An update on reconfiguring $10$-colorings of planar graphs

The reconfiguration graph $R_k(G)$ for the $k$-colorings of a graph $G$ has as vertex set the set of all possible proper $k$-colorings of $G$ and two colorings are adjacent if they differ in the color of exactly one vertex. A result of Bousquet and Perarnau (2016) regarding graphs of bounded degeneracy implies that if $G$ is a planar graph with $n$ vertices, then $R_{12}(G)$ has diameter at most $6n$. We improve on the number of colors, showing that $R_{10}(G)$ has diameter at most $8n$ for every planar graph $G$ with $n$ vertices.


Introduction and result
Let G be a graph, and let k be a non-negative integer. A (proper) k-coloring of G is a function ϕ : V (G) → {1, . . . , k} such that ϕ(u) = ϕ(v) whenever uv ∈ E(G). The reconfiguration graph R k (G) of the k-colorings of G has as vertex set the set of all k-colorings of G, with two colorings adjacent if they differ in the color of exactly one vertex. That is, two k-colorings ϕ 1 and ϕ 2 are joined by a path in R k (G) if and only if we can transform ϕ 1 into ϕ 2 by recoloring vertices one by one, always keeping the coloring proper, and the number of recolorings needed is equal to the distance between ϕ 1 and ϕ 2 in graph with n vertices, then R 8 (G) has diameter O(n(log n) 7 ) and R 12 (G) has diameter at most 6n. This motivates the following question.
Problem 3. What is the minimum integer κ such that for every planar graph G with n vertices, R κ (G) has diameter O(n)?
The object of this paper is to show κ ≤ 10, improving on the bound 12 following from Theorem 2.
Theorem 4. Let G be a planar graph on n vertices. Then R 10 (G) has diameter at most 8n.
Consider the coloring of the icosahedron graph D where the opposite vertices get the same color. This gives a 6-coloring of D where the closed neighborhood of each vertex contains all 6 colors, and hence this 6-coloring forms an isolated vertex in R 6 (D). Consequently, R 6 (G) does not even need to be connected for planar graphs, implying κ ≥ 7. However, not much is known about R 7 (G) for planar graphs G. The 5-degenerate graphs for which R 7 (G) has quadradic diameter constructed in [3] (paths with four apex vertices) are non-planar. A natural candidate for a planar graph G with R 7 (G) of quadratic diameter is as follows: Consider the drawing of K 7 on the torus. Cut this drawing along a non-contractible triangle and glue together many copies of the resulting cylinder. We obtain a planar graph with a 7-coloring such that the closed neighborhood of all but six vertices contains all 7 colors, so to recolor this graph, one has to "propagate" from the ends of the cylinder. However, this graph G is 3-degenerate and chordal, and thus R 7 (G) in fact has linear diameter by the aforementioned result of Bartier and Bousquet [4]. Hence, we cannot exclude the possibility that the answer to Problem 3 is κ = 7.

Outline of the proof
In this section, we lay out our strategy for proving Theorem 4. Let us start off by noting that Theorem 4 will follow as an immediate consequence to the following theorem.
Theorem 5. Let G be a planar graph. Let α be a 10-coloring of G. Then there exists a sequence of recolorings from α to some 9-coloring of G that recolors every vertex either at most once, to a color distinct from 10, or exactly twice, first to the color 10 and then to a color distinct from 10.

Theorem 4 follows by a standard argument.
Proof of Theorem 4. Let α and β be 10-colorings of G. To prove the theorem, it suffices to show that we can recolor α to β by at most 8n recolorings.
By Theorem 5, we can recolor α to some 9-coloring α 1 of G by at most 2n recolorings and β to some 9-coloring β 1 by at most 2n recolorings. By [16], there exists a partition of V (G) into an independent set I and a 3-degenerate graph D. From α 1 and β 1 recolor the vertices in I to color 10 (the color that is not used in α 1 and β 1 ). Let α 2 and β 2 denote the restrictions of α 1 and β 1 to D. Applying Theorem 2, the distance between α 2 and β 2 in R 9 (D) is at most 4|V (D)|, and thus we can recolor α 2 to β 2 by at most 4|V (D)| recolorings without using the color 10. This completes the proof.
