A Formal Proof of R(4,5)=25

Abstract

In 1995, McKay and Radziszowski proved that the Ramsey number R(4,5) is equal to 25. Their proof relies on a combination of high-level arguments and computational steps. The authors have performed the computational parts of the proof with different implementations in order to reduce the possibility of an error in their programs. In this work, we prove this theorem in the interactive theorem prover HOL4 limiting the uncertainty to the small HOL4 kernel. Instead of verifying their algorithms directly, we rely on a verified SAT solver to prove gluing lemmas. To reduce the number of such lemmas and thus make the computational part of the proof feasible, we implement a generalization algorithm and verify that its output covers all the possible cases.

1 Introduction↩︎

Formalizations are useful to verify that there are no bugs in some software and also that there are no errors in a mathematical proof. Researchers write their formalizations in an interactive theorem prover also called a proof assistant. An interactive theorem prover transforms high-level proof steps, written by its user in the language of the interactive theorem prover, into low-level proof steps at the level of logical rules and axioms. These low-level steps are then verified by the kernel of the proof assistant. Formalizations are thus doubly appropriate when a proof combines advanced human-written arguments and computer-generated lemmas. This is the case for the four-color theorem [1] which was proved by Appel and Haken in 1976 and the Kepler conjecture [2] which was proved by Hales and Ferguson in 1998.

In those two cases, a human argument is used to reduce a potentially infinite number of cases to a finite number. Then, a computer algorithm is used to generate a proof for each of these cases. The generated proofs are too numerous to be verified manually and so the generating code, which is in those cases quite complicated, had to be trusted. To avoid trusting that the generating code fits together with the human argument, a formalization of the four-color theorem [3] was completed in the Coq proof assistant [4] by Gonthier in 2005 and a formalization of the Kepler conjecture [5] was completed in the HOL Light proof assistant [6] by a team led by Hales in 2014.

The case of $$R(4,5) \leq 25$$ is different. Since it is a finite problem, one could prove it by considering a finite number of cases. Since there are $$\frac{25\times24}{2}=300$$ edges in a graph with 25 vertices, a naive proof would consist of checking the presence of a 4-clique or a 5-independent set in all graphs of size 25 which would amount to $$2^{300}\approx 10^{90}$$ graphs. Another approach would be to encode the clique constraints into a SAT solver. This reduces the search space dramatically but so far no proof of $$R(4,5) \leq 25$$ relying only on calls to SAT solvers has been found.

The proof of McKay and Radziszowski [7] first uses a high-level argument and then relies on a pre-processing algorithm to reduce the number of cases to a manageable number. Each of those cases consists of a pair of graphs for which it has to be proved that they cannot occur together in an $$\mathcal{R}(4,5,25)$$-graph. These kinds of problems are called gluing problems. Our formalization of $$R(4,5)=25$$ in the HOL4 theorem prover [8] will mostly follow the initial splitting argument. We construct a slightly different pre-processing algorithm that uses gray edges instead of removing vertices. We also make use of the HOL4 interface [9] to the SAT solver MiniSat [10], instead of re-using the custom-built solver of McKay and Radziszowski, to prove that the gluing problems are unsatisfiable. This greatly simplifies our proof as we do not need to trace the proof steps of their optimized solver and we do not have to replay those proof steps in HOL4. Additional differences between our formal proof and the original proof are discussed in Section 8.

We now explain in more detail the different components of our formal proof. To conclude that $$R(4,5) = 25$$, we prove that $$R(4,5) \leq 25$$ and that $$R(4,5) > 24$$. The statement $$R(4,5) > 24$$ can simply be proved by exhibiting an $$\mathcal{R}(4,5,24)$$-graph. The existence of such graph has been known since 1965 thanks to a construction by Kalbfleisch [11]. The formal proof of the existence of an $$\mathcal{R}(4,5,24)$$-graph is given in Section 6.2. The other parts of this paper describe how to formally prove the more challenging statement $$R(4,5) \leq 25$$. In Section 2, we first give three important definitions. In particular, we define the Ramsey number $$R(4,5)$$ which is necessary to state the final theorem. In Section 3, we prove that in an $$\mathcal{R}(4,5,25)$$-graph there exists a vertex of degree $$d\in \{8, 10 ,12 \}$$ and that the neighbors of that vertex form an $$\mathcal{R}(3,5,d)$$-graph and the anti-neighbors form an $$\mathcal{R}(4,4,24-d)$$-graph as illustrated in Figure [fig:split]. This vertex is referred to in other parts of the proofs as the splitting vertex. In Section 4, we enumerate all possible $$\mathcal{R}(3,5,d)$$-graph and $$\mathcal{R}(4,4,24-d)$$-graph modulo isomorphism. We then regroup similar graphs together in what we call generalizations. In Section 5, we prove that there is no way to glue an $$\mathcal{R}(3,5,d)$$-graph and an $$\mathcal{R}(4,4,24-d)$$-graph while respecting the clique constraints. This is achieved by encoding the gluing into SAT and calling the HOL4 interface to MiniSat.

Not every algorithm needs to have its computation steps verified in a formal manner. Sometimes, it is enough to verify that the mathematical object produced by the algorithm satisfies the desired properties. For example, we did not verify every step of the nauty algorithm [12] which we rely on to normalize graphs in Section 4. Indeed, it is sufficient to save the witness permutations used during graph normalization to show that two graphs are isomorphic.

2 Preliminaries↩︎

Throughout our proof, we rely on the following definitions.

Definition 1. (neighbors, anti-neighbors)
Given a graph $$(V,E)$$:

• the set of neighbors (blue-neighbors) of a vertex $$x\in V$$ is $$\lbrace y\in V\;|\;y \not= x \wedge (x,y) \in E \rbrace$$,

• the set of anti-neighbors (red-neighbors) of a vertex $$x\in V$$ is $$\lbrace y\in V\;|\;y \not= x \wedge (x,y) \not\in E \rbrace$$.

Definition 2. (Ramsey property)
The Ramsey property $$\mathcal{R}(n,m,k)$$ holds for a graph $$(V,E)$$ if:

• $$V$$ has size $$k$$,

• $$(V,E)$$ does not contain a clique (blue-clique) of size n,

• $$(V,E)$$ does not contain an independent set (anti-clique, red-clique) of size m.

We also use $$\mathcal{R}(n,m,k)$$ to refer to the set of graphs with property $$\mathcal{R}(n,m,k)$$.
A graph for which the property $$\mathcal{R}(n,m,k)$$ holds is called a $$\mathcal{R}(n,m,k)$$-graph.

