Characterization of ( 2 m , m )-paintable graphs

In this paper, we prove that for any graph G and any positive integer m, G is (2m,m)-paintable if and only if G is 2-paintable. It was asked by Zhu in 2009 whether k-paintable graphs are (km,m)-paintable for any positive integer m. Our result answers this question in the affirmative for k = 2.


Introduction
Graphs considered in this paper are finite.Suppose G is a graph, b is a positive integer.A b-fold colouring c of G assigns to each vertex v a set c(v) of b colours, and colour sets assigned to adjacent vertices are disjoint.If, for a ∈ Z + , we have c(v) ⊆ {1, 2, . . ., a} for all vertices v of G, then c is called a b-fold a-colouring of G; we then say that G is (a, b)colourable.A 1-fold a-colouring of G is also called an a-colouring of G.The chromatic number χ(G) of G is the minimum a such that G is a-colourable.The fractional chromatic number χ f (G) of G is defined as It is well-known that "inf" in the definition of χ f (g) can be replaced by "min." A list assignment of G is a mapping L which assigns to each vertex v of G a set L(v) of permissible colours.If |L(v)| = a for all vertices v of G, then L is called an a-list assignment.A b-fold L-colouring of G is a b-fold colouring c of G such that c(v) ⊆ L(v) for each vertex v. Similarly, a 1-fold L-colouring of G is called an L-colouring of G.A graph G is called a-choosable if there is an L-colouring for any a-list assignment L of G; and is called (a, b)-choosable if there is a b-fold L-colouring for any a-list assignment L of G.The choice number ch(G) of G is the minimum a for which G is a-choosable, and the fractional choice number ch f (G) is defined as List colouring of graphs was introduced independently in the 1970s by Vizing [8] and by Erdős, Rubin, and Taylor [3]; it has been studied extensively in the literature [7].By definitions, we have χ(G) ch(G) and χ f (G) ch f (G) for any graph G.It is easy to see (cf.[3]) that the choice number of bipartite graphs can be arbitrarily large.In particular, ch(G) − χ(G) (as well as ch(G)/χ(G)) is not bounded.On the other hand, it was proved by Alon, Tuza, and Voigt [1] that for any graph G, χ f (G) = ch f (G).Moreover, the infimum in the definition of ch f (G) is attained, and hence can be replaced by minimum.
In this paper, we consider the on-line version of list colouring of graphs.The on-line list colouring of a graph is defined through a two-person game.At the beginning of the game, instead of assigning to each vertex v a set of permissible colours, each vertex v is assigned a set of tokens.In the process of the game, each token is replaced by a permissible colour.The coloring algorithm, which we call Painter, needs to decide right away which independent set of vertices with this permissible colour will receive the colour.We now give a price definition.Let V (G) denote the vertex set of a graph G. Definition 1. Suppose G = (V, E) is a graph and f, g : V → N are functions that assign nonnegative integers to vertices of G.The (f, g)-painting game on G is played by two players: Lister and Painter.Initially, each vertex v has f (v) tokens, and no colours.In the ith step, Lister chooses a nonempty subset M i of the vertices v that have received less than g(v) colours and takes away one token from each chosen vertex.Painter chooses an independent set X i in G contained in M i and assigns colour i to each vertex of X i .If at the end of some step, there is a vertex v which has no tokens left, and received less than g(v) colours, then Lister wins the game.Otherwise, each vertex v receives g(v) colours at some step, and Painter wins the game.
