Plane Graphs are Facially-non-repetitively 10^{4 \cdot10^7}-Choosable
aa r X i v : . [ m a t h . C O ] J un PLANE GRAPHS ARE FACIALLY-NON-REPETITIVELY · -CHOOSABLE GRZEGORZ GUTOWSKI
Abstract.
A sequence ( x , x , . . . , x n ) of even length is a repetition if ( x , . . . , x n ) = ( x n +1 , . . . , x n ) . We prove existence of a constant C < · such that given any planar drawing of a graph G , and a list L ( v ) of C permissible colors for each vertex v in G , there is a choice of a permissiblecolor for each vertex such that the sequence of colors of the vertices on anyfacial simple path in G is not a repetition. Introduction
For two real-valued functions f and g whose domains are cofinite subsetsof N , we write f ( k ) = O ( g ( k )) if there exist constants n and c such that | f ( k ) | c | g ( k ) | for all k > n . All graphs considered in this paper are finite,undirected and contain no loops nor multiple edges. Given a graph G , a planardrawing of G is a geometric representation of G in the plane such that: • each vertex v is drawn as a distinct point p v , • each edge { u, v } is drawn as a simple curve connecting p u and p v , • no two edges intersect except at their common endpoints.A graph is planar if it admits a planar drawing. A planar drawing of G is a straight-line drawing of G if each edge is drawn as a segment. Let G be a connected planar graph. A planar drawing of G divides the plane intotopologically connected regions, called faces . Exactly one face is an infiniteregion, and is called the external face . Each face F is described by the cyclicorder of vertices of F as they are visited when the boundary of F is traversedin the clockwise direction. The description of all the faces determined by aplanar drawing of G and choice of the external face is a planar embedding of G . Two planar drawings of G are equivalent if they both determine the sameplanar embedding. A plane graph is a non-empty connected planar graph witha fixed planar drawing.For a sequence X = ( x , . . . , x n ) , and any i j n , a sequence X [ i, j ] = ( x i , . . . , x j ) is a block of X . A sequence X = ( x , x , . . . , x n ) of evenlength is a repetition if X [1 , n ] = X [ n + 1 , n ] . A sequence X is non-repetitive if no block of X is a repetition. The study of non-repetitive sequences was Partially supported by Polish National Science Center grant 2016/21/B/ST6/02165. initiated by Thue [13] who proved that there are arbitrarily long non-repetitivesequences with only three different elements.A proper coloring of a graph G is a coloring of the vertices of G such that notwo endpoints of an edge of G are colored the same. A non-repetitive coloring of G is a coloring of the vertices of G such that the sequence of colors of verticeson any simple path in G is not a repetition. In particular, any non-repetitivecoloring of G is a proper coloring of G . The study of non-repetitive coloringswas initiated by Alon et al. [1]. They conjectured that every planar graphis non-repetitively O (1) -colorable. Currently, the best result supporting thisconjecture is by Dujmović et al. [5] who proved that every planar graph on n vertices is non-repetitively O (log n ) -colorable.A facial walk in a plane graph G is a walk that traverses consecutive verticesof the boundary of some face of G , and that traverses each edge at most once.A facial path in G is a facial walk which is a simple path. A facially-non-repetitive coloring of a plane graph G is a coloring of the vertices of G suchthat the sequence of colors of vertices on any facial path in G is not a repetition.Barát and Czap [4] proved that every plane graph is facially-non-repetitively -colorable.A list assignment of a graph G is a mapping L which assigns to each vertex v of G a set L ( v ) of permissible colors. For two list assignments L , and M ofthe graph G , we write M ⊆ L to denote that M ( v ) ⊆ L ( v ) , for each vertex v in G . An L -coloring of G is a coloring c of vertices of G such that c ( v ) ∈ L ( v ) for every vertex v in G . We say that a list assignment M of G is proper (ornon-repetitive, or facially-non-repetitive) if any M -coloring of G is a proper,(or non-repetitive, or facially-non-repetitive) coloring of G .A list assignment L of G is a k -list assignment of G if | L ( v ) | > k , for eachvertex v in G . A graph G is properly (or non-repetitively, or facially-non-repetitively) k -choosable if for any k -list assignment L of G there is a proper(or non-repetitive, or facially-non-repetitive) L -coloring of G . Fiorenzi et al. [6]showed that for any constant C there is a tree which is not non-repetitively C -choosable. Przybyło et al. [10] proved that every plane graph of maximumdegree ∆ is facially-non-repetitively O (∆) -choosable. In this paper we improvethis result and prove that every plane graph is facially-non-repetitively O (1) -choosable.We say that a graph G is properly (or non-repetitively, or facially-non-repetitively) ( k : m ) -choosable if for any k -list assignment L of G there is aproper, (or non-repetitive, or facially-non-repetitive) m -list assignment M ⊆ L of G .Main result proved in this paper is the following. Theorem 1.
