Uniqueness for the 3-State Antiferromagnetic Potts Model on the Tree

The antiferromagnetic $q$-state Potts model is perhaps the most canonical model for which the uniqueness threshold on the tree is not yet understood, largely because of the absence of monotonicities. Jonasson established the uniqueness threshold in the zero-temperature case, which corresponds to the $q$-colourings model. In the permissive case (where the temperature is positive), the Potts model has an extra parameter $\beta\in(0,1)$, which makes the task of analysing the uniqueness threshold even harder and much less is known. In this paper, we focus on the case $q=3$ and give a detailed analysis of the Potts model on the tree by refining Jonasson's approach. In particular, we establish the uniqueness threshold on the $d$-ary tree for all values of $d\geq 2$. When $d\geq3$, we show that the 3-state antiferromagnetic Potts model has uniqueness for all $\beta\geq 1-3/(d+1)$. The case $d=2$ is critical since it relates to the 3-colourings model on the binary tree ($\beta=0$), which has non-uniqueness. Nevertheless, we show that the Potts model has uniqueness for all $\beta\in (0,1)$ on the binary tree. Both of these results are tight since it is known that uniqueness does not hold in the complementary regime. Our proof technique gives for general $q>3$ an analytical condition for proving uniqueness based on the two-step recursion on the tree, which we conjecture to be sufficient to establish the uniqueness threshold for all non-critical cases ($q\neq d+1$).


Introduction
The q-state Potts model is a fundamental spin system from statistical physics that has been thoroughly studied in probability and computer science. The model has two parameters q and β, where q ≥ 3 is the number of the states, and β > 0 is a parameter which corresponds to the temperature of the system 1 . The set of states is given by [q] = {1, . . . , q} and we will usually refer to them as colours. The case q = 2 is known as the Ising model, and the Potts model is the generalisation of the Ising model to multiple states. When β = 0, the Potts model is known as the q-colourings model.
A configuration of the Potts model on a finite graph G = (V, E) is an assignment σ : V → [q]. The weight of the configuration σ is given by w G (σ) = β m(σ) , where m(σ) denotes the number of monochromatic edges in G under the assignment σ. The Gibbs distribution of the model, denoted by Pr G [·], is the probability distribution on the set of all configurations, where the probability mass of each configuration σ is proportional to its weight w G (σ). Thus, for any σ : V → [q] it holds that where Z G = σ:V →[q] w G (σ) is the so-called partition function. Note that in the case β = 0 the Gibbs distribution becomes the uniform distribution on the set of proper q-colourings of G. The Potts model is said to be ferromagnetic if β > 1, which means that more likely configurations have many monochromatic edges. It is said to be antiferromagnetic if β < 1, which means that more likely configurations have fewer monochromatic edges. This paper is about the antiferromagnetic case.
For spin systems like the Ising model and the Potts model, one of the most well-studied subjects in statistical physics is the so-called uniqueness phase transition on lattice graphs, such as the grid or the regular tree. Roughly, the uniqueness phase transition on an infinite graph captures whether boundary configurations can exert non-vanishing influence on far-away vertices. In slightly more detail, for a vertex v and an integer n, fix an arbitrary configuration on the vertices that are at distance at least n from v. Does the influence on the state of v coming from the boundary configuration vanish when n → ∞? If yes, the model has uniqueness, and it has non-uniqueness otherwise. 2 (See Definition 1 for a precise formulation in the case of the tree.) Note that uniqueness is a strong property, which guarantees that the effect of fixing an arbitrary boundary configuration eventually dies out. As an example, for the antiferromagnetic Ising model on the d-ary tree it is well-known that uniqueness holds iff β ≥ d−1 d+1 ; the value d−1 d+1 is a point of a phase transition and is also known as the uniqueness threshold because it is the point at which the uniqueness phase transition occurs. The uniqueness phase transition plays a prominent role in connecting the efficiency of algorithms for sampling from the Gibbs distribution to the properties of the Gibbs distribution itself. One of the first examples of such a connection is in the analysis of the Gibbs sampler Markov chain for the Ising model on the 2-dimensional lattice, where the uniqueness phase transition marks the critical value of β where the mixing time switches from polynomial to exponential (see [16,15,25]).
From a computational complexity perspective, it is the uniqueness phase transition on the regular tree which is particularly important. For many 2-state spin models, including the antiferromagnetic Ising model and the hard-core model, it has been proved [23,24,8,12] that the uniqueness phase transition on the tree coincides with a more general computational transition in the complexity of approximating the partition function or sampling from the Gibbs distribution. In the case of the antiferromagnetic Ising model for example, the problem of approximating the partition function on (d + 1)-regular graphs undergoes a computational transition at the tree uniqueness threshold: it admits a polynomial-time algorithm when β ∈ ( d−1 d+1 , 1) and it is NP-hard for β ∈ (0, d−1 d+1 ). This connection has been established in full generality for antiferromagnetic 2-state systems . For antiferromagnetic multi-state systems, the situation is much less clear and, in fact, even understanding the uniqueness phase transition on the tree poses major challenges. One of the key reasons behind these difficulties is that certain monotonicities that hold for two-state systems simply do not hold in the multi-state setting, which therefore necessitates far more elaborate techniques. For analysing the uniqueness threshold on the tree, this difficulty has already been illustrated in the case of the q-colourings model, where Jonasson [11], building upon work of Brightwell and Winkler [2], established via a painstaking method that the model is in uniqueness on the d-ary tree iff q > d + 1. The goal of this paper is to extend this analysis 2 The terminology comes from the theory of Gibbs measures, where the interest is in examining whether there is a unique infinite-volume measure whose marginals on finite regions is given by the Gibbs distribution (it can be shown that an infinite-volume measure always exists). See [10,6] for a thorough exposition of the theory. The two formulations of uniqueness/non-uniqueness that we have described, i.e., examining infinite-volume measures and examining the limit of marginals in growing finite regions, turn out to be equivalent. to the Potts model (beyond the zero-temperature case).
There are several reasons for focusing on establishing uniqueness on the tree. For the colourings model and the antiferromagnetic Potts model, it is widely conjectured that the uniqueness phase transition on the d-ary tree captures the complexity of approximating the partition function on graphs with maximum degree d + 1, as is the case for antiferromagnetic 2-state models. It has been known since the 80s that non-uniqueness holds for the colourings model when q ≤ d + 1 and for the Potts model when β < 1 − q/(d + 1), see [18]. More recently, it was shown in [7] that the problem of approximating the partition function is NP-hard when q < d + 1 for the colourings model and when β < 1 − q/(d + 1) for the Potts model (for q even). It is not known however whether efficient algorithms can be designed in the complementary regime; for correlation decay algorithms in particular (see [9,14,13]), it has been difficult to capture the uniqueness threshold in the analysis -this becomes even harder in the case of the Potts model where uniqueness is not known. For a more direct algorithmic consequence of uniqueness, it has been demonstrated that, on sparse random graphs, sampling algorithms for the Gibbs distribution can be designed by exploiting the underlying tree-like structure and the decay properties on the tree guaranteed by uniqueness. In particular, in the G(n, d/n) random graph, Efthymiou [4] developed a sampling algorithm for q-colourings when q > (1 + ǫ)d, based on Jonasson's uniqueness result. Related results on G(n, d/n) appear in [27,5,22,17]. Also, after presenting our main result, we will describe an application on random regular graphs, appearing in [1].