The game was called the on-line list colouring game in [10], because the sets of permissible colours are given on-line, and the colouring of the graph is constructed on-line.The (f, 1)-painting game was originally described by Schauz [6] as a game between Mr. Paint and Mrs. Correct, and in [2], it was described as a game between Marker and Remover (as in the (f, 1)-painting game, once a vertex v is coloured, then it requires no further attention, and we can consider v to be removed from G). Definition 2. A graph G is (f, g)-paintable when Painter has a winning strategy for the (f, g)-painting game on G.If f (v) = a and g(v) = b are constant functions, then an (f, g)-paintable graph is called (a, b)-paintable.If G is (a, 1)-paintable, then we say G is a-paintable.The paint number χ p (G) of G is the minimum a such that G is a-paintable.The fractional paint number χ f p (G) of G is defined as By definition, χ(G) ch(G) χ p (G) and χ f (G) ch f (G) χ f p (G) for every graph G.It is known (see [6], [10]) that there are graphs G for which χ p (G) > ch(G).However, Gutowski [4] proved that for any graph G, χ f p (G) = ch f (G) = χ f (G).Nevertheless, example graphs are given in [4] to show that the infimum in the definition of χ f p (G) may not be attained, and hence cannot be replaced by the minimum.
If G is a nonempty bipartite graph, then -choosable , we know that for some integer m, G is (2m, m)choosable.Although the fractional paint number of G is also 2, in the definition of χ f p (G), the infimum cannot be replaced by the minimum.Thus χ f p (G) = 2 does not imply that G is (2m, m)-paintable for some integer m.A natural question is which graphs are (2m, m)paintable for some integer m.
It was conjectured in [3] that if a graph G is k-choosable, then for any positive integer m, G is (km, m)-choosable.More generally, it was conjectured that if G is (a, b)-choosable then for any positive integer m, G is (am, bm)-choosable.We propose the on-line version of the above conjecture.
It was asked in [10] (Question 24) whether k-paintable graphs are (km, m)-paintable for any positive integer m.The main result of this paper is the following theorem, which answers this question in affirmative for k = 2, and surprisingly, in this case the converse is also true.Theorem 4. For every graph G and every positive integer m, G is (2m, m)-paintable if and only if G is 2-paintable.
In the remaining paper, we prove Theorem 4. Section 2 contains some easy lemmas that we use to make later proofs more efficient.In Section 3, we prove the forward direction of the implication in Theorem 4. In Section 4, we prove the reverse direction of Theorem 4.

Lemmas
Denote by N G (v) (respectively, N G [v]) the neighborhood of v (respectively, the closed neighborhood of v).We say v is an (x, y)-vertex when f (v) = x and g(v) = y.
We begin with an observation about necessary conditions on f and g for a graph to be (f, g)-paintable.We call the pair (f, g) the token-colour functions of G.If g(v) = 0 for some v, then the game is equivalent to the game restricted to G − v. Using a term introduced in [2], we call the set M i chosen by Lister the set marked at step i.When the step number is clear from the context or is irrelevant, we simply call M a marked set.
The following proposition is usually used as an equivalent, recursive definition of a graph to be (f, g)-paintable.Let δ X be the characteristic function of a set X, defined as δ X (v) = 1 for v ∈ X and δ X (v) = 0 otherwise.Proposition 5. Assume G is a graph and (f, g) are token-colour functions with g(v) > 0 for all v. Then G is (f, g)-paintable if and only if any subset U of V (G) contains an independent set X such that G is (f − δ U , g − δ X )-paintable.Proposition 6. Assume G is a graph and (f, g) are token-colour functions with g(v) > 0 for all v.If G is (f, g)-paintable, then , then Lister wins by marking {v} until it has no more tokens, but still needs to be coloured.If uv ∈ E(G) and max{f (u), f (v)} < g(u) + g(v), then Lister wins by marking {u, v} as long as f (u), f (v), g(u), and g(v) are nonnegative.Each round, g(u) + g(v) decreases by at most 1, so some vertex still needs to be coloured after losing all of its tokens.
We say that a vertex v is forced when f (v) = g(v), and an edge uv is tight when max{f To make studying the painting game more efficient, we make use of the following observations about Painter's responses on bipartite graphs: Corollary 8. Assume G is a bipartite graph and the set of tight edges induces a connected spanning subgraph of G.If Lister marks V (G), then Painter must colour all vertices in one of the partite sets.