Every plane graph is facially-non-repetitively ( O ( m ) : m ) -choosable. LANE GRAPHS ARE FACIALLY-NON-REPETITIVELY · -CHOOSABLE 3 The proof gives an explicit polynomial f ( m ) of degree such that everyplane graph is facially-non-repetitively ( f ( m ) : m ) -choosable. When we com-pute the value of this polynomial for m = 1 we get that every plane graph isfacially-non-repetitively · -choosable.The proof is based on the following earlier results. First ingredient is thefamous Four Color Theorem. Theorem 2 (Appel, Haken, Koch [2, 3]) . Every plane graph is properly -colorable. Second tool is a beautiful technique by Thomassen [12] who showed thatevery plane graph is properly -choosable. We use the following stronger state-ment which can be obtained using the same proof. The modified proof ispresented in Appendix A. Theorem 3 (Thomassen [12]) . Every plane graph is properly (5 m : m ) -choosable. Grytczuk et al. [8] showed that every path is non-repetitively -choosable.In our proof, we use an even stronger result by Gągol et al. [7, Lemma 6]. Theproof of the following theorem uses the entropy compression method and worksfor graphs more general than simple paths. Theorem 4 (Gągol, Joret, Kozik, Micek [7]) . Every simple path is non-repetitively (32 m + 1 : m ) -choosable. The last ingredient is the study of bipolar orientations of planar graphs.This notion was first used by Lempel et al. [9] to develop planarity testingalgorithm. Let G be a plane graph with a planar straight-line drawing suchthat no two vertices have the same y -coordinate. Let ~G denote an orientationof G that arises from directing each edge upward, i.e. towards a vertex withbigger y -coordinate. A source in a directed graph is a vertex with no incomingedges. Similarly, a sink is a vertex with no outgoing edges. We say that thedrawing of G is bipolar if ~G is acyclic, has a single source s , and a single sink t . In order to easily distinguish vertices s and t , we call such a drawing to be ( s, t ) -bipolar . Theorem 5 (Lempel, Even, Cederbaum [9]) . For every 2-connected plane graph G , and two vertices s , and t on the externalface of G , there is an equivalent ( s, t ) -bipolar drawing of G . For a face F of an ( s, t ) -bipolar drawing of G , let s ( F ) – source of F , and t ( F ) – sink of F be the vertex with respectively the minimal, and the maximal y -coordinate among vertices of F . Observe, that the source of the external faceof G is s , and the sink of the external face of G is t . Tamassia and Tollis [11]showed that the boundary of any face F of a bipolar orientation consists oftwo directed paths from s ( F ) to t ( F ) . As a result, we get that any vertex v isa source or sink of all but at most two faces. See Figure 1. G. GUTOWSKI s ( F ) t ( F ) F vU U D D D L R
Figure 1.
