Three-edge-colouring doublecross cubic graphs

In this paper, all graphs are finite and simple. The four-colour theorem (4CT), that every planar graph is four-colourable, was proved by Appel and Haken [1, 2] in 1977, and a simplified proof was given in 1997 by three of us, with Neil Robertson [4]. The 4CT is equivalent to the statement that every two-edge-connected cubic planar graph is three-edge-colourable, and a strengthening was proposed as a conjecture by Tutte in 1966 [9]; that every two-edge-connected cubic graph with no Petersen graph minor is three-edge-colourable. This paper is a step in the proof of Tutte’s conjecture.

A graph $G$ is apex if $G\setminus v$ is planar for some vertex $v$ (we use $\setminus$ to denote deletion); and a graph $G$ is doublecross if it can be drawn in the plane with only two crossings, both on the infinite region in the natural sense.

It is easy to check that apex and doublecross graphs do not contain the Petersen graph as a minor; but there is also a converse. Let us say a graph $G$ is theta-connected if $G$ is cubic and has girth at least five, and $|\delta_{G}(X)|\geq 6$ for all $X\subseteq V(G)$ with $|X|,|V(G)\setminus X|\geq 6$. ($\delta_{G}(X)$ denotes the set of edges of $G$ with one end in $X$ and one end in $V(G)\setminus X$.) Two of us (with Robertson) proved in [6] that every theta-connected graph with no Petersen graph minor is either apex or doublecross (with one exception, that is three-edge-colourable); and in [7] that every minimal counterexample to Tutte’s conjecture was either apex or theta-connected. It follows that every minimal counterexample to Tutte’s conjecture is either apex or doublecross, and so to prove the conjecture in general, it suffices to prove it for apex graphs and for doublecross graphs. Two of us proved in [8] that every two-edge-connected apex cubic graph is three-edge-colourable, so all that remains is the doublecross case, which is the objective of this paper. Our main theorem is:

The proof method is by modifying the proof of the 4CT given in [4]. Again we give a list of reducible configurations (the definition of “reducible” has to be modified to accommodate the two pairs of crossing edges), and a discharging procedure to prove that one of these configurations must appear in every minimal counterexample (and indeed in every non-apex theta-connected doublecross graph). This will prove that there is no minimal counterexample, and so the theorem holds. Happily, the discharging rules given in [4] still work without any modification.

We are only concerned with graphs that can be drawn in the plane with two crossings, and one might think that these are not much different from planar graphs, and perhaps one could just use the 4CT rather than going to all the trouble of repeating and modifying its proof. For graphs with one crossing this is true: here is a pretty theorem of Jaeger [3] (we include a proof because we think it is of interest):

We have tried (hard!) to do something similar to handle the doublecross case, but failed; it seems necessary to do it the long way, modifying the proof of the 4CT. Fortunately that is not as difficult as it was for the apex case in [8].

Our proof is a modification of the proof of the 4CT, so let us begin with that. We can work in terms of planar cubic graphs or in terms of planar triangulations, but we found it most convenient to use triangulations (because for instance, it is much easier to present long lists of reducible configurations as subgraphs of triangulations than as subgraphs of cubic graphs). Thus we need to show that every planar triangulation is 4-colourable. If there is a counterexample, the smallest one is “internally 6-connected”; that is, every vertex-cutset of order at most five separates at most one vertex from the rest of the graph. Now the average degree of vertices is just less than six, and so there are vertices of degree five; but we need to show that there is a small clump of vertices with small degrees. (We listed explicitly exactly what counts as a “clump”; there were 633 of them in that paper. They have somewhere between four and twelve vertices all with small degree). To show this we use a “discharging procedure”; we initially assign “charges” (integers) to the vertices, depending on their degrees and with positive total; and then move small amounts among neighbouring vertices according to some carefully chosen rules; and prove a theorem that any vertex that ends up with positive charge (and there must be such a vertex) belongs to or neighbours one of the clumps we are hoping for.

Next we show that for any counterexample to the 4CT, there must be a smaller counterexample. We just showed that it would contain one of the 633 clumps, and the clumps were carefully chosen with a special property: for every clump $C$ in a internally 6-connected triangulation $T$, we can excise $C$ from $T$ and replace it by something smaller, so that $T$ is changed into a smaller triangulation that is no easier to 4-colour. This would show that the 4CT is true.