The rest of this paper will be devoted to the proof of Theorem 5. In order to prove the theorem, we must first make a few definitions. A scene is a pair (G, α), where G is a plane graph and α is a 10-coloring of G. We say that a sequence of recolorings from α to some coloring γ of G is valid if γ uses only colors 1, . . . , 9 and every vertex v of G is recolored either at most once (to the color γ(v)) or exactly twice, first to the color 10 and then to the color γ(v). We say that the scene (G, α) is recolorable if G admits a valid sequence of recolorings starting from α.
The scene (G, α) is said to be a minimal counterexample if (G, α) is not recolorable and all scenes (G , β) such that Our aim will be to exclude the existence of a minimal counterexample, which will prove Theorem 5. We begin with an easy proposition. Lemma 6. If (G, α) is a minimal counterexample, then G is a triangulation and the color 10 appears in the closed neighbourhood of every vertex of G under α.
Proof. Suppose that G is not a triangulation; then for some face f of G, there exist distinct non-adjacent vertices u and v incident with f . If α(u) = α(v), we insert the edge uv. If α(u) = α(v) we identify u and v into a new vertex u . The resulting graph G is planar and, by minimality, (G , α) is recolorable (we consider α to be a coloring of G by defining α(u ) = α(u) = α(v)). As any valid sequence of recolorings in G easily translates into a valid sequence of recolorings in G, this shows that G must be a triangulation.
Suppose that the color 10 does not appear on some vertex v of G or any of its neighbors. Recolor v to the color 10 and let α denote the resulting coloring. By minimality, (G, α ) is recolorable. It follows, by definition, that (G, α) is recolorable.
We now analyze the structure of a minimal counterexample (G, α) by showing that G cannot contain a number of induced subgraphs whose vertices are of prescribed degrees (here and in Section 3). Afterwards, we will show that no such minimal counterexample exists using the discharging method (see Section 4).
Let H be an induced subgraph of G. By the minimality of (G, α), there exists a valid sequence of recolorings in G − V (H) from the restriction of α to G − V (H) to some coloring γ of G − V (H). Let us define a list assignment L H for H by setting for each v ∈ V (H). We say that L H is an assignment of available colors to H in (G, α); let us remark that there may be several different assignments of available colors, corresponding to different colorings of G − V (H).
We have the following proposition. A sequence of recolorings of H is said be a once-only recoloring if every vertex of H is recolored at most once. The induced subgraph H of G is said to be reducible in (G, α) if for every assignment of available colors L H to H, there exists a once-only recoloring of H from the restriction of α to some L H -coloring of H. Lemma 7. In a minimal counterexample (G, α), no induced subgraph of G is reducible.
Proof. Let H be an induced subgraph of G. By minimality, G − V (H) has a valid sequence of recolorings σ to some coloring γ. Let L H be the corresponding assignment of available colors to H. Suppose for a contradiction H is reducible. Then there exists a once-only recoloring σ of H from the restriction of α to some L H -coloring γ H of H. But σ followed by σ is a valid sequence of recolorings in G. Indeed, recoloring of a vertex v ∈ V (H) according to σ does not conflict with the colors of its neighbors in G−V (H), since if u ∈ V (G) \ V (H) and uv ∈ E(G), then α(u) ∈ L H (v). Afterwards, recolorings of u ∈ V (G) \ V (H) do not conflict with the color of its neighbors v ∈ V (H), since u can only be recolored to 10 or γ(u) and neither of these colors belongs to L H (v). This is a contradiction.