We also use the following two results, which were first proven by Zhu [10]. and the electronic journal of combinatorics 22(2) (2015), #P2.14 Proof.If G is (f, g)-paintable and Lister marks N [v] for g(v) consecutive rounds, then Corollary 7 implies Painter colours v each time.After these moves, each w of )-paintable since Painter had no other possible responses against this Lister strategy.If G − v is (f , g )-paintable, then in G, Painter "reserves" g(v) tokens at each neighbor of v. Anytime v is marked, Painter colours v, and uses up at most one of the reserved tokens for each u ∈ N (v).Since this happens at most g(v) times and G − v is (f , g )-paintable, Painter has a winning strategy in G.
For a subset where f and g are the restrictions of f and g to G − v.
Proof.If G is (f, g)-paintable, then Painter has a winning strategy on every subgraph of G with restrictions of f and g.Suppose G − v is (f , g )-paintable.Painter wins in G by following a winning strategy for G − v, colouring v when it is marked and none of its neighbors are coloured by that strategy.At most g(N G (v)) tokens of v are used without colouring it, so v will receive enough colours.
The following proposition is obvious.In our later proofs, we implicitly use it to restrict our attention to Painter responses that are maximal independent subsets of Lister's moves.
Proposition 11.If G is (f, g)-paintable, then Painter has a winning strategy in which on each round, the vertices coloured form a maximal independent subset of the marked set.
We now prove that Lister has a winning strategy in a particular position on C 4 .
Proof.We use induction on the total number of tokens , and g is the restriction of g to {v 1 , v 2 , v 3 }.However, with respect to (f , g ), both v 1 and v 3 are forced.Lister wins the game by applying Proposition 9 again.
We may now assume that v 0 is not forced.As (v 1 , v 0 ) and (v 3 , v 0 ) are strictly tight pairs, i.e, 1 f Lister marks {v 2 , v 3 }.By Corollary 7, Painter colours v 3 ; now (v 1 , v 2 ) is strictly tight.Then Lister marks {v 2 , v 1 }.By Corollary 7 again, Painter colours v 1 ; now (v 3 , v 2 ) is strictly tight and (v 1 , v 2 ) remains strictly tight.Since g(v i ) > 0 for all v i and f (v i ) is smaller, the induction hypothesis implies that Lister wins.
the electronic journal of combinatorics 22(2) (2015), #P2.14When a graph G is not (2m, m)-paintable, we wish to conclude that "similar" graphs are also not (2m, m)-paintable.We make this intuitive idea more precise with the following definition.Note that every (2m, m)-colouring of a path assigns the same set of b colours to every other vertex along the path.We now apply Lemma 15 with a = 2m and b = m in the following corollary.In Section 3, we use it to reduce the number of cases that we must consider in Theorem 17.
Corollary 16.Given a graph G and an edge uv of G, if G is not (2m, m)-paintable, then the graph obtained by replacing uv with a path of odd length is not (2m, m)-paintable.
Proof.Assume G is obtained from G by replacing uv with a path of length 2r + 1.Let H be a path of length 2r with vertices {v 1 , v 2 , . . ., v 2r+1 } taken in order along the path.Then (H, {v 1 , v 2r+1 }) is a (2m, m)-gadget, and we obtain G as in Definition 14 by taking the disjoint union of G and H, splitting u into two vertices u 1 and u 2 , and let u 1 be adjacent to v and u 2 adjacent to the rest of N G (u), and finally, u 1 is identified with v 1 and u 2 with v 2r+1 .Using this (H, U )-augmentation of G, Lemma 15 implies that G is not (2m, m)-paintable.

Non-(2m, m)-paintable graphs
Our goal in this section is to prove the following theorem: for all e ∈ E(G).If G is not 2-paintable, then it must contain a 3-paint-critical subgraph.Thus to prove Theorem 17, it suffices to show that every 3-paint-critical graph is not (2m, m)-paintable.Riasat and Schauz [5], and independently Carraher et al. [2], characterized 3-paint-critical graphs.The characterization first requires the following definition: the theta-graph Θ r,s,t consists of two vertices joined by internally disjoint paths of lengths r, s, t.
