On the Running Time of Hypergraph Bootstrap Percolation

Given $r\geq2$ and an $r$-uniform hypergraph $F$, the $F$-bootstrap process starts with an $r$-uniform hypergraph $H$ and, in each time step, every hyperedge which"completes"a copy of $F$ is added to $H$. The maximum running time of this process has been recently studied in the case that $r=2$ and $F$ is a complete graph by Bollob\'as, Przykucki, Riordan and Sahasrabudhe [Electron. J. Combin. 24(2) (2017), Paper No. 2.16], Matzke [arXiv:1510.06156v2] and Balogh, Kronenberg, Pokrovskiy and Szab\'o [arXiv:1907.04559v1]. We consider the case that $r\geq3$ and $F$ is the complete $r$-uniform hypergraph on $k$ vertices. Our main results are that the maximum running time is $\Theta\left(n^r\right)$ if $k\geq r+2$ and $\Omega\left(n^{r-1}\right)$ if $k=r+1$. For the case $k=r+1$, we conjecture that our lower bound is optimal up to a constant factor when $r=3$, but suspect that it can be improved by more than a constant factor for large $r$.

Our focus in this paper is on hypergraph bootstrap percolation which generalizes the notion of graph bootstrap percolation introduced by Bollobás [16] in 1968, originally under the name "weak saturation." For a set S and r ⩾ 0, let S (r) denote the collection of all subsets of S of cardinality r. Given an r-uniform hypergraph F , the F -bootstrap process starts with an r-uniform hypergraph H 0 . Then, at each time step t ⩾ 1, the hypergraph H t consists of all hyperedges of H t−1 as well as each e ∈ V (H 0 ) (r) for which there exists a copy F ′ of F containing e such that E (F ′ ) ⊆ E (H t−1 ) ∪ {e} (that is, e "completes" a copy of F when added to H t−1 ). The hypergraph H 0 is referred to as the initial infection and, for t ⩾ 0, the hyperedges of H t are said to be infected at time t.
The most well studied problem related to the F -bootstrap process is to determine the minimum number of hyperedges in an initial infection H 0 on n vertices such that every element of V (H 0 ) (r) is eventually infected. When F is the complete r-uniform hypergraph on k vertices, which we denote by K r k , this is equivalent to the famous Skew Two Families Theorem proved by Frankl [25] using a beautiful exterior algebraic approach of Lovász [41]; see also [37,4]. Alon [3] proved a multipartite version of the Skew Two Families Theorem, which was extended further by Moshkovitz and Shapira [46]; the results of both of these papers can be thought of in terms of hypergraph bootstrap percolation in a multipartite "host" hypergraph. Pikhurko [47,48] solved the problem for several classes of non-complete hypergraphs F using the exterior algebraic method of [41,25] and a related matroid-theoretic approach of Kalai [37]. For results on graphs, i.e. the case r = 2, see [16,40,15,45,36,37].
Another major focus in bootstrap percolation has been on analyzing the maximum number of time steps that a given bootstrap percolation model can take before it stabilizes [17,42,14,13,49,31,5]. In the context of the F -bootstrap process with initial infection H 0 , the running time is M F (H 0 ) := min {t : H t = H t+1 } and the maximum running time among all hypergraphs H 0 with n vertices is denoted by M F (n). If F = K r k for some k, then we write M F (H 0 ) and M F (n) as M r k (H 0 ) and M r k (n), respectively. We omit the superscript in the case r = 2.
Our goal is to bound M r k (n) for fixed r and k asymptotically as a function of n. This problem was first studied in the case r = 2 independently by Bollobás, Przykucki, Riordan and Sahasrabudhe [17] and Matzke [42]. Clearly, M r k (n) ⩽ n r for all n, r and k as at least one hyperedge must become infected in each step. It is an easy exercise to show that the maximum running time for the K 3 -bootstrap process, i.e. M 3 (n), is precisely ⌈log 2 (n − 1)⌉. Interestingly, for k = 4, the maximum running time jumps from logarithmic to linear.
Bollobás et al. [17] conjectured that M k (n) = o (n 2 ) for all k ⩾ 5. This was disproved for all k ⩾ 6 by Balogh, Kronenberg, Pokrovskiy and Szabó [11]. Since the inequality M k+1 (n + 1) ⩾ M k (n) can be shown by simply taking a construction on n vertices, adding a vertex and infecting every edge incident to it (see [11,Proposition 10] or Lemma 8 in this paper), the critical case is k = 6.
