On the number of solutions in random hypergraph 2-colouring

We determine the limiting distribution of the logarithm of the number of satisfying assignments in the random $k$-uniform hypergraph 2-colouring problem in a certain density regime for all $k\ge 3$ . As a direct consequence we obtain that in this regime the random colouring model is contiguous wrt. the planted model, a result that helps simplifying the transfer of statements between these two models.

In this paper we consider random k-uniform hypergraphs H k (n, m) on the vertex set [n] = {1, . . . , n} with exactly m hyperedges each comprising of k distinct vertices and chosen uniformly at random from all possible subsets of [n] of size k. The hypergraph 2-colouring problem is a random constraint satisfaction problem where one is interested in the number Z(H k (n, m)) of 2-colourings (also called solutions) of H k (n, m), which are maps σ : [n] → {0, 1} that generate no monochromatic edges (i.e. edges e such that |σ(e)| = 1).
In the following we only consider sparse random hypergraphs, meaning that m = O(n) as n → ∞. We call the parameter d = km/n the hyperedge density.
As for many other random constraint satisfaction problems, there is a conjecture as to a sharp threshold for the existence of solutions in terms of the hyperedge density. The best currently known bounds on this threshold d k,col are from Achlioptas and Moore [3] and Coja-Oghlan and Zdeborová [10]. Furthermore, there was a prediction for this density by statistical physicists [11,13] suggesting that d k,col /k = 2 k−1 ln 2 − ln 2 2 − 1 4 + ε k with lim k→∞ ε k → 0.
This prediction was proved by Coja-Oghlan and Panagiotou [8] for the problem of NAE-k-SAT which is almost equivalent to hypergraph 2-colouring and it should be possible to transfer the result without major difficulties.
For a long period of time, the best rigorous upper and lower bounds on d k,col were based on the non-constructive first and second moment method applied to the random variable Z. Achlioptas and Moore [3] proved that there is a critical density d k,sec such that E Z 2 ≤ C · E [Z] 2 for some constant C = C(d, k) > 0 for all d < d k,sec but is violated for d > d k,sec . Via the Paley-Zygmund inequality it can be established that d k,sec is a non-constructive lower bound on d k,col . In [3] it was shown that For d ′ , k covered by Corollary 1.3 we have ln Z k (G(n, m)) = Θ(n) w.h.p.. Thus, it would be reasonable to expect that ln Z k (G(n, m)) has fluctuations of order e. g. √ n. However, the first part of Corollary 1.3 shows that actually ln Z k (G(n, m)) fluctuates w.h.p. by no more than ω = ω(n) for any ω(n) → ∞. Moreover, the second part shows that this is best possible.

Planted model and silent planting.
When proving results about random hypergraphs and investigating their properties, it turns out very useful and often essential to have the notion of typical 2-colourings at hand. By a typical 2-colouring of a random hypergraph we mean a 2-colouring chosen uniformly at random from all its 2-colourings. To make this formal, let Λ k,n,m be the set of all pairs (H k , σ) with H k = H k (n, m) and σ a 2-colouring of H k . Now we define a probability distribution π rc k,n,m [H k , σ] on Λ k,n,m by letting We call this distribution the random colouring model or Gibbs distribution. It can also be described as the distribution produced by the following experiment. RC1: Generate a random hypergraph H k = H k (n, m) provided that Z(H k ) > 0. RC2: Choose a 2-colouring σ of H k uniformly at random. The result of the experiment is (H k , σ). However, up to now there is no known method to implement this experiment efficiently for a wide range of hyperedge densities. In fact, the first step RC1 is easy to process, because we are only interested in values of d where H k (n, m) is 2-colourable w.h.p. and thus the conditioning on Z(H k ) > 0 does not cause problems (the probability P [H k (n, m) is 2-colourable] is close to 1). But what turns the direct study of the distribution π rc k,n,m into a challenge is step RC2, because in the interesting density regimes we cannot even find one 2-colouring algorithmically let alone sample one uniformly: The currently best-performing algorithms for sampling a 2-colouring of H k (n, m) are known to succeed up to density d ′ /k = c2 k−1 /k for some constant c > 0 [2], which is about a factor of k below the colourability threshold.