How can we show that for each of the 633 clumps, if it occurs in a triangulation, it can be replaced by something smaller without making the triangulation easier to colour? The idea is, we show this clump by clump, using a computer. Let $C$ be one of our clumps, in an internally 6-connected triangulation $T$ that is not 4-colourable. (Since we are working through the clumps one at a time, at this stage we know $C$ completely, but have no knowledge of the remainder of $T$, which is some hypothetical triangulation.) Let $R$ be the cycle of $T$ (the “ring” of $C$) that bounds the region that we obtain when $C$ is deleted. (It might not actually be a cycle, but ignore that. We do know its length, because it is determined by the degrees in $T$ of the vertices in $V(C)$, and we know these numbers.) Some 4-colourings of the ring can be extended to the part of $T$ outside of $R$, and some extend inwards to $C$, but none of them extends in both directions because $T$ is not 4-colourable. Since we know $C$, we can compute exactly which 4-colourings of $R$ extend inwards to $C$; so we know that the set of ring-colourings that extend outwards is a subset of the complement of this. Also, the set that extend outwards has to be closed under Kempe chain recolouring (we call this “consistent”); and sometimes we find that this is impossible, that there is no such consistent set (except the null set, which is consistent but cannot be the right answer). In this case, deleting $C$ still gives a planar graph that cannot be 4-coloured; we don’t need to substitute anything for it (unless we want to make a triangulation, when we had better put something in, just to make all the regions triangles). For other clumps, we find that there is a nonempty consistent set of 4-colourings of $R$ none of which extends inwards, and we figure out the maximal such set $S$, which is unique (fortunately); and $S$ is still very restricted, so much so that we can replace $C$ by some smaller gadget, with the crucial property that no colourings in $S$ extend inwards to the gadget.

That is roughly the idea of the proof of the 4CT. How can we adapt it to handle doublecross graphs? The first issue is that we can’t take the planar dual any more to make a triangulation. But we could if we modify the four crossing edges somehow to make the graph planar first; eg subdivide all four of them and identify their midpoints. So we have something that is “almost” a planar dual of the cubic graph that we want to three-edge-colour. It is a planar triangulation except for one region of length eight. This is worth doing, because it is easier to work with triangulations than with cubic graphs.

Now we have $T$, which is almost a triangulation. If it really were a triangulation, it would have to contain one of the 633 clumps that we used for the 4CT, but

• •

  it isn’t really a triangulation;

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  if it did contain one of those clumps, we need to make sure that the region of length eight is not involved;

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  even if we find a clump far away from the big region, it will not necessarily do what we want.

The third problem above arises because we have to work with Kempe chains in the original cubic graph, which is not planar; now Kempe chains might use the four crossing edges, and so we can’t prove that the set of colourings of the ring that extend outwards is consistent in the old planar sense. All we know is, it must be consistent in a slightly weaker sense that allows for a little bit of crossing between the various Kempe chains. Call this new sense “XX-consistent”. The maximal XX-consistent subset of ring-colourings disjoint from those that extend inwards gets bigger than in the planar case; and it might now contain a colouring that extends inwards to the gadget we were planning to replace the clump by. We need to search for another set of clumps to replace the 633, that work with this weaker definition of consistent. We did this; so the third problem goes away, if we replace the 633 with the 756 clumps described at the end of the paper. And one really nice feature; the same discharging procedure as for the 4CT still proves that one of these clumps must always be present. But we still have the first and second problems to handle.

Here is a way to handle them. At the moment our “almost”-triangulation $T$ has one special region of length eight, inside which the four crossing edges were drawn. Fill in this region with a dense triangulation, so that $T$ is extended to something that really is a triangulation, $T^{\prime}$; and do it in such a way that all the new vertices close to the eight boundary vertices have big degree (at least 12). We know that $T^{\prime}$ must contain one of the 756 clumps; but we want to get one of them inside of $T$. The idea is, the discharging argument tells us a little more; that every vertex that ends up with positive charge belongs to or neighbours one of the 756 clumps. So if only we can prove that some vertex of $T$ itself ends up with positive charge, we would have one of the 756 at least neighbouring a vertex of $T$; and it cannot be part in $T$ and part sticking out, because all the new vertices close to the boundary have big degree. So it would be completely in $T$, which is what we want.

How can we show that some vertex in $T$ ends up with positive charge? We need to show that the total initial charge assigned to the interior of $T$ is big, and not too much gets shipped over the border out of $T$. This boils down to something easy. Take the initial doublecross cubic graph that we want to three-edge-colour, delete the four crossing edges, and look at the length of the cycle $Z$ that bounds the region in which the four crossing edges were drawn. Provided $Z$ has length at least 21, the discharging argument will give us what we want, one of the 756 clumps in the right place where it can be used. (We are back-and-forth between the cubic graph and its “dual” here, unfortunately.)