Definition 3. (Ramsey number)
The Ramsey number $$R(n,m)$$ is the least $$k\in \mathbb{N}$$ such that $$\mathcal{R}(n,m,k)$$ is empty.

In our formalization, a set of vertices $$V$$ will be represented by a subset of nonnegative integers. Moreover, we often use an equivalent formulation of graphs when discussing algorithms on graphs. In the equivalent formulation, all graphs are complete graphs but their edges are colored either blue or red. The correspondence between the two formulations is straightforward. There is a blue edge in the second formulation if and only if there is an edge in the first and there is a red edge in the second formulation if and only if there is no edge in the first.

3 Degree Constraints↩︎

As an intermediate concept, we define $$R'(r,s,n)$$ to hold if $$\mathcal{R}(r,s,n)$$ is empty. It is easy to see $$R(r,s) \leq n$$ iff $$R'(r,s,n)$$. In our formalization, we are primarily interested in proving $$R'(r,s,n)$$ for values of $$r$$, $$s$$ and $$n$$. In this section our focus is on reducing the goal of proving $$R'(4,5,25)$$ to ruling out vertices of degrees 8, 10 or 12.

Given a graph $$(V,E)$$1 and a vertex $$v\in V$$, we write $$\mathcal{N}^{(V,E)}(v)$$ for the set of neighbors of $$v$$ and $$\mathcal{A}^{(V,E)}(v)$$ for the set of antineighbors of $$v$$. We will almost always omit the superscript and write $$\mathcal{N}(v)$$ and $$\mathcal{A}(v)$$. The degree of $$v$$ is defined to be the cardinality of $$\mathcal{N}(v)$$. Likewise, the antidegree of $$v$$ is the cardinality of $$\mathcal{A}(v)$$.

Several relevant smaller Ramsey numbers are well-known: $$R(2,s) = s$$, $$R(3,3) = 6$$, $$R(3,4) = 9$$, $$R(3,5) = 14$$ and $$R(4,4) = 18$$. In our formalization we prove the $$R'$$ variant, only proving the known values are upper bounds. We begin by sketching a description of these results as well as some of the preliminary results used to obtain them.

By considering complements of graphs we know that $$R'(r,s,n)$$ implies $$R'(s,r,n)$$. If $$(V,E)\in\mathcal{R}(r+1,s,n)$$ and $$v\in V$$ is a vertex of degree $$d$$, then $$(\mathcal{N}(v),E)\in\mathcal{R}(r,s,d)$$. Likewise, if $$(V,E)\in\mathcal{R}(r,s+1,n)$$ and $$v\in V$$ is a vertex with antidegree $$d$$, then $$(\mathcal{A}(v),E)\in\mathcal{R}(r,s,d)$$.

Every graph in $$\mathcal{R}(2,s,m)$$ has no edges (since an edge would be a 2-clique). Thus every graph in $$\mathcal{R}(2,s,m)$$ is an independent set of size $$m$$. This is impossible if $$s\leq m$$, and so we conclude $$R'(2,m,m)$$. Likewise, $$R'(m,2,m)$$.

We next prove a well-known result that provides upper bounds for values of $$R(r,s)$$.

Lemma 1. If $$R'(r+1,s,m+1)$$ and $$R'(r,s+1,n+1)$$, then $$R'(r+1,s+1,m+n+2)$$.

Proof. Assume we have a graph $$(V,E)$$ in $$\mathcal{R}(r+1,s+1,m+n+2)$$. We choose a vertex $$v\in V$$ with degree $$d$$ and antidegree $$d'$$. We know $$(\mathcal{N}(v),E)\in \mathcal{R}(r,s+1,d)$$ and $$(\mathcal{A}(v),E)\in \mathcal{R}(r+1,s,d')$$. We obtain a contradiction using $$d + d' = m+n+1$$, $$d < m+1$$ (since $$R'(r+1,s,m+1)$$) and $$d' < n+1$$ (since $$R'(r,s+1,n+1)$$). ◻

Applying the previous results, we immediately obtain $$R'(3,3,6)$$. We also obtain $$R'(3,4,10)$$, but need the stronger result $$R'(3,4,9)$$.

There is an easy informal argument for why $$\mathcal{R}(3,4,9)$$ is empty. Assume $$(V,E)$$ is a graph in $$\mathcal{R}(3,4,9)$$. The results above ensure every vertex $$v\in V$$ must have degree $$d < 4$$ (since $$R'(2,4,4)$$) and antidegree $$d' < 6$$ (since $$R'(3,3,6)$$). Since $$d+d'=8$$, we must have $$d=3$$ and $$d'=5$$. We now consider the sum of the degrees of each vertex. Since the relation is symmetric, the sum must be even, as each edge is counted as part of the degree of each of the vertices of the edge. However, the sum is also $$9\cdot 3 = 27$$, which is odd. Hence no such graph exists. Below we describe our formalization of general results allowing us to prove $$R'(3,4,9)$$. The results will also allow us to later prove every graph in $$\mathcal{R}(4,5,25)$$ must have a vertex with even degree.

Lemma 2. Let $$(V,E)$$ be a graph in which every vertex has odd degree. For each $$U\subseteq V$$, $$U$$ has odd cardinality if and only if $$\Sigma_{u\in U}|\mathcal{N}(u)|$$ is odd.

Proof. The proof follows by an induction over the finite set $$U$$. ◻

Applying Lemma 2 with $$U=V$$, we obtain that if $$V$$ has odd cardinality and every vertex has odd degree, then $$\Sigma_{v\in V}|\mathcal{N}(v)|$$ is odd. In particular for a hypothetical graph $$(V,E)\in\mathcal{R}(3,4,9)$$, $$\Sigma_{v\in V}|\mathcal{N}(v)|$$ is odd since $$9$$ and $$3$$ are odd.

On the other hand we can prove $$\Sigma_{v\in V}|\mathcal{N}(v)|$$ is always even, though this requires two inductions on finite sets. We first prove that if we extend a graph with a new vertex, the neighbors of the new vertex in the larger graph contribute twice to the sum.

Lemma 3. Let $$V$$ be a finite set, $$u\notin V$$ and $$E$$ be a symmetric relation (on $$V\cup\{u\}$$).2 For every finite set $$U$$, if $$\mathcal{N}^{(V\cup\{u\},E)}(u) = U$$, then $\Sigma_{w\in V\cup\{u\}} |\mathcal{N}^{(V\cup\{u\},E)}(w)| = \Sigma_{v\in V} |\mathcal{N}^{(V,E)}(v)| + 2|U|.$

Proof. This is proved by induction on the finite set $$U$$. ◻

We can now prove the sum is even by induction on the finite set of vertices $$V$$.

