Flexibility of Planar Graphs -- Sharpening the Tools to Get Lists of Size Four
Ilkyoo Choi, Felix Christian Clemen, Michael Ferrara, Paul Horn, Fuhong Ma, Tomáš Masařík
FFlexibility of Planar Graphs—Sharpening the Tools to Get Lists ofSize Four
Ilkyoo Choi , Felix Christian Clemen , Michael Ferrara ,Paul Horn , Fuhong Ma , and Tomáš Masaˇrík Hankuk University of Foreign Studies [email protected] University of Illinois at Urbana-Champaign [email protected] University of Colorado Denver [email protected] University of Denver [email protected] Shandong University, Jinan, China [email protected] Charles University, Prague, Czech Republic University of Warsaw, Poland Simon Fraser University, Burnaby, Canada [email protected]
Abstract
A graph where each vertex v has a list L ( v ) of available colors is L-colorable if there is a propercoloring such that the color of v is in L ( v ) for each v . A graph is k-choosable if every assignment L of atleast k colors to each vertex guarantees an L -coloring. Given a list assignment L , an L-request for a vertex v is a color c ∈ L ( v ). In this paper, we look at a variant of the widely studied class of precoloring extensionproblems from [Z. Dvoˇrák, S. Norin, and L. Postle: List coloring with requests. J. Graph Theory 2019],wherein one must satisfy "enough", as opposed to all, of the requested set of precolors. A graph G is ε -flexible for list size k if for any k -list assignment L , and any set S of L -requests, there is an L -coloringof G satisfying an ε -fraction of the requests in S .It is conjectured that planar graphs are ε -flexible for list size 5, yet it is proved only for list size 6 andfor certain subclasses of planar graphs. We give a stronger version of the main tool used in the proofsof the aforementioned results. By doing so, we improve upon a result by Masaˇrík and show that planargraphs without K − are ε -flexible for list size 5. We also prove that planar graphs without 4-cycles and3-cycle distance at least 2 are ε -flexible for list size 4. Finally, we introduce a new (slightly weaker) formof ε -flexibility where each vertex has exactly one request. In that setting, we provide a stronger tool andwe demonstrate its usefulness to further extend the class of graphs that are ε -flexible for list size 5. A proper (vertex) coloring of a graph G is an assignment of colors to the vertices of G such that adjacentvertices receive distinct colors. A widely studied class of problems in numerous branches of chromatic graph1 a r X i v : . [ c s . D M ] A p r heory is the family of precoloring extension problems, wherein the goal is to extend some partial coloring ofthe graph to a coloring with a desired property. This general notion was introduced in [6, 15, 16], and hasbeen studied in a breadth of settings (see, for instance, [1, 2, 3, 4, 5, 7, 13, 20, 24]).Motivated by the work of Dvoˇrák and Sereni in [12], Dvoˇrák, Norin, and Postle [11] introduced therelated concept of flexibility . The question of interest is the following: if some vertices of the graph havea preferred color, then is it possible to properly color the graph so that at least a constant fraction of thepreferences are satisfied? While this is not a precoloring extension problem in the classical sense, the idea ofretaining a set of preferred, as opposed to prescribed, colors establishes a clear link between these problems.This question is trivial in the usual proper coloring setting with a fixed number of colors, as we can permutethe colors in a proper k -coloring of a graph in order to satisfy at least a 1 / k -fraction of the requests [11]. Onthe other hand, in the setting of list coloring, the concept of flexibility gives rise to a number of interestingproblems.Before continuing, we present some formalities necessary for the results that follow. A list assignment L for a graph G is a function that assigns a set L ( v ) of colors to each vertex v ∈ V ( G ), and an L-coloring is aproper coloring ϕ such that ϕ ( v ) ∈ L ( v ) for all v ∈ V ( G ). A graph G is k-choosable if G is L -colorable fromevery list assignment L where each vertex receives at least k colors. The choosability of a graph G is theminimum k such that G is k -choosable. A weighted request is a function w that assigns a non-negative realnumber to each pair ( v , c ) where v ∈ V ( G ) and c ∈ L ( v ). Let w ( G , L ) = (cid:88) v ∈ V ( G ) , c ∈ L ( v ) w ( v , c ). For ε >
0, we saythat w is ε -satisfiable if there exists an L -coloring ϕ of G such that (cid:88) v ∈ V ( G ) w ( v , ϕ ( v )) ≥ ε · w ( G , L ) . An important special case is when at most one color can be requested at each vertex and all such colorshave the same weight. A request for a graph G with list assignment L is a function r with dom( r ) ⊆ V ( G )such that r ( v ) ∈ L ( v ) for all v ∈ dom( r ). If each vertex requests exactly one color, i.e., dom( r ) = V ( G ), thensuch a request is widespread . For ε >