Our result
In this paper, we study the uniqueness threshold for the antiferromagnetic Potts model on the tree. We establish the uniqueness threshold for q = 3 for every d ≥ 2. Our proof technique, which is a refinement of Jonasson's approach, also gives, for general q > 3, an analytical condition for proving uniqueness, which we conjecture to be sufficient for establishing the uniqueness threshold whenever q = d + 1. As we shall discuss shortly, the case q = d + 1 is special, since it incorporates the critical case for the colourings model. To formally state our result, we will need a few definitions.
Given a graph G = (V, E), a configuration σ : V → [q], and a subset U of V , we use σ(U ) to denote the restriction of the configuration σ to the vertices in U . For a vertex v ∈ V and a colour c ∈ [q], we denote by Pr G [σ(v) = c] the probability that v takes the colour c in the Gibbs distribution. Let T d,n be the d-ary tree with height n (i.e., every path from the root to a leaf has n edges, and every non-leaf vertex has d children). 3 Let Λ T d,n be the set of leaves of T d,n and let v d,n be its root. The following definition formalises uniqueness on the d-ary tree. (See also [2] for details about how to translate Definition 1 to the Gibbs theory formalisation.) It has non-uniqueness otherwise.
Equation 1 formalises the fact that the correlation between the root of a d-ary tree and vertices at distance n from the root vanishes as n → ∞. We are now ready to state our main result.
Theorem 2. Let q = 3. When d ≥ 3, the 3-state Potts model on the d-ary tree has uniqueness for all β ∈ [ d−2 d+1 , 1). When d = 2, the 3-state Potts model on the binary tree has uniqueness for all β ∈ (0, 1). Theorem 2 precisely pinpoints the uniqueness threshold for the 3-state Potts model since it is known that the model is in non-uniqueness in the complementary regime. When d ≥ 3, nonuniqueness for β < d−2 d+1 follows from the existence of multiple semi-translation-invariant Gibbs measures 4 . When d = 2, the 3-state Potts model for β = 0 corresponds to the 3-colouring model, and non-uniqueness holds in this case because of the existence of so-called frozen 3-colourings; in these colourings, the configuration on the leaves determines uniquely the colour of the root, see [2].
Interestingly, our result and proof technique for the 3-state Potts model suggests that the only obstruction to uniqueness in the 3-colouring model on the binary tree are the frozen colouring configurations. It is reasonable to believe that this critical behaviour in the colourings model happens more generally whenever q = d + 1. For comparison, note that the colourings model has non-uniqueness when q < d + 1 ([2], see also footnote 4) and it has uniqueness when q > d + 1 [11].
This critical behaviour for the colourings model when q = d + 1 arises in the context of the Potts model as well, and, as we shall see in the next section, it causes complications in the proof of Theorem 2. Nevertheless, we formulate a general condition for all non-critical cases (q = d + 1) which will be sufficient to establish the uniqueness threshold. We conjecture that the condition holds whenever q = d + 1 (see Conjecture 17). The condition is tailored to the Potts model on a tree, unlike other known sufficient criteria for uniqueness (see for example [3,26]). Our condition reduces to single-variable inequalities and can be verified fairly easily for small values of q, d. Since Theorem 2 includes the critical case (q, d) = (3,2), our proof of the theorem necessarily goes a slightly different way (as we explain below), so in Section 2, we give a more detailed outline of our proof approach.