Lemma 4. For every finite set $$V$$ and symmetric relation $$E$$, $$\Sigma_{v\in V} |\mathcal{N}^{(V,E)}(v)|$$ is even.

With Lemmas 2 and 4 we can conclude $$R'(3,4,9)$$ since the sum of the degrees of the vertices in a hypothetical graph $$(V,E)\in\mathcal{R}(3,4,9)$$ would be both odd and even.

Using Lemma 1 we now immediately obtain $$R'(3,5,14)$$ and $$R'(4,4,18)$$, giving us all the upper bounds for small Ramsey numbers we will need.

We now turn to the consideration of $$R'(4,5,25)$$. Assume $$(V,E)\in\mathcal{R}(4,5,25)$$. Let $$v\in V$$ with degree $$d$$ and antidegree $$d'$$ be given. Since $$(\mathcal{N}(v),E)\in\mathcal{R}(3,5,d)$$ and $$(\mathcal{A}(v),E)\in\mathcal{R}(4,4,d')$$ we know $$d < 14$$ and $$d' < 18$$. Since $$d+d'=24$$, we must have $$d > 6$$. This provides our basic upper and lower bounds on degrees of vertices in $$(V,E)$$.

These same degree bounds are, of course, given in [7]. The argument in [7] considers graphs in $$\mathcal{R}(3,5,d)$$ and corresponding graphs in $$\mathcal{R}(4,4,24-d)$$ that could hypothetically correspond to $$\mathcal{N}(v)$$ and $$\mathcal{A}(v)$$ for a vertex $$v\in V$$. The case with $$d=11$$ is ruled out since if every vertex had degree $$11$$, the sum of degrees would be odd, giving a contradiction. That is, we can be assured of the existence of a vertex $$v\in V$$ with degrees $$d\in\{7,8,9,10,12,13\}$$. We apply Lemmas 2 and 4 more generally to conclude that there must be a vertex $$v\in V$$ of even degree. Thus, we can be assured there is a $$v\in V$$ with degree $$d\in\{8,10,12\}$$. Combining this result with the main results described in the paper, we obtain $$R'(4,5,25)$$.

4 Enumeration of $$\mathcal{R}(3,5,d)$$ and $$\mathcal{R}(4,5,24-d)$$ Graphs and Construction of Covers↩︎

From the result mentioned at the end of Section 3, we know that there must exist a vertex $$v$$ of degree $$d \in \lbrace 8,10,12\rbrace$$. So, if we prove that for all $$d \in \lbrace 8,10,12 \rbrace$$ and for all pair of graphs $$G\in \mathcal{R}(3,5,d)$$ and $$H\in \mathcal{R}(4,4,24-d)$$, there is no way to color edges connecting G and H without creating a 4-blue or a 5-red clique, then we would have proved that $$R'(4,5,25)$$ (i.e. $$R(4,5) \leq 25$$).

Here is a simple approach. First, enumerate all the graphs in $$\mathcal{R}(3,5,d)$$ and in $$\mathcal{R}(4,4,24-d)$$. And then to prove the absence of gluing between each pair of graphs (see Section 5). This however not efficient enough given our computational means. To save time in both algorithms, we regroup graphs that are similar to each other, differing only by a few edges, in what we call generalizations. This way, our proofs will avoid repeating the same proving arguments in similar situations.

From a set of graphs $$\mathcal{G}$$, we will construct a set of generalizations $$\mathcal{G}^*$$ (this is a set of set of graphs) with the following properties. Every graph in $$\mathcal{G}$$ is a member of a generalization in $$\mathcal{G}^*$$ (we are not missing any case) and every graph in a generation $$G^* \in \mathcal{G}^*$$ is in $$\mathcal{G}$$ (we are not covering extra cases).

Definition 4. (cover,exact cover)
A set of generalizations $$\mathcal{G}^*$$ is a cover of a set of graphs $$\mathcal{G}$$ if $$\mathcal{G} \subseteq \bigcup_{G^*\in \mathcal{G}^*} G^*$$.
A set of generalizations $$\mathcal{G}^*$$ is an exact cover of a set of graphs $$\mathcal{G}$$ if $$\mathcal{G} = \bigcup_{G^*\in \mathcal{G}^*}G^*$$.

In a cover the generalizations do not need to be disjoint. Furthermore, our proof does not fundamentally require the constructed covers to be exact and better covers may be obtained by dropping this requirement. Yet, having exact covers simplify our presentation of the gluing algorithm as it enable us to ignore all clique constraints containing the splitting vertex (see Section 5).

4.1 Algorithm for Constructing an Exact Cover↩︎

In the following, we describe our base algorithm for constructing an exact cover for a set of graphs $$\mathcal{G}$$. This will be achieved by incrementally growing a set of generalizations $$\mathcal{G}^*_\mathit{partial}$$. We refer to the set of graphs $$G\in\mathcal{G}$$ that are not currently covered by $$\mathcal{G}^*_\mathit{partial}$$ as $$\mathcal{G}_\mathit{uncovered}$$. Initially, $$\mathcal{G}^*_\mathit{partial}$$ is empty and thus $$\mathcal{G}_\mathit{uncovered}$$ is equal to $$\mathcal{G}$$.

At each iteration of our algorithm, we randomly pick a graph $$G$$ from $$\mathcal{G}_\mathit{uncovered}$$ constructing a singleton generalization $$G^*_0 = \lbrace G \rbrace$$. Then, we color one of the edges of $$G$$ gray. This represents a generalization $$G^*_1$$ that contains the two graphs obtained by coloring the gray edge red or blue. Note that one of this graph is $$G$$ and thus $$G \in G^*_1$$ and $$G^*_0 \subseteq G^*_1$$. In general, the process starts from a generalization $$G^*_n$$ represented as a graph with $$n$$ gray edges. By definition, the generalization $$G^*_n$$ is defined to be the set of all graphs that can be obtained by coloring its $$n$$ gray edges red or blue in its representation. Then, the algorithm selects randomly one edge to gray among edges respecting the following conditions: the produced generalization $$G_{n+1}^*$$ must only contain graphs that are in $$\mathcal{G}$$ and $$(G_{n+1}^*\setminus G_{n})\;\cap\; \mathcal{G}_\mathit{uncovered}$$ must contain at least $$\lceil 2^{n-3} \rceil$$ graphs. The first condition makes the cover exact, the second condition prevents the construction of unnecessary large generalizations. This process is repeated graying one more edge per generalization step, as illustrated in Figure [fig:gen]. It stops when the number of gray edges exceeds a user-given limit or when there are no more edges respecting the conditions.