Theorem 2 (Balogh et al. [11]). M 6 (n) ⩾ n 2 2500 for all n sufficiently large. The growth rate of the maximum running time for the K 5 -bootstrap process is still unknown. In fact, it is an interesting open problem to determine whether or not it is quadratic [17,11]. To date, the best known lower bound is given by Balogh et al. [11] by exploiting connections to additive combinatorics; their result improved on a bound of n 13/8−o(1) due to [17]. Let r 3 (n) be the largest cardinality of a subset of [n] without a 3-term arithmetic progression. While the asymptotics of r 3 (n) are not known precisely, Roth's Theorem [50] implies that r 3 (n) = o (n) (see [51,38] for recent quantitative bounds) and the Behrend Construction [12] yields r 3 (n) ⩾ n Theorem 3 (Balogh et al. [11]). M 5 (n) ⩾ nr 3 (n) 1200 . Here, we initiate the study of M r k (n) for r ⩾ 3. Our main contributions are constructions of initially infected hypergraphs yielding lower bounds. Our first result concerns the case that k = r + 1. It would be interesting to know whether this bound is tight up to a constant factor, especially in the case r = 3 and k = 4; see Conjecture 25 and Question 26.
Theorem 4. Let r ⩾ 3. If k = r + 1, then M r k (n) = Ω (n r−1 ). In contrast, for k ⩾ r + 2, we show that the trivial upper bound n r is tight up to a constant factor. Theorem 5. Let r ⩾ 3. If k ⩾ r + 2, then M r k (n) = Θ (n r ). The rest of the paper is organized as follows. In the next section, we build up some basic notation and terminology and establish a few preliminary lemmas. In particular, these lemmas will be used to reduce Theorems 4 and 5 to finding constructions with some additional properties for r = 3 and k ∈ {4, 5}. In Section 3, we introduce a key construction, which we call the "beachball hypergraph". This hypergraph has running time that is only linear with respect to its number of vertices, but plays a key role in the proofs of both of our main theorems. In the same section, we show how linearly many beachball hypergraphs can be "chained together" to prove Theorem 4. In Section 4, we use the beachball construction in a different way to prove Theorem 5. We conclude the paper in Section 5 with two open problems.

Preliminaries
Given r-uniform hypergraphs F and H, say that a copy F ′ of F is susceptible to H if there exists a hyperedge e / ∈ E (H) and e ∈ E (F ′ ) such that E (F ′ ) ⊆ E (H) ∪ {e}. Say that H is F -stable if there are no copies of F that are susceptible to H. We make the following simple observation. Observation 6. Let H 0 be an initial infection for the F -bootstrap process. Then, for t ⩾ 1, E (H t ) is the union of E (H t−1 ) and the hyperedge sets of all copies of F which are susceptible to H t−1 .
The following very straightforward definition and the lemma that follows it allow us to use a lower bound on M r k (n) to get a lower bound on M r k+1 (n) which is only slightly worse; c.f. [11,Proposition 10].

Lemma 8.
Let H 0 be an initial infection for the K r k -bootstrap process and, for w / Proof. Suppose not and let t be the minimum time that the equality is violated. Since Observe that no hyperedge containing w becomes infected in any step t ⩾ 1 since all such hyperedges are already in H 0 .
First, suppose that there is e ∈ E (H t ∨ w) such that e / ∈ E (H ′ t ). As noted above, we may assume that e ∈ E (H t ) since w / ∈ e. Let F be the corresponding copy of K r k that is susceptible to H t−1 . Then, by minimality of t, all hyperedges of F ∨ w except for e are present in H ′ t−1 . Thus, e ∈ E (H ′ t ), which is a contradiction. The proof of the other direction is similar.
The analysis of the running time of H 0 tends to become unwieldy if there are time steps t in which there is more than one copy of F that is susceptible to H t . For this reason, the constructions in this paper will be designed with a specific goal of avoiding this situation; a similar approach is taken for graphs in [11,17]. This motivates the following definitions. Definition 9. Let H 0 be an initial infection for the F -bootstrap process. Say that H 0 is F -tame if there is at most one copy of F which is susceptible to H t for all t ⩾ 0.
where T = M F (H 0 ) and, for all 0 ⩽ t ⩽ T − 1, F t is the unique copy of F that is susceptible to H t and e t+1 is the unique hyperedge of E (F t ) \ E (H t ).
The following proposition is an easy consequence of Lemma 8; we omit the proof.
In our constructions, it will be especially useful to restrict our attention to F -tame hypergraphs with limited interaction between the elements of their trajectories.
Definition 12. Let (F 0 , e 1 , . . . , F T −1 , e T ) be the trajectory of an F -tame hypergraph H 0 and let e 0 ∈ E (F 0 ). We say that H 0 is F -civilized with respect to e 0 if the following two conditions hold: In other words, an F -tame hypergraph H 0 is F -civilized with respect to e 0 if every copy of F in its trajectory has exactly two hyperedges missing from H 0 \{e 0 } and H 0 \{e 0 } is F -stable. A useful property of an F -civilized hypergraph is that its trajectory can be reversed by swapping e 0 and e T . Lemma 13. Let H 0 be a hypergraph which is F -civilized with respect to e 0 with trajectory (F 0 , e 1 , . . . , F T −1 , e T ) and let Proof. To show that H ′ 0 is F -civilized with respect to e T , it suffices to show that it is F -tame with the correct trajectory since the additional conditions of Definition 12 will follow from the fact that H 0 is F -civilized with respect to e 0 .