0, a request r is ε -satisfiable if there exists an L -coloring ϕ of G suchthat at least (cid:15) | dom( r ) | vertices v in dom( r ) receive color r ( v ). In particular, a request r is 1-satisfiable if andonly if the precoloring given by r extends to an L -coloring of G .We say that a graph G with list assignment L is ε -flexible , weakly ε -flexible , and weighted ε -flexible if every request, widespread request, and weighted request, respectively, is ε -satisfiable. Note that weakflexibility does not make sense in the weighted setting since one can set up the weights to turn o ff the requestsfor some of the vertices. If G is (weighted / weakly) ε -flexible for every list assignment with lists of size k ,then G is (weighted / weakly) ε -flexible for lists of size k. The main meta-question is whether there exists a universal constant ε > k -choosablegraphs are (weighted) ε -flexible for lists of size k . In the paper that introduced flexibility, Dvoˇrák, Norin,and Postle [11] established some basic properties and proved several theorems in terms of degeneracy andmaximum average degree. A graph G is d-degenerate if every subgraph has a vertex of degree at most d , andthe degeneracy of G is the minimum d such that G is d -degenerate. Theorem 1 (Dvoˇrák, Norin, and Postle [11]) . For every d ≥ , there exists ε > such that d-degenerategraphs are weighted ε -flexible for lists of size d + . Compare this result with the corresponding trivial greedy bound for choosability; namely, d -degenerategraphs are d + ε > ε -flexible with lists of size 7. In the same paper [11], two bounds in2erms of the maximum average degree were developed, one of which implies that there exists ε > ε -flexible with lists of size 6. Since planar graphs are 5-choosable [22] and there existsa planar graph that is not 4-choosable [25], the following question is natural: Question 2.
Does there exist an ε > such that every planar graph is (weighted) ε -flexible for lists of size 5? Dvoˇrák, Masaˇrík, Musílek, and Pangrác [10] showed that planar graphs of girth at least 4 are weighted ε -flexible for lists of size 4. This is tight as planar graphs of girth at least 4 are 3-degenerate, and hence4-choosable, and there exists a planar graph of girth 4 that is not 3-choosable [14, 26]. They also showed thatplanar graphs of girth at least 6 are weighted ε -flexible for lists of size 3 [9]. There is still a gap, in terms ofgirth constraints, since planar graphs of girth at least 5 are 3-choosable [23].Masaˇrík made further progress towards Question 2 by showing that planar graphs without 4-cycles are ε -flexible for lists of size 5 [19]. This raises the natural corresponding question for planar graphs without4-cycles, which are known to be 4-choosable [18]. Question 3.
Does there exist an ε > such that every planar graph without C is (weighted) ε -flexible forlists of size 4? We prove a strengthening of the result in [19] towards solving Question 2. Wang and Lih [28] conjectured thatplanar graphs without K − ( K without an edge, also known as a diamond) are 4-choosable. This conjectureremains open. Improving the result in [19], we prove that planar graphs without K − are weighted ε -flexiblefor lists of size 5. This is the largest subclass of planar graphs that is known to be weighted ε -flexible for listsof size 5. Theorem 4.
There exists ε > such that every planar graph without K − is weighted ε -flexible for lists ofsize 5. We also investigate Question 3; ε -flexibility for two subclasses of planar graphs without 4-cycles, andshow that lists of size 4 are actually su ffi cient. Theorem 5.
There exists ε > such that every planar graph without C and with C distance at least 2 isweighted ε -flexible for lists of size 4. Theorem 6.
There exists ε > such that every planar graph without C , C , C is weighted ε -flexible forlists of size 4. In order to prove the Theorems 4, 5, and 6, we strengthened the main tool from [11], which was explicitlypresented in [10]. This tool, Lemma 13 in Section 2, allows us to construct more fine-tuned reducibleconfigurations, thereby reducing the complexity of the discharging argument.The concept of weak flexibility enables stronger reducible configurations when widespread requests areconsidered. To identify such configurations, we adapted Lemma 13 for widespread requests. Using the notionof weak flexibility, we derive the following result. For an example of a configuration that is possible in thissetting, but not in that of general flexibility, see (RC3) in Section 7.As planar graphs without C are not 3-degenerate, the following result further extends our attempts toattack Question 2. In this case K − is allowed and we instead forbid the “house" C + (a C and a C sharing anedge) and K , . 3 heorem 7. There exists ε > such that every planar graph with neither C + nor K , is weakly ε -flexible forlists of size 5. Table 1 compares known related results in terms of flexibility and choosability regarding planar graphswith forbidden structures. The results proved in this paper are highlighted in green.Planar graphs C C , C C , C , C Degeneracy 5 C C , C dist. ≤ C , C C , C , C K − C + , K , Degeneracy 4 , [29] 3 Choosability 4 [18] 4 4 [27] 4 5 5Weak Flexibility 5 4 5 4 5 5 T7Flexibility 5 4 5 4 5 6Weighted Flexibility 5 [19] 4 T5 5 4 T6 5 T4 6Table 1: Known results in terms of degeneracy, flexibility, and choosability, including known lower boundsfor choosability. Each graph class is a subclass of planar graphs. The first row indicates the forbiddensubgraphs for each column. Results without a reference follow from some other entry in the table, includingthe degeneracy arguments implied by Theorem 1. Entries in blue are prior results that are not known to betight, and entries in turquoise are from this paper and are not known to be tight.In Section 2, we develop the notation and prove Lemma 13, our main tool. We lay out our proof strategyin Section 3 and the proofs of Theorems 4, 5, 6, and 7 are in Sections 4, 5, 6, and 7, respectively.