To circumvent these difficulties, we consider an alternative probability distribution on Λ k,n,m called the planted model, which is much easier to approach. To describe this experiment, for σ : We observe that in contrast to the "difficult" step RC2, step PL2 is much easier to implement. Of course, the two probability distributions π rc k,n,m and π pl k,n,m differ. Under π rc k,n,m , the hypergraph is chosen uniformly at random, whereas under π pl k,n,m it comes up with a probability that is proportional to its number of solutions, meaning that hypergraphs exhibiting many 2-colourings are "favoured" by the planted model.
However, the two models are related if m = m(n) is such that Coja-Oghlan an Achlioptas showed in [1] (where it was actually formulated for the problem of graph k-colouring, but the authors asserted that it also holds for hypergraph 2-colouring), that if (1.2) is satisfied, then the following is true.
3) The statement (1.3) was baptised "quiet planting" by Krzakala and Zdeborová [14] and has ever since been used to study the behaviour of the set of colourings and its geometrical structure in various random constraint satisfaction problems [1,5,15,16,17]. Yet a significant complication in the use of (1.3) is that E n must not only have a small probability but is required to be exponentially unlikely in the planted model. This has caused substantial difficulties in several applications (e.g., [6,5,15]). Theorem 1.1 enables us to establish a very strong connection between the random colouring model and the planted model. To state this, we recall the following definition. Suppose that µ = (µ n ) n≥1 , ν = (ν n ) n≥1 are two sequences of probability measures such that µ n , ν n are defined on the same probability space Ω n for every n. Then (µ n ) n≥1 is contiguous with respect to (ν n ) n≥1 , in symbols µ ⊳ ν, if for any sequence (E n ) n≥1 of events such that lim n→∞ ν n (E n ) = 0 we have lim n→∞ µ n (E n ) = 0.
We show that as a consequence of (1.1) the statement (1.3) can be sharpened in the strongest possible sense. Roughly speaking, we are going to show that in a density regime nearly up to the second moment lower bound the random colouring model is contiguous with respect to the planted model, i.e., in (1.3) it suffices that π pl k,n,m [E n ] = o(1).
1.3. Discussion and further related work. The ideas for the proofs follow the way beaten in [4], where statements analogue to Corollary 1.3 and Corollary 1.4 are shown for the problem of k-colouring random graphs. However, Theorem 1.1 is stronger than the results obtained in [4] because we determine the exact distribution of ln Z − E [ln Z] asymptotically. The main observation used in the proofs is that the variance in the logarithm of the number of 2colourings can be attributed to the fluctuations of the number of cycles of bounded length. The same phenomenon was observed in [4] and also in [9], where a combination of the second moment method and small subgraph conditioning was applied to derive a result similar to ours for the problem of random regular k-SAT.
Small subgraph conditioning was originally developed by Robinson and Wormald in [20] to investigate the Hamiltonicity of random regular graphs of degree at least three. Janson showed in [12] that the method can be used to obtain limiting distributions. Neeman und Netrapalli [19] used the method to obtain a result on non-distinguishability of the Erdős-Rényi model and the stochastic block model. Moore [18] used the method to determine the satisfiability threshold for positive 1-in-k-SAT, a Boolean satisfiability problem where each clause contains k variables and demands that exactly one of them is true.
Similar to [12], we aim at obtaining a limiting distribution. Unfortunately, Jansons result does not apply directly in our case for the following reason. Since verifying the required properties to apply small subgraph conditioning directly for the random variable Z is very intricate, we break Z down into Z = s Z s for some number s > 0 and determine the first and second moment of these smaller variables Z s . However, it is not evident how to apply Jansons 3 result to that growing number of variables simultaneously. Instead, we choose to perform a variance analysis along the lines of [20]. The same approach was pursued in [9], and thus our proof technique is similar to theirs in flavour, but we have the advantage of only having to deal with a very moderately growing number s of variables, which simplifies matters slightly.
1.4. Preliminaries and notation. We always assume that n ≥ n 0 is large enough for our various estimates to hold and denote by [n] the set {1, ..., n}.