For 0 ⩽ t < T , let F ′ be a copy of F which is susceptible to H ′ t and let e ′ be the unique hyperedge of F ′ that is not in H ′ t . We claim that F ′ = F T −t−1 and e ′ = e T −t−1 . For the sake of contradiction, suppose that at least one of these equalities does not hold and let t be the minimum time for which such an F ′ and e ′ exist. By minimality of t, we have that In particular, H ′ t is a subhypergraph of H T . Now, observe that e ′ must be contained in H T ; if not, then F ′ would be susceptible to H T which contradicts the assumption that (F 0 , e 1 , . . . , F T −1 , e T ) is the trajectory of H 0 . Next, we claim that F ′ is not a subhypergraph of H 0 . If it were, then, since e ′ is not contained in H ′ t , we must have that e ′ = e 0 and all other hyperedges of F ′ are contained in H 0 . However, this would contradict condition 12 of Definition 12. Therefore, F ′ is not a subhypergraph of H 0 .
So, H 0 does not contain every hyperedge of F ′ but H T does. Thus, we can let 0 ⩽ j < T be the maximum index such that there is a hyperedge of F ′ that is not contained in H j . So, F ′ is susceptible to H j , which, by the assumption of the lemma, implies that F ′ = F j . Since H 0 is F -civilized with respect to e 0 , the only hyperedges of F j missing from H 0 \{e 0 } are e j and e j+1 . Since F j is susceptible to . . , e T −1 } and also to H j , but not to H ′ t−1 , the only possibility is that e T −t = e j+1 and e ′ = e j . So, j + 1 = T − t which implies that F ′ = F j = F T −t−1 and e j = e T −t−1 , as we wanted.
Thus, F T −t−1 is the only copy of F that can be susceptible to H ′ t . The last thing to show is that F T −t−1 is, indeed, susceptible to H ′ t . To see this, note that H 0 is F -civilized with respect to e 0 , and so e T −t−1 is the unique hyperedge of F T −t−1 that is not in H ′ t . Thus, H ′ 0 is F -tame with the desired trajectory and the proof is complete. Next, we define a natural "step up" construction for converting constructions for the K r k -bootstrap process into constructions for the K r+1 k+1 -bootstrap process. Later, in Lemma 16, we will see how Lemma 13 can be applied to "chain together" these step up constructions to transform a K r k -civilized hypergraph into a K r+1 k+1 -civilized hypergraph with a linear number of extra vertices in such a way that the running time is boosted by a linear factor. Definition 14. Let r ⩾ 2, let H be an r-uniform hypergraph and let w / ∈ V (H). Define H +w to be the (r + 1)-uniform hypergraph with vertex set V (H) ∪ {w} and hyperedges Next, we show that H +w 0 behaves in essentially the same way with respect to the K r+1 k+1 -bootstrap process as H 0 does with respect to the K r k -bootstrap process.
Proof. Suppose that the lemma is false and let t be the smallest index such that H ′ t ̸ = (H t ) +w . By definition, we have H ′ 0 = H +w 0 and so t ⩾ 1. First, suppose that there is a hyperedge which is contained in (H t ) +w but not in H ′ t . Since both of these hypergraphs contain all of V (H 0 ) (r+1) , this hyperedge must have the form e ∪ {w} where e ∈ E (H t ). By minimality of t, we must have e / ∈ E (H t−1 ) and so there must be a copy F of K r k containing e which is susceptible to H t−1 . As F is a copy of K r k containing e, this means that F +w must be a copy of K r+1 k+1 containing e ∪ {w}. However, by minimality of t, we have that H ′ t−1 = (H t−1 ) +w , and hence e ∪ {w} / ∈ H ′ t−1 . Since H t−1 contains every hyperedge of F except e, we conclude that H ′ t−1 must contain every hyperedge of F +w except for e ∪ {w}. Therefore, F +w is susceptible to H ′ t−1 and so e ∪ {w} is contained in H ′ t , which is a contradiction. Now, suppose that there is a hyperedge e ′ that is contained in H ′ t but not in (H t ) +w . As in the previous case, this hyperedge must contain w. By minimality of t, we must have e ′ / ∈ E H ′ t−1 and so there must be a copy This contradiction completes the proof.
As alluded to earlier, we now apply Lemmas 13 and 15 to show that step up constructions for K r k -civilized hypergraphs can be chained together to yield K r+1 k+1 -civilized hypergraph with very long running time.
Lemma 16. For r ⩾ 3 and k ⩾ r + 1, let H 0 be a K r k -civilized hypergraph. Then, for any Proof. Let e 0 be a hyperedge of H 0 such that H 0 is K r k -civilized with respect to e 0 and let (F 0 , e 1 , . . . , F T −1 , e T ) be the trajectory of H 0 . Let us describe the construction of be a set of vertices disjoint from the set of vertices introduced so far.
containing all hyperedges in this set, except for e +j f (j) and e . Finally, Intuitively, the way to think of this is as follows. For each 1 ⩽ j ⩽ m, the hypergraph H j 0 will emulate the K r k -bootstrap process with initial infection H 0 , either in the forward or backwards (as in Lemma 13) direction, depending on the parity of j. The hypergraph C j for 1 ⩽ j ⩽ m − 1 is used to link the process on H j 0 to the process H j+1 0 in such a way that the termination of the former triggers the start of the latter.