In this section, we strengthen and modify the main tool used in [10, 11], in order to better understand thestructural properties of a hypothetical minimum counterexample. Let 1 I denote the characteristic function of I , meaning that 1 I ( v ) = v ∈ I and 1 I ( v ) = H ,we define addition and subtraction in the natural way, adding and subtracting, respectively, their values ateach vertex. Given a function f : V ( H ) → Z and a vertex v ∈ V ( H ), let f ↓ v denote the function such that( f ↓ v )( w ) = f ( w ) for w (cid:44) v and ( f ↓ v )( v ) =
1. A list assignment L is an f -assignment if | L ( v ) | ≥ f ( v ) for all v ∈ V ( H ). Given a set of graphs F and a graph H , a set I ⊂ V ( H ) is F -forbidding if the graph H togetherwith one additional vertex adjacent to all of the vertices in I does not contain any graph from F . Definition 8 (( F , k )-boundary-reducibility) . A graph H is an ( F , k ) -boundary-reducible induced subgraphof G if there exists a set B ⊂ V ( H ) such that Note that: The dodecahedron is a planar graph without C , C that is not 2-degenerate. The icosidodecahedron is a planargraph without C that is not 3-degenerate. The truncated cube is a planar graph without C , C , C , C that is not 2-degenerate. Theicosahedron is a planar graph that is not 4-degenerate. We have not found a reference for this result and therefore include a proof for completeness as Observation 20 in Section 5 v ∈ V ( H ) \ B , H − B is L -colorable for every (( k − deg G + deg H − deg B ) ↓ v )-assignment L ,and(FORB) for every F -forbidding set I ⊂ V ( H ) \ B of size at most k − H − B is L -colorable for every( k − deg G + deg H − deg B − I )-assignment L .We define an additional reducibility condition wherein we weaken the (FIX) property in order to considera new, weaker variation of flexibility. Definition 9 (weak ( F , k )-boundary-reducibility) . A graph H is a weakly ( F , k ) -boundary-reducible inducedsubgraph of G if there exists a set B ⊂ V ( H ) and a non-empty set Fix( H ) ⊆ ( V ( H ) \ B ) such that(FIX) for every v ∈ Fix( H ), H − B is L -colorable for every (( k − deg G + deg H − deg B ) ↓ v )-assignment L , and(FORB) for every F -forbidding set I ⊂ V ( H ) \ B of size at most k − H − B is L -colorable for every( k − deg G + deg H − deg B − I )-assignment L .In general, we may allow Fix( H ) to be only one vertex, which clearly establishes this notion as a weakerone than that given in Definition 2. In both of the preceding definitions, we will sometimes refer to the set B as the boundary of the configuration. Definition 10 (( F , k , b )-resolution) . Let G be a graph with lists of size k that does not contain any graph in F as an induced subgraph. We define ( F , k , b ) -resolution of G as a set G i of nested subgraphs for 0 ≤ i ≤ M ,such that G : = G and G i : = G − i (cid:91) j = (cid:16) H j − B j (cid:17) , where each H i is an induced ( F , k )-boundary-reducible subgraph of G i − with boundary B i such that | V ( H i ) \ B i | ≤ b and G M is itself ( F , k )-boundary-reducible graph with empty boundary and size at most b . Fortechnical reasons, let G M + : = ∅ .A weak ( F , k , b ) -resolution is defined anaologously to an ( F , k , b )-resolution; it uses weak ( F , k )-boundary-reducibility in the place of ( F , k )-boundary-reducibility.It is our goal to show that every graph that does not contain any graph from F as a subgraph contains a(weakly) reducible subgraph. Conceptually, we then think of a (weak) resolution as an inductively-definedobject obtained by iteratively identifying some (weakly) reducible subgraph H with boundary B and deleting H − B until V ( G ) is exhausted.Going forward, when considering weak reducibility, let Fix( G ) denote the union of each Fix( H ) overthe weak ( F , k )-boundary-reducible subgraphs H in some resolution of G . While Fix( G ) depends on theparticular resolution under consideration, we will generally omit mention of the resolution when the contextis clear. By definition, the following property holds:Fix( G ) ∩ ( G i − G i + ) (cid:44) ∅ . When considering ( F , k )-boundary-reducibility, we take Fix( H ) = V ( H ) \ B where B is the boundary of H .To prove weighted ε -flexibility, we use the following observation made by Dvoˇrák et al. [11]. Lemma 11 ([11]) . Let G be a graph and let L be a list assignment on V ( G ) . Suppose G is L-colorable andthere exists a probability distribution on L-colorings ϕ of G such that for every v ∈ V ( G ) and c ∈ L ( v ) , Prob [ ϕ ( v ) = c ] ≥ ε. Then G with L is weighted ε -flexible.