It is often convenient to focus just on the sizes of the lists. For a function s : X → N with V (H) ⊆ X, we say that a list assignment L for H is an s-list  In the next section, we show a number of motifs that are oo-recolorable, and thus they cannot be induced in a minimal counterexample. Before we do that, let us point out the aspects of our argument that we consider to be novel: Our original plan was to restrict ourselves to once-only recolorings; this enables us to apply the method of reducible configurations which has not been previously used in the area, since we only need to forbid two colors (the initial and the final color) per neighbor outside of the configuration. A bit of a breakthrough for us then was the seemingly counterintuitive notion of valid sequences of recolorings, where we introduce new vertices of color 10 in order to eventually eliminate the color 10. This enables us to assume that color 10 appears in the closed neighborhood of every vertex, which is extremely useful in proving the reducibility of configurations.

Structure of minimal counterexample
In this section, we show in a series of lemmas that if (G, α) is a minimal counterexample, then G has minimum degree at least five and does not contain any of the graphs in Figure 1 as induced subgraphs with the prescribed degrees of vertices. Let us start with a trivial observation. Corollary 11. If (G, α) is a minimal counterexample, then G has minimum degree at least five.
Proof. Consider a vertex v ∈ V (G). By Lemma 9, there exists a motif M induced by v that is not oo-colorable, and thus |L M (v)| = 0 by Observation 10.
In order to facilitate the proofs that the graphs in Figure 1 Hence, we can first recolor v to c and then perform the recolorings according to σ, showing that M is oo-recolorable.
Similarly, we obtain the following observation.
Proof. By assumptions, M − v is oo-recolorable to some coloring γ, via a sequence σ of recolorings. We can first perform the recolorings σ in M , as they do not conflict with the color 10 of v. Finally, we can recolor v to a color in Proof. By assumptions, M − (v → c) is oo-recolorable via a sequence σ of recolorings. We can first recolor v to c (since no neighbor of v has color c) and then perform the recolorings σ in C. For a neighbor u of v, the recoloring of u according to σ does not conflict with the color c, since c ∈ L M −(v→c) (u). This shows M is oo-recolorable. Lemma 14 has the following useful consequence. For a motif M and In particular, repeatedly applying Lemma 15 until a motif with single vertex is obtained and using Observation 10, we have the following consequence.
Proof. By assumptions, M (v → c) is oo-recolorable via a sequence σ of recolorings. This sequence of recolorings can also be performed in M , since no neighbor of v can be assigned the color α M (v). Finally, we can recolor v to c, since no neighbor may end up with the color c. This shows M is oo-recolorable.
We will generally repeatedly use the preceding claims to simplify the motif obtained by Lemma 9, often to one contradicting Corollary 16. For brevity, let us introduce a notation for this kind of arguments. Suppose vertices of . . , s m ) ∼ M , we mean the following: The motif M is described by the vector (s 1 , . . . , s m ), and applying Lemma n with v = v i , we obtain a motif M described by (s 1 , . . . , s m ), such that if M is not oo-colorable, then M also is not oocolorable. In case Lemma 15 or Lemma 17 with more than one color choice is applied, we also specify the color c over the arrow. In case the resulting motif M is not further discussed (e.g., a contradiction with Corollary 16 is obtained), the ∼ M part is omitted. We can also chain several such statements in the natural way. In all the arguments, we without loss of generality assume that |L(v i )| = s i , implicitly removing extra colors from the lists if needed.
Recall that by Lemma 6, the color 10 appears in the closed neighbourhood of every vertex of a minimal counterexample.
If neither u nor v has color 10, then since the color 10 appears in the closed neighbourhood of every vertex, we have However, this contradicts Corollary 16. We also need the following three easy observations.
We now make two observations about triangles in a minimal counterexample.
Lemma 22. Let (G, α) be a minimal counterexample. If G contains a triangle T with vertices v 1 , v 2 , and v 3 such that v 1 has degree five and v 2 and v 3 have degree at most six, then α −1 (10) ∩ V (T ) = ∅.