Let us formalize this. First, we show that, if x ∈ X j for some 1 ⩽ j ⩽ m − 1, then, for all t ⩾ 0, every hyperedge e of H ′ t containing x is contained in V (C j ). Suppose that this is not the case, let t be the minimum time for which it fails and let z be a vertex of e that is not in V (C j ). Clearly, by construction of H ′ 0 , we must have t ⩾ 1. By minimality of t, there must be a copy F of K r+1 k+1 which is susceptible to H ′ t−1 and contains e. Since k + 1 > r + 1, we can let u be a vertex of V (F ) \ e. Since r + 1 ⩾ 4, we can choose y to be a vertex of e \ {x, z}. Now, consider the set e ′ = e \ {y} ∪ {u}. This is a hyperedge of F which contains both of x and z. However, all hyperedges of F other than e are in H ′ t−1 , and so e ′ is in H ′ t−1 which contradicts the minimality of t. Next, observe that, for any t ⩾ 0, the hypergraph H ′ t does not contain any hyperedge which includes w j and w j ′ for |j − j ′ | > 1. Note that H ′ 0 has no such hyperedge. Consider the first time t ⩾ 1 that such a hyperedge appears, and observe that the relevant copy of K r+1 k+1 that was allegedly susceptible to H ′ t−1 is missing all other hyperedges containing w j and w j ′ by minimality of t, which is a contradiction. Now, let us show that, for 1 ⩽ j ⩽ m − 1, for every t ⩾ 0, every hyperedge e of H ′ is violated and let z be the offending vertex of e. As usual, t ⩾ 1 by construction of H ′ 0 . Take a copy F of K r+1 k+1 which is susceptible to H ′ t−1 and contains e. Since k + 1 > r + 1, we can let u be a vertex of V (F ) \ e. Since r + 1 ⩾ 4, we can pick y ∈ e \ {w j , w j+1 , z}. Now, consider e ′ = e \ {y} ∪ {u}. This is a hyperedge of F other than e which contains w j , w j+1 and z, which contradicts the minimality of t.
Putting all of this together, we see that the only hyperedges that can become infected during the K r+1 k+1 -bootstrap process starting with H ′ 0 are those which are contained within V H j 0 for some 1 ⩽ j ⩽ m or within V (C j ) for some 1 ⩽ j ⩽ m − 1. The hyperedges of H ′ 0 within V (C j ) form nothing more than a copy of K r+1 k+1 with two hyperedges removed. The same is true for the hyperedges within V (F i ) ∪ {w j } for any copy F i of K r k in the trajectory of H 0 and any 1 ⩽ j ⩽ m with the exception of the case i = 0 and j = 1, in which case the hyperedges form a copy of K r+1 k+1 with one hyperedge removed. Therefore, the trajectory of H ′ 0 follows that of H +w 1 0 for the first M r k (H 0 ) steps. At that point, the hyperedge e +1 T has become infected, which causes C 1 to become susceptible. Thus, e +2 T becomes infected. This triggers the process in H 2 0 to start running in the reverse direction as in Lemma 13, and so on. Putting all of this together, the fact that H 0 is K r k -civilized with respect to e 0 now implies that H ′ 0 is K r+1 k+1 -civilized with respect to e +1 0 . The running time includes M r k (H 0 ) steps for every 1 ⩽ j ⩽ m, plus one extra step for each 1 ⩽ j ⩽ m − 1. This completes the proof.
Using the results of this section, we show that, in order to prove Theorems 4 and 5, it suffices to find K r k -civilized constructions with long running time for r = 3 and k ∈ {4, 5}. Proof. For any fixed r ⩾ 3 and k such that k = r + 1, applying Lemma 16 with m = Θ (n) exactly r − 3 times to the hypergraph H 0 in the statement of the corollary yields an r-uniform hypergraph with O (n) vertices whose running time with respect to the K r kbootstrap process is Ω (n r−1 ). For r ⩾ 3 and k ⩾ r + 2, apply Proposition 11 to H ′ 0 exactly k − r − 2 times, and then Lemma 16 with m = Θ (n) exactly r − 3 times to get an r-uniform hypergraph with O (n) vertices whose running time with respect to the K r kbootstrap process is Ω (n r ). Trivially, M r k (n) ⩽ n r = O (n r ) and so M r k (n) = Θ (n r ).