5n light of lemma 11, we can derive a similar lemma for weak ε -flexibility, provided that the assumptionshold for su ffi ciently large and evenly distributed set of vertices. Lemma 12.
Let b be an integer. Let G be a graph with a widespread request and let L be a list assignmenton V ( G ) . Suppose G is L-colorable with a weak ( F , k , b ) -resolution and there exists a probability distributionon L-colorings ϕ of G such that for every v ∈ Fix ( G ) and c ∈ L ( v ) , Prob [ ϕ ( v ) = c ] ≥ ε . Then G with L isweakly (cid:16) ε · b (cid:17) -flexible. The proof is very similar to proof of Lemma 11, but it makes use of the fact that requests are made for allvertices of G . Proof.
Let r be a widespread request for G and L . Let φ be chosen at random based on the given probabilitydistribution. By assumption | Fix(G) | ≥ | V | b . By linearity of expectation: E (cid:34) (cid:88) v ∈ Fix( G ) ,φ ( v ) = r ( v ) (cid:35) = (cid:88) v ∈ Fix( G ) Prob [ φ ( v ) = r ( v )] ≥ ε · | V | b , and thus there exists an L -coloring φ with (cid:16) ε · | V | b (cid:17) satisfied requests. (cid:3) Now, we are ready to strengthen the key lemma implicitly presented by Dvoˇrák, Norin, and Postle in [11],and explicitly formulated as Lemma 4 in [10]).
Lemma 13.
For all integers k ≥ and b ≥ and for all sets F of forbidden subgraphs, let G be a graphwith an ( F , k , b ) -resolution. Then, there exists an ε > such that G with any assignment of lists of size k isweighted ε -flexible. Furthermore, if the request is widespread and G has weak ( F , k , b ) -resolution, then Gwith any assignment of lists of size k is weakly (cid:16) ε · b (cid:17) -flexible. Even though the statement of the lemma is noticeably stronger, its proof remains quite similar to theoriginal formulation. We include the proof for the sake of completeness.
Proof.
Let p = k − b and ε = p k − . For a graph G satisfying the assumptions and an assignment L of lists ofsize k , we prove the following claim by induction on the ( F , k , b )-resolution:There exists a probability distribution on L -colorings ϕ of G such that(i) for every v ∈ Fix( G ) and a color c ∈ L ( v ), the probability that ϕ ( v ) = c is at least ε , and(ii) for every color c and every F -forbidding set I in G of size at most k −
2, the probability that ϕ ( v ) (cid:44) c for all v ∈ I is at least p | I | .Part (i) with Fix( G ) = V ( G ) implies that G with L is weighted ε -flexible by Lemma 11. Part (i) with assumedwidespread request implies that G with L is weakly (cid:16) ε · b (cid:17) -flexible by Lemma 12.The claim clearly holds for a graph with no vertices, the base case of the induction. Hence, suppose V ( G i ) (cid:44) ∅ . By the assumptions, there exists a subgraph H of G such that H is ( F , k )-boundary-reducible.By definition, there exists a boundary set B ⊂ V ( H ), and let Q : = H − B . Moreover, by assumption, weknow that the order of Q is at most b . By the induction hypothesis, there exists a probability distributionon L -colorings of G i + satisfying (i) and (ii). Choose an L -coloring ψ from this distribution and let L (cid:48) be the list assignment on V ( G [ Q ]) defined by L (cid:48) ( y ) = L ( y ) \ { ψ ( v ) : v ∈ V ( G ) \ V ( Q ) , vy ∈ E ( G ) } . Notethat | L (cid:48) ( y ) | ≥ k − deg G ( y ) + deg G [ Q ] ( y ) for all y ∈ V ( H ), and thus G [ V ( Q )] has an L (cid:48) -coloring by (FORB)6pplied with I = ∅ . Among all L (cid:48) -colorings of G [ V ( Q )], choose one uniformly at random, extending ψ to an L -coloring ϕ of G .Let us first argue (ii) holds. Let I = I \ V ( Q ) and I = I ∩ V ( Q ). By the induction hypothesis,we have ϕ ( v ) (cid:44) c for all v ∈ I with probability at least p | I | . If I = ∅ , then this implies (ii). Hence,suppose | I | ≥
1. For y ∈ I , let L c ( y ) = L (cid:48) ( y ) \ { c } ; and for y ∈ V ( Q ) \ I , let L c ( y ) = L (cid:48) ( y ). Note that | L c ( y ) | ≥ k − deg G ( y ) + deg G [ V ( Q )] ( y ) − I ( y ) for all y ∈ V ( Q ). Thus by (FORB), G [ V ( Q )] has an L c -coloring.Since G [ V ( Q )] has at most k b L (cid:48) -colorings, we conclude that the probability that ϕ ( y ) (cid:44) c for all y ∈ I is atleast 1 / k b = p ≥ p | I | . Hence, the probability that ϕ ( y ) (cid:44) c for all y ∈ I is at least p | I | + | I | ≥ p | I | , implying (ii).Next, let us argue (i) holds. For v ∈ V ( G ) \ V ( Q ), this is true by the induction hypothesis. Hence, supposethat v ∈ V ( Q ), and let I be the set of neighbors of v in V ( G ) \ V ( Q ). Since G does not contain any graph from F but does contain the boundary B , and all vertices in I have a common neighbor, the set I is F -forbiddenin G − V ( Q ). Furthermore, (FORB) implies 1 ≤ k − deg G ( v ) + deg G [ V ( Q )] ( v ) − { v } ( v ) = k − | I | −