Bipolar drawing. On the left, boundary of face F consists of two directed paths from s ( F ) to t ( F ) . On the right,vertex v is a source/sink of all but two faces, i.e. v = s ( U ) = s ( U ) = t ( D ) = t ( D ) = t ( D ) .2. Result
The main idea behind the coloring algorithm is the following. We say thata vertex v on a face F is either regular or special for F . We divide occurrencesof the vertices on the faces so that: • there are two special vertices for any face, • each vertex is regular for at most two faces.Lemma 6 gives such a division. The construction is based on bipolar orienta-tions of -connected plane graphs.Our coloring algorithm first filters list assignment so that any two vertices indistance at most two on any face have disjoint lists of colors. This is obtainedby applying Theorem 3 a few times and is described in Lemma 7. Then, we in-troduce a technique that chooses colors for a single face in a slightly augmentedsetting. Intuitively, each face F "controls" the colors of the regular verticesfor F , but have to "accept" any colors of the special vertices for F . This way,large faces "control" the colors of most of its vertices. On the other hand, listof colors for any vertex is "controlled" by at most two faces. Lemma 10 givesa method to filter lists of the regular vertices for a single face F so that nomatter how the special vertices for F are colored there is no repetition on anyfacial path of F . The proof covers the face with paths and uses Theorem 4.Last observation that completes the proof is that we can introduce an auxil-iary planar graph on the faces of G , color them with four colors, and applyLemma 10 simultaneously for all faces of the same color. LANE GRAPHS ARE FACIALLY-NON-REPETITIVELY · -CHOOSABLE 5 Lemma 6 (Drawing Lemma) . Let G be a plane graph, and s , t be two verticeson the external face of G . There is a way to name each occurrence of a vertexon a face either normal or special and satisfy the following conditions: • there are two special vertices for any face, • each vertex is regular for at most two faces, • vertices s and t are not regular for any face.Proof. By induction on the number of vertices of G . If G has two vertices s and t , then both vertices are special for the only face of G .If G is 2-connected, we apply Theorem 5 and get an ( s, t ) -bipolar drawing of G . For each face F , we choose the special vertices for F to be s ( F ) and t ( F ) .Properties of bipolar drawings, see Figure 1, guarantee that each vertex v isspecial for all but at most two faces. Vertices s and t are special for all thefaces.If G is not 2-connected, let v be a cut-point of G and G , . . . , G k be thecomponents of G r { v } . Without loss of generality, assume that s is in G ,and that t is either in G , or in G . Let G ′ denote the graph G r G k , and G ′′ denote the graph G r G , . . . , G k − . Let F be the only face of G such that theboundary of F has edges both in G ′ and G ′′ . Let F ′ be the face of G ′ suchthat G ′′ is drawn inside face F ′ . Let F ′′ be the external face of G ′′ . Observe,that vertex v is in F ∩ F ′ ∩ F ′′ and that the boundary of F is a boundary of F ′ with "inserted" boundary of F ′′ . See Figure 2.First, assume that t is in G ′ . Apply induction for G ′ , s , and t . Choose anyvertex w other than v in F ′′ . Apply induction for G ′′ , v , and w . For each faceof G different than F , special vertices are determined by induction. We setthe special vertices for F to be the special vertices for F ′ in G ′ . Observe thatvertex w is special for all faces of G ′′ , and thus w is normal only for one face F of G . Every vertex other than w that is normal for F is either normal for F ′ , or normal for F ′′ .Now, assume that t is in G ′′ . This means that k = 2 , t is in G and F is theexternal face of G . Apply induction for G ′ , s , and v . Apply induction for G ′′ , v , and t . For each face of G different than F , special vertices are determinedby two induction calls. We set the special vertices for F to be s and t . Vertex v is special for all faces of G ′ , and for all faces of G ′′ , and thus v is normal onlyfor one face F of G . Every vertex other than v that is normal for F is eithernormal for F ′ , or normal for F ′′ . (cid:3) Now, we prove several lemmas, that allow us to filter list assignments ofdifferent structures. Each of those lemmas tells us that some plane graphs are ( f ( m ) : m ) -choosable for a polynomial function f . For example, set f ( m ) = 5 m .Theorem 3 gives that every planar graph is properly ( f ( m ) : m ) -choosable. G. GUTOWSKI st s ( F ) t ( F ) F vw svt F
Figure 2.