Getting the Beachball Rolling
The following definition provides a simple gadget that will be used in both of our main constructions. See Figure 1 for an illustration.  We call u 1 and u 2 the top and bottom vertices of B (v 1 , . . . , v n , u 1 , u 2 ), respectively. The vertices v 1 , v 2 , . . . , v n are the middle vertices. A key property of beachball hypergraphs is that they behave in a very straightforward way with respect to the K 3 4 -process.
Proof. For the purposes of this proof, denote B (v 1 , . . . , v n , u 1 , u 2 ) by B n 0 . Say that a hyperedge e is flat if it is contained in {v 1 , . . . , v n } and wide if it contains exactly one of u 1 or u 2 and is not an element of E (B n 0 ). We claim that B n t contains no flat or wide hyperedges for any t ⩾ 0. For t = 0, this is true by definition of B n 0 . Now, consider the minimum t for which the statement does not hold and let e ∈ B n t be a hyperedge which is either flat or wide. Then there must exist a copy F of K 3 4 with e ∈ E (F ) which is susceptible to B n t−1 . Let v be the unique vertex of V (F ) \ e. If e is flat, then we can write e = {v i , v j , v k } for 1 ⩽ i < j < k ⩽ n.
If v ∈ {v 1 , . . . , v n }, then B n t−1 must contain the hyperedge e \ {v i } ∪ {v}. However, this hyperedge is flat, which contradicts the minimality of t. Likewise, if v ∈ {u 1 , u 2 }, then B n t−1 contains the hyperedge {v i , v k , v}, which is wide, and so we get another contradiction. Similarly, if e is wide, then, depending on whether v ∈ {v 1 , . . . , v n } or v ∈ {u 1 , u 2 }, we find either a flat or a wide hyperedge in B n t−1 , respectively. This proves the claim. Thus, the only hyperedges which can become infected after any number of steps are those of the form {u 1 , u 2 , v i } for some i ∈ {2, . . . , n}. This implies that the only copies of K 3 4 that can be susceptible at any time are those whose vertex sets have the form Now, we claim that, for all 0 ⩽ t ⩽ n − 2, a hyperedge of the form {u 1 , u 2 , v i } for 1 ⩽ i ⩽ n is contained in B n t if and only if i ⩽ t + 1. This is clearly true for t = 0 by construction. Consider the minimum t ⩾ 1 for which this statement is false and let j > t + 1 such that e = {u 1 , u 2 , v j } ∈ B n t . Then, by the result of the previous paragraph, the only possible copies of K 3 4 containing e which could be susceptible at time t − 1 are the ones with vertex set {u 1 , u 2 , v j−1 , v j } or {u 1 , u 2 , v j , v j+1 }. However, this implies that either {u 1 , u 2 , v j−1 } or {u 1 , u 2 , v j+1 } is in E B n t−1 ; in either case, this contradicts the minimality of t.
Therefore, B n 0 is K 3 4 -tame with the trajectory described in the statement in the lemma. Let us argue that B n 0 is K 3 4 -civilized with respect to e 0 = {u 1 , u 2 , v 1 }. Condition 12 of Definition 12 follows from the construction of the beachball hypergraph, since all of the sets {u 1 , u 2 , v i+1 , v i+2 } for 0 ⩽ i ⩽ n − 2 form a copy of K 3 4 with two hyperedges missing from B n 0 \{e 0 }. For the same reason, condition 12 also holds. This completes the proof. In our construction for the K 3 4 -bootstrap process, we will also require the hypergraph B ′ (v 1 , . . . , v n , u 1 , u 2 ) obtained from B (v 1 , . . . , v n , u 1 , u 2 ) by adding the hyperedge {u 1 , v 1 , v n }. We call this an augmented beachball hypergraph. Next, we show that the addition of this hyperedge extends the running time by one step without disrupting any of the steps that came before it. In fact, this hypergraph is still K 3 4 -civilized, albeit with respect to a different hyperedge e ′ 0 ̸ = e 0 . at time zero, as is the case in B ′ 0 n , then, for 0 ⩽ t ⩽ n − 2, the copy of K 3 4 with this vertex set has at most two infected hyperedges at time t, and is therefore not susceptible at that time. Any other copy of K 3 4 containing {u 1 , v 1 , v n } has at most one infected hyperedge. Thus, we have that B ′ t n = B n t ∪ {{u 1 , v 1 , v n }} for 0 ⩽ t ⩽ n − 1. Given that the hyperedge {u 1 , v 1 , v n } is infected in B ′ 0 n , we see that the copy of K 3 4 with vertex set {u 1 , u 2 , v 1 , v n } is susceptible to B ′ n n−1 . Note that, by Lemma 19, {u 1 , v 1 , v n } is the unique wide hyperedge (to borrow terminology from the proof of Lemma 19) that is infected at time n − 1. Thus, there is only one