If p = (p 1 , . . . , p l ) is a vector with entries p i ≥ 0, then we let Here and throughout, we use the convention that 0 ln 0 = 0. Hence, if l i=1 p i = 1, then H(p) is the entropy of the probability distribution p. Further, for a number x and an integer h > 0 we let For the sake of simplicity we choose to prove Theorem 1.1 using the random hypergraph model H(n, m). This is a random k-uniform (multi-)hypergraph on the vertex set [n] obtained by choosing m hyperedges e 1 , . . . , e m of the complete hypergraph on n vertices uniformly and independently at random (i.e., with replacement). In this model we may choose the same edge more than once, however, the following statement shows that this is quite unlikely. All statements and all proofs are valid for any k ≥ 3.

OUTLINE OF THE PROOF
We classify the 2-colourings according to their proportion of assigned colours: For a map σ : [n] → {0, 1} we define and call this value the colour density of σ. We let A(n) signify the set of all possible colour densities ρ(σ) for σ : [n] → {0, 1}. We will later show that when bounding the moments of Z(H(n, m)) we can confine ourselves to colourings such that the proportion of the two colours does not deviate too much from 1/2. Formally, we say that and we denote by A ω (n) the set of all (ω, n)-balanced colour densities ρ ∈ A(n). For a hypergraph H on [n] we let Z ω (H) signify the number of (ω, n)-balanced colourings, which are 2-colourings σ such that ρ(σ) ∈ A ω (n). As we will see, it will turn out useful to split up the set A ω (n) into smaller sets in the following way. For ν ∈ N and s ∈ [ων] let Let A s ω,ν (n) be the set of all colour densities ρ ∈ A(n) such that For a hypergraph H let Z s ω,ν (H) denote the number of 2-colourings σ of H such that ρ(σ) ∈ A s ω,ν (n). We are going to apply small subgraph conditioning to Z s ω,ν rather than directly to Z. We observe that for each fixed ν we have Z ω = ων s=1 Z s ω,ν . In Section 3 we will calculate the first moments of Z and Z ω to obtain the following.
As outlined in Section 1.3, our basic strategy is to show that the fluctuations of ln Z can be attributed to fluctuations in the number of cycles of a bounded length. Hence, for an integer l ≥ 2 we let C l,n denote the number of cycles of length (exactly) l in H(n, m). Let We will see that λ l denotes the expected number of cycles of length l in a random k-uniform hypergraph, whereas δ l is a correction factor taking into account that we only allow for bichromatic edges. It is well-known that C 2,n , . . . are asymptotically independent Poisson variables [7,Theorem 5.16]. More precisely, we have the following.
Next, we investigate the impact of the cycle counts C l,n on the first moment of Z s ω,ν . In Section 4 we prove the following.
Moreover, let ω, ν ∈ N. If c 2 , . . . , c L are non-negative integers, then for any s ∈ [ων]: Additionally, we need to know the second moment of Z s ω,ν very precisely. The following proposition is the key result of our approach and the one that requires the most technical work. Its proof can be found at the end of Section 5. Proposition 2.4. Assume that k ≥ 3 and d ′ /k < 2 k−1 ln 2 − 2 and let ω, ν ∈ N. Then for every s ∈ [ων] we have We now derive Theorem 1.1 from Propositions 2.1-2.4. The key observation we will need is that the variance of the random variables Z s ω,ν can almost entirely be attributed to the fluctuations of the number of short cycles. As done in [9], the arguments we use are similar to the small subgraph conditioning from [12,20]. But we do not refer to any technical statements from [12,20] directly because instead of working only with the random variable Z we need to control all Z s ω,ν for fixed ω, ν ∈ N simultaneously. In fact, ultimately we have to take ν → ∞ and ω → ∞ as well. Our line of argument follows the path beaten in [9] and the following three lemmas are an adaption of the ones there.
