (G 4 ) → 1. Now, it follows from the same proof as Lemma 4.2 that P p (R i is internally spanned) → 1.
Let N be the random number of subgraphs among R 1 , . . . , R m/m that are internally spanned, and observe that |ω ∞ | ≥ m n 2 N . Fix > 0. We have which proves the claim.
Lemma 5.1. Assume that Z m × K 2 n is spanned. Then both of the following conditions hold: (1) There is no blocking interval.
(2) There is at least one copy of K 2 n that is not θ-inert. n Proof. The necessity of (2) is trivial. Necessity of (1) is also easy, because if (1) fails and [i 1 , i 2 ] is a blocking interval, no point in [i 1 , i 2 ] × K 2 n gets added even if ([i 1 , i 2 ] × K 2 n ) c is completely occupied.
Proof. If γ < 1/ , then (2) fails with probability tending to 1 as n → ∞ by Lemmas 3.5 and 3.7. For the remainder of the proof, we assume γ > 1/ and a / ! < (γ + 1/ )/2. This time call a Hamming plane K 2 n exceptional if it is either not θ-II or it is (θ − 1)-II, and let p good be the probability that K 2 n is not θ-II, and p bad the probability that K 2 n is (θ − 1)-II. Embed the random configuration on Z m ×K 2 n into a random configuration on Z + ×K 2 n .
For any i ≥ 0, let ξ i be the random configuration on [0, i] × K 2 n . Let I 1 and I 2 be the smallest (random) indices of two consecutive exceptional planes that are (θ − 1)-II. Fix i 1 ≥ 0 and k ≥ 1. Then, conditioned on ξ i1−1 , the event {I 1 = i 1 , I 2 − I 1 ≤ k} is a decreasing function of the configuration on [i 1 , ∞) × K 2 n .
The key to the argument that follows is the event G i1,i1+k , that there is no neighboringplane interference in [i 1 , i 1 + k] × K 2 n . To be more precise, this is the event that no vertex in {i 1 } × K 2 n has an initially occupied neighbor in {i 1 + 1} × K 2 n together with at least θ − 2 initially occupied neighbors in {i 1 }×K 2 n ; and that no vertex in {j}×K 2 n , i 1 +1 ≤ j ≤ i 1 +k, has an initially occupied neighbor in {j − 1} × K 2 n together with at least θ − 2 initially occupied neighbors in {j} × K 2 n . Note that G i1,i1+k is also decreasing, and independent of the configuration on ξ i1−1 . Observe also that We will denote by ξ a generic realization of ξ i1−1 , and let k = log n/(p good + p bad ). Conditioned on ξ i1−1 = ξ and I 1 = i 1 , I 2 − i 1 is a geometric random variable with success probability p good + p bad , and therefore P p (I 2 − i 1 > k | ξ i1−1 = ξ, I 1 = i 1 ) ≤ e −k(pgood+pbad) = n −1 . (5.1) We proceed by a similar argument as in the proof of Lemma 4.2. Given γ > 1/ and a / ! < (γ + 1/ )/2, if T is the number of exceptional planes in [0, m/2] × K 2 n , then by Lemma 2.2, P p T ≥ 1 4 (p good + p bad )m → 1.
This completes the proof of the lemma.
Proof. The argument is similar to that of the previous lemma, and is somewhat simpler, so we omit the details.
Proof. Choose constants α, β such that a / ! < α < β < 1/ , and let m = n β . As in the proof of Lemma 4.4, divide the cycle Z m into m/m intervals of length m , and denote the resulting subgraphs of Z m × K 2 n by R 1 , . . . , R m/m . Call a site in R i Z-assisted if it has an initially occupied Z-neighbor in R i and at least 2 − 1 initially occupied K-neighbors Call a subgraph R i almost inert if the following conditions are met.
(1) All Hamming planes in R i are θ-II.
(2) No sites in R i are Z-assisted.
(3) There is a (θ − 1)-II plane among the first n α Hamming planes and among the last n α Hamming planes in R i .
If R i is almost inert, then it contains an interval of Hamming planes of length at least m − 2n α = m (1 − o(1)), in which the initial configuration remains unchanged by the bootstrap dynamics, even if every site outside of the interval becomes occupied. The probability that R i satisfies condition (1) converges to 1 by Lemma 3.5, since β < 1/ . The probability that R i does not satisfy condition (2) is, by Lemma 3.8, O(n β−1 (log n) 2 ) = o(1). Finally, the probability that R i satisfies condition (3) tends to 1 by Lemma 3.4. So we have P p (R i is almost inert) → 1.
Let N be the number of subgraphs among R 1 , . . . , R m/m that are not almost inert, and observe that |ω ∞ | ≤ m n 2 (N + 1) + 2n α (m/m )n 2 + |ω 0 | , n and that m n 2 = o(mn 2 ), n α (m/m )n 2 = o(mn 2 ). Fix > 0. Then, for large enough n, where the first term goes to zero because each subgraph R 1 , . . . , R m/m is independently almost inert, and the second term goes to zero because p → 0. This completes the proof of the lemma.