Division of a graph G in Lemma 6 into G ′ (darkgray) and G ′′ (light gray). On the left, vertex t is in G ′ . On theright, vertex t is in G ′′ .For a plane graph G , a facial-square of G is a graph on the same vertexset in which two vertices u and v are connected by an edge when u and v areconnected in G by a facial path of length at most two. We say that a coloringof a plane graph G is a facially-square-proper if it is a proper coloring of thefacial-square of G .For the next lemma, set f ( m ) = f ( f ( f ( m ))) = 125 m . Lemma 7 (Facial-Square Filtering Lemma) . Facial-square of a plane graph isproperly ( f ( m ) : m ) -choosable.Proof. Let G be a plane graph, and let L be an f ( m ) -list assignment of G .We show that edges of facial-square of G can be decomposed into threesets, say red, green, and blue so that any two edges of the same color arenon-crossing. See Figure 3 for an example of the following decomposition.First, color red all edges of G . Then, for each face F , we color the edgesthat correspond to pairs of vertices in distance two on F . Let l denote thenumber of vertex occurrences on the boundary of F , and let v , v , . . . , v l − be the vertices of F in the clockwise order. Color green every edge { v i , v j } ,where j = (( i + 2) mod l ) and both i and j are odd numbers. Similarly, colorgreen every edge { v i , v j } , where j = (( i + 2) mod l ) and both i and j are evennumbers. If l is odd, color green the edge { v , v l − } , and color red the edge { v , v l − } . LANE GRAPHS ARE FACIALLY-NON-REPETITIVELY · -CHOOSABLE 7 We apply Theorem 3 to the red graph and obtain a f ( f ( m )) -list assignmentof the facial-square of G in which any two vertices connected by a red edgehave disjoint lists of permissible colors. We repeat the same two more times,for green, and for blue edges. In the end we obtain a facially-square-proper m -list assignment of G . (cid:3) Figure 3.
Coloring edges of a facial-square in Lemma 7.For the next lemma, set f ( m ) = 32 m + 1 , f ( m ) = f ( m ) + m = 32 m + m + 1 , f ( m ) = f ( f ( m )) + m + f ( m ) = O ( m ) .Theorem 4 gives that a path is non-repetitively ( f ( m ) : m ) -choosable. Lemma 8 (Path Filtering Lemma) . Let m > . Let P be a simple path ( v , . . . , v n ) , and v s be a selected vertex in P . Let L be a proper list assignmentof P such that: • | L ( v i ) | = f ( m ) , for i s , • | L ( v i ) | = f ( m ) , for i > s .There is a non-repetitive list assignment M ⊆ L of P such that: • | M ( v i ) | = m , for i < s , • | M ( v i ) | = f ( m ) , for i > s .Proof. Set M ( v s ) = L ( v s ) . If s > , then choose any m colors for M ( v s − ) from L ( v s − ) . Next, for each i < s − , set L ′ ( v i ) = L ( v i ) r M ( v s − ) . L ′ is a proper f ( m ) -list assignment of ( v , . . . , v s − ) . Apply Theorem 4 to L ′ .The resulting m -list assignment is non-repetitive and uses the set of colors G. GUTOWSKI disjoint with M ( v s − ) . Thus, we get a non-repetitive list assignment M of ( v , . . . , v s − ) .Then, for each i > s , set L ′′ ( v i ) = L ( v i ) r ( M ( v s ) ∪ M ( v s − )) , or L ′′ ( v i ) = L ( v i ) r M ( v s ) if s = 1 . L ′′ is a proper f ( f ( m )) -list assignment of ( v s +1 , . . . , v n ) .Apply Theorem 4 to L ′′ . The resulting f ( m ) -list assignment is non-repetitiveand uses the set of colors disjoint with M ( v s ) . Thus, we get a non-repetitivelist assignment M of ( v s , . . . , v n ) .Suppose, that there is a repetition in M . Such a repetition must includevertex v s − . Observe that M ( v s − ) and M ( v s ) are disjoint as L is a proper listassignment. The set of colors M ( v s − ) was removed from the list of permissiblecolors for every other vertex. Thus, the color of v s − is not repeated. Acontradiction. (cid:3) For a walk W in a graph, a simple W -block is a block of W that is a simplepath. For the next lemma, set f ( m ) = f ( f ( m )) = O (cid:0) m (cid:1) . Lemma 9 (Walk Filtering Lemma) . Let W be a facial walk in the graph, andlet L be a proper f ( m ) -list assignment of W . There is an m -list assignment M ⊆ L of W such that for any simple W -block P , M is a non-repetitive listassignment of P .Proof. Observe, that the repeated occurrences of vertices in W have the fol-lowing laminar structure. Let w i = w j = v , and let u be any vertex in W otherthan v . Then, either all occurrences of u are in W [ i +1 , j − , or all occurrencesof u are in W [1 , i − ∪ W [ j + 1 , n ] . Assume to the contrary w