Theorem 18 ( [5,2]).A graph is 3-paint-critical if and only if it is one of the following: • An odd cycle.
• Two vertex-disjoint even cycles connected by a path.
• Two edge-disjoint even cycles having exactly one vertex in common.
• Θ r,s,t where r, s, t have the same parity and max {r, s, t} > 2.
To prove Theorem 17, it suffices by Corollary 16 to show that the following seven graphs in Figure 1 are not (2m, m)-paintable for any positive integer m.Since C 3 is not (2m, m)-colourable, it is not (2m, m)-paintable for any positive integer m.We now reduce this family further by applying Lemma 15 to F 0 .
Proof.Let G = F 0 , let u be the vertex of degree 4, and suppose that G is not (2m, m)paintable.Let H = P 3 and V (H) = {v 1 , v 2 , v 3 } with v 1 , v 3 as the endpoints.Note that (H, {v 1 , v 3 }) is a (2m, m)-gadget.We split u into two copies u 1 , u 2 and partition the edges incident to u so that u 1 is incident to the two edges in the copy of C 4 on the left and u 2 is incident to the two edges in the copy of C 4 on the right.Identifying u 1 with v 1 and u 2 with v 3 yields the graph F 2 .Therefore, F 2 is an (H, {v 1 , v 3 })-augmentation of F 0 , and by Lemma 15, F 2 is not (2m, m)-paintable.
It remains to show that each of K 2,4 , F 0 , F 1 , Θ 2,2,4 , and Θ 1,3,3 is not (2m, m)-paintable for any m ∈ Z + .All these graphs are bipartite graphs.We use A and B to denote the two partite sets.Vertices in A are named a 1 , . . ., a |A| and vertices in B are b 1 , . . ., b |B| .
Theorem 20.Let G = K 2,4 and f, g be positive on every vertex.If every edge of G is tight, then G is not (f, g)-paintable.) is strictly tight).Similarly as above, it is easy handle the case when g(a 2 ) = 1 and we can assume g(a 2 ) 2. Now g = g − δ {a 1 ,a 2 } is positive on every vertex v and every edge is tight, so Lister wins by induction hypothesis.
Theorem 22.Let G = F 0 and f, g be positive on every vertex.If every edge of G is tight, then G is not (f, g)-paintable.Proof.We prove by induction on the total number of tokens.Let F 0 be labelled as in Figure 2 and let A = {a 1 , a 2 , a 3 , a 4 }.If there is a forced vertex x, then let C = (x, y, z, w) be a 4-cycle in G containing x.We know that (w, x) and (y, x) are strictly tight pairs.It follows from Lemma 12 that Lister has a winning strategy.Thus we assume that there is no forced vertex.
Suppose there is a strictly tight pair (x, y).Note since y is not forced, g(y) < f (y) < g(y) + g(x), and thus g(x) 2. Also, since no vertex is forced, f (v) 2 for all v. Lister marks N G [x] − {y}, and by Corollary 7, Painter colours x.After this move, f, g are still positive on every vertex.By the induction hypothesis, Lister has a winning strategy.Thus we assume that there is no strictly tight pair.This implies that for each edge xy of G, f (x) = f (y) = g(x) + g(y).Hence, f is constant on V (G), and g is constant on each partite set of G. Theorem 27.Let G = Θ 1,3,3 and f, g be positive on every vertex.If every edge of G is tight, then G is not (f, g)-paintable.
Proof.Let u, v be adjacent vertices of degree 2. Lister repeatedly marks {u, v}.Since uv is a tight edge, eventually, one vertex, say u, becomes a forced vertex.Lister marks u and its neighbor u in G − {u, v}.Painter must colour u, as u is forced.Lemma 12 implies that Lister wins in the remaining graph.

(2m, m)-paintable graphs
The core of a connected graph G is the graph obtained from G by successively deleting vertices of degree 1; it is unique up to isomorphism.The following theorem characterizes 2-paintable graphs.