When the generalization algorithm stops, it creates a maximal generalization $$G^*_{\mathit{max}}$$ which is added to the set of generalizations $$\mathcal{G}^*_\mathit{partial}$$ and the instantiations of $$G^*_{\mathit{max}}$$ are removed from the set $$\mathcal{G}_\mathit{uncovered}$$ We keep adding new generalizations to $$\mathcal{G}^*_\mathit{partial}$$ by the same procedure until the set $$\mathcal{G}_\mathit{uncovered}$$ is empty and therefore $$\mathcal{G}^*_\mathit{partial}$$ is an exact cover of $$\mathcal{G}$$.

We can reduce the size of the final cover $$\mathcal{G}^*$$ by sampling multiple graphs in $$\mathcal{G}_\mathit{uncovered}$$ at each iteration of the algorithm. In our implementation, we sample 1000 graphs when $$\mathcal{G}=\mathcal{R}(4,4,k)$$ and all graphs when $$\mathcal{G}=\mathcal{R}(3,5,k)$$. This produces one maximal generalization for each of those graphs. We then select among them a generalization $$G^*$$ that contains a maximum number of uncovered graphs. That is to say one for which $$|G^* \cap \mathcal{G}_\mathit{uncovered}|$$ is maximum. We call this strategy for selecting generalization the best cover strategy.

The nauty algorithm [12] is called to normalize graphs in $$\mathcal{G}$$ and in each generalization $$G^*$$. By normalizing all graphs, we can check that two graphs are isomorphic by simply checking if their normalizations are equal.

Table 1: Number of $$\mathcal{R}(3,5,k)$$-graphs and $$\mathcal{R}(4,4,k)$$-graphs up-to-isomorphism together with the number of generalizations in the respective covers. All the covers were initially constructed with a maximum of 10 gray edges. We later updated the cover for the bold cases using an edge selection algorithm and an improved selection algorithm for generalizations (see Section 5.2).
k $$\mathcal{R}(3,5,k)$$ $$\mathcal{R}^*(3,5,k)$$ $$\mathcal{R}(4,4,k)$$ $$\mathcal{R}^*(4,4,k)$$
1 1 1
2 2 2
3 3 4
4 7 9 1
5 13 3 24 3
6 32 3 84 6
7 71 5 362 11
8 179 27 2079 47
9 290 11 14701 271
10 313 43 103706 1669
11 105 12 546356 7919
12 12 12 1449166 26845
13 1 1 1184231 13078
14 130816 11752
15 640 67
16 2 2
17 1 1

Computing the lists of $$\mathcal{R}(3,5,k)$$-graphs and $$\mathcal{R}(4,4,k)$$-graphs up-to-isomorphism can be done efficiently by simply repeatedly extending graphs in $$\mathcal{R}(3,5,k)$$ (resp.$$\mathcal{R}(4,4,k)$$) by one vertex while respecting the clique constraints to produce $$\mathcal{R}(3,5,k+1)$$ (resp.$$\mathcal{R}(4,4,k+1)$$). Such lists have been repeatedly compiled as mentioned in [7] and therefore we will not discuss in more detail how to construct them. From those lists, we construct corresponding covers $$\mathcal{R}^*(3,5,k)$$ and $$\mathcal{R}^*(4,4,k)$$. The size of those constructed covers for the sets of graphs $$\mathcal{R}(3,5,k)$$ with $$5 \leq k \leq 13$$ and the sets of graphs $$\mathcal{R}(4,4,k)$$ with $$4 \leq k \leq 17$$ is presented in Table 1.

4.2 Proof that any Graph with Property $$\mathcal{R}(3,5,k)$$ (resp.$$\mathcal{R}(4,4,k)$$) is a Member of One of the Element of $$\mathcal{R}^*(3,5,k)$$ (resp.$$\mathcal{R}^*(4,4,k)$$)↩︎

In this section, we only present the proof for the covers $$\mathcal{R}^*(3,5,k)$$ since the proof for the covers $$\mathcal{R}^*(4,4,k)$$ follows by an analog argument. Given the result presented in Section 3, it is enough to consider the cases of a splitting vertex with degree degree $$d \in \{8, 10, 12\}$$. Therefore, it would be enough to prove that $$\mathcal{R}^*(3,5,d)$$ covers the set of graphs with property $$\mathcal{R}(3,5,d)$$. However, to do so, we found it easier to prove the stronger result: $\forall\;5 \leq k \leq 13.\; G \;has\;property\; \mathcal{R}(3,5,k) \Rightarrow \exists G^* \in \mathcal{R}^*(3,5,k).\;G \in G^*$

We prove the result by a finite induction over the number of vertices $$k$$.
The base case $$k=5$$ consists of searching for all the possible graphs with property $$\mathcal{R}(3,5,5)$$ and show that they appear modulo isomorphism in one of the generalizations in $$\mathcal{R}*(3,5,5)$$.
The inductive case is similar. The main difference is that we start the search from a generalization $$G^*$$ instead of the empty generalization. We prove that for all generalizations $$G^*$$ in $$\mathcal{R}^*(3,5,k)$$, any extension of $$G^*$$ by one vertex that respects the property $$\mathcal{R}^(3,5,k+1)$$ is isomorphic to an element of $$cR^*(3,5,k+1)$$. This is achieved by exploring all possible colorings (in blue or in red) of edges that are either gray or contain the new vertex. This extension process is depicted in Figure [fig:ext].