For L > 2 let F L = F L,n (d, k) be the σ-algebra generated by the random variables C l,n with 2 ≤ l ≤ L. For each L ≥ 2 the standard decomposition of the variance yields The term Var E Z s ω,ν (H(n, m))|F L accounts for the amount of variance induced by the fluctuations of the number of cycles of length at most L. The strategy when using small subgraph conditioning is to bound the second summand, which is the expected conditional variance In the following lemma we show that in fact in the limit of large L and n this quantity is negligible. This implies that conditioned on the number of short cycles the variance vanishes and thus the limiting distribution of ln Z s ω,ν is just the limit of ln E Z s ω,ν |F L as n, L → ∞. This limit is determined by the joint distribution of the number of short cycles. Lemma 2.5. For d ′ ∈ (0, ∞) and any ω, ν ∈ N and s ∈ [2ων] we have Proof. Fix ω, ν ∈ N and set Z s = Z s ω,ν (H(n, m)). Using Fact 2.2 and equation (2.5) from Proposition 2.3 we can choose for any ε > 0 a constant B = B(ε) and L ≥ L 0 (ε) large enough such that for each large enough n ≥ n 0 (ε, B, L) we have for any s ∈ [ων]: The tower property for conditional expectations and the standard formula for the decomposition of the variance yields and thus, using (2.6) we have Finally, the estimate exp[−x] ≥ 1 − x for |x| < 1/8 combined with (2.7) and Proposition 2.4 implies that for large enough ν, n, L and each s ∈ [ων] we have As this holds for any ε > 0 and by equation (2.4) from Proposition 2.3 the expression exp ∞ l=2 δ 2 l λ l is bounded, the proof of the lemma is completed by first taking n → ∞ and then L → ∞. Lemma 2.6. For d ∈ (0, ∞) and any α > 0 we have Proof. To unclutter the notation, we set Z = Z(H(n, m)) and Z ω = Z ω (H(n, m)). First we observe that Proposition 2.1 implies that for any α > 0 we can choose ω ∈ N large enough such that (2.8) We let ν ∈ N. To prove the statement, we need to get a handle on the cases where the random variables Z s ω,ν (H(n, m)) deviate strongly from their conditional expectation E Z s ω,ν (H(n, m))|F L . We let Z s = Z s ω,ν (H(n, m)) and define By the definition of the X s 's and Chebyshev's inequality it is true for every s that Hence, using that with Proposition 2.1 there is a number and n large enough, we have Taking expectations, choosing ε = ε(α, β, ω) small enough and applying Lemma 2.5, we obtain (2.10) Using (2.9), Markov's inequality, (2.10) and (2.8), it follows that Finally, the triangle inequality combined with Markov's inequality and equations (2.8) and (2.11) yields which proves the statement. Proof. In a first step we show that E [|U L |] is uniformly bounded. As x − x 2 ≤ ln(1 + x) ≤ x for |x| ≤ 1/8 we have for every l ≤ L:

Lemma 2.7. Let
(2.14) Proposition 2.3 ensures that l δ 2 l λ l < ∞. Furthermore, as we are in the regime d ′ /k ≤ 2 k−1 ln 2, we have l δ l √ λ l ≤ l k l 2 −(k−1)l/2 < ∞ and thus (2.14) shows that E [|U L |] is uniformly bounded. To prove (2.13), for given n and a constant B > 0 we let C B be the event that C l,n < B for all l ≤ L. Referring to Fact 2.2, we can find for each L, ε > 0 a B > 0 such that (2.15) To simplify the notation we set Z = Z(H(n, m)) and Z ω = Z ω (H(n, m)). By Proposition 2. On the other hand, for α sufficiently small and large enough n we have Thus, the proof of (2.13) is completed by combining (2.15), (2.16), (2.17) and taking logarithms.
Proof of Theorem 1.1. For L ≥ 2 we define Then Fact 2.2 implies that for each L the random variables U L defined in (2.12) converge in distribution to W L as n → ∞. Furthermore, because l δ l √ λ l , l δ 2 l λ l < ∞, the martingale convergence theorem implies that W is well-defined and that the W L converge to W almost surely as L → ∞. Therefore, from Lemmas 2.7 and 2.6 it follows that ln Z(H(n, m)) − ln E [Z(H(n, m))] converges to W in distribution, meaning that for any ε > 0 we have To prove the second part, we construct an event whose probability is bounded away from 0 and that is such that conditioned on this event, the number of solutions of the random hypergraph H k (n, m) is not concentrated very strongly. We consider the event T t that the random hypergraph H k (n, m) contains t isolated triangles, i. e. t connected components such that each component consists of 3k − 3 vertices and 3 edges and the intersection of each pair of edges contains exactly one vertex. It is well-known that for t ≥ 0 there exists ε = ε(d, t) > 0 such that (2.20) Given T t , we let H * k (n, m) denote the random hypergraph obtained by choosing a set of t isolated triangles randomly and removing them. Then H * k (n, m) is identical to H k (n − (3k − 3)t, m − 3t) and with Proposition 2.1 there exists a constant C = C(d, k) such that A very accurate calculation of the number of 2-colourings of a triangle in a hypergraph yields that this number is given by 2 k−2 − 1 2 2k−1 − 2 k + 2 . Thus, we obtain implying that for any ω > 0 we can choose t large enough so that E [Z(H k (n, m))|T t ] ≤ E [Z(H k (n, m))] /(2 exp [ω]). Using Markov's inequality, we obtain Thus, combining (2.20) and (2.21) yields that for any finite ω > 0 there is ε > 0 such that for large enough n we have thereby completing the proof of the second claim.