In the (2m, m)-painting game, vertices of degree 1 are degenerate.Thus for any positive integer m, a graph G is (2m, m)-paintable if and only if its core is (2m, m)paintable.Thus to prove Theorem 4, it suffices to show that each of K 1 , C 2n , K 2,3 is (2m, m)-paintable for all m ∈ Z + .Given m ∈ Z + , it is obvious that K 1 is (2m, m)paintable, so we move on to considering even cycles.An oriented graph G is kernel perfect if every induced subgraph G[M ] has a kernel, i.e., M contains an independent set I such that every v ∈ M − I has an out-neighbour in I.For an oriented graph D, N + D (v) is the set of out-neighbours of v, and Theorem 30.Assume G has an orientation D which is kernel perfect and g : V (G) → N is a mapping which assigns to each vertex x a non-negative integer.If f (v) Proof.We prove this theorem by induction on the total number of tokens.For any subset Proof.Orient C 2n as a directed cycle.This orientation is kernel perfect.By Theorem 30, C 2n is (2m, m)-paintable.
Let G = K 2,3 be labeled as in Figure 4. Let (f, g) be token-colour functions on G.For each edge a i b j of G, let For a set D of edges, we let w A,f,g (D) = e∈D w A,f,g (e) and w B,f,g (D) = e∈D w B,f,g (e).An edge set D is special if |D| 2 and there is an edge e ∈ D such that every other edge e ∈ D has no common endpoint with e. Observe that a special edge set D contains either two or three edges.Indeed, up to isomorphism, there are only two special edge sets {a 1 b 1 , a 2 b 2 } and {a 1 b 1 , a 2 b 2 , a 2 b 3 }.
(2) For any special edge set D, w A,f,g (D) 0 and w B,f,g (D) 0.
Proof.The proof is by induction on the total number of tokens.Assume (f, g) has Property ( ).First we consider the case that there exists a forced vertex.
Assume a 1 is forced, say a 1 is an (a, a)-vertex.Let a 2 be a (c + d, d)-vertex, and each b i be a (x i + y i , y i )-vertex.Then w A,f,g ({a 2 b 1 , a 1 b 2 , a 1 b 3 }) 0 implies that c y 1 + y 2 + y 3 , i.e., a 2 is degenerate.By Proposition 10, G is (f, g)-paintable if and only if G−a 2 is (f, g)paintable.By (1) of Property ( ), Assume b 1 is forced, say b 1 is a (b, b)-vertex.Let b j be a (c j + d j , d j )-vertex for j ∈ {2, 3}, and let a i be a (x i + y i , y i )-vertex for i ∈ {1, 2}.Then w B,f,g ({a 1 b 2 , a 2 b 1 }) 0 implies that c 2 y 1 + y 2 .and w B,f,g ({a 1 b 3 , a 2 b 1 }) 0 implies that c 3 y 1 + y 2 .Thus both b 2 are b 3 are degenerate, and Proposition 10 implies that G is (f, g)-paintable if and Assume there are no forced vertices.We shall prove that for any set M of vertices, there is an independent set X contained in M such that (f − δ M , g − δ X ) has Property ( ).
Assume f (a 1 ) < g(a 1 ) + g(b 1 ).Then f (b 1 ) g(a 1 ) + g(b 1 ), and f (b 1 ) < g (a 1 ) + g (b 1 ) implies that b 1 ∈ M, a 1 ∈ M and X = M ∩ A. So w A,f,g (a 1 b 1 ) < 0 and yet (R2) is not applied.This implies that (R1) is applied.Without loss of generality, we may assume that a Next we show (2) of Property ( ) holds for (f , g ), i.e., for any special edge set D, w A,f,g (D) 0 and w B,f,g (D) 0.