Application
We have already discussed some results in the literature where the uniqueness of spin-models on trees enables fast algorithms for sampling from these models on bounded-degree graphs and sparse random graphs. It turns out that Theorem 2 can also be used in this way. In particular, Blanca et al. have obtained the following theorem.
Theorem 3 (Theorem 8 of [1]). Let q ≥ 3, d ≥ 2, and β ∈ (0, 1) be in the uniqueness regime of the d-ary tree with β = (d + 1 − q)/(d + 1). Then, there exists a constant δ > 0 such that, for all sufficiently large n, the following holds with probability 1 − o(1) over the choice of a random There is a polynomial-time algorithm which, given the graph G as input, outputs a random assignment σ : V → [q] from a distribution which is within total variation distance O(1/n δ ) from the Gibbs distribution of the Potts model on G with parameter β.
Thus, Theorem 2 has the following corollary.
There is a polynomial-time algorithm which, given the graph G as input, outputs a random assignment σ : V → [q] from a distribution which is within total variation distance O(1/n δ ) from the Gibbs distribution of the Potts model on G with parameter β.
We next discuss our approach for proving Theorem 2.

Proof Approach
In this section, we outline the key steps of our proof approach for proving uniqueness for the antiferromagnetic Potts model on the tree. As mentioned in the Introduction, the model does not enjoy the monotonicity properties which are present in two-state systems (or the ferromagnetic case) 5 , so we have to establish more elaborate criteria to resolve the uniqueness threshold.
We first review Jonasson's approach for colourings [11]. One of the key insights there is to consider the ratio of the probabilities that the root takes two distinct colours and show that this converges to 1 as the height of the tree grows large. Jonasson analysed first a one-step recursion to establish bounds on the marginals of the root and used those to obtain upper bounds on the ratio. Then, he bootstrapped these bounds by analysing a more complicated two-step recursion and showed that the ratio converges to 1. Our approach refines Jonasson's approach in the following way; we jump into the two-step recursion and analyse the associated optimisation problem by giving an explicit description of the maximisers for general q and d (see Lemma 10). It turns out that the maximisers change as the value of the ratio gets closer to 1, so to prove the desired convergence to 1, we need to account for the roughly q d possibilities for the maximiser. This yields an analytic condition that can be checked easily for small values of q, d and thus establish uniqueness. In the context of Theorem 2 where q = 3, most of the technical work is to deal analytically with the potentially large values of the arity d of the tree.
A further complication arises in the case q = 3 and d = 2 (and more generally q = d + 1), since this incorporates the critical behaviour for colourings described in Section 1. 1. This manifests itself in our proof by breaking the (global) validity of our uniqueness condition. We therefore have to use an analogue of Jonasson's approach to account for this case by first using the one-step recursion to argue that the ratio gets sufficiently close to 1 and then finishing the argument with the two-step recursion.
Our proofs are computer-assisted but rigorous -namely we use the (rigorous) Resolve function of Mathematica to check certain inequalities. We also provide Mathematica code to assist the reader with tedious-but-straightforward calculations (such as differentiating complicated functions). The Mathematica code is in Section 7.
Note that if β > 0 and n > 0, then for every τ : Suppose, for fixed q, β and d, that lim n→∞ γ(q, β, d, n) = 1. This implies that the limsup in the uniqueness definition (Definition 1) is zero. Thus, Theorem 2 is an immediate consequence of the following theorem.
In Section 3 we obtain Theorem 2 by proving Theorem 5.