Proof of Corollary 1.4. This proof is nearly identical to the one in [4]. Assume for contradiction that (A n ) n≥1 is a sequence of events such that for some fixed number 0 < ε < 1/2 we have

THE FIRST MOMENT CALCULATION
The aim in this section is to prove Proposition 2.1 and a result that we need for Proposition 2.4. For a hypergraph H let Z ρ (H) be its number of 2-colourings with colour density ρ. We setρ = 1 2 . For ρ ∈ [0, 1] we define The next lemma shows that f 1 (ρ) is the function we need to analyse in order to determine the expectation of Z ρ .
Lemma 3.1. Let d ′ ∈ (0, ∞). There exist numbers C 1 = C 1 (k, d), C 2 = C 2 (k, d) > 0 such that for any colour density ρ: Proof. The edges in the random hypergraph H(n, m) are independent by construction, so the expected number of solutions with colour density ρ can be written as Further, the number of "forbidden" edges is given by To proceed we observe that ln x + y n = ln(x) + ln 1 + y xn for x > 0, y < xn and consequently The following corollary states an expression for E [Z (H(n, m))]. Additionally, it shows that when ω → ∞, this value can be approximated by E [Z ω (H(n, m))].

COUNTING SHORT CYCLES
We recall that for l ∈ {2, . . . , L} we denote by C l,n the number of cycles of length l in H(n, m). Further we let c 2 , . . . , c L be a sequence of non-negative integers and S be the event that C l,n = c l for l = 2, . . . , L. Additionally, for an assignment σ : [n] → {0, 1} we let V(σ) be the event that σ is a colouring of the random graph H(n, m). We also recall λ l , δ l from (2.3).
Proof of Proposition 2.3. First observe that from the definition of λ l and δ l in (2.3) and the fact that Together with (4.1), Proposition 2.3 readily follows from the following lemma about the distribution of the random variables C l,n given V(σ).
Before we establish Lemma 4.1, let us point out how it implies Proposition 2.3. By Bayes' rule, we have We consider the number of sequences of m 2 + · · · + m L distinct cycles such that m 2 corresponds to the number of cycles of length 2, and so on. Clearly this number is equal to (C 2,n ) m2 · · · (C L,n ) mL .
We call a cycle good, if it does not contain edges that overlap on more than one vertex. We call a sequence of good cycles good sequence if for any two cycles C and C ′ in this sequence, there are no vertices v ∈ C and v ′ ∈ C ′ such that v and v ′ are contained in the same edge. Let Y be the number of good sequences andȲ be the number of sequences that are not good. Then it holds that (4.4) The following claim states that the contribution of E[Ȳ |V(σ)] is negligible. Its proof follows at the end of this section.
Thus it remains to count good sequences given V(σ). We let σ ∈ A ω (n) and first consider the number D l,n of rooted, directed, good cycles of length l. This will introduce a factor of 2l for the number of all good cycles of length l, thus D l,n = 2lC l,n . For a rooted, directed, good cycle of length l we need to pick l vertices (v 1 , ..., v l ) as roots, introducing a factor (1 + o(1)) n 2 l , and there have to exist edges between them which generates a factor To choose the remaining vertices in the participating edges we have to distinguish between pairs of vertices (v i , v i+1 ) that are assigned the same colour and those that are not, because if σ(v i ) = σ(v i+1 ) we have to make sure that at least one of the other k − 2 vertices participating in this edge is assigned the opposite colour. This gives rise to the third factor in the following calculation.