1, and thus | I | ≤ k −
2. Hence, by the induction hypothesis, we have ψ ( u ) (cid:44) c for all u ∈ I with probability at least p k − .Assuming this is the case, (FIX) implies there exists an L (cid:48) -coloring of G [ V ( Q )] that gives v the color c . Since G [ V ( Q )] has at most k b L (cid:48) -colorings, we conclude that the probability that ϕ ( v ) = c is at least p k − / k b = ε .Hence, (i) holds. (cid:3) We gather common definitions and reducible configurations for the forthcoming proofs. We also give anoverview of the discharging method.A d-vertex , a d + -vertex , and a d − -vertex are a vertex of degree d , at least d , and at most d , respectively. A d-face , a d + -face , and a d − -face are defined analogously. A ( d , d , d ) -face is a 3-face where the degreesof the vertices on the face is d , d , d . Throughout all the figures in the paper, black vertices have all theirincident edges drawn, whereas a white vertex may have more edges incident than drawn since white verticesare in the boundary. Lemma 14. A ( k − -vertex is ( ∅ , k ) -boundary-reducible with empty boundary.Proof. Let H be the graph induced by a ( k − − -vertex v and set the boundary to be empty.(FIX) holds since H has only one vertex v .(FORB) also holds since there is still one available color for v . (cid:3) Lemma 15.
For k ≥ , two ( k − -vertices u , v on a -cycle uvw are ( { K − } ) , k ) -boundary-reducible withboundary { w } .Proof. Let H be a 3-cycle uvw such that deg( u ) = deg( v ) = k − B = { w } .(FIX): Each of u , v has two available colors. Fix a color ϕ ( u ) and choose an available color for ϕ ( v ) thatis not ϕ ( u ) to extend the coloring. Fixing a color for v is symmetric.(FORB): Let I ⊂ V ( H ) \ B where | I | ≤ k −
2. If | I | ≥
2, then I is not { K − } -forbidding since connectinga new vertex to both u and v always creates a K − . It remains to consider the cases where | I | = (cid:3) Going forward, we will take to the number of colors “available" to a vertex v in a configuration H to meanthe maximum number of colors remaining in L ( v ) after coloring vertices exterior to the configuration. Whenconsidering (FIX) we reduce the number of available colors on a “fixed" vertex to 1, and when considering(FORB), we reduce the number of available colors on the vertices of a candidate set I by 1.7e use the discharging method for the proofs of our theorems. We end this section with a brief overviewof the method: Given a theorem we aim to prove, let H be a counterexample with the minimum number ofvertices. Fix a plane embedding of H and let F ( H ) denote the set of faces of H . To each z ∈ V ( H ) ∪ F ( H ),assign an initial charge ch ( z ) so that the total sum is negative. We then redistribute the charge according tosome discharging rules , which will preserve the total charge sum. Let ch ∗ ( z ) denote the final charge at each z ∈ V ( H ) ∪ F ( H ). We recount the charge at this point and show that the final charge is non-negative for eachvertex and face to conclude that the sum of the final charge is non-negative. This is a contradiction since theinitial charge sum is negative and the discharging rules preserve the total charge sum. We conclude that acounterexample could not have existed. In this section we prove Theorem 4. Let F = { K − } and let G be a counterexample to Theorem 4 withthe minimum number of vertices. Fix a plane embedding of G and note that by minimality, G must beconnected. Let L be a list assignment on V ( G ) where each vertex receives at least five colors. The followingconfigurations cannot appear in G by Lemmas 14 and 15:(RC1) A 3 − -vertex.(RC2) A 3-cycle incident with two 4-vertices. (RC1) (RC2) Figure 1: Reducible configurationsFor each vertex v and each face f , let ch ( v ) = deg( v ) − ch ( f ) = | f | −
