Isomorphism for Random $k$-Uniform Hypergraphs

We study the isomorphism problem for random hypergraphs. We show that it is polynomially time solvable for the binomial random $k$-uniform hypergraph $H_{n,p;k}$, for a wide range of $p$. We also show that it is solvable w.h.p. for random $r$-regular, $k$-uniform hypergraphs $H_{n,r;k},r=O(1)$.


Introduction
In this note we study the isomorphism problem for two models of random k-uniform hypergraphs, k ≥ 3. A hypergraph is k-uniform if all of its edges are of size k. The graph isomporphism problem for random graphs is well understood and in this note we extend some of the ideas to hypergraphs.
The first paper to study graph isomorphism in this context was that of Babai, Erdős and Selkow [12]. They considered the model G n,p where p is a constant independent of n. They showed that w.h.p. 1 G = G n,p has a canonical labelling and that this labelling can be constructed in O(n 2 ) time. In a canonical labelling we assign a unique label to each vertex of a graph such that labels are invariant under isomorphism. It follows that two graphs with the same vertex set are isomorphic, if and only if the labels coincide. (This includes the case where one graph has a unique labeling and the other does not. In which case the two graphs are not isomorphic.) The failure probability for their algorithm was bounded by O(n −1/7 ). Karp [9], Lipton [11] and Babai and Kucera [3] reduced the failure probability to O(c n ), c < 1. These papers consider p to be constant and the paper of Czajka and Pandurangan [13] allows p = p(n) = o(1). We use the following result from [13]: the notation A n ≫ B n means that A n /B n → ∞ as n → ∞.
Theorem 1. Suppose that p ≫ log 4 n n log log n and p ≤ 1 2 . Then G n,p has a canonical labeling q.s. 2 Our first result concerns the random hypergraph H n,p;k , the random k-uniform hypergraph on vertex set [n] in which each of the possible edges [n] k occurs independently with probability p. We say that two k-uniform hypergraphs . . , f (x k )} is an edge of H 2 . We extend the notion of canonical labelling to hypergraphs.
Theorem 2. Suppose that k ≥ 3 and p, 1 − p ≫ n −(k−2) log n then H n,p;k has a canonical labeling w.h.p.
Bollobás [1] and Kucera [10] proved that random regular graphs have canonical labelings wh.p. We extend the argument of [1] to regular hypergraphs.
A hypergraph is regular of degree r if every vertex is in exactly r edges. We denote a random r-regular, k-uniform hypergraph on vertex set [n] by H n,r;k .
Explanation: There are n 2 choices for i, j. There are at most n 2 choices for y = f (i), x = f −1 (j) in an isomorphism f between H i and H j . This accounts for the n 4 term. There are (n − 3)! < n! possible isomorphisms between H i − {y, j} and H j − {x, i}. Then for every (k − 1)-set of vertices S that includes none of i, j, x, y, the probability for there to be an edge or non-edge in both H i and H j is given by the expression The above estimation shows that even disregarding edges containing i, j, Let G k be the event that H n,p;k has a canonical labeling and that it can be constructed in O(n 2k ) time. Now assume inductively that The base case, k = 2, for (1) are given by the results of [13], [9] and [11].
Indeed, if none of the events in (2) occur then in time O(n 2 ×n 2(k−1) ) = O(n 2k ) we can by induction uniquely label each vertex via the canonical labeling of its link. After this we can confirm that E k has occurred. This confirms the claimed time complexity. Given that E k does not occur, this will determine the only possible isomorphism.
Going back to (2) we see by induction that This completes the proof of Theorem 2.