The formalization and efficiency of the previous arguments rely on our custom-made solver for labeled graphs. Our solver mostly works like a DPLL SAT solver [13]. The principal difference is that it represents clauses as essentially first-order formulas. We show how we represent the property $$\mathcal{R}^*(3,5,k+1)$$ in first-order. Given a graph $$G$$ of size $$k+1$$, represented by a binary relation $$E$$ over a set of vertices $$V=[|0,k|]=\{ 0,1,\ldots,k \}$$, our first-order representation of the statement “$$G$$ has property $$\mathcal{R}^*(3,5,k+1)$$” is given by the two formulas: \begin{align} \forall\underset{\mathit{distinct}}{x_0 x_1 x_2} < k \!+\! 1.\;& \neg E x_0 x_1 \vee \neg E x_0 x_2 \vee \neg E x_1 x_2\\ \forall\underset{\mathit{distinct}}{x_0 x_1 x_2 x_3 x_4} < k \!+\! 1.\;& E x_0 x_1 \vee E x_0 x_2 \vee E x_0 x_3 \vee E x_0 x_4 \vee E x_1 x_2 \vee \\ &E x_1 x_3 \vee E x_1 x_4 \vee E x_2 x_3 \vee E x_2 x_4 \vee E x_3 x_4 \end{align}

where $$\mathit{distinct}$$ means that we add inequalities between each of pair of quantified variables. For example, $$\forall\underset{\mathit{distinct}}{x_0 x_1 x_2} < k+1.\;P[x_1,x_2,x_3]$$ stands for: $\forall x_0 x_1 x_2.\; (x_0 \!<\! k \!+\! 1 \wedge x_1 \!<\! k \!+\! 1 \wedge x_2 \!<\! k \!+\! 1 \wedge x_0 \!\neq\! x_1 \wedge x_0 \!\neq\! x_2 \wedge x_1 \!\neq\! x_2) \Rightarrow P[x_1,x_2,x_3]$

Our solver is designed to work exclusively on graphs where edges are represented by first-order literals and the coloring of an edge $$(i,j)$$ to blue (resp.red) corresponds to assuming the literal $$E x_i x_j$$ in the branch (resp.the negation of the literal $$\neg E x_i x_j$$). A gray edge just indicates that no color for that edge is currently assumed. The prover starts by assuming all literals corresponding to colored edges in a generalization $$G^*$$ with $$k$$ vertices. It then explores all possible colorings of the gray edges and then all the possible coloring of the edges containing a new vertex $$x_k$$. At the leaf, this produces a graph of size $$k+1$$ represented by all the literals assumed in the branch.

By representing vertices by variables $$x_0,\ldots,x_k$$ instead of the concrete value $$0,\ldots,k$$, it is easier to prove that the graphs in the leaves are indeed isomorphic to one of the elements of one of the generalizations $$G' \in \mathcal{R}^*(3,5,k+1)$$. Instead of proving that every generalization in $$\mathcal{R}^*(3,5,k)$$ extends into graphs belonging to a generalization in $$\mathcal{R}^*(3,5,k+1)$$, we will instead forbid the creation of a graph that is a member of a generalization in $$\mathcal{R}^*(3,5,k+1)$$. Given a generalization $$G' \in \mathcal{R}^*(3,5,k+1)$$ with a set of blue edges $$B$$ and a set of red edges $$R$$, we use the following formula to forbid the creation of an element of $$G'$$: $\forall \underset{\mathit{distinct}}{x_0 \ldots x_i \ldots x_j \ldots x_k} < k+1.\;\; ((\bigwedge\limits_{(i,j) \in B} E x_i x_j)\; \wedge \; (\bigwedge\limits_{(i,j) \in R} \neg E x_i x_j)) \Rightarrow \bot$ Note that this implies that all permutations of graphs that are members of this generalization are forbidden. The first reason is that the formula does not assume any constraints on the gray edges of $$G'$$ therefore forbids all members of $$G'$$. The second reason is one can permute the indices of variables by a simple instantiation of the variables with a permutation being given by the nauty algorithm to make the labeled graph on the branch match with one of the labeled generalizations.

5 Gluing↩︎

In the previous section, we constructed covers for $$\mathcal{R}(3,5,d)$$-graphs and $$\mathcal{R}(4,4,24-d)$$-graphs. The next step of our proof is to prove that given a generalization $$G^*$$ in $$\mathcal{R}^*(3,5,d)$$ and a generalization in $$H^*$$ in $$\mathcal{R}^*(3,5,24-d)$$, there is no way to extend color gray edges and transverse edges to form an $$\mathcal{R}(4,5,24)$$-graph (see Figure [fig:hi]) and thus an $$\mathcal{R}(4,5,25)$$-graph by adding the splitting vertex. In the rest of this section, we can ignore clique constraints that include the splitting vertex as they are already satisfied. This is a consequence of the fact that our covers are exact covers. All our gluing problems are formulated at the propositional level and contain the following clauses representing the property $$\mathcal{R}(4,5,24)$$. Let us number the vertices of an $$\mathcal{R}^*(3,5,d)$$-generalization $$G^*$$ from $$0$$ to $$d-1$$, and the vertices of a $$\mathcal{R}^*(4,4,24-d)$$-generalization graph $$H^*$$ from $$d$$ to $$23$$.
$For each subset S \subset [|0,23|] of size 4, we create the clause\bigvee\limits_{a,b \in S \wedge a < b}\neg E_{a,b}\;.$ $For each subset T \subset [|0,23|] of size 5, we create the clause \bigvee\limits_{a,b \in T \wedge a < b} E_{a,b}\;.$

In all these propositional clauses, $$E_{a,b}$$ is a propositional variable that is true if there is a blue edge between $$a$$ and $$b$$ and that is false if there is a red edge between $$a$$ and $$b$$. In each gluing problem ($$G^*$$,$$H^*$$), we add unit clauses for each colored edge (red or blue) of $$G^*$$ and $$H^*$$. If an edge $$(a,b)$$ with $$a<b$$ is blue then we add the unit clause $$E_{a,b}$$, if it is red then we add the unit clause $$\neg E_{a,b}$$. If an edge is gray we do not add a unit clause. Together, with the clique clauses this forms our SAT problem that is sent to the verified MiniSat interface. In practice, we had to perform unit propagation to reduce the number of clauses before sending the problem to the interface. This is due to some limitations in the interface as this does not happen when we call the SAT solver directly.