4. The following is the onlydischarging rule:(D1) Every 5 + -vertex sends 1 / − -vertex. If v is a 4-vertex, then ch ∗ ( v ) = ch ( v ) = deg( v ) − =
0, since v is not involvedin the discharging rules. If v is a 5 + -vertex, then ch ∗ ( v ) ≥ deg( v ) − − (cid:98) deg( v )2 (cid:99) · ≥ v is incident with at most (cid:98) deg( v )2 (cid:99) K − . Hence, each vertex has non-negativefinal charge.We next turn our attention to faces, again with the goal of showing that their final charge is non-negative.If f is a 4 + -face, then ch ∗ ( v ) = ch ( v ) = | f | − ≥
0, since f is not involved in the discharging rules. If f is a3-face, then by (RC1) and (RC2), f is incident with at least two 5 + -vertices. By (D1), ch ∗ ( f ) ≥ − + · = Proof of Theorem 5
In this section, we prove Theorem 5. Let F consist of C and C at distance at most 1 and let G be acounterexample to Theorem 5 with the minimum number of vertices. Fix a plane embedding of G and notethat by minimality, G must be connected. Let L be a list assignment on V ( G ) where each vertex receives atleast four colors. The following configurations cannot appear in G :(RC1) A 2 − -vertex.(RC2) A 3-vertex adjacent to two 3-vertices.(RC3) A d -vertex v adjacent to d − v is either on a 3-cycle or is adjacent to avertex on a 3-cycle. (RC1) (RC2) (RC3) Figure 2: Reducible configurations: dotted lines indicate that there is a 3-cycle somewhereFirst, observe that (RC1) is reducible by Lemma 14.
Lemma 16. (RC2) is ( { C } , -boundary-reducible with empty boundary . Proof.
Let a , b , c be consecutive 3-vertices on a path. If ac is an edge, then this configuration is reducible byLemma 15, so we may assume a and c are non-adjacent. Hence b has three available colors and a and c bothhave two available colors. Let H be the graph on a , b , c and set the boundary B = ∅ .(FIX): When the color of b is fixed as ϕ ( b ), then the coloring can be extended by choosing colors for a and c that are di ff erent from ϕ ( b ), which is possible since a and c have two available colors. When the colorof a is fixed as ϕ ( a ), then the coloring can be extended by choosing a color ϕ ( b ) for b that is not ϕ ( a ), andthen choosing a color for c that is not ϕ ( b ). Fixing the color of c is symmetric.(FORB): Let I ⊂ V ( H ) \ B where | I | ≤
2. Up to symmetry, the only { C } -forbidding set is { a , b } , sinceconnecting a new vertex to { a , c } creates a C . When I = { a , b } , a , b , c have one, two, two, respectively,available colors. An L -coloring of H can be obtained by choosing an available color ϕ ( a ) for a , choosing anavailable color ϕ ( b ) for b that is not ϕ ( a ), and then choosing an available color for c that is not ϕ ( b ). (cid:3) Lemma 17. (RC3) is ( F , -boundary-reducible.Proof. Let v be a d -vertex adjacent to d − v is either on a 3-cycle or is adjacentto a vertex on a 3-cycle. Denote the 3-cycle as w w w , let A be the set of 3-vertices in N ( v ). Let H be thegraph on A ∪ { v } ∪ { w , w , w } and set the boundary B = V ( H ) \ ( A ∪ { v } ).Note that H − B is isomorphic to either K , d − or K , d − with one additional edge. Observe that everyvertex x of H − B has two available colors if x ∈ A ∪ { v } and has three available colors if x ∈ { w , w , w } \ { v } .(FIX): If H − B is a star, then fixing a color of an arbitrary vertex of H − B can be extended to all of H − B since all vertices have at least two available colors. If H − B contains a C , then fixing a color of an9rbitrary vertex of the C can be extended to all of H since the vertices in { w , w , w } \ { v } have at least threeavailable colors.(FORB): Let I ⊂ V ( H ) \ B where | I | ≤
2. If | I | =
2, then I is not F -forbidding since connecting a newvertex to any pair of vertices in H always creates either two C at distance at most 1 or a 4-cycle. It remainsto consider the cases where | I | = (cid:3) For each vertex v and each face f , let ch ( v ) = deg( v ) − ch ( f ) = −
2. Consider a vertex v that isadjacent to a 3-vertex w . If w is on a 3-face f that is not incident with v , then f is a pendent -face of v . Thedischarging rules are the following:(D1) Every 3-vertex sends charge 1 / + -vertex v sends 2 / / + -faces.We now check that each vertex and each face has non-negative final charge. Lemma 18.
Each vertex has non-negative final charge.Proof.