susceptible copy of Let v 0 , . . . , v n−1 be distinct vertices whose indices are viewed "cyclically" modulo n. In particular, for any i, we have v −i = v n−i = v 2n−i , and so on. Let u 0 , . . . , u m−1 be m additional vertices which are distinct from one another, and from v 0 , . . . , v n−1 . For The construction can be thought of intuitively as follows. Start with the hypergraph B ′ (v 0 , . . . , v n−1 , u 0 , u 1 ). Then, we take the same beachball hypergraph again, except that we delete one specific hyperedge, rotate the middle vertices by one, move the bottom vertex to the top, and insert a new vertex on the bottom. Repeat until there are no remaining vertices to add. See Figure 3 for an illustration. First, we claim that the hyperedges of H 0 within S 0 are precisely the hyperedges of the augmented beachball hypergraph on its vertices. Every hyperedge of the augmented beachball hypergraph on S 0 was added, by definition of H 0 , and so it suffices to show that no additional hyperedges are present. For i > 1, all of the hyperedges of B ′ (v −i , v 1−i , . . . , v n−1−i , u i , u i+1 ) contain u i or u i+1 and thus are not contained in S 0 . Moreover, the only hyperedge of B ′ (v n−1 , v 0 , . . . , v n−2 , u 1 , u 2 ) contained in S 0 that is not a hyperedge of the augmented beachball on its vertices is {v n−1 , v 0 , u 1 } which was deleted during the construction of H 0 . This proves the claim. The same argument also proves that, for all i ⩾ 1, the hyperedges within the vertices S i are precisely those of the augmented beachball hypergraph on these vertices, except that the hyperedge {v −i , v 1−i , u i } is missing. Now, we also observe that, for any t ⩾ 0, the hypergraph H t does not contain any hyperedge which includes u i and u i ′ with |i − i ′ | > 1. To see this, consider the first time that it fails and consider the susceptible K 3 4 which caused it; as H 0 contains no such hyperedges containing u i and u i ′ , we get a contradiction.
By Lemma 20, every augmented beachball in the chain is K 3 4 -civilised, and the corresponding trajectory implies that the hyperedge So we see that, for any i ⩾ 1, the first hyperedge in S i that becomes infected does so due to a susceptible copy of K 3 4 with precisely one vertex outside of S i (since it cannot be contained entirely inside S i ); until such a copy appears, no additional hyperedges in S i become infected. Note that this statement is not true for i = 0, as the first hyperedge in S 0 becomes infected due to a susceptible copy of K 3 4 contained within S 0 . Putting this together, we see that the only hyperedges that become infected in the first n steps are exactly those which become infected in the K 3 4 -bootstrap process in B ′ (v 0 , v 1 , . . . , v n−1 , u 0 , u 1 ). By Lemma 20, all of these hyperedges are disjoint from S i for all i ⩾ 1, except for the last one, namely {u 1 , v 0 , v n−1 }, which happens to be the unique hyperedge of B ′ (v n−1 , v 0 , . . . , v n−2 , u 1 , u 2 ) that is missing from H 0 . From this point, the infection follows the K 3 4 -bootstrap process in B ′ (v n−1 , v 0 , . . . , v n−2 , u 1 , u 2 ) for n steps. This pattern repeats itself m − 1 times. Thus, H 0 is K 3 4 -tame. The proof that the additional conditions of Definition 12 hold with respect to e ′ 0 = {u 0 , v 0 , v 1 } is analogous to the arguments given in the proofs of the lemmas in this section, and so we omit it.
This completes the proof.
Proof of Theorem 4. For each n ⩾ 8, Theorem 21 gives us a K 3 4 -civilized hypergraph H 0 on n vertices such that The result now follows by Corollary 17.
4 Cubic Time Construction for K 3

5
Our final task is to use the beachball hypergraph again to obtain a construction for the K 3 5 -bootstrap process with cubic running time.
Theorem 22. For any n ⩾ 2 and m ⩾ 2, there exists a K 3 5 -civilized hypergraph H 0 with 2n + 4m + 4 vertices such that Proof. The vertices of the construction are naturally divided into two "halves" which share exactly two special vertices, w 1 and w 2 . For i ∈ {1, 2}, the vertices that are "exclusive" to the ith half are v i 1 , . . . , v i n , u i 1 , . . . , u i m , z i 1 , . . . , z i m and x i . Each half of the construction has n + 2m + 1 exclusive vertices. Together with w 1 and w 2 , this makes 2n + 4m + 4 vertices in total.