a = w b = u and i < a < j < b . Each of the three walks W [ i, a ] , W [ a, j ] , W [ j, b ] connects u and v . These three walks divide the plane into at least three regions, and thus W is not a facial walk.We say that a path P is a maximal simple W -block if P is a simple W -block and P cannot be extended into either direction in W and remain asimple block. Let P denote the set of all maximal simple W -blocks. Weshow that P can be decomposed into two sets, say red and green so thatany two paths of the same color are non-overlapping W -blocks. Let P = { P = W [ l , r ] , . . . , P k = W [ l k , r k ] } and without loss of generality l < l <. . . < l k . It follows from the maximality of each path that r < r < . . . < r k .See Figure 4.For a path P i , we know that P i cannot be extended to the right, so either i = k and r i = n , or the vertex w r i +1 is already on path P i . Assume the secondcase and let j be such that w j = w r i +1 and l i j < r i . The fact that P i is asimple path, and the laminar structure of repeated occurrences guarantee thateach vertex w j +1 , . . . , w r i has no other occurrences in W . Thus, any simple W -block that includes w j does not include w r i +1 and any simple W -block thatincludes w r i +1 can be extended to the left so that it includes w j +1 . Hence, LANE GRAPHS ARE FACIALLY-NON-REPETITIVELY · -CHOOSABLE 9 l i +1 = j + 1 . Further, if i + 2 k , we can repeat the same reasoning for P i +1 and get that l i +2 is to the right of j + 1 and to the right of some vertex withrepeated occurrence. As vertices w j +1 , . . . , w r i have no other occurrences in W ,we get that l i +2 > r i . Thus, paths P i and P i +2 are non-overlapping W -blocks.We color red each path P i for i odd, and color green each path P i for i even.Let t = f ( m ) . Now, we filter list assignment L so that we get a t -listassignment in which there is no repetition on any subpath of a red path. Weprocess paths P , P , . . . , one by one in this order. We will keep the followinginvariants that are satisfied after processing each path P i : • each vertex v that has no occurrences on paths P , . . . , P i has at least f ( m ) = f ( t ) permissible colors; • each vertex v that has at least one occurrence on paths P , . . . , P i and atleast one occurrence on paths P i +1 , . . . , P k has at least f ( t ) permissiblecolors; • each vertex v that has no occurrences on paths P i +1 , . . . , P k has at least t permissible colors.When processing path P i = ( w l i , . . . , w r i ) , we divide vertices in P i into fourdisjoint sets: • Only – a vertex v is in Only if there are no occurrences of v on pathsother than P i . • F irst – a vertex v is in F irst if there are no occurrences of v on paths P , . . . , P i − , and at least one on paths P i +1 , . . . , P k . • M iddle – a vertex v is in M iddle if there is at least one occurrence of v on paths P , . . . , P i − , and at least one on paths P i +1 , . . . , P k . • Last – a vertex v is in Last if there is at least one occurrence of v onpaths P . . . , P i − , and no occurrences on paths P i +1 , . . . , P k .Our invariants guarantee that each vertex in Only ∪ F irst has at least f ( t ) permissible colors, and that each vertex in M iddle ∪ Last has at least f ( t ) permissible colors. The fact that P i is a simple path, and the laminar structureof repeated occurrences guarantee that: • There is at most one vertex in
M iddle . • Every vertex in
Last is before any vertex in
M iddle ∪ F irst . • Every vertex in
F irst is after any vertex in
M iddle ∪ Last .Set vertex s to be the last vertex in M iddle ∪ Last ∪ { w l i } . Our invariantsguarantee that: • each vertex to the left of s has at least f ( t ) colors. • s has at least f ( t ) colors. • each vertex to the right of s has at least f ( t ) colors.We apply Lemma 8 to path P i with special vertex s and get new list assignment.We get that each vertex in Last keeps at least t colors. Vertex in M iddle , if itexists, keeps at least f ( t ) colors. Any vertex in F irst keeps at least f ( t ) colors. Thus, our invariants are satisfied after P i is processed. When we process allred paths, we get an f ( m ) -list assignment of W in which there is no repetitionon any simple path, a subpath of a red maximal simple W -block. To finishthe proof, we repeat the same process for t = m and green maximal simple W -blocks. (cid:3) W : a b c d e c f g c h b i j k l j m n j b o a p q r s t q u aP P P P P P Figure 4.