5.1 Simplicity Heuristic↩︎

Before sending the problem to the SAT solver, we will try to construct covers that create easier problems. Let $$b_{k}$$ represent the number of blue $$k$$-cliques in $$G^*$$. Let $$r_{k}$$ represent the number of red $$k$$-cliques in $$G^*$$. Let $$b'_{k}$$ represent the number of blue $$k$$-cliques in $$H^*$$. Let $$r'_{k}$$ represent the number of red $$k$$-cliques in $$H^*$$. We use the following formula to estimate the difficulty of a gluing problem $$(G^*,H^*)$$ $\mathit{simplicity}(G^*,H^*) =\frac{1}{2^3} b_1 b'_3 + \frac{1}{2^4}b_2 b'_2 + \frac{1}{2^6} r_2 r'_3 + \frac{1}{2^6}r_3 r'_2 + \frac{1}{2^4} r_4 r'_1$ Our formula is originally designed to estimate the simplicity of a problem of gluing an $$\mathcal{R}(3,5,d)$$-graph $$G$$ with an $$\mathcal{R}(4,4,24-d)$$-graph $$H$$. There, 5 different types of configurations may create a blue 4-clique or a 5 red-clique as illustrated in Figure [fig:config]. These correspond to the 5 terms in our formula. Two types of configurations may create a blue 4-clique: a blue 1-clique in $$G$$ and a blue 3-clique in $$H$$ forming a constraint on 3 transverse edges, a blue 2-clique in $$G$$ and a blue 2-clique in $$H$$ forming a constraint on 4 transverse edges. Three types of configurations may create a red 5-clique: a red 2-clique in $$G$$ and a red 3-clique in $$H$$ forming a constraint consisting of 6 transverse edges, a red 3-clique in $$G$$ and a red 2-clique in $$H$$ forming a constraint on 6 transverse edges, a red 4-clique in $$G$$ and a red 1-clique in $$H$$ forming a constraint on 4 transverse edges.

Other potential configurations never occur because $$G$$ is a $$\mathcal{R}(3,5,d)$$-graph and $$H$$ is an $$\mathcal{R}(4,4,24-d)$$-graph. For example, we know that there are no red 4-clique in $$H$$ so a configuration consisting of a red 1-clique in $$G$$ and a red 4-clique in $$H$$ cannot happen.

In the simplicity formula, each of the 5 terms counts the corresponding number of configurations and multiply them by a weight before adding them up. The choice of the weight is based on the number of transverse edges in the configuration. For example, if 3 transverse edges must not all be blue (e.g. in the first configuration) then we can rule out $$1/2^3$$ of the cases in our proof. Thus, our formula computes cases covered as a proportion of the total number of cases, sometimes counting the same cases multiple times. If there were no overlap between the different configurations, we could prove the absence of gluing as soon as the simplicity score exceeds 1.

We test how good the simplicity score is at predicting the run time of our SAT solver on 200 gluing problems between $$\mathcal{R}(3,5,10)$$-graphs and $$\mathcal{R}(4,4,14)$$-graphs in Figure [fig:simp]. The results reveal that our simplicity score is a good predictor in this setting. When we apply the formula to predict the run-time for gluing generalizations, we are therefore ignoring the contributions of configurations (clauses) containing gray edges. This is desirable as we would prefer the problem to be simple without requiring a case analysis (splitting) on the color of gray edges.

Finally, during the construction of a cover for $$\mathcal{G}$$ we are not aware of the corresponding cover for $$\mathcal{H}$$. The covers would have to be built simultaneously making the algorithm more complicated. To avoid those complications, we devise a measure to predict if a generalization will create difficult problems on its own without depending on the possible counterparts. To this end, we chose to estimate the simplicity of a generalization $$G^*$$ by how difficult it is glue it with an average counterpart graph $$\bar{H}$$. Let $$\bar{b}'_{k}$$ represent the average number of blue $$k$$-cliques per graph in $$\mathcal{H}$$ and $$\bar{r}'_{k}$$ represent the average number of red $$k$$-cliques per graph in $$\mathcal{H}$$, the simplicity of $$G^*$$ is: $\mathit{simplicity}(G^*) = \frac{1}{2^3} b_1 \bar{b}'_3 + \frac{1}{2^4}b_2 \bar{b}'_2 + \frac{1}{2^6} r_2 \bar{r}'_3 + \frac{1}{2^6}r_3 \bar{r}'_2 + \frac{1}{2^4} r_4 \bar{r}'_1$ Similarly, let $$\bar{b}_{k}$$ represent the average number of blue $$k$$-cliques per graph in $$\mathcal{G}$$ and $$\bar{r}_{k}$$ represent the average number of red $$k$$-cliques per graph in $$\mathcal{G}$$, the simplicity of $$H^*$$ is: $\mathit{simplicity}(H^*) = \frac{1}{2^3} \bar{b}_1 b'_3 + \frac{1}{2^4}\bar{b}_2 b'_2 + \frac{1}{2^6} \bar{r}_2 r'_3 + \frac{1}{2^6}\bar{r}_3 r'_2 + \frac{1}{2^4} \bar{r}_4 r'_1$

5.2 Creation of Better Covers by Parameter Search↩︎

We now improve the exact cover algorithm by relying on our simplicity heuristic in two places. The first one is during the selection of edges. Previously, the edges were selected at random as long as the produced generalization satisfied some conditions. Now, we will select among the possible edges allowed by the conditions the one that produces the generalization with the highest simplicity score. We call this strategy fastest in Table 2. The second place where the simplicity score will influence the algorithm is during the selection of generalizations. Previously, we selected maximal generalizations $$G^*$$ with highest coverage value $$n_\mathit{cover}(G^*) = | G^* \cap \mathcal{G_\mathit{uncover}} |$$. This strategy was called the best cover strategy. Now, we will also prefer generalizations with higher simplicity scores. Since we want to optimize for both objectives at the same time, we will select the graph $$G^*$$ with the highest combined score $$\mathit{simplicity}(G^*)^{c} \times n_\mathit{cover}(G^*)$$ where $$\mathit{c}$$ is a real number parameter influencing how much one heuristic is preferred over the other. In Table 2, we call this selection strategy mixed-c.

Although the simplicity score is important to reduce the difficulty of the problems, the most important parameter in reducing the total computation time is the number of maximum allowed gray edges in each generalization. In Table 2, we optimize those parameters for the three relevant cases $$d=8,10,12$$ for the gluing. Each experiment consists of a line in Table 2. There, we compute new covers with different parameters. To figure out the best parameters, we estimate the run time of an average problem by sampling 200 random problems from a pair of covers. We then multiply this estimate by the number of problems this pair of covers would create to get an estimated total run time for this pair of covers. In the end, we decided not to go with the best parameters according to the estimated times given in Table 2. The reason is that by increasing the number of gray edges, the total number of problems is reduced but the difficulty of each problem is increased. This makes the problems harder and they would have taken more memory than we had available. That is why we chose to compromise and instead use the fastest parameter settings, shown in bold in Table 2, that would not use more memory than available on our machines. Table 3 gives a comparison of the total run time for our final problems with the total runtime that we would get by simply gluing pairs of graphs instead of generalizations. For degree $$d = 10$$, the number of gluing problems is reduced by a factor of $$81.0$$ and the estimated total time by a factor $$14.6$$. For degree $$d = 12$$, the number of gluing problems is reduced by a factor of $$53.9$$ and the estimated total time by a factor of $$20.6$$.