Observe that if X = M ∩A, then for any edge e = ab, w A,f,g (e) = w A,f ,g (e).Indeed, if a ∈ M , then f (a) = f (a)−1, g (a) = g(a)−1, and g (b) = g(b).So w A,f,g (e) = w A,f ,g (e).If a ∈ M , then f (a) = f (a), g (a) = g(a), and g (b) = g(b).Again w A,f,g (e) = w A,f ,g (e).Similarly, if X = M ∩ B, then for any edge e, we have w B,f,g (e) = w B,f ,g (e).
Since X = M ∩ A, by the observation above, for any special edge set D, w A,f ,g (D) w A,f,g (D) 0. It remains to show that w B,f ,g (D) 0.
As (R1) applies, there is an edge, say e = a 1 b 1 , such that w B,f,g (a 1 b 1 ) < 0, a 1 ∈ M and b 1 ∈ M .
The proof of this case is the same as that of Case 1.One simply needs to interchange A and B in the subscripts and the roles of a 1 and b 1 in the marked set M .

Corollary 7 .
If the marked set contains a forced vertex v, then Painter must colour v; if the marked set contains u and v where uv is a tight edge, then Painter must colour one of u and v; if the marked set contains u and not v where (u, v) is a strictly tight pair, then Painter must colour u.

Definition 13 .
Assume H is a graph and U is a subset of vertices of H and a b are positive integers.We say (H, U ) is an (a, b)-gadget if H is (a, b)-colourable, and in any (a, b)-colouring of H, all vertices in U are coloured by the same b-set.Definition 14. Assume G and H are graphs, v ∈ V (G) and U ⊆ V (H).If G is obtained from the disjoint union of G and H by splitting v into |U | copies, arbitrarily partitioning the edges incident to v among those copies, and identifying the |U | copies of v with the vertices of U in H, then we say G is an (H, U )-augmentation of G.We use the gadget from Definition 13 and the augmentation described in Definition 14 to build many non-(a, b)-paintable graphs G from a single non-(a, b)-paintable graph G.The folowing lemma makes this idea precise by showing how "non-(a, b)-paintability" can be preserved when augmenting G to form G .Lemma 15.If G is not (a, b)-paintable, (H, U ) is an (a, b)-gadget, and G is an (H, U )augmentation of G, then G is not (a, b)-paintable.Proof.Since G is not (a, b)-paintable, Lister has a winning strategy S. Each round, Lister obtains a marked set M ⊆ V (G) according to S. If v ∈ M , then in G , Lister marks V (H) ∪ (M − v); otherwise Lister marks M as a subset of V (G ).Let D be the set that Painter colours.If 0 < |D ∩ U | < |U |, then Lister marks V (H) in every remaining round.Since every (a, b)-colouring of H assigns vertices in U the same set of b colours, Painter will not be able to colour each vertex of H by a set of b colours.So Lister wins the game.If D ∩ U = ∅, then Lister views Painter's response as D − V (H) in the game on G.If D ∩ U = U , then Lister views Painter's response as D ∪ {v} − V (H) in the game on G. Thus Lister can continue using strategy S and eventually wins the game.

Figure 1 :
Figure 1: Family of graphs for Theorem 17

Proof.
We prove by induction on the total number of tokens.Let A = {a 1 , a 2 } be the set of vertices of degree 4, and let B = {b 1 , . . ., b 4 } be the set of vertices of degree 2. Lister marks L 1 = {a 1 , b 1 , b 2 }.Painter must colour a 1 , for otherwise, (b 3 , a 1 ) and (b 4 , a 1 ) would be strictly tight pairs and Lister wins at the 4-cycle C 4 = (a 1 , b 3 , a 2 , b 4 ) by Lemma 12.If g(a 1 ) = 1, then after the first move, b 1 , b 2 become forced vertices, and it is easy to check that Lister wins on the subgraph induced by {a 2 , b 1 , b 2 }.Assume g(a 1 ) 2. Next Lister marks {a 2 , b 3 , b 4 }, and Painter must colour a 2 by Corollary 7 (as (a 2 , b 1

3 .