The two-step recursion
In this section, we formulate an appropriate recursion on the infinite d-ary tree, which will be one of our main tools for tracking the ratio γ(q, β, d, n).
We denote the set of q-dimensional probability vectors by △, i.e., Suppose that c 1 and c 2 are two colours in [q]. We define two functions g c 1 ,c 2 ,β and h c 1 ,c 2 ,β , indexed by these colours. The argument of each of these functions is a tuple (p (1) , . . . , p (d) ) where, for each j ∈ [d], p (j) ∈ △. The functions are defined as follows.
The following proposition, proved in Section 4, shows the relevance of these functions for analysing the tree.
Since △ α is compact and h c 1 ,c 2 ,β is continuous, the maximisation in (5) is well-defined.
Definition 7 ensures that △ α is the subset of △ induced by probability vectors whose entries are within a factor of α > 1 of each other. M α,c 1 ,c 2 ,β is the maximum of the two-step recursion function h c 1 ,c 2 ,β when each of its arguments are from △ α . The following proposition gives a preliminary condition for establishing uniqueness when β ∈ (0, 1) -it is proved in Section 3.
In the next section, we will show how to simplify the condition in Proposition 8.

A simpler condition for uniqueness
Proposition 8 gives a sufficient condition on the two-step recursion that is sufficient for establishing uniqueness based on the maximisation of h c 1 ,c 2 ,β . Due to the many variables involved in the maximisation, this is rather complicated for any direct verification. We will simplify this maximisation signifantly by showing that it suffices to consider very special vectors whose entries are either equal to α or 1. We start with the following definition of "extremal tuples".
One of the consequences of Lemma 10 is that the validity of the inequality in Proposition 8 is monotone with respect to β. In particular, we have the following lemma (also proved in Section 6.1).
Lemma 11. Let q ≥ 3, d ≥ 2 and β ′ , β ′′ ∈ [0, 1) with β ′ ≤ β ′′ . Then, for all α > 1 and any colours c 1 , c 2 ∈ [q], it holds that Another consequence of Lemma 10 is that, combined with the scale-free property, it reduces the verification of the condition in Proposition 8 to the verification of single-variable inequalities in α. These inequalities are obtained by trying all d-tuples of q-dimensional vectors whose entries are as follows.
Ex c (α) = (p 1 , . . . , p q The following simplified condition will be our main focus henceforth. If C(α) holds, we say that the pair (q, d) satisfies Condition 12 for α. Now, to verify the inequality in Proposition 8, we will show shortly that it suffices only to establish Condition 12 for all α > 1, which turns out to be a much more feasible task because of the very explicit form of the set Ex c 2 (α). In the next section, we discuss how to do this in detail, but for now let us state a proposition which asserts that this is indeed sufficient.
Then by Lemma 11, we obtain that, for all β ∈ [β * , 1) it holds that M α,c 1 ,c 2 ,β < α 1/d as well for all α > 1 and therefore, by Proposition 8, the Potts model has uniqueness for all such β.
For c ∈ [q] and k ∈ [d], denote by p (k) c the entry of p (k) corresponding to colour c and letp (k) be the vector t k p (k) where t k = 1/p (k) c 2 . By the definition of an (α, c 2 )-extremal tuple, we have thatp (1) , . . . ,p (d) ∈ Ex c 2 (α).

Verifying the Condition
In this section, we give more details on how to verify Condition 12. To apply Proposition 13, we will need to verify Condition 12. The latter is fairly simple to verify for small values of q, d since it reduces to single-variable inequalities in α. We illustrate the details when (q, d) = (3,3) and (q, d) = (4,4).
Combining Lemma 15 with Proposition 13 we get the following immediate corollary. Corollary 16 establishes the uniqueness threshold for (q, d) = (3,3) and (q, d) = (4,4). More generally, we are interested in the following question: When is Condition 12 satisfied for all α > 1? We conjecture the following.
We have only been able to verify Conjecture 17 for specific values of q, d (with methods similar to those used in the proof of Lemma 15). However, it is important to note that the restriction q = d + 1 in the conjecture cannot be removed. For example, the pair (q, d) = (3,2) does not satisfy Condition 12 for all α > 1 -it only satisfies the condition for α fairly close to 1. Thus, to prove Theorem 2 we need a different argument to account for the case (q, d) = (3,2).
Thus, instead of trying to prove Conjecture 17 for all values of α (which wouldn't be enough for our theorem), we follow Jonasson's approach and use the one-step recursion to argue that the ratio γ(q, β, d, n) gets moderately close to 1; close enough that we can then use the two-step recursion to finish the proof of uniqueness. Note that, in contrast to the two-step recursion, the one-step recursion is not sufficient on its own to obtain tight uniqueness results for any values of q, d (this was also observed by Jonasson [11] in the case of colourings).
First, we state the one-step recursion that we are going to use on the tree. This recursion, as well as the two-step recursion of Proposition 6, are well-known, but we prove them explicitly in Section 4 for completeness.
be an arbitrary configuration. Let , let T i be the subtree of T rooted at v i and let Λ i denote the set of leaves of the subtree T i . Then, for any colour c ∈ [q], it holds that Tracking the one-step recursion relatively accurately requires a fair amount of work, and to aid the verification of Condition 12 in the case q = 3, we do this for general values of d. In particular, we prove the following lemma in Section 5.
Lemma 19. Let q = 3 and c ∈ [3] be an arbitrary colour. For d ≥ 2, consider the d-ary tree T d,n with height n and let τ : Λ T d,n → [3] be an arbitrary configuration on the leaves.
When d = 2, for all β ∈ (0, 1), for all sufficiently large n it holds that , there exist sequences {L n } and {U n } (depending on d and β) such that for all sufficiently large n The following corollary is an immediate consequence of Lemma 19.