Table 2: Tested parameters for creating exact covers. The columns titled 3,5 and 4,4 show the maximum number of gray edges allowed during the construction of the cover.
Gluing 3,5 4,4 Edge sel. Gen. sel. CPU-days
3,5,8-4,4,16 0 0 none none 0.055
4 0 fastest mixed-0.5 0.018
3,5,10-4,4,14 0 0 none none 8373
3 4 fastest mixed-1.0 725
4 3 fastest mixed-1.0 689
4 4 random best cover 734
4 4 fastest mixed-10.0 625
4 4 fastest mixed-2.0 595
4 4 fastest mixed-1.0 658
4 4 fastest mixed-0.5 572
4 4 fastest mixed-0.1 706
5 4 fastest mixed-1.0 547
4 5 fastest mixed-1.0 586
5 5 fastest mixed-0.5 396
3,5,12-4,4,12 0 0 none none 7702
2 6 fastest mixed-0.5 641
3 6 fastest mixed-0.5 782
4 6 fastest mixed-0.5 784
0 8 fastest mixed-0.5 374
1 8 fastest mixed-0.5 353
2 8 fastest mixed-0.5 538
3 8 fastest mixed-0.5 360
4 8 fastest mixed-0.5 419
Table 3: Reduction of the number of SAT solver calls and faster estimated times
Graphs Generalizations
2-5 (lr)6-9 d 3,5,d 4,4,24-d problems days 3,5,d 4,4,24-d problems days
8 179 2 358 0.055 27 2 54 0.018
10 313 130816 40945408 8373 43 11752 505336 572
12 12 1449166 17389992 7702 12 26845 322140 374

6 Combining the Different Parts of the Proof↩︎

Our proof is expressed using three different formal representations of mathematical statements. In Section 3, we express our statements in a higher-order form allowing us to make counting arguments. In Section 4.2, we rely on an almost first-order representation to implement a custom theorem prover for graphs and in particular to prove isomorphism between graphs. In Section 5, the problems are stated at the propositional level. Here, we first describe how we connect the different representations and as a consequence prove that $$R(4,5) \leq 25$$. Then, we give the proof of the existence of an $$\mathcal{R}(4,5,24)$$-graph and show that $$R(4,5) > 24$$.

6.1 Connecting Representations↩︎

We will start by translating propositional gluing lemmas to first-order formulas. The SAT problems do not explicitly mention on which vertices the graphs are lying on since they are only constraining SAT variables that represent edges. Surprisingly, one can instantiate the SAT variables $$E_{i,j}$$ by the atom $$E\;x_i\;x_j$$ in the gluing problem. As a consequence, gluing problems for all permutations of edges are proved at once. We can also freely add the following additional constraints. All variables $$x_i$$ must be distinct and variables with indices less than the degree $$d$$ must have a value less than $$d$$ and other variables must have a value greater or equal to $$d$$. This ensures that our generalizations $$G^*$$ and $$H^*$$ have distinct sets of vertices. We then prove that for each $$\mathcal{R}^*(3,5,d)$$-generalization a single theorem stating that this particular generalization can not be glued to any of the corresponding generalizations in $$\mathcal{R}^*(4,4,24-d)$$ by regrouping gluing theorems. This step constructs $$27$$ theorems for degree $$d=8$$, $$43$$ theorems for degree $$d=10$$, and $$12$$ theorems for degree $$d=12$$. These numbers correspond to the number of $$\mathcal{R}^*(3,5,d)$$-generalizations presented in Table 1. Together, these 27 theorems (respectively 43 and 12) can be used to prove a theorem stating that the splitting edge, represented by the vertex number 24, cannot have degree 8 (respectively 10 and 12). The higher-order version of these three final theorems do not state on which set of vertices the neighbors and anti-neighbors should lie although it requires them to form sets of nonnegative integers of size $$d$$ and $$24-d$$ respectively. To prove the more general higher-order formulations, we rely on the fact that there is a bijection from $$[|0,d-1|]$$ to sets of vertices of size $$d$$ and a bijection from sets $$[|d,24|]$$ to sets of vertices of size $$24-d$$. These three theorems together are enough to prove that $$R(4,5) \leq 25$$ according to the proof given in Section 3.

6.2 Existence of an $$R(4,5,24)$$-graph↩︎

To prove the existence of an $$R(4,5,24)$$-graph, we pick a graph from the full list of $$R(4,5,24)$$-graphs compiled in 2017. This was necessary step to prove that $$R(5,5)\leq 48$$ as described in [14]. For our purpose, we only need one witness graph $$G_0$$. Let $$B$$ be the set of blue edges in $$G_0$$, we represent the graph $$G_0$$ as the relation: $E_0 =_{\mathit{def}} \lambda i j.\; \bigvee\limits_{(a,b) \in B \wedge a < b} (i = a \wedge j = b) \vee (j = a \wedge i = b)$ We prove on the first-order level that this graph does not contain any blue 4-cliques or any red 5-cliques which can be stated as: \begin{align} &\vdash \forall\underset{\mathit{distinct}}{x_0 x_1 x_2 x_3} < 24.\; \neg E_0 x_0 x_1 \vee \neg E_0 x_0 x_2 \vee \neg E_0 x_0 x_3 \vee \neg E_0 x_1 x_2 \vee \neg E_0 x_1 x_3 \vee \neg E_0 x_2 x_3\\ &\vdash \forall\underset{\mathit{distinct}}{x_0 x_1 x_2 x_3 x_4} < 24.\; E_0 x_0 x_1 \vee E_0 x_0 x_2 \vee E_0 x_0 x_3 \vee E_0 x_0 x_4 \vee E_0 x_1 x_2 \vee \\ & E_0 x_1 x_3 \vee E_0 x_1 x_4 \vee E_0 x_2 x_3 \vee E_0 x_2 x_4 \vee E_0 x_3 x_4 \end{align} This was achieved by repeatedly applying the following lemma to eliminate the quantified variables $$\vdash (\bigwedge\limits_{x<24} P(x)) \Rightarrow (\forall x<24. P(x))$$. To speed up the process, we stop applying the lemma as soon as we were able to prove the goal on the branch either because we find a red edge in blue clique (or a blue edge in a red clique) or because we have selected the same vertex twice in the clique. The connection with the higher-order formulation can be obtained because by proving that the set $$[|0,23|]$$ has cardinality 24. And thus we get $$R(4,5) > 24$$ which together with $$R(4,5) \leq 25$$ gives: $\vdash R(4,5)=25$

7 Reproducibility↩︎

We provide instructions on how to reproduce the proof in the README.md of our repository https://github.com/barakeel/ramsey.