Scaling in the gradual regime
Lemma 5.2 shows that the magnitude of p of the form (3.1) is too small when γ < 1/ and θ = 2 + 1. In this case, we need to scale p so that a θ-IS plane has a chance to appear, so we let p = a · 1 n 1+1/( +1) m 1/( +1) , where a > 0 is a constant.
Lemma 6.2. Assume that θ = 2 + 1 and γ < 1/ , and that p is of the form (6.1). Then As the size of p given by (6.1) is much larger than that given by (3.1) we can, by monotonicity in p, take the scaling constant a in Lemma 3.4 to be arbitrarily large, which implies that So, by FKG inequality, by Lemma 6.1 and elementary Poisson convergence.
n Call a site Z-assisted if one of the following holds: the site has an initially occupied Z-neighbor and at least 2 initially occupied K-neighbors, or the site has two initially occupied Z-neighbors and at least 2 − 1 initially occupied K-neighbors. The probability that there exists a Z-assisted site is bounded above by a constant times mn 2 p(np) 2 + p 2 (np) 2 −1 = O(m −1 /(n 2 p)) = o(1).
If an initially empty site becomes occupied in the first step of bootstrap percolation, then either that site is in a plane that is not (2 + 1)-II, or that site is Z-assisted. Therefore, P p (there exists a site that becomes occupied in the first step) again by Lemma 6.1. The two asymptotic bounds (6.2) and (6.3) establish the desired convergence.

Threshold 2
We now show that when θ = 2, the critical a is twice as large as obtained by taking = 1 in Lemmas 4.3 and 5.3. Thus we will still assume (in this and in the next subsection) that p has the scaling given by (3.1), that is, p = a log n n 2 .
(7.1) Lemma 7.1. Assume that p is of the form (7.1). Then P p K 2 n is not 2-IS ∼ P p K 2 n is 2-II ∼ a log n n a .
Proof. Let G k be the event that K 2 n contains exactly k initially occupied points. Then P p K 2 n is 2-II ≥ P p (G 1 ) ∼ n 2 pe −n 2 p = an −a log n.
Lemma 7.2. Suppose θ = 2, and p has the form (7.1). Then P p (Span) converges to 0 when a < γ/2 and to 1 when a > γ/2. n Proof. If two neighboring copies of K 2 n are initially empty then spanning cannot occur.
By Lemma 2.1, the probability of this is close to 0 (resp., 1) if m(1 − p) 2n 2 goes to ∞ (resp., to 0), which happens when a > γ/2 (resp., when a < γ/2). Thus, when a < γ/2, P p (Span) → 0. On the other hand, a sufficient condition for spanning is that there are no initially empty neighboring copies of K 2 n and there is at least one 2-IS copy of K 2 n .

Threshold 3
We now handle the case θ = 3, beginning with the restatement of Lemma 3.5 for this case.
Lemma 7.3. Assume p has the form (7.1). Then P p K 2 n is 3-IS ∼ P p K 2 n is not 3-II ∼ a 2 · (log n) 2 n .
The rest of the proof of Lemma 5.2 needs to be slightly adapted, as now the probability in Lemma 3.8 is only O(P p K 2 n is 3-IS ), by Lemma 7.3. However, this means that P p (G i1,i1+k ) ≥ α > 0 for some fixed number α > 0 (instead of converging to 1), and as a result the probability of occurrence of a blocking interval is at least (1 − o(1))α. This is still true if we replace m by m = n γ , where γ < γ and a / ! < (γ + 1/ )/2. But this means that we have have m/m 1 independent possibilities for a blocking interval to occur, which is sufficient.

A boundary case
Here, we provide an example with θ = 2 + 1 whereby the transition is neither sharp nor gradual. This occurs at a boundary case γ = 1/ ; more precisely, we assume that m = n 1/ (log n) 1+1/ , n which proves that P p (Span) → 0 when a < λ d−θ+1 . To get the upper bound, observe that P p (z is initially black) = Ω P p (z is initially grey) θ . (8.1) Next, observe that if a point z ∈ Z d m has θ − 1 black points in its neighborhood and an additional grey point (that could be z itself), then z eventually becomes black. The local growth of black points can then be constructed in the same way as in [BBM], and due to (8.1) has the same leading order asymptotics as the local growth probability of grey points, as the main contribution comes from a lower-dimensional process where, apart from an initial black nucleus, only grey points are used. This shows that P p (spanning) → 1 when a > λ d−θ+1 .