What do we do if $Z$ has length less than 21? In this case we need a completely different approach; we look for a “clump” in the cubic graph (perhaps we should call this a dual clump) containing the four crossing edges and some edges of $Z$. We prove that there must always be one that can be replaced by something smaller without making the graph 3-edge-colourable. Again, it is an argument using consistency under Kempe chains, but here consistency is back to its old sense from the proof of the 4CT, because all the nonplanarity is inside the clump itself, and the Kempe chains are out in the planar part of the graph. And it works, just – we could do this up to length 20, but not for length 21, which is exactly the cutoff for the discharging procedure to work.

It is most convenient to present the reducible configurations as subgraphs of triangulations, even though most of the argument is done in terms of cubic graphs. A drawing is defined as in [4], and therefore has no “crossings”. (Sometimes we need to allow crossings, but then we say so.) A triangulation $T$ means a non-null drawing in a 2-sphere such that every region is bounded by a cycle of length three (briefly, “is a triangle”), and a near-triangulation is a non-null connected drawing in the plane such that every finite region is a triangle. If $T$ is a near-triangulation, its infinite region is bounded by a cycle if and only if $T$ is two-connected; and if so, we denote this cycle by $T_{\infty}$. A configuration $K$ consists of a near-triangulation $G_{K}$ together with a map $\gamma_{K}:V(G_{K})\rightarrow\mathbb{Z}$ ($\mathbb{Z}$ denotes the set of all integers) with the following properties, where $d(v)$ is the degree of $v$ in $G_{K}$:

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  $|V(G_{K})|\geq 2$;

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  for every vertex $v$, $G_{K}\setminus v$ has at most two components, and if there are two, then $\gamma_{K}(v)=d(v)+2$;

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  for every vertex $v$, if $v$ is not incident with the infinite region then $\gamma_{K}(v)=d(v)$; and otherwise $\gamma_{K}(v)>d(v)$, and in either case $\gamma_{K}(v)\geq 5$;

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  $K$ has ring-size at least two, where the ring-size of $K$ is defined to be $\sum_{v}(\gamma_{K}(v)-d(v)-1)$, summed over all vertices $v$ incident with the infinite region such that $G_{K}\setminus v$ is connected. (In fact, all configurations used in this paper have ring-size at least six.)

We use the same conventions as in [4] to describe configurations, and in particular, we use the same vertex shapes in drawings to represent the numbers $\gamma_{K}(v)$. Thus, a vertex $v$ with $\gamma_{K}(v)=5$ is represented by a solid circle; 7 by a hollow circle; 8 by a square; 9 by a triangle; 10 by a pentagon; and 6 and 11 by a point (there is only one instance of 11, in the last configuration of the list).

Two configurations $K,L$ are isomorphic if there is a homeomorphism of the plane mapping $G_{K}$ to $G_{L}$ and mapping $\gamma_{K}$ to $\gamma_{L}$. In the appendix to this paper there are 756 configurations. Any configuration isomorphic to one of these is called an XX-good configuration. Note that every XX-good configuration $K$ has the property that $\gamma_{K}(v)\leq 11$ for every $v\in V(G_{K})$.

A triangulation $T$ (in a 2-sphere $\Sigma$) or a near-triangulation (in a plane $\Sigma$) is internally six-connected if for every cycle $C$ of $T$ with length at most five, either some open disc in $\Sigma$, bounded by $C$, contains no vertex of $T$, or $C$ has length five and some such open disc contains a unique vertex of $T$.

We say a configuration $K$ appears in a triangulation $T$ if

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  $G_{K}$ is an induced subgraph of $T$; and

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  for each $v\in V(G_{K})$, $\gamma_{K}(v)$ equals the degree of $v$ in $T$.

Let $T$ be a two-connected near-triangulation. We say a configuration $K$ appears internally in $T$ if

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  $G_{K}$ is an induced subgraph of $T\setminus V(T_{\infty})$;

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  for each $v\in V(G_{K})$, $\gamma_{K}(v)$ equals the degree of $v$ in $T$; and

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  every vertex or edge of $T$ that does not belong to $G_{K}$ is drawn within the infinite region of $G_{K}$.

The main result of this section is the following.