Here are the computational resources necessary to run our proof. The final gluing step was run on 4 different machines allowing us to finish the gluing phase in less than 9 days. This is slightly longer than what we expected according to the estimated times. Two machines were used for the $$d=10$$ case and the other 2 were used for the $$d=12$$ case. Three of them have 512 GB of RAM and one of them has 1024 GB of RAM. All of those machines have 64 hyper-threaded CPU cores for a total of 128 available CPUs. However, we only used 40 cores per machine as our main limitation was memory. In all those machines, the same copy of HOL4 was used to get compatible timestamps in the produced gluing theories. The other phases were much faster and required less memory and but were still run on the machine with more memory.

Here are other issues one could encounter when trying to reproduce the proof. As expected, all the technical problems come during or after the expensive gluing phase. The first issue we discovered is that the communication files between HOL4 and MiniSat are stored in the temporary directory of the system. Since the proof file produced by MiniSat can be up to 2GB and our temporary directory sits in a partition of only 32GB, we ran out of memory in that partition because we were running 40 processes in parallel. So, we changed the temporary directory used by the HOL4 interface to MiniSat by modifying the file dimacsTools.sml. We changed he temporary directory used by MiniSat using the TMPDIR bash variable. The second issue is that the reconstruction of a SAT proof in HOL4 can require a lot of memory (in the order of 20GB per problem) and time (about 3 times longer than the SAT solver call). We found out that creating theories with one theorem per theory diminished memory consumption. To guarantee that the memory consumption did not exceed a threshold, we also ran the scripts using buildheap instead of Holmake as the latter does not provide a way to limit the memory consumption per core. Finally, we were not able to load all the gluing theories together into HOL4 in a reasonable time. Indeed, we observed a slowdown of time taken to load one theory as more and more theories were added making it impossible for us to load more than 200,000 theories. Thus, we used the following workaround instead. We loaded the theorems produced by the 43 theorems for degree $$d=10$$ and the 12 theorems for degree $$d=12$$, mentioned in Section 6.1, without specifying in which theory they were proved. This was achieved by creating an alternative theory loader where we omitted the call to the function link_parents. To ensure the safety of this procedure, we externally check that constructing the complete theory graph for our proof without any broken dependencies is possible in the directory theorygraph . There, we make sure that the time stamps are coherent and that there are no cycles. In the complete theory graph, the final theory r45_equals_24 has 828857 ancestor theories.

8 Related Works↩︎

Our formalization is based on the work of McKay and Radziszowski [7]. Their proof already contains the three steps performed in our formalization namely: the degree constraints for a splitting, the creation of covers, and the proof of the absence of satisfiable gluing problems. We explain here how our proof differs in each of those steps. In the first step, their proof mentions that it is not possible for all vertices to have degree 11. We realized during the formalization that this argument can be applied to prove that it is not possible for all vertices to have odd degrees. This allows us to save time during the formalization compared to their original proof by additionally ignoring the cases of a splitting vertex of degree $$d\in\{7,9,13\}$$. In the second step, their proof removes vertices from graphs to create generalizations which would correspond to graying all edges connected to the removed vertices in our formalization. We tried this approach but the large number of grayed edges made the problems very difficult for the SAT solvers. Moreover, our tests gave an estimated time to completion much larger than with our more parsimonious approach to graying edges. In the final step, they rely on a custom prover for gluing generalizations and spend probably half of the paper describing how they optimize its different components. In contrast, we perform the gluing step with an existing SAT solver. Furthermore, they state that they prefer sparser (redder) generalizations when constructing a cover of $$\mathcal{R}(3,5,10)$$-graphs and denser (bluer) generalizations when constructing a cover of $$\mathcal{R}(4,4,12)$$-graphs. Our approach instead rely on a much more involved heuristics. We try to estimate the simplicity of a generalization $$G^*$$ by understanding which type of clauses that would appear in a gluing problem where one of the generalization is $$G^*$$.

Proving mathematical theorems in combinatorics with the help of SAT solvers is not a new phenomenon. For instance in [15], the authors prove using SAT solvers that the 3-color Ramsey number R(4,3,3) is equal to 30 . This proof contains consideration about the degrees of vertices, a symmetry-breaking component to avoid considering isomorphic graphs and an abstraction component relying on degree matrices to represent sets of graphs. Ultimately, the approach relies on encoding each of these steps into SAT clauses and calling a SAT solver. A more famous example is the proof of the Pythagorean Triple theorem [16]. In that paper, the authors rely on cube-and-conquer and look-ahead methods implemented in a SAT solver to create a single large proof of the theorem in the DRAT format [17]. This proof was later verified independently by the DRAT proof checker [18]. In our proof, we could not use this approach for multiple reasons. First of all, we rely on multiple independent calls to a SAT solver. It would be hard to regroup them in a single call. Second of all, we formalize high-level counting arguments as well as symmetry-breaking arguments that are much easier to encode in a more expressive logic than propositional logic. Finally, we believe that it is better to express the mathematical statement $$R(4,5)=25$$ in higher-order logic and our formula is essentially identical to the original mathematical statement (up to the representation of vertices by nonnegative integers). In contrast, a propositional statement would consist of a large number of clauses. Thus, it would be harder for the reader to be convinced that those clauses correspond to the original problem. Therefore, we believe that to minimize the possibility of an error, we should in general try to formalize theorems in a form that most closely matches its original mathematical form.

9 Conclusion↩︎

We have created the first formalization of the theorem $$R(4,5)=25$$. This verification was performed within the HOL4 theorem prover. During this process, we have realized that we can generalize the argument given for degree $$d=11$$ to eliminate all odd degree cases. We have designed a verified algorithm for regrouping similar graphs, creating generalizations and ultimately speeding up our subsequent proofs. Finally, we have created and tested a heuristic for predicting the relative run-time of a SAT solver on gluing problems. This helped us choose generalizations that create easier gluing problems.

In the future, we would like to investigate if a proof of $$R(4,5)=25$$ is intrinsically computational or if there exist additional high-level arguments that could eliminate the need for a large computation.

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1. We always implicitly assume the set $$V$$ is finite.↩︎

2. In the formalization, $$E$$ is assumed to be symmetric on the relevant type, ignoring $$V\cup\{u\}$$.↩︎