A facial walk W . Above the walk, the laminarstructure of repeated occurences of vertices. Below the walk,maximal simple W -blocks colored red and green.For the next lemma, set f ( m ) = f ( m ) + 10 m = O (cid:0) m (cid:1) . Lemma 10 (Face Filtering Lemma) . Let L be a facially-square-proper f ( m ) -list assignment of a face F , and let s and t be the special vertices for F . Thereis a non-repetitive list assignment M ⊆ L of F such that: • M ( s ) = L ( s ) , and M ( t ) = L ( t ) , • | M ( v ) | > m , for each v in F .Proof. Let A denote the set of all regular neighbors of s and all regular neigh-bors of t in F . For any vertex v in F let A ( v ) denote the five vertices in A encountered first when traversing F in clockwise, and in counter-clockwisedirection from v . Each set A ( v ) has at most ten elements. See Figure 5.First we define M for vertices in A . For each v ∈ A we choose any m colorsfrom L ( v ) so that M ( v ) is disjoint with M ( w ) for any w ∈ A ( v ) . As for each v ∈ A , the size of A ( v ) is at most ten, and | L ( v ) | > m , this can be easilydone. Now, for each regular vertex v not in A , remove from L ( v ) the colors M ( w ) for each w ∈ A ( v ) . This removes at most m colors from each list L ( v ) . The resulting list assignment L ′ has f ( m ) colors for each regular vertexnot in A . Further, set L ′ ( v ) = M ( v ) for v in A ∪ { s, t } .Let P be a facial path in F such that there exists an L ′ -coloring c of P suchthat the sequence of colors of vertices in P is a repetition. Observe that, as L is a facially-square-proper list assignment of F , we have that P has at least six LANE GRAPHS ARE FACIALLY-NON-REPETITIVELY · -CHOOSABLE 11 vertices. Let A ( P ) denote the vertices in P ∩ A , and let a ( P ) = | A ( P ) | . First,observe that P is a simple path and that between any three vertices in A ( P ) there is at least one special vertex. Thus we have that a ( P ) . Furthermore,if a ( P ) = 6 then both endpoints of P are in A . Thus, we have A ( P ) ⊆ A ( v ) for any regular vertex v in P . Hence, only a special vertex can match the colorof a vertex in A ( P ) and we have that a ( P ) , as there are only two specialvertices.If a ( P ) = 2 , then P contains exactly two vertices a , a in A , and exactlytwo special vertices. If both special vertices are in the first half of P , then, asthere are at least three vertices in the first half, there is at least one vertex a in A in the first half of P and colors of a cannot be matched in the secondhalf. Thus, in each half of P there is exactly one special vertex, and exactlyone vertex in A. Without loss of generality, assume that vertices s and a arein the first half of P . If s is neither the first, nor the last vertex in the firsthalf of P , then there are two vertices in A in that half of P – one to the left,and one to the right of s . Similarly, t is either the first, or the last vertex inthe second half of P . On the other hand, we get that a , which is next to s in the first half, is neither the first, nor the last vertex in the first half of P .Thus, color of a cannot be matched by color of t . A contradiction.If a ( P ) = 1 , then let A = { a } and assume that a is in the first half of P .Color of a is matched by the color of a special vertex w in the second halfof P . As the last vertex in P before w is in A , we have that a and w areneighbors in G . As L is a proper list assignment of F , we have that L ( a ) and L ( w ) are disjoint. A contradiction.Thus, we have that any simple path with a repetition of colors in L ′ does notinclude neither a vertex in A , nor a special vertex. We will further filter listassignment L ′ to remove all such repetitions. Let W , . . . , W k be the connectedcomponents of the boundary of F with vertices in A ∪ { s, t } removed. Observethat walks W , . . . , W k are pairwise vertex disjoint. List assignment L ′ is aproper f ( m ) -list assignment of each walk W , . . . , W k . We apply Lemma 9to L ′ and each walk W , . . . , W k independently. We obtain a facially-non-repetitive list assignment M of F . (cid:3) Now we combine Lemmas 6, 7, and 10 to prove the main theorem of thispaper. For the proof of the main result, set f ( m ) = f ( f ( f ( f ( f ( m ))))) = O (cid:0) m (cid:1) . Theorem 1.