To prove this, we use the discharging procedure from [4], so now we turn to that. A discharging function in a triangulation $T$ means a map $\phi$ from the set of all ordered pairs of adjacent vertices of $T$ into $\mathbb{Z}$, such that $\phi(u,v)+\phi(v,u)=0$ for all adjacent $u,v$. In [4], we defined an explicit discharging function, for every internally six-connected triangulation. Since it is rather complicated, we refer the reader to [4] for the details of its definition. We need the following two properties; the first can be verified by hand, but the second needs a computer.

Both of these statements are minor variants of theorems proved in [4] (theorems 4.7 and 4.4 of that paper, respectively) and the methods of proof are unchanged. The proof of the first statement is virtually identical with the proof of theorem 4.7 of [4], because all the “good configurations” used in that proof are also XX-good, except for one, the configuration called conf(2,10,6) in that paper, which is not XX-good. This is only needed at one step of the proof, and at that step we can use a different configuration instead, which is XX-good, the second on line 5 of page 3 of the Appendix. In the proof in [4] of theorem 4.7 of that paper, we did not include in the statement of the theorem that the good configuration we find contains $u$, as we are claiming now; but in fact the proof shows that.

The proof of the second needs analogues of theorems 4.5, 4.6, 4.8, 4.9 of the same paper. Again, in [4] we did not include in the statement of the theorem that the good configuration we find contains $u$ or one of its neighbours; but this is implied by the fact that the configuration is always found within a “cartwheel” centred on $u$. The proofs of the analogues of 4.5, 4.6 and 4.8 are unchanged, because all the good configurations used for those proofs in [4] are also XX-good. The analogue of 4.9 is proved by computer. The computer program just checks a machine-readable proof of unavoidability, and is the same as was used in [4]; we just changed its two inputs, the list of configurations we want to prove unavoidable, and the files containing the machine-readable proofs. We are making the program and the computer-readable proofs available on the arXiv [5].

Let there be $k$ edges between $V(J)\setminus V(C)$ and $V(C)$. It follows that $\sum_{v\in V(C)}d_{J}(v)=k+2|V(C)|$. Consequently

and so

Now extend the drawing of $J$ to a drawing of an internally six-connected triangulation $T$, by adding more vertices and edges drawn within $r_{0}$, in such a way that every vertex in $V(C)$ has degree in $T$ at least 12. (It is easy to see that this is possible, since $J$ is internally six-connected.)

(1) If there is an XX-good configuration $K$ that appears in $T$ such that some vertex in $V(J)\setminus V(C)$ either belongs to $G_{K}$ or has a neighbour in $G_{K}$, then $K$ appears internally in $J$.

No vertex in $V(C)$ belongs to $V(G_{K})$, since $\gamma_{K}(v)\leq 11$ for every vertex $v$ of $K$, and $d_{T}(v)\geq 12$ for every $v\in V(C)$. From the hypothesis, it follows that some vertex of $G_{K}$ belongs to $V(J)\setminus V(C)$, and hence $V(G_{K})\subseteq V(J)\setminus V(C)$ since $G_{K}$ is connected. Moreover, every finite region of $G_{K}$ is a finite region of $J$, since $J$ is internally six-connected. But every vertex in $V(J)\setminus V(C)$ has the same degrees in $T$ and in $J$, and so $K$ appears internally in $J$. This proves (1).

Let $\phi$ be the discharging function on $T$ defined in [4]. Suppose first that $\phi(u,v)>5$ for some edge $uv$ of $T$ with $u\in V(J)\setminus V(C)$ and $v\in V(C)$. By the first statement of 4.2, some XX-good configuration $K$ appears in $T$ and contains $u$, and the result follows from (1).

Thus we may assume that there is no such edge $uv$. Let there be $k$ edges $uv$ in $J$ with $u\in V(J)\setminus V(C)$ and $v\in V(C)$. Consequently the sum of $\phi(u,v)$, over all edges $uv$ of $T$ with $u\in V(J)\setminus V(C)$ and $v\in V(C)$, is at most $5k$. But this equals the sum over all $u\in V(J)\setminus V(C)$, of the sum of $\phi(u,v)$ over all neighbours $v$ of $u$, since $\phi(u,v)=-\phi(v,u)$ for all $u,v$. Therefore the sum over all $u\in V(J)\setminus V(C)$ of

is at least $10(k+6-2|V(C)|)-5k$. Since $k\geq 4|V(C)|-11$ by hypothesis, the last is positive, and so there exists $u\in V(J)\setminus V(C)$ such that

For such a vertex $u$, its degrees in $J$ and $T$ are the same; and so by the second assertion of 4.2, there is an XX-good configuration $K$ that appears in $T$, such that either $u$ or some neighbour of $u$ is a vertex of $G_{K}$. But then again, the result follows from (1). This proves 4.1.     