If an edge d is incident to neither a 1 nor b 1 , i.e., d ∈ {a 2 b 2 , a 2 b 3 }, then w B,f ,g (d) w B,f,g (d)−1.However, for such an edge d, by (2) of Property ( ), we have w B,f,g (d) −w B,f,g (a 1 b 1 ) 1, which implies that w B,f ,g (d) 0. First we assume that a 1 b 1 ∈ D. If D contains at most one of a 2 b 2 and a 2 b 3 , then w B,f ,g (D) w B,f,g (D) 0 by the observations above.If D contains both a 2 b 2 , a 2 b 3 , then D − {a 2 b 3 } is also special, and hence w B,f ,g (D − {a 2 b 3 }) 0. As w B,f ,g (a 2 b 3 ) 0, we have w B,f ,g (D) = w B,f ,g (D − {a 2 b 3 }) + w B,f ,g (a 2 b 3 ) w B,f ,g (D − {a 2 b 3 }) 0. Next assume D does not contain a 1 b 1 .Then D contains at most one of the edges a 2 b 2 , a 2 b 3 (for otherwise, D is not special).If D contains none of a 2 b 2 , a 2 b 3 , then by the observations above, w B,f ,g (D) w B,f,g (D) 0. Thus we may assume that a 2 b 2 ∈ D and a 2 b 3 ∈ D. If every other edge of D are non-adjacent to a 2 b 2 , then D ∪ {a 1 b 1 } is also special and wB,f ,g (D) = w B,f ,g (D ∪ {a 1 b 1 }) − w B,f ,g (a 1 b 1 ) 0 (as w B,f ,g (a 1 b 1 ) 0). Assume D contains another edge adjacent to a 2 b 2 .The only possible special edge set is D = {a 2 b 1 , a 2 b 2 , a 1 b 3 }.In this case, w B,f ,g (D) = w B,f ,g (D − {a 2 b 2 }) + w B,f ,g (a 2 b 2 ) w B,f,g (D−{a 2 b 2 })+w B,f ,g (a 2 b 2 ).Since D−{a 2 b 2 }is special, we have w B,f,g (D−{a 2 b 2 }) 0. As w B,f ,g (a 2 b 2 ) 0, we conclude that w B,f ,g (D) 0. This completes the proof of Case 1.

Case 3 (
R3) applies.If M ∩ A = A, then w A,f ,g (D)w A,f,g (D) 0 for any special edge set D. If |M ∩ A| = 2, then w B,f ,g (e)w B,f,g (e) for every edge e. Assume that|M ∩ A| = 1.Then |M ∩ B| 1.The case M ∩ B = ∅ is trivial.Thus we may assume M = {a 1 b 1 }.Then w B,f ,g (a 2 b 1 ) = w B,f,g (a 2 b 1 ) − 1, w B,f ,g (a 1 b 2 ) = w B,f,g (a 1 b 2 ) + 1, w B,f ,g (a 1 b 3 ) = w B,f,g (a 1 b 3 ) + 1,and for every other edge e, w B,f ,g (e) = w B,f,g (e).If a special edge set D does not contain a 2 b 1 , then w B,f ,g (D) w B,f,g (D) 0. If D contains a 2 b 1 , then D contains at least one of a 1 b 2 and a 1 b 3 .In this case, we also have w B,f ,g (D) w B,f,g (D) 0. Case 4 (R4) applies.As (R3) does not apply, |M ∩ B| 2. If |M ∩ B| = 3, then w A,f ,g (e) w A,f,g (e) and w B,f ,g (e) w B,f,g (e) for every edge e.Thus we may assume M = {a 1 , b 1 , b 2 }.If D is a special edge set not containing a 1 b 3 , then w A,f ,g (e) w A,f,g (e) for each edge e ∈ D, and hence w A,f ,g (D) w A,f,g (D) 0. If a 1 b 3 ∈ D, then D contains at least one of a 2 b 1 , a 2 b 2 .As in Case 3, w B,f ,g (D) w B,f,g (D) 0. Corollary 33.K 2,3 is (2m, m)-paintable for all positive integer m.