By (RC1), there is no 2 − -vertex. If v is a 3-vertex, then ch ∗ ( v ) = − · = v is a 4 + -vertex. If v is on a 3-face, then v has no pendent 3-faces since there are no 3-cycleswith distance at most 1.Since ch ( v ) = deg( v ) − ≥ v has non-negative final charge. If v is not on a 3-face, then it has at mostdeg( v ) − v ) − ≥ deg( v ) − , v has non-negative final charge. (cid:3) Lemma 19.
Each face has non-negative final charge.Proof.
Note first that there are no 4-faces since there are no 4-cycles.Suppose f is a 3-face. Each 4 + -vertex on f sends to f by (D2). If v is a 3-vertex on f , then theneighbor w of v not on f is a 4 + -vertex by (RC3), so f is a pendent 3-face of w . Thus, v and w send and to f by (D1) and (D2), respectively. Therefore, each vertex on f guarantees at least to be sent to f , so ch ∗ ( f ) ≥ − + · = f is a 5-face, and let v , v , v , v , v be the vertices on f in clockwise ordering. By (RC2), f cannot be incident with four 3-vertices. If f is incident with at most one 3-vertex, then f is incident with atleast four 4 + -vertices. Since each 4 + -vertex on f sends at least to f by (D2), ch ∗ ( f ) ≥ − + · + = f is incident with either two or three 3-vertices.Suppose v , v are 3-vertices on f , so v and v are 4 + -vertices by (RC2). Since two 3-vertices v and v are adjacent to each other, every vertex at distance at most 1 from either v or v cannot be on a 3-cycle by(RC3). Moreover, for i ∈ { , } , if v i is a 4-vertex and 5-vertex, then v i has no and at most one pendent 3-faceby (RC3), so it sends and , respectively, to f by (D2). If v i is a 6 + -vertex, then it sends to f by (D2).Thus, each of v and v sends at least to f , so ch ∗ ( f ) ≥ − + · + · = v , v are 3-vertices on f , so all other vertices on f are 4 + -vertices. Note that a 4 + -vertexis guaranteed to send at least to f . If v is on a 3-face, then it must be a 5 + -vertex by (RC3). Thus, ch ( f ) ≥ − + + · + · = >
0. Now assume v is not on a 3-face. By (RC3), if v is a 4-vertex and 5-vertex, then v has no and at most one pendent 3-face, so it sends and , respectively, to f by (D2). If v is a6 + -vertex, then it sends to f by (D2). Thus, v sends at least to f , so ch ∗ ( f ) ≥ − + + · + · = = f is a 6 + -face, then each vertex on f sends at least to f by (D1) and (D2), so ch ∗ ( f ) ≥− + · = (cid:3) Observation 20.
Every planar graph without C and with C distance at least 2 is 3-degenerate.Proof. Suppose to the contrary that there exists a planar graph with minimum degree at least 4 but neither C nor C distance at most 1. We use a simple discharging argument without any reducible configurations.Let the initial charge of each vertex v and each face f be deg( v ) − | f | −
4, respectively. Therefore only3-faces have negative initial charge. The discharging rules are the following: • each 5 + -face sends 1 / • each vertex sends all its charge to its incident 3-face if it exists.The final charge of each vertex remains non-negative. Each 5 + -face f has non-negative final charge since( | f | − / | f | ) ≥ /
5. Let f be a 3-face and let v be a vertex on f . Each face incident with v except f isa 5 + -face, since there are neither C nor C distance at most 1. Since the minimum degree is at least 4, v receives charge at least , all of which is sent to f . Thus, the final charge of f is at least − + / > (cid:3) In this section, we prove Theorem 6. Let F = { C , C , C } and let G be a counterexample with the minimumnumber of vertices. Fix a plane embedding of G and note that by minimality, G must be connected. Let L be a list assignment on V ( G ) where each vertex receives at least four colors. The following configurationscannot appear in G :(RC1) A 2 − -vertex.(RC2) A 3-cycle incident with two 3-vertices.(RC3) A 4-vertex v on two (3 , , , , (RC1) (RC2) (RC3) (RC4) Figure 3: Reducible configurationsFirst, observe that (RC1) is reducible by Lemma 14 and (RC2) is reducible by Lemma 15.