By a minimal counterexample, we mean a cubic two-edge-connected doublecross graph $G$ which is not three-edge-colourable, and such that every cubic two-edge-connected minor of $G$ is three-edge-colourable except $G$ itself.

To prove 1.1, we use the same approach as the proof of the four-colour theorem, proving the existence of an unavoidable set of reducible subgraphs. Some of these reducible subgraphs use all four of the edges $g_{1},\ldots,g_{4}$, and the others use none of them. The length of the cycle $Z$ of 5.1 is the deciding factor here; if $|E(Z)|\leq 20$ we will show the existence of a reducible subgraph using $g_{1},\ldots,g_{4}$, while if $|E(Z)|\geq 21$ we will show the presence of one of the other kind. In this section we handle the case when $|E(Z)|\leq 20$.

A choice of the 8-tuple $(L_{1},\ldots,L_{8})$ is called plausible if it satisfies the conditions just listed. Suppose that $|E(Z)|\leq 20$; then there are only finitely many plausible choices for $(L_{1},\ldots,L_{8})$, and we handle them one at a time. Now, therefore, we assume that we are dealing with some such plausible choice, and so we know the lengths $L_{1},\ldots,L_{8}$. Let $G^{-}$ be the graph obtained from $G$ by deleting the four crossing edges $g_{1}=z_{1}z_{3}$, $g_{2}=z_{2}z_{4}$, $g_{3}=z_{5}z_{7}$ and $g_{4}=z_{6}z_{8}$. Then $Z$ is a cycle of $G^{-}$, bounding a region in a planar drawing of $G^{-}$. Every vertex of $Z$ different from $z_{1},\ldots,z_{8}$ is incident with an edge of $G^{-}$ that does not belong to $E(Z)$. Let the vertices of $Z$ different from $z_{1},\ldots,z_{8}$ be $v_{1},\ldots,v_{k}$ say, numbered in circular order (starting from some arbitrary first vertex), and for $1\leq i\leq k$ let $f_{i}$ be the edge of $G$ incident with $v_{i}$ and not in $E(Z)$. Note that $f_{1},\ldots,f_{k}$ might not all be distinct (because for instance some $f_{i}$ might be incident with a vertex of the interior of $Z_{2}$ and incident with a vertex of the interior of $Z_{6}$). Let $F=\{f_{1},\ldots,f_{k}\}$, and let $\mathcal{D}$ be the set of all maps from $F$ to $\{1,2,3\}$.

A subset $\mathcal{C}\subseteq\mathcal{D}$ is said to be consistent if it has the following property. For all distinct $x,y\in\{1,2,3\}$, and each $\phi\in\mathcal{C}$, let $F_{x,y}$ be the set of all $f\in F$ with $\phi(f)\in\{x,y\}$; then there is a partition $\Pi$ of $F_{x,y}$ into sets of size one and two, with the following properties:

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  for $f\in F_{x,y}$, the member of $\Pi$ containing $f$ has size one if and only if both ends of $f$ belong to $V(Z)$;

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  for $1\leq a<b<c<d\leq k$, not both $\{f_{a},f_{c}\},\{f_{b},f_{d}\}\in\Pi$; and

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  $\phi^{\prime}\in\mathcal{C}$ for every subset $F^{\prime}\subseteq F_{x,y}$ which is expressible as a union of members of $\Pi$, where $\phi^{\prime}$ is defined by

  $$\phi^{\prime}(f)=\begin{cases}\phi(f)&\mbox{if }f\in F\setminus F^{\prime}\\ y&\mbox{if }f\in F^{\prime}$ and $\phi(f)=x\\ x&\mbox{if }f\in F^{\prime}$ and $\phi(f)=y.\end{cases}$$

For any graph $H$ with $F\subseteq E(H)$, we denote by $\mathcal{C}_{H}$ the set of all members of $\mathcal{D}$ that can be extended to a three-edge-colouring of $H$. Let $J=G^{-}\setminus E(Z)$, and let $K$ be the subgraph of $G$ formed by the edges in $F\cup E(Z)\cup\{g_{1},g_{2},g_{3},g_{4}\}$ and their incident vertices. Since $|E(Z)|\leq 20$, it follows that $|F|\leq 12$.

For each plausible choice of $(L_{1},\ldots,L_{8})$ (except one, that we handle separately), there are three steps to be carried out on a computer, which we explain now. All three involve computation with subsets of $\mathcal{D}$, but since $|F|\leq 12$ all three steps are easily implemented.