Proof of Theorem 3
We extend the analysis of Bollobás [1] to hypergraphs. We use the configureation model for hypergraphs, which is a simple generalisation of the model in Bollobás [2]. We let W be a set of size rn where m = rn/k is an integer. Assume that it is partitioned into sets W 1 , W 2 , . . . , W n of size r. We define f : . . , f (w k )} for i = 1, 2, . . . , m. It is known that if γ(F ) has a graph property w.h.p. then H n,r;k will also have this property w.h.p., see for example [4]. Let ρ = (r − 1)(k − 1).
For a vertex v we let d ℓ (v) denote the number of vertices at hypergraph distance ℓ from v in H n,r;k . We show that if ℓ * = 3 5 log ρ n then w.h.p. no two vertices have the same sequence (d ℓ (v), ℓ = 1, 2, . . . , ℓ * ). In the following H = H n,r;k . For a set S ⊆ [n], we let e H (S) denote the number of edges of H that are contained in S.
Proof. We have that Let E denote the high probability event in Lemma 5. We will condition on the occurrence of E. Now for v ∈ [n] let S ℓ (v) denote the set of vertices at distance ℓ from v and let S ≤ℓ (v) = j≤ℓ S j (v). We note that Furthermore, Lemma 5 implies that there exist b r,k < a r,k < (k − 1)r such that w.h.p., we have for all v, w ∈ [n], 1 ≤ ℓ ≤ ℓ 0 , This is because there can be at most one cycle in S ≤ℓ 0 (v) and the sizes of the relevant sets are reduced by having the cycle as close to v, w as possible.
Now consider ℓ > ℓ 0 . Consider doing breadth first search from v or v, w exposing the configuration pairing as we go. Let an edge be dispensable if exposing it contains two vertices already known to be in S ≤ℓ . Lemma 5 implies that w.h.p. there is at most one dispensable edge in S ≤ℓ 0 . Lemma 6. With probability 1 − o(n −2 ), (i) at most 20 of the first n 2 5 exposed edges are dispensable and (ii) at most n 1/4 of the first n 3 5 exposed edges are dispensable.
Now let ℓ 1 = log ρ n 2/5 and ℓ 2 = log ρ n 3/5 . Then, we have that, conditional on E, with probability 1 − o(n −2 ), We deduce from this that if ℓ 3 = log r−1 n 4/7 and ℓ = ℓ 3 + a, a = O(1) then with probability 1 − o(n −2 ), Suppose now that we consider the execution of breadth first search up until we have exposed S k (v). Let d ℓ (v) denote the number of vertices at distance ℓ from v. Then in order to have d ℓ (v) = d ℓ (w), conditional on the history of the search, there has to be an exact outcome for |S ℓ (w)\S ℓ (v)|. Now consider the pairings of the W x , x ∈ S ℓ (w) \ S ℓ (v). Now at most n 1/4 of these pairings are with vertices in S ≤ℓ (v) ∪ S ≤ℓ (w). Condition on these. There must now be s = Θ(n 4/7 ) pairings between W x , x ∈ S ℓ (w)\S ℓ (v) and W y , y / ∈ S ℓ (v)∪S ℓ (w). Furthermore, to have d ℓ (v) = d ℓ (w) these s pairings must involve exactly t of the sets W y , y / ∈ S ℓ (v) ∪ S ℓ (w), where t is determined before the choice of these s pairings. The following lemma will easily show that w.h.p. H has a canonical labeling defined by the values of Lemma 7. Let R = µ i=1 R i be a partitioning of an rµ set R into µ subsets of size r. Suppose that S is a random s-subset of R, where µ 5/9 < s < µ 3/5 . Let X S denote the number of sets R i intersected by S. Then for some constant c 0 .
Proof. We may assume that s ≥ µ 1/2 . The probability that S has at least 3 elements in some set R i is at most So the lemma will follow if we prove that for every j, for some constant c 1 .
Clearly, P j = 0 if j < s/2 and otherwise Now for s/2 ≤ j < s we have We note that if s − j ≥ 10s 2 µ then P j+1 P j ≥ 10r 3(r−1) ≥ 2 and so the j maximising P j is of the form s − αs 2 µ where α ≤ 10. If we substitute j = s − αs 2 µ into (8) then we see that P j+1 P j ∈ 2αr r − 1 1 ± c 2 s µ for some absolute constant c 2 > 0.
It follows that if j 0 is the index maximising P j then Furthermore, if j 1 = j 0 − s µ 1/2 then P j+1 P j ≤ 1 + c 3 µ 1/2 s for j 1 ≤ j ≤ j 0 , for some absolute constant c 3 > 0.
We apply Lemma 7 with µ = n, s = ρ = Θ(n 4/7 ) to show that This completes the proof of Theorem 3.