Hence, recalling that C l,n = 1 2l D l,n , we get (4.5) In fact, since Y considers only good sequences and l, m 2 , . . . , m L remain fixed as n → ∞, (4.5) yields Plugging the above relation and Claim 4.2 into (4.4) we get (4.3). The proposition follows.

Proof of Claim 4.2:
The idea of the proof is to find an event, namely that there exists an induced subgraph with too many edges, that always occurs ifȲ > 0 and whose probability we can bound from above. To this aim let A = {i ∈ R|i = (l − 1)(k − 1) + j for some l ≤ L, j ∈ {0, ..., k − 2}}. For every subset R of (l − 1)(k − 1) + j vertices, where l ≤ L and j ∈ {0, ..., k − 2} let I R be equal to 1 if the number of edges that only consist of vertices in R is at least l. Let the H L be the event that R:|R|∈A I R > 0. It is direct to check that ifȲ > 0 then H L occurs. This implies that For any set R such that |R| = (l − 1)(k − 1) + j, we can put l edges inside the set in at most ( (l−1)(k−1)+j k ) l ways, which obviously gets largest if j = k − 2 and thus (l − 1)(k − 1) + j = l(k − 1) − 1. Clearly conditioning on V(σ) can only reduce the number of different placings of the edges. 13 We observe that for a colouring σ and two fixed vertices v and v ′ with σ(v) = σ(v ′ ) the probability that e(v, v ′ ) . Using inclusion/exclusion and the binomial theorem, with N = n k and F (σ) ∼ 2 1−k N , for a fixed set R of cardinality (l − 1)(k − 1) + j we get that With m = dn k and since i j ≤ (ie/j) j , it holds that where the last equality follows since L is a fixed number.

THE SECOND MOMENT CALCULATION
In this section we prove Proposition 2.4. To this end, we need to derive an expression for the second moment of the random variables Z s ω,ν for s ∈ [ων] that is asymptotically tight. As a consequence, we need to put more effort into the calculations than done in prior work on hypergraph-2-colouring (e.g. [10]), where the second moment of Z is only determined up to a constant factor. Part of the proof is based on ideas from [4], but as we aim for a stronger result, the arguments are extended and adapted to our situation.
If we further remember the definition from (2.1), we can alternatively represent ρ(σ, τ ) as .
To simplify the notation, for a 2 × 2-matrix ρ = (ρ ij ) we introduce the shorthands We  ρ (H) be the number of pairs (σ, τ ) of 2-colourings of H whose overlap matrix is ρ. Analogously to (3.1), we define the functions f 2 , g 2 : B → R as The following lemma states a formula for E Z (2) ρ (H(n, m)) for ρ ∈ B(n) in terms of f 2 (ρ).
Then for ρ ∈ B(n) we have Proof. Let ρ = ρ 00 ρ 10 ρ 01 ρ 11 ∈ B(n). Then where N = n k and F (σ, τ ) is the total number of possible monochromatic edges under either σ or τ . In the last line, σ and τ are just two arbitrary fixed 2-colourings with overlap ρ and the equation is valid because the following computation shows that F (σ, τ ) only depends on ρ: 15 We proceed as in the proof of Lemma 3.1 by using that ln x − y n = ln(x) + ln 1 − y xn for x > 0, y n < x and consequently As F (σ, τ ) does only depend on ρ, (5.4) becomes Using Stirling's formula, we get the following approximation for the number of colour assignments with overlap ρ: n ρ 00 n, ρ 01 n, ρ 10 n, ρ 11 n

Dividing up the interval.
Let ω, ν ∈ N and s ∈ [ων]. Analogously to the notation in Section 2 we introduce the sets and B s ω,ν (n) = ρ ∈ B ω (n) : ρ i⋆ , ρ ⋆i ∈ ρ s ω,ν − imposing constraints on the overlap matrix ρ insofar as the colour densities resulting from its projection on each colouring must not deviate too much from 1/2 in the set B ω (n) and from ρ s ω,ν in the set B s ω,ν (n). By the linearity of expectation, for any s ∈ [ων] we have E Z s ω,ν (H(n, m)) 2 = ρ∈B s ω,ν (n) E Z (2) ρ (H(n, m)) .
To prove this proposition, we need the following lemma.