Concluding uniqueness
In this section, we prove Proposition 8 and also conclude the proof of Theorem 5 (assuming for now Lemmas 19 and 21 and also Lemma 10, which we have already used). Recall that We will need the following proposition.
Proposition 22. Let q ≥ 3, d ≥ 2 and β ∈ (0, 1). Suppose that, for some integer n ≥ 3 and some α > 1, it holds that γ(q, β, d, n − 2) = α and M α, be an arbitrary configuration. As in Proposition 6, let {z i,j } i,j∈ [d] denote the grandchildren of the root, let T i,j be the subtree of T rooted at z i,j , and let Λ i,j be the set of leaves of T i,j . Further, let r (i,j) be the marginal distribution at z i,j in the subtree T i,j , conditioned on the configuration τ (Λ i,j ). By the assumption γ(q, β, d, n − 2) = α and the definition (2) of the ratio γ(q, β, d, n − 2), we have that r (i,j) ∈ △ α for all i, j ∈ [d]. Proposition 6 also guarantees that for colours c 1 ∈ [q] and c 2 ∈ [q] we have where the strict inequality follows by the assumption that M α,c 1 ,c 2 ,β < α 1/d . Since τ was an arbritrary configuration on the leaves Λ, we obtain that γ(q, β, d, n) < γ(q, β, d, n − 2) as needed.
We start with Proposition 8, which we restate here for convenience.
Proposition 8. Let q ≥ 3, d ≥ 2 and β ∈ (0, 1). Suppose that for all α > 1 and any colours Then, it holds that γ(q, β, d, n) → 1 as n → ∞, i.e., the q-state Potts model with parameter β has uniqueness on the d-ary tree. Proof. Fix q, d and β as in the statement. For all n ≥ 1, let α n := γ(q, β, d, n). We may assume that α n > 1 for all n ≥ 1 (otherwise, uniqueness follows trivially by choosing n 0 such that α n 0 = 1 and then applying Proposition 18 repeatedly to show α n = 1 for all n ≥ n 0 .) Using Proposition 22 and the assumption that M α,c 1 ,c 2 ,β < α 1/d for all α > 1 and colours This implies that both of the sequences {α 2n } and {α 2n+1 } are decreasing. Since both of these sequences is bounded below by 1, we obtain that for n → ∞ it holds that 6 α 2n ↓ α ev , α 2n+1 ↓ α odd for some α ev , α odd ≥ 1. We claim that in fact both of α ev , α odd are equal to 1, which proves that γ(q, β, d, n) → 1 as n → ∞.
Assuming Lemmas 19 and 21, we can also conclude the proof of Theorem 5 in a similar way.
Proof. We first consider the case d ≥ 3. Let β ∈ [1 − 3 d+1 , 1) and for all n ≥ 1, set α n = γ (3, d, β, n). By Lemma 19, we have that there exists n 0 such that for all n ≥ n 0 , it holds that α n ∈ (1, 53/27]. (The reason that the left end-point of the interval is open is that we finish if α n = 1, as in the proof of Proposition 8.) By Lemma 21, the pair (q, d) satisfies Condition 12 for α n (for n ≥ n 0 ). By the definition of Condition 12, for all c 1 , c 2 ∈ [q], and (p (1) , . . . , p (d) Using Proposition 22 we obtain that for all n ≥ n 0 + 2, it holds that This implies that both of the sequences {α 2n } n≥n 0 and {α 2n+1 } n≥n 0 are decreasing, and since both are bounded below by 1, they converge. We now use an argument that is almost identical to the one used in the proof of Proposition 8. The only difference is that now the sequences start from 2n 0 and 2n 0 + 1 instead of from n = 2 and n = 3, respectively. Using this argument, we obtain that the limits of {α 2n } n≥n 0 and {α 2n+1 } n≥n 0 must be equal to 1, thus proving that γ(3, d, β, n) → 1 as n → ∞.
The argument for the case d = 2 and β ∈ (0, 1) is actually the same; the only difference to the case d ≥ 3 is that β lies in an open interval instead of a half-open interval.