For 1 ⩽ j ⩽ m 2 , let (a j , b j ) be the jth element of {1, 2, . . . , m} × {1, 2, . . . , m} in lexicographic order. For 1 ⩽ j ⩽ m 2 , let C 1 j be the copy of K 3 5 with vertex set u 1 a j , z 1 b j , u 2 a j , z 2 b j , w 2 and, if j ⩽ m 2 − 1, then we additionally define C 2 j to be the copy of K 3 5 with vertex set u 2 a j , z 2 b j , u 1 a j+1 , z 1 b j+1 , w 1 . Let for all 1 ⩽ j ⩽ m 2 . Define H 0 to be the union of the following five hypergraphs Intuitively, the way that the infection evolves can be described as follows. It starts by propagating through B 1 1 , the first beachball hypergraph on the first half, in the way described by Lemma 19 and Proposition 11. This takes n + 1 steps. The hyperedge that gets infected in the last step is u 1 Consequently, now all of the hyperedges within V (C 1 1 ) except for u 2 a 1 , z 2 b 1 , w 2 are infected, and so this hyperedge gets infected in the next step; at this point, we have done n + 2 steps in total. This was the only hyperedge of B 2 1 that is not in H 0 , and so its infection triggers the beachball hypergraph B 2 1 on the right half of the construction. The process inside B 2 1 ends with the infection of the hyperedge u 2 a 1 , z 2 b 1 , w 1 , which is in V (C 2 1 ). So now all the hyperedges of V (C 2 1 ) except u 1 a 2 , z 1 b 2 , w 1 are infected, and hence this hyperedge gets infected in the next step. This transfers the infection back over to the left half of the construction by triggering B 1 2 , and so on. If the process does indeed progress in this manner, then, in total, it will take n + 2 steps for every 1 ⩽ j ⩽ m 2 and every i ∈ {1, 2}, with the exception of j = m 2 and i = 2, which only contributes n + 1 steps (since C 2 m 2 has not been defined). Thus, there are (2n + 4) m 2 − 1 steps.
Let us now make this rigorous. Our goal is to show that H 0 is K 3 5 -civilized with respect to e 0 = u 1 a 1 , z 1 b 1 , w 1 ; on the way to that goal, we will need to establish several claims. First, we observe that any pair of vertices which are not contained together in a hyperedge of H 0 are not contained together in any hyperedge of H t for any t ⩾ 0; to see this, consider the first time that such a hyperedge becomes infected, look at the susceptible copy of K 3 5 which caused it and get a contradiction. This argument immediately yields the following claim. Throughout the statement of the claim, keep in mind that w 1 and w 2 are regarded as being on both halves of the construction, but all other vertices are exclusive to one half or the other. The purpose of the next claim is similar to the previous one; i.e. to rule out certain types of hyperedges from becoming infected.
Claim 24. For any t ⩾ 0, every hyperedge of H t containing both w 1 and w 2 must also contain one of x 1 or x 2 .
Proof of Claim 24. Consider the first time t such that H t contains a hyperedge e that contains both of w 1 and w 2 but neither of x 1 nor x 2 . Note that t ⩾ 1 by construction of H 0 . Let F be the copy of K 3 5 containing e which is susceptible to H t−1 . By minimality of t, the only two hyperedges of H t−1 containing w 1 and w 2 are {w 1 , w 2 , x 1 } and {w 1 , w 2 , x 2 }. So, the vertices of F must consist of w 1 , w 2 , x 1 , x 2 and a fifth vertex, say y. However, y is exclusive to one of the two halves of the construction, and so, regardless of which half it is, H t−1 contains several hyperedges which satisfy condition 23 of Claim 23, which is a contradiction.
Let us now show that H 0 is K 3 5 -tame with the trajectory that was described earlier in the proof. Suppose that this is not the case, let t ⩾ 1 be the minimum time in which there is an unexpected infected hyperedge, say e. Let F be a copy of K 3 5 containing e which is susceptible to H t−1 . By Lemma 19 and minimality of t, we cannot have V (F ) ⊆ B i j for any 1 ⩽ j ⩽ m 2 and i ∈ {1, 2}. Now, any five vertices on the ith half of the construction are either contained in B i j for some j or contain two of u i 1 , . . . , u i m or two of z i 1 , . . . , z i m . So, as there are no hyperedges satisfying condition 23 of Claim 23, V (F ) must contain at least one vertex that is exclusive to each half of the construction. But now, since no hyperedges satisfy Claim 23 23 or 23, we get that V (F ) cannot contain any of the vertices x i or v i j for i ∈ {1, 2} and 1 ⩽ j ⩽ n. This also means that it cannot contain both of w 1 and w 2 by Claim 24. On the other hand, it must contain at least one of w 1 or w 2 ; if not, then three vertices of F are exclusive to one of the sides of the partition, and we get a contradiction from Claim 23 23. Using Claim 23 23 one more time, we see that V (F ) has the form {u 1 a , z 1 b , w ℓ , u 2 c , z 2 d } for some 1 ⩽ a, b, c, d ⩽ m and ℓ ∈ {1, 2}. We now divide the proof into cases. Case 1. e does not contain w ℓ .
We assume that e = {u 1 a , z 1 b , u 2 c } and note that the other three cases follow from similar arguments. Let 1 ⩽ j ⩽ m 2 be chosen so that a = a j and b = b j ; such a j exists by construction of H 0 . The hypergraph H t−1 contains the hyperedge e ′ = e \ {u 2 c } ∪ {z 2 d } and so, by minimality of t, we get that e ′ is contained in either V C 1 j or V C 2 j−1 . Indeed, under the trajectory described at the beginning of the proof, every hyperedge containing two vertices exclusive to one side and one exclusive to the other had this property. The fact that e ′ is contained in V C 1 j or V C 2 j−1 implies that e is as well. However, every hyperedge of C 1 j or C 2 j−1 not containing w 1 or w 2 was added to H 0 originally, which contradicts our choice of e. Case 2. e contains w ℓ .