Step 1: Compute $\mathcal{C}_{K}$.

Step 2: Compute the maximal consistent subset $\mathcal{C}$ of $\mathcal{D}\setminus\mathcal{C}_{K}$. (The union of any two consistent sets is consistent, and so there is a unique maximal consistent subset of any set.)

Step 3: Verify that there is a graph $K^{\prime}$, obtained from $K$ by deleting one or two of the edges $g_{1},\ldots,g_{4}$, and suppressing the vertices of degree two that arise, such that $\mathcal{C}\cap\mathcal{C}_{K^{\prime}}=\emptyset$. (This involves looking at the handful of possibilities for $K^{\prime}$ and computing $\mathcal{C}_{K^{\prime}}$ for each of them.)

The exception is the 8-tuple $(4,1,4,1,4,1,4,1)$, which is plausible but for which there is no $K^{\prime}$ as in step 3 above. For this 8-tuple let $K^{\prime}$ be obtained from $K$ by deleting the unique edge in $Z_{2}$ and suppressing the vertices of degree two that arise. Again we check that $\mathcal{C}\cap\mathcal{C}_{K^{\prime}}=\emptyset$.

In all cases, since $G$ is not three-edge-colourable, and $J\cup K=G$, it follows that $\mathcal{C}_{J}$ and $\mathcal{C}_{K}$ are disjoint. The planarity of $J$ implies that $\mathcal{C}_{J}$ is consistent (this is easy to see, and is a standard argument in proving the four-colour theorem – see [4]). Consequently $\mathcal{C}_{J}\subseteq\mathcal{C}$, and so $\mathcal{C}_{J}\cap\mathcal{C}_{K^{\prime}}=\emptyset$. It follows that $J\cup K^{\prime}$ is not three-edge-colourable. But $J\cup K^{\prime}$ can be obtained from $G$ by deleting either one or two disjoint edges, and suppressing the vertices of degree two that arise; and so $J\cup K^{\prime}$ is two-edge-connected (since $G$ is theta-connected), doublecross, and smaller than our supposedly minimal counterexample, which is impossible. This proves 5.2.     

We have omitted the details of the computer checking; this is all straightforward. (The program is available on the arXiv.) There are 2957 plausible 8-tuples to check, up to symmetry, but the program only takes about a minute to do them all, so we were content with that. If desired, running through all possible choices of $(L_{1},\ldots,L_{8})$ could be made more efficient at the cost of complicating the proof. For instance, we could quickly dispose of the case when $\min(L_{1},L_{2},L_{3})=\min(L_{5},L_{6},L_{7})=1$, because in this case $G$ contains a “C-reducible” subgraph (of a different kind), no matter what the other six lengths are. But we are aiming for simplicity rather than speed here.

An island means a graph $I$ drawn in the plane, with the following properties:

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  $I$ is two-connected;

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  every vertex has degree two or three; and

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  every vertex of degree two is incident with the infinite region.

Let $I$ be an island, and $J$ be a geometric dual, where $j\in V(J)$ corresponds to the infinite region of $I$. For each $v\in V(J)\setminus\{j\}$, let $\gamma(v)$ be the length of the region of $I$ that corresponds to $v$. Then the pair $(J\setminus\{j\},\gamma)$ might or might not be a configuration; but more important, for every configuration $K$, there is an island that gives rise to it in this way, unique up to homeomorphism of the plane. (We leave checking this to the reader. One way is to go to the “free completion” of $K$ defined in [4], take a dual, and delete the vertex corresponding to the infinite region.) We call this the island of $K$, and denote it by $I(K)$.

We need to work mostly with the islands of the XX-good configurations, but it is more compact to draw the configurations themselves. Sometimes we need to refer to an edge $e$ of one of the islands, say of $I(K)$. Now $e$ corresponds to some edge $f$ of $J$ under the duality (where $J$ is as before), and if $f\in E(G_{K})$ then we can refer to $e$ by defining it as the edge dual to $f$. But sometimes the edge $f$ is not an edge of $G_{K}$. For this reason, in the list of XX-good configurations, some vertices are drawn with extra “half-edges”. These indicate some of the edges of $J$ that are not edges of $G_{K}$, for convenience in referring to certain edges of $I(K)$.