Proving Tree Recursions
In this section, we give proofs of the (standard) tree recursions, which we have already used. We first prove Proposition 18 for the one-step recursion.
Proposition 18. Suppose q ≥ 3, d ≥ 2 and β ∈ (0, 1). For an integer n ≥ 1, let T be the tree T d,n with root v = v d,n and leaves Λ = Λ T d,n . Let τ : Λ → [q] be an arbitrary configuration. Let , let T i be the subtree of T rooted at v i and let Λ i denote the set of leaves of the subtree T i . Then, for any colour c ∈ [q], it holds that Proof. For any graph G, we use V (G) to denote the vertex set of G. Recall that, for any configuration σ : V (G) → [q], its weight in the Potts model with parameter β is given by We will typically be interested in the case where G is a sub-tree of T . We have Combining this with (15) and (16), we obtain the statement of the lemma.
We next prove Proposition 6 for the two-step recursion of Section 2.2.
Proof of Proposition 6. For i ∈ [d] and c ∈ [q], let T i be the subtree of T rooted at z i with leaves Λ T i and let r (i) We have r (i) c > 0 for every c ∈ [q], hence we can apply Proposition 18 to T to obtain Analogously, for every c 1 , c 2 ∈ [q], we have Plugging (17) into (18), and using the definition of h c 1 ,c 2 ,β from (3), we obtain the statement of the lemma.

Bounds from the one-step recursion -Proof of Lemma 19
In this section, we prove Lemma 19.

Bounding the marginal probability at the root by the one-step recursion
We begin by giving an upper and a lower bound for the marginal probability that the root is assigned a colour c via the one-step recursion (see the upcoming Lemma 24). First we define two functions. Let We will use the following lemma.
Lemma 23. Let f be a convex function on an interval I = [a, b].

Given
Proof. We first prove Item 1. Suppose y, z ∈ J satisfy y < z, we will show that g(z) ≤ g(y).
We have y < z ≤ ρ − z < ρ − y and y ≥ a, ρ − y ≤ b (using ρ ≤ a + b). It follows that all of y, z, ρ − y, ρ − z belong to I. Moreover, by the convexity of f on I, we conclude that the slope of f in the interval [ρ − z, ρ − y] is greater or equal to the slope of f in the interval [y, z], i.e., Re-arranging, we obtain g(z) ≤ g(y). The proof of Item 2 is analogous. For y, z ∈ J satisfying y < z, we have that ρ − z < ρ − y ≤ y < z and all of y, z, ρ − y, ρ − z belong to I (using ρ ≥ a + b). By the convexity of f on I, we conclude that the slope of f in the interval [ρ − z, ρ − y] is less or equal to the slope of f in the interval [y, z], which gives that g(z) ≥ g(y).
The following lemma gives recursively-generated bounds on the probability that the root of T d,n is a given colour.
Next, we show that p ≤ f u (d, β, U, L). To give an upper bound on p, it suffices to lower bound R. Since p i,1 ≥ L for every i ∈ [d], we obtain the lower bound Using the arithmetic-mean geometric-mean inequality we have . Let x = U and y = max{p i,2 , p i,3 }. Since p i,2 and p i,3 are at most U , x and y are in J and satisfy x ≥ y. Therefore, g(x) ≥ g(y), which gives that ln 1 −βy + ln 1 −β(ρ − y) ≥ ln 1 −βx + ln 1 −β(ρ − x) .
Thus, by substituting in the values of x, y and ρ and exponentiating, we have Using the inequality p i,2 + p i,3 = 1 − p i,1 ≤ 1 − L in the right-hand side, we get Plugging this into (23) for each i ∈ [d] and then into (22), we obtain that Therefore, using (19), we obtain the upper bound p ≤ f u (d, β, U, L).

Properties of the functions f u and f ℓ
In this section, we establish useful monotonicity properties of the functions f u and f ℓ that will be relevant later.
The derivative of f u with respect to β is given by (Obviously, this can be checked directly, but the reader may prefer to use the Mathematica code in Section 7.2 to check this and the derivative of f ℓ with respect to β, which appears below.) Using the conditions on x and y in the statement of the lemma, we find that We conclude that ∂fu ∂β ≤ 0 so f u (d, β, x, y) is a decreasing function of β on the interval [0, 1]. Similarly, is an increasing function of β on the interval (0, 1).