Proof. By Lemma 9, there exists a motif M induced by T in (G, α) that is not oo-recolorable. Suppose for a contradiction no vertex of T has color 10. Since the color 10 appears in the closed neighbourhood of every vertex, we have Lemma 23. Let M be a motif such that H M is a triangle with vertices v 1 , v 2 , and v 3 . If M is described by (4, 3, 1), then M is oo-recolorable, and if M is described by (3, 3, 1) or (3, 3, 2), then M is oo-recolorable un- Proof. Suppose first M is described by (3,3,1) or (3,3,2), and that M is not oo-recolorable. If there exists contradicting Observation 10. This gives the characterization of non-oorecolorable motifs described by (3,3,1) or (3,3,2). Suppose now M is described by (4, 3, 1); then we can delete a color from L M (v 1 ) to obtain a motif M described by (3,3,1), but with L M (v 1 ) = L M (v 2 ). The motif M is oo-recolorable by the previous paragraph, and thus M is oo-recolorable as well.
We also require the following observation on diamonds in a minimal counterexample.
Proof. By Lemma 9, there exists a motif M induced by F in (G, α) that is not oo-recolorable. Suppose for a contradiction no vertex of F has color 10. Since the color 10 appears in the closed neighbourhood of every vertex, M is described by (2,4,4,2). If there exists a color We first recolor v 2 to c 2 , then v 3 to α(v 2 ), then v 1 to α(v 3 ), and finally v 4 to α(v 1 ).
We are now ready to demonstrate that the graphs in Figure 1 are reducible. Figure 1: Reducible induced subgraphs, where denotes a vertex of degree at most seven, • denotes a vertex of degree five and · denotes a vertex of degree at most six.
Lemma 25. If (G, α) is a minimal counterexample, then G contains none of the induced subgraphs with prescribed vertex degrees depicted in Figure 1.
Proof. Suppose for a contradiction C is one of the graphs depicted in Figure 1 and contained in G as an induced subgraph with the prescribed degrees of vertices. By Lemma 9, there exist a motif M induced by C in (G, α) that is not oo-recolorable. We prove that each of the cases are reducible separately, starting with C 1 and working our way towards C 16 . We fix the labelling of vertices as indicated in Figure 1.

Discharging phase
Consider a plane triangulation G, a vertex v ∈ V (G) of degree k ≥ 3, and its neighbors v 1 , . . . , v k in the clockwise order around G. We say that the subgraph of G consisting of the cycle v 1 . . . v k , the vertex v, and the edges vv i for i = 1, . . . , k is a wheel, v is its center and v 1 , . . . , v k its rim. Note that a wheel is not necessarily an induced subgraph of G. Let T be the triangle bounding the outer face of G. Let C be a graph and d : V (C) → N a function assigning a prescribed degree to each vertex of C. We say that C with the prescribed degrees d appears in G if there exists a wheel W in G and an injective function f : , and Hence, C is an induced subgraph of W , but not necessarily of G (since W may not be an induced subgraph of G). Let us remark that the last technical condition from the definition of appearance will be later used to deal with this issue.
Lemma 26. Suppose G is a plane triangulation such that every vertex not incident with the outer face of G has degree at least five. If |V (G)| ≥ 4, then one of the graphs with prescribed degrees depicted in Figure 1 appears in G.
Proof. Suppose for a contradiction none of these graphs appears in G. We assign the initial charge ch 0 (v) = 10 · deg v − 60 to each vertex v of G. Since G is a triangulation, we have |E(G)| = 3|V (G)| − 6 by Euler's formula, and thus A vertex is big if it either has degree at least 7 or it is incident with the outer face of G, medium if it has degree six and is not incident with the outer face of G, and small if it has degree five and is not incident with the outer face of G. Next, we redistribute the charges according to the following rules. For accounting purposes, for a rule sending some amount of charge from a vertex v to another vertex u, we also specify faces incident with v through which the charge leaves v, and an edge e incident with u along which the charge arrives to u. Additionally, we specify a face incident with e through which the charge passes. Note that (R2) applies in addition to the two units of charge sent by v to u by (R1), but the charge arrives to u along a different edge. Furthermore, if x is small, the charge is also being sent from v to x by (R2) with the roles of u and x exchanged. Furthermore, note that (R3) may possibly send charge from v m to v 1 twice around the same vertex x, once in the clockwise direction, once in the counterclockwise one (when x is the center of a wheel whose rim contains v 1 and v m and every other vertex of the rim is medium). We now analyze the final charge ch(v) of each vertex v of G after the redistribution of the charge. Clearly, for a medium vertex v, we have ch(v) = ch 0 (v) = 0.