Again, let 1 ⩽ j ⩽ m 2 be chosen so that a = a j and b = b j . The hyperedge is contained in H t−1 and so, by minimality of t, it must be contained in either V C 1 j or V C 2 j−1 . By definition of V C 1 j and V C 2 j−1 , and our specific choice of lexicographic order in the construction of H 0 , this implies that either c = a j or c = a j −1 and b j = 1.
Suppose first that c = a j . Then we get that H t−1 contains z 1 b j , u 2 a j , z 2 d which, by minimality of t and construction of H 0 , implies that either d = b j or d = b j − 1. If d = b j , then we must have that ℓ = 2 by minimality of t since z 1 b j , w 1 , z 2 b j is not contained in any of the hypergraphs C 1 1 , . . . , C 1 m 2 or C 2 1 , . . . , C 2 m 2 −1 . So, in the case that c = a j and d = b j , we get that V (F ) = V C 1 j . This implies that e is one of the only two hyperedges of C j that are missing from H 0 . If e = u 1 a j , z 1 b j , w 2 , then the hyperedge w 2 , u 2 a j , z 2 b j is present before e, which contradicts minimality of t. If e = w 2 , u 2 a j , z 2 b j , then e is being infected due to C 1 j being susceptible, which fits the description of the trajectory from earlier in the proof, and so this contradicts the definition of e. Now, suppose that c = a j and d = b j − 1. In this case, we must have ℓ = 1 by minimality of t since z 1 b j , w 2 , z 2 b j −1 is not contained in any of the sets C 1 1 , . . . , C 1 m 2 or C 2 1 , . . . , C 2 m 2 −1 . So, what we end up with is V (F ) = V C 2 j−1 and we get a contradiction similar to the case that we just analyzed. Now, suppose that c = a j − 1 and b j = 1. Since H t−1 contains z 1 1 , u 2 a j −1 , z 2 d , we must have either d = 1 or d = m. If d = 1, we get a contradiction, since u 1 a j , u 2 a j −1 , z 2 1 is not contained in H t−1 by minimality of t and the fact that m ⩾ 2 and so, under lexicographic order, (a j − 1, 1) does not immediately precede or follow any pair involving a j in the first coordinate. So, d = m. Now, we get that ℓ = 1 by minimality of t since {z 1 1 , w 2 , z 2 m } is not contained in any of the sets C 1 1 , . . . , C 1 m 2 or C 2 1 , . . . , C 2 m 2 −1 . So, we get that V (F ) = C 2 j−1 which leads us to a contradiction, as in the previous paragraph. Similar arguments also show that, for e 0 = u 1 a 1 , z 1 a 1 , w 1 , the hypergraph H 0 \ {e 0 } is K 3 5 -stable; thus, condition 12 of Defintion 12 holds. Clearly each of the copies of K 3 5 in the trajectory of H 0 , as described above, has precisely two hyperedges missing from H 0 \ {e 0 } and so 12 holds, too. This completes the proof.
Proof of Theorem 5. For n sufficiently large and n ≡ 4 mod 6, Theorem 22 provides a K 3 5 -civilized hypergraph H ′ 0 with n vertices such that The result follows by Corollary 17.

Open Problems
While Theorem 5 determines M r k (n) up to a constant factor for all r ⩾ 3 and k ⩾ r + 2, it would be interesting to have a better understanding of the growth rate in the case k = r + 1. The most interesting question here seems to be whether there is a non-trivial upper bound on the running time of the K 3 4 -bootstrap process; we conjecture that the quadratic lower bound from Theorem 4 is tight up to a constant factor. In contrast, we do not believe that Theorem 4 is tight for all r ⩾ 3. It is even conceivable that, for large enough r, the maximum running time of the K r r+1 -process on n vertices eventually becomes Θ (n r ). We ask whether this is, indeed, the case. Question 26. Does there exist an integer r 0 such that, if r ⩾ r 0 , then M r r+1 (n) = Θ (n r )?

Acknowledgements
We would like to thank the anonymous referee for their helpful comments that improved particular details as well as the overall coherence of the paper.

Remarks
After completing this work, Hartarsky and Lichev [32] and Espuny Díaz, Janzer, Kronenberg and Lada [23] independently disproved 25 by showing that M 3 4 (n) = Θ(n 3 ), and consequently answered Question 26 in the affirmative in a rather strong sense with r 0 = 3. Hartarsky and Lichev [32] also determine the leading asymptotics of the prefactor when r → ∞. Additionally, Espuny Díaz et al. [23] provide the first nontrivial exact result about the maximum running times of hypergraph bootstrap percolation by showing that the maximum running time for the H-bootstrap process on n vertices when H is K 3