Now take a geometric dual $T$ of $G^{+}$, such that $z_{\infty}$ belongs to the infinite region of $T$. It follows that:

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  $T$ is a near-triangulation;

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  the cycle $T_{\infty}$ is an induced subgraph of $T$;

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  $T$ is internally six-connected (since $G$ is theta-connected); and

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  the number of edges of $T$ between $V(T_{\infty})$ and $V(T)\setminus V(T_{\infty})$ is at least 21, since $|E(Z)|\geq 21$ by 5.2, and for each $e\in E(Z)$, the corresponding edge of $T$ has one end in $V(T_{\infty})$ and the other in $V(T)\setminus V(T_{\infty})$.

Since $|V(T_{\infty})|=8$ and so $4|V(T_{\infty})|-11=21$, 4.1 implies that some XX-good configuration $K$ appears internally in $T$. Consequently the union of the closures of the regions of $G^{-}$ that correspond to vertices in $G_{K}$ is a closed disc that defines an island satisfying the theorem. This proves 6.1.     

It remains to show that the outcome of 6.1 is impossible, but for that we need to discuss reducibility further.

If $a,b,c,d$ are integers and $1\leq a<b<c<d$, we call $\{\{a,c\},\{b,d\}\}$ a cross. Let $\Pi$ be a finite set of finite sets of positive integers, each of cardinality two and pairwise disjoint. We say that $\Pi$ is doublecross if the following conditions hold:

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  at most two subsets of $\Pi$ are crosses

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  if $A,B,C,D\in\Pi$ are distinct, and $\{A,B\},\{C,D\}$ are crosses, let $X=A\cup B\cup C\cup D$ (so $|X|=8$). Then for all $P\in\Pi$ with $P\cap X=\emptyset$ there do not exist $x_{1},x_{2}\in X$ such that $\{P,\{x_{1},x_{2}\}\}$ is a cross.

This is equivalent to the following geometric condition, which may be easier to grasp: choose $k\geq 3$ such that $A\subseteq\{1,\ldots,k\}$ for each $A\in\Pi$. Take a regular $k$-vertex polygon in the plane, with vertices $v_{1},\ldots,v_{k}$ in order. For each $A\in\Pi$ draw a line segment $L_{A}$ between $v_{i},v_{j}$, where $A=\{i,j\}$. Then we ask that

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  no point of the plane belongs to more than two of $L_{A}\;(A\in\Pi)$;

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  at most two points of the plane belong to more than one of these lines;

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  if there are two points $x,y$ each belonging to two of the lines $L_{A}\;(A\in\Pi)$ say, then either some $L_{A}$ contains them both, or no $L_{A}$ intersects the interior of the line segment between $x,y$.

We leave the equivalence to the reader.

Let $k\geq 1$, and let $\mathcal{D}$ be the set of all maps $\phi:\{1,\ldots,k\}\rightarrow\{1,2,3\}$. We say a subset $\mathcal{C}$ of $\mathcal{D}$ is XX-consistent if it has the following property. For all distinct $x,y\in\{1,2,3\}$, and each $\phi\in\mathcal{C}$, let $R_{x,y}$ be the set of all $i\in\{1,\ldots,k\}$ with $\phi(i)\in\{x,y\}$; then there is a doublecross partition $\Pi$ of $R_{x,y}$, such that $\phi^{\prime}\in\mathcal{C}$ for every subset $F^{\prime}\subseteq R_{x,y}$ which is expressible as a union of members of $\Pi$, where $\phi^{\prime}$ is defined by

Let $I$ be the island of a configuration $K$. We say $F\subseteq E(I)$ is a matching if no two edges in $F$ have a common end, and $V(F)$ denotes the set of vertices incident with an edge in $F$. A three-edge-colouring modulo $F$ of $I$ means a map $\phi:\;E(I)\setminus F\rightarrow\{1,2,3\}$, such that for all distinct edges $e,f\in E(I)\setminus F$ with a common end $v$ say, $\phi(e)=\phi(f)$ if and only if $v\in V(F)$.

Let the vertices with degree two in $I$ be $v_{1},\ldots,v_{k}$ in order on the boundary of the infinite region. With $\mathcal{D}$ as before, let $\mathcal{C}_{K}$ be the set of all $\psi\in\mathcal{D}$ such that there is a three-edge-colouring $\phi$ of $I$ with $\phi(e)\neq\psi(i)$ for $1\leq i\leq k$ and for each edge $e$ of $I$ incident with $v_{i}$. We say that $K$ is XXD-reducible if there is no non-null XX-consistent subset of $\mathcal{D}\setminus\mathcal{C}_{K}$. We say that $K$ is XXC-reducible if there is a matching $F$ of $I$ with the following properties:

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  $1\leq|F|\leq 4$.