Proof. Let
We first prove Item 2. Let a = 0, b = 1, ρ = 1 − x and consider the interval It follows that, for fixed x, R is a decreasing function of y on J and therefore f ℓ is increasing in y. It remains to observe that, for x ∈ [0, 1], the condition y ∈ J is equivalent to the condition x + 2y ≤ 1 and y ∈ [0, 1] in the statement. For is an increasing nonnegative function of x and 1 − (1 − β)x is a decreasing nonnegative function of x, so R is an increasing function of x. Thus, f ℓ is a decreasing function of x.

Proof. The proof is analogous to that of Lemma 26. Let
We first prove Item 1 It follows that the function is a decreasing function of x on J, and therefore R has the same property as well. Thus, f u is an increasing function of x on the interval J. It remains to observe that, for y ∈ [0, 1], the condition x ∈ J is equivalent to the condition 1 ≤ 2x + y and x ∈ [0, 1] in the statement. For Item 2, note that 1 − (1 − β)(1 − x − y) is an increasing nonnegative function of y and 1 − (1 − β)y is a decreasing nonnegative function of y, so R is an increasing function of y. Thus, f u is a decreasing function of y.

Proof.
Using the assumptions in the statement of the lemma, we obtain 1 ≤ 2x 1 + y 2 and 2y 1 + x 2 ≤ 1. Therefore, by Lemmas 26 and 27, we obtain

Bounding the marginal probability at the root by two sequences
For any β > 0 and d ≥ 2, we define two sequences: and for every non-negative integer n, Our interest in the sequences u n (d, β) and ℓ n (d, β) is that they give upper and lower bounds on the probability Pr T d,n [σ(v d,n ) = c], respectively (subject to any boundary configuration at the leaves).
Lemma 30. Suppose that q = 3, d ≥ 2, and β ∈ (0, 1). For any n ≥ 0, for the d-ary tree T d,n with depth n and root v d,n , for any configuration τ : Λ T d,n → [q] on the leaves and any colour c ∈ [q], it holds that Proof. We prove the lemma by induction on n. For the base case n = 0, note that T d,n has a single vertex. Thus, for every c ∈ [q] and every τ assigning a colour to this vertex,

By Lemma 24 and (24), we conclude that
The following lemma will be used to show that the sequences u n (d, β) and ℓ n (d, β) converge.
By Lemma 31, we have that the sequences {u n (d, β}) and {ℓ n (d, β)} are bounded and monotonic, so they both converge. Let We have the following characterisation of the limits u ∞ (d, β), ℓ ∞ (d, β).

Bounding the maximum ratio
In this section, we place the final pieces for the proof of Lemma 19. The first lemma accounts for the d = 2 case of Lemma 19.
Note that β * (d) > 0. The following lemma shows that u n (d, β) and ℓ n (d, β) are bounded by the values corresponding to the critical parameter.
Our next goal is to prove Lemma 46 below, which will help us to obtain an upper bound on the ratio u ∞ (d, β * (d))/ℓ ∞ (d, β * (d)) when d is sufficiently large. In order to do this, we first define some useful re-parameterisations of f u and f ℓ , and establish some properties of these.
Note that the argument µ in g u , g ℓ corresponds to the ratio x/y of the arguments x, y of f u , f ℓ .
The following lemma is analogous to Lemma 37, but for the function g ℓ .

Proof. Let
Since µ ≥ 1 and d ≥ 3 and y ≤ 1/(µ + 2) all of the factors in W are positive, so W > 0. The derivative of g ℓ with respect to y (see Section 7.5 for Mathematica assistance) is given by the following.
We've already seen that W > 0 and the denominator is greater than 0 since it is a square. Since 3µy < 3 < d + 1, the remaining term is also positive, so ∂g ℓ ∂y < 0, as required.
Definition 39. Define the quantity y µ as follows.
Then we have the following lemmas.