Consider now a small vertex z. We claim that for each edge e = wz incident with z and each face f = wzx incident with e, a unit of charge passes through f to arrive to z along e, and thus ch(z) = ch 0 (z) + 10 × 1 = 0. Indeed, if w is big, then this is the case by (R1). If w is not big and x is big, then a unit of charge passing through f arrives to z along e from x by (R2). If neither w nor x is big, then since C 2 does not appear in G, both of them are medium. Since C 4 does not appear in G, x has a neighbor distinct from z that is not medium. Let v 1 = z, v 2 = w, v 3 , . . . , v m be the neighbors of x in order, where v 3 , . . . , v m−1 are medium and v m is not medium. Since C 3 , C 6 , C 8 and C 2 do not appear in G, the vertex v m is not small, and thus v m is big. Consequently, a unit of charge passing through f arrives to z along e from v m by (R3).
Suppose now v is a vertex of degree d ≥ 7 not incident with the outer face of G. For a face f = vxy, let t(f ) denote the total amount of charge that leaves v through f . If both x and y are small, then t(f ) = 4 since two units leave through f by (R1), one along the edge vx and the other along vy, and two by (R2), both along the edge xy. If x is small and y is medium or vice versa, then t(f ) = 2 since one unit leaves through f by (R1) and one by (R2). If both x and y are medium, then t(f ) ≤ 2, since at most two units leave through f by (R3). If x is small and y is big or vice versa, then t(f ) = 1, since only one unit leaves through f by (R1). Otherwise, t(f ) = 0.
Furthermore, consider the faces f 1 and f 2 following f in the clockwise order around f . Since C 1 does not appear in G, if t(f ) = 4, then t(f 1 ) ≤ 2 and t(f 2 ) ≤ 2. Consequently, there are at most d/3 faces f incident with v such that t(f ) = 4. If d ≥ 8, this implies Hence, we can assume d = 7, and thus ch 0 (v) = 10. Let v 1 , . . . , v 7 be the neighbors of v in the clockwise order, and for i = 1, . . . , 7, let f i be the face i=1 t(f i ) be the total amount of charge sent by v. We argue that s ≤ 10, and thus ch(v) = ch 0 (v) − s ≥ 0. To do so, we discuss several cases.
• v is adjacent to two consecutive small vertices in the cycle on neighbors of v. Thus v is incident with a face f such that t(f ) = 4. By symmetry, we can assume t(f 1 ) = 4, and thus v 1 and v 2 are small. Since C 1 does not appear in G, v 3 and v 7 are not small.
Suppose v 6 and v 7 are both medium. Since C 13 does not appear in G, v 5 is big, and thus t(f 7 ) + t(f 6 ) + t(f 5 ) ≤ 2 + 2 + 0 = 4. Since C 14 and C 10 do not appear in G, v 4 is not small and v 3 and v 4 are not both medium, respectively, implying t(f 3 ) = 0 and t(f 4 ) = 0. Consequently, s ≤ 4 + 2 + 0 + 0 + 4 = 10. Hence, assume v 6 and v 7 are not both medium, and symmetrically, that v 3 and v 4 are not both medium.
• small vertices are not consecutive in the cycle on neighbors of v. Consequently, t(f ) ≤ 2 for each face incident with v and v is adjacent to at most three small vertices.