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  If $|F|=4$, then either some finite region of $I$ is incident with at least three members of $F$, or there are two finite regions of $I$, say $r,s$, such that some edge of $I$ is incident with both $r,s$, and every edge of $F$ is incident with one of them.

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  Let $\mathcal{C}_{F}$ be the set of all $\psi\in\mathcal{D}$ such that there is a three-edge-colouring modulo $F$ of $I$, say $\phi$, such that for $1\leq i\leq k$ and every edge $e\in E(I)\setminus F$, $\phi(e)=\psi(i)$ if and only if $v_{i}\in V(F)$. Then every XX-consistent subset of $\mathcal{D}\setminus\mathcal{C}_{K}$ is disjoint from $\mathcal{C}_{F}$.

We call such a set $F$ a reducer for $K$. (We will show that if $K$ appears in a minimal counterexample $G$, then deleting from $G$ the edges in $F$ and suppressing the resultant vertices of degree two will make a smaller counterexample, which is impossible; and so $K$ cannot appear.) We need:

This is just the same process as in the proof of the four-colour theorem [4], and is carried out on a computer the same way; we omit further details. (Again, we are making the program available on the arXiv [5].)

Now we combine these various lemmas to prove 1.1.

Now let $\mathcal{D}$ be the set of all maps from $\{1,\ldots,k\}$ to $\{1,2,3\}$. We say a map $\phi:E(G)\setminus E(I)\rightarrow\{1,2,3\}$ is a three-edge-colouring of $\overline{I}$ if $\phi(e)\neq\phi(f)$ for every two distinct edges $e,f\in E(G)\setminus E(I)$ with a common end in $G$. Let $\mathcal{C}$ be the set of $\psi\in\mathcal{D}$ such that there is a three-edge-colouring $\phi$ of $\overline{I}$ with $\psi(i)=\phi(e_{i})$ for $1\leq i\leq k$. We claim:

(1) $\mathcal{C}$ is non-null and XX-consistent.

Clearly it is non-null, from the minimality of $G$. Let $\psi\in\mathcal{C}$, and choose a three-edge-colouring $\phi$ of $\overline{I}$ with $\psi(i)=\phi(e_{i})$ for $1\leq i\leq k$. Let $x,y\in\{1,2,3\}$ be different, and let $H$ be the subgraph of $G$ formed by the edges $e\in E(G)\setminus E(I)$ with $\phi(e)\in\{x,y\}$ and their ends. It follows that every component of $H$ is either a cycle, or a path with distinct ends both in $V(D)$. Let $\Pi$ be the set of all $\{i,j\}$ such that $1\leq i<j\leq k$ and some component of $H$ is a path with end-edges $e_{i},e_{j}$. (Possibly $e_{i}=e_{j}$, and this path has only one edge.) Then we can switch colours $x,y$ on any subset of $\{1,\ldots,k\}$ that is expressible as a union of members of $\Pi$, by exchanging the colours $x,y$ on the corresponding components of $H$. It remains to show that $\Pi$ is doublecross. To see this, note first that if $\{\{a,b\},\{c,d\}\}\subseteq\Pi$ is a cross, and $P,Q$ are the components of $H$ with end-edges $e_{a},e_{b}$ and $e_{c},e_{d}$ respectively, then either $P$ contains one of $g_{1},g_{2}$ and $Q$ contains the other, or $P$ contains one of $g_{3},g_{4}$ and $Q$ contains the other. Since $G$ can be drawn with no crossing pairs of edges except $g_{1},g_{2}$ and $g_{3},g_{4}$, and the two crossings they form are on a common region, it follows that $\Pi$ is doublecross. This proves (1).

Now let $\mathcal{C}_{K}$ be as in 7.1; that is, $\mathcal{C}_{K}$ is the set of all $\psi\in\mathcal{D}$ such that there is a three-edge-colouring $\phi$ of $I$ with $\phi(e)\neq\psi(i)$ for $1\leq i\leq k$ and for each edge $e$ of $I$ incident with $v_{i}$.

(2) $\mathcal{C}_{K}\cap\mathcal{C}=\emptyset$.

For suppose that $\psi\in\mathcal{C}_{K}\cap\mathcal{C}$. Choose a three-edge-colouring $\phi_{1}$ of $\overline{I}$ such that $\psi(i)=\phi_{1}(e_{i})$ for $1\leq i\leq k$. Choose a three-edge-colouring $\phi_{2}$ of $I$ such that $\phi_{2}(e)\neq\psi(i)$ for $1\leq i\leq k$ and each edge $e$ of $I$ incident with $v_{i}$. For each edge $e$ of $G$, let