Proof.
Since d is fixed in the proof of this lemma, we will drop it as an argument of β * , h u , g u . We will useβ * to denote 1 − β * = 3/(d + 1). We will drop d and β * as an argument of f u . So, plugging in Definitions 36 and 39, we get h u (µ) = g u (µ, y µ ) = f u (x µ , y µ ) − x µ . We have Let The derivatives of f u (x, y) with respect to x and y are as follows (see Section 7.7 for Mathematica assistance).
The Mathematica code in Appendix 7.7 uses Resolve to show rigorously that there is no µ > 1 satisfying (34).
We will use the following function in several of the remaining lemmas.
where ψ is the function defined in Definition 42. The derivative of h u (d, µ) with respect to d is given as follows (see Appendix 7.8 for the Mathematica code).
Facts 1 and 2 guarantee that, for all d ≥ 23, ζ(d, x µ , y µ ) > 0. Lemma 40 guarantees that all other factors in (35) are also positive. Thus, ∂hu(d,µ) ∂d is positive for all d ≥ 23. Together with Fact 3, this proves the first part of the lemma, that h u (d, 157/80) < 0. The three facts are proved in the Mathematica code in Section 7. 8. Finally, fix µ = 32. Lemma 40 again guarantees that all factors in (35) other than ζ(d, x µ , y µ ) are positive. Thus, it suffices to prove the three facts for µ = 32, and this is done in the Mathematica code in Section 7.8.
Lemmas 41 and 43 have the following corollary.
The Mathematica code in Appendix 7.9 verifies that h ℓ (23, µ) > 0 for all µ ≥ 157/80. Together with (36), this proves the lemma. Therefore, in the rest of the proof, we prove (36). Using Definitions 39 and 36, we have h ℓ (d, µ) = f ℓ (d, β * (d), x µ , y µ ) − y µ . We use the following definitions in order to describe ∂h ℓ (d,µ) Then the derivative of h ℓ (d, µ) with respect to d is given as follows (see Appendix 7.9 for Mathematica assistance).
Lemma 46. If d ≥ 23 then there is no solution to the system of equations which satisfies x ≥ 157y/80 ≥ 0 and 2x + y ≥ 1 ≥ 2y + x.
Corollary 47. For every integer d ≥ 23, there exists a positive integer n 0 such that for all n ≥ n 0 , Proof. Fix d ≥ 23. For simplicity, we will write β * instead of β * (d).
Corollary 47 accounts for integers d ≥ 23. To account for integers 3 ≤ d ≤ 22, we define the following two sequences.
u ′ 0 (d) = 1 ℓ ′ 0 (d) = 0 and for every non-negative integer n, We have the following lemma, which is proved by brute force.
By the induction hypothesis, we have Using Lemma 29, we therefore obtain that This completes the proof.
Corollary 50. For every integer d ∈ {3, . . . , 22} and every integer n ≥ 60, Proof. Fix an arbitrary integer d between 3 and 22. We have the following chain of inequalities (see below for explanation): The first inequality holds by Lemma 31, since the sequence {u n (d, β * (d))} is increasing and the sequence {ℓ n (d, β * (d))} is decreasing. The second inequality holds by Lemma 49. Finally, the third inequality holds by Lemma 48.
We can now prove Lemma 19, which we restate here for convenience.
Lemma 19. Let q = 3 and c ∈ [3] be an arbitrary colour. For d ≥ 2, consider the d-ary tree T d,n with height n and let τ : Λ T d,n → [3] be an arbitrary configuration on the leaves. When d = 2, for all β ∈ (0, 1), for all sufficiently large n it holds that When d ≥ 3, for all β ∈ [1 − 3 d+1 , 1), there exist sequences {L n } and {U n } (depending on d and β) such that for all sufficiently large n L n ≤ Pr T d,n [σ(v d,n ) = c | σ(Λ T d,n ) = τ ] ≤ U n and U n /L n ≤ 53/27.

Proof.
The statement for d = 2 follows directly from Lemma 33.
To prove Lemma 10, it will be helpful in this section to consider the set of maximisers of h c 1 ,c 2 ,β .
The following lemmas give properties of the maximisers in M α,c 1 ,c 2 ,β .
To proceed, we will need the following technical fact.
Lemmas 54 and 56 yield Lemma 10 as an immediate corollary.
We also now prove Lemma 11.
In turn, using the definition (3) of h c 1 ,c 2 ,β , we obtain from this that (43) holds, as wanted.
We then have the following lemma.
Lemma 58. Suppose q = 3, d ≥ 2 and β ∈ [0, 1]. Fix α > 1 and distinct colours c 1 , c 2 ∈ [q], and let p (1) , p (2) , . . . , p (d) ∈ Ex c 2 (α). Then there are nonnegative integers d 0 and d 1 with The following definition applies Definition 57 to the critical value of β for the special case where d 0 + d 1 = d; we will see that this special case is all that we need to consider to verify Condition 12 for α ∈ (1, 2). Proof. This is rigorously verified using the Resolve function of Mathematica in Section 7.11.
Now note that the first parenthesised expression in (47) is positive since X, Y > 0 for all α > 1. Thus, to show ∂φ ∂α < 0, it suffices to show that the second parenthesised expression in (47) is less than 0. To do this, we can apply the strict upper bound on g X /g Y from (48), and show that the resulting expression, which is ξ 1 − ξ 2 , is at most 0. Thus, we have completed Step 1.

Lemma 26
Both of the queries in the following code give the output False.

Lemma 48
The code checks that all of the desired inequalities are satisfied. The output is True.