Every plane graph is facially-non-repetitively ( O ( m ) : m ) -choosable.Proof. Let G be a plane graph and L be an f ( m ) -list assignment of G . First,we use Lemma 6 to divide occurrences of the vertices on the faces into regu-lar and special. Next, we apply Lemma 7 to obtain a facially-square-proper f ( f ( f ( f ( m )))) -list assignment of G . Then, we construct an auxiliary graph Fs ( F ) t ( F ) Figure 5.
Filtering of list assignment of face F in Lemma 10.Special vertices marked with red circles. Vertices in A markedwith blue squares. Walks W , . . . , W k colored green. H on faces of G in which we put an edge between two faces F , F of G whenthere is a vertex v in G that is regular both for F and for F . Observe thatgraph H is planar. Indeed, we can construct a planar drawing of H from pla-nar drawing of G by placing a vertex corresponding to face F anywhere in theface F and routing an edge { F , F } through the vertex that is regular bothfor F and F . Theorem 2 gives a proper coloring of H with colors red, green,blue and yellow. Observe that, as no two red faces share a common regularvertex, we can apply Lemma 10 to L ′ and all red faces of G simultaneously. Weobtain an f ( f ( f ( m ))) -list assignment of G in which there is no repetition onany facial path of a red face. We repeat the same three more times, for green,blue, and yellow faces. In the end we obtain a facially-non-repetitive m -listassignment M of G . (cid:3) Discussion
In order to get the final result, we need to compute value f (1) . A computercalculation gave us the value with approximately 33 million decimal digits.Thus, we get that any plane graph is facially-non-repetitively C -choosable for C = 10 · . The presented proof is far from optimal and the polynomialsin some of the lemmas can be improved at the expense of a more technicalargument. Nevertheless, these improvements do not lead to any reasonablevalue C . We do not know any non-trivial lower bounds for C . References [1] Noga Alon, Jarosław Grytczuk, Mariusz Hałuszczak, and Oliver Riordan. Nonrepetitivecolorings of graphs.
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Appendix A. Thomassen’s proof
We present a slightly modified version of Thomassen’s [12] proof that givesthe following statement. We use almost the exact same wording as in theoriginal proof to make it obvious that the same proof works.
Theorem.
Let G be a near-triangulation; i.e., G is a planar graph which hasno loops or multiple edges and which consists of a cycle C : v v · · · v p v , andvertices and edges inside C such that each bounded face is bounded by a tri-angle. Assume that v and v are colored { , . . . , m } and { m + 1 , . . . , m } ,respectively, and that L ( v ) is a list of at least m colors if v ∈ C − { v , v } andat least m colors if v ∈ G − C . Then the coloring of v and v can be extendedto a list m -coloring of G .Proof. (by induction on the number of vertices of G ). If p = 3 and G = C there is nothing to prove. So we proceed to the induction step.If C has a chord v i v j where i j − p − ( v p +1 = v ), then we applythe induction hypothesis to the cycle v v · · · v i v j v j +1 · · · v and its interior andthen to v j v i v i +1 · · · v j − v j and its interior. So we can assume that C has nochord. Let v , u , u , . . . , u m , v p − be the neighbors of v p in that clockwise orderaround v p . As the interior of C is triangulated, G contains the path P : v u u · · · u m v p − . As C is chordless, P ∪ ( C − v p ) is a cycle C ′ . Let X be aset of m distinct colors in L ( v p ) r { , . . . , m } . Now define L ′ ( u i ) = L ( u i ) r X for i m and L ′ ( v ) = L ( v ) if v is a vertex of G not in { u , u , . . . , u m } .Then we apply the induction hypothesis to C ′ and its interior and the new list L ′ . We complete the coloring by assigning m colors from X to v p such that v p and v p − get disjoint sets of colors. (cid:3) (G. Gutowski) Theoretical Computer Science Department, Faculty of Math-ematics and Computer Science, Jagiellonian University, Kraków, Poland
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