Before we proceed, let us make a useful observation: This is clearly the case unless v b , v b+1 , and v b+2 are all medium and leaves v through f b and f b+1 twice by (R3), and thus either both z 1 and z 3 are small, or none of z 1 , z 2 , and z 3 is big and at least one of them is small. But then either C 5 or C 4 appears in G, which is a contradiction.
Let us now continue with the case analysis.
v is adjacent to three small vertices. By symmetry we can assume v 1 , v 3 , and v 5 are small. Since C 12 does not appear in G, we can by symmetry assume v 2 is big hence t(f 1 ) = t(f 2 ) = 1. If v 4 is big, then t(f 3 ) = t(f 4 ) = 1 implying s ≤ 4 × 1 + 3 × 2 = 10. Thus, since C 1 does not appear in G, we can assume v 4 is medium. Since C 9 does not appear in G, v 6 and v 7 cannot both be medium, and thus t(f 6 ) = 0. Consequently, s ≤ 1 + 1 + 2 + 2 + 2 + 0 + 2 = 10.
v is adjacent to two small vertices, at distance two in the cycle on neighbors of v. By symmetry we can assume v 1 and v 3 are small.
v is adjacent to two small vertices, at distance three in the cycle on neighbors of v. By symmetry we can assume v 1 and v 4 are small.
v is adjacent to at most one small vertex. By symmetry we can assume no neighbor of v other than v 1 is small. If v i is big for some i ∈ {1, 3, 4, 5, 6}, then t(f i−1 ) = t(f i ) = 0 (where f 0 = f 7 ) and s ≤ 5 × 2 = 10. Hence, we can assume v i is medium for i ∈ {3, 4, 5, 6} and v 1 is medium or small. Since C 15 does not appear in G, v 2 and v 7 are not both medium; by symmetry, we can assume v 2 is big, and thus t(f 1 ) + t(f 2 ) ≤ 1. By ( ), we have t(f 3 ) + t(f 4 ) ≤ 3, and thus s ≤ 1 + 3 + 3 × 2 = 10.
We conclude that every vertex not incident with the outer face of G has non-negative final charge. Finally, let us consider a vertex v incident with the outer face of G. Since |V (G)| ≥ 4 and G is a triangulation, we have deg(v) ≥ 3. Furthermore, the outer face f of G is incident only with big vertices by definition, and thus t(f ) = 0. In the utmost case, t(f ) ≤ 4 for every face f = f incident with v and hence ch(v) ≥ ch 0 (v) − (deg v − 1) × 4 = 6 deg v − 56 ≥ −38. Therefore, (1) together with the fact that no charge is created or lost in the redistribution process gives Corollary 27. If G is a plane triangulation of minimum degree at least five, then one of the graphs depicted in Figure 1 is an induced subgraph of G with prescribed vertex degrees.
Proof. If G contains a separating triangle, then let T be a separating triangle in G such that the open disk in the plane bounded by T is minimal; otherwise, let T be the triangle bounding the outer face of G. Let G be the induced subgraph of G drawn in the closed disk bounded by T . By Lemma 26, one of the graphs C with prescribed degrees depicted in Figure 1 appears in G, via a map f : V (C) → V (W ) for a wheel W in G . By the choice of G , observe that G does not contain any separating triangle, and thus W is an induced subgraph of G , and thus also of G. Since C is an induced subgraph of W , it follows that C is an induced subgraph of G. Furthermore, V (C) ∩ V (T ) = ∅ by the last condition from the definition of appearance, and thus the vertices of f (V (C)) have the same degree in G and in G.
The proof of the main result is now straightforward.
Proof of Theorem 5. Suppose for a contradiction that there exists a nonrecolorable scene (G, α). Choose such a scene with the smallest number of vertices, among those with the largest number of edges, and among those with the largest number of vertices of color 10. Then (G, α) is a minimal counterexample, and thus G is a triangulation by Lemma 6, has minimum degree at least five by Corollary 11, and does not contain any of the induced subgraphs with prescribed vertex degrees depicted in Figure 1. However, this contradicts Corollary 27.