Lemma 21. (RC3) is ( { C } , -boundary-reducible with empty boundary. roof. Let v be a 4-vertex on two 3-cycles vu u and vw w , where deg( u ) = deg( w ) = u ) = deg( w ) =
4. Let H be the graph on u , u , v , w , w and set the boundary B = ∅ .(FIX): If vertex v is fixed, then the coloring can be extended as u and w have lists of size 3 and u and w have lists of size 2. Fixing any other vertex, say u , allows us to first color u , which then the leaves v with at least two available colors. Therefore the coloring can be extended as the list sizes of w , w , v are2 , ,
3. Fixing any other vertex in H − v is handled in a similar fashion.(FORB): Let I ⊂ V ( H ) \ B where | I | ≤
2. If | I | =
2, then I is not { C } -forbidding since connecting a newvertex to any pair of vertices in H always creates a C . It remains to consider the cases where | I | = (cid:3) Lemma 22. (RC4) is ( { C , C } , -boundary-reducible with empty boundary.Proof. Let v , v be adjacent 4-vertices on disjoint 3-cycles v u u and v w w , where deg( u ) = deg( w ) = u ) = deg( w ) =
4. Let H be the graph on u , u , v , v , w , w and set the boundary B = ∅ .(FIX): If vertex v is fixed, then we are left with two 3-cycles. The coloring can be extended since thetwo 3-cycles have lists of sizes 1 , , , ,
3. If we fix any other vertex x , then the v i that lies on the same3-cycle as x has at least one available color and therefore the coloring can be extended.(FORB): Let I ⊂ V ( H ) \ B where | I | ≤
2. Only subsets of { v , v } are { C , C } -forbidding. In that case,we assign v and v distinct colors, leaving the two vertices on each 3-cycle with one and two available colors.The coloring can be extended by coloring the vertices with one available color first, then the remaining vertexon each 3-cycle.As was the case previously, the case where | I | = (cid:3) For each vertex v and each face f , let ch ( v ) = − ch ( f ) = | f | −
2. Note that there are no 2 − -verticesby (RC1), so there are no 4-faces and no 5-faces. We remark that even though C is forbidden, there mightstill be 6-faces; these can appear only in the form of two embedded 3-cycles.The discharging rules are as follows:(D1) Every 6 + -face uniformly distributes its initial charge to every incident vertex.(D2) Let f be a 3-face.(D2A) If f is incident with a 3-vertex, then f sends 4 / f is not incident with a 3-vertex, then f sends 3 / , , / , , + )-face, and sends 2 / + , + , + )-faces.Note that a 6-face and a 7 + -face sends 2 / /
7, respectively, to each incident vertex by (D1).Also, since each 3-face is incident with at most one 3-vertex by (RC2), it sends at least 1 / Lemma 23.
Each vertex has non-negative final charge.Proof. If v is a 5 + -vertex, then it is incident with at least three 6 + -faces, so ch ∗ ( v ) ≥ − + · = v is a 3-vertex. If v is on a 3-face, then v is also on two 7 + -faces, so ch ∗ ( v ) ≥ − + · + = v is not on a 3-face, then v is on three 6 + -faces, so ch ∗ ( v ) ≥ − + · = v is a 4-vertex. If v is on at most one 3-face, then it is on at least three 6 + -faces, so ch ∗ ( v ) ≥− + · = v is on two 3-faces f and f . Note that v is also on two 7 + -faces, whicheach sends 5 / v by (D1). If both f and f are (4 + , + , + )-faces, then each of f and f sends 2 / v , so ch ∗ ( v ) ≥ − + · + · =
0. If f is a (3 , , + )-face, which sends 3 / v by (D2A), then f sends at least1 / v by (D2), so ch ∗ ( v ) ≥ − + · + + =
0. By (RC3), the only remaining case to consider is when f is a (3 , , f is a (4 + , + , + )-face. In this case, f and f send 1 / /
7, respectively, to v ,so ch ∗ ( v ) ≥ − + · + + = (cid:3) Lemma 24.
Each face has non-negative final charge.Proof.
The final charge of every 6 + -face is 0 since its initial charge is positive and it distributes its initialcharge uniformly to all incident vertices by (D2A). Recall that there are no 4-faces and no 5-faces.Let f be a 3-face, whose initial charge is 1. Recall that f is incident with at most one 3-vertex by (RC2).If f is incident with a 3-vertex, then it has non-negative final charge by (D2A). If f is not incident with a3-vertex, then by (RC4), it sends 3 / ch ∗ ( v ) ≥ − − · = (cid:3) In this section, we prove Theorem 7. Let F = { C + , K , } and let G be a counterexample with the minimumnumber of vertices. Fix a plane embedding of G and note that by minimality, G must be connected. Let L bea list assignment on V ( G ) where each vertex receives at least five colors. The following configurations cannotappear in G :(RC1) A 3 − -vertex.(RC2) A 4-cycle with two adjacent 4-vertices.(RC3) A 4-cycle with consecutive vertices of degrees 4 , , (RC1) (RC2) (RC3) (RC4) (RC5) Figure 4: Reducible configurationsFirst, observe that (RC1) is reducible by Lemma 14. We establish the reducibility of the remainingconfigurations in the following lemmas. Note that (RC4) in Figure 4 is only weakly ( F , F , Lemma 25.
A 4-cycle v v v v with two adjacent 4-vertices v and v , (RC2), is ( { C + } , { v , v } . 13 roof. Let v v v v be a 4-cycle with deg( v ) = deg( v ) =
4. Let H be the graph on v , v , v , v and set theboundary B = { v , v } .(FIX): The only F -forbidding sets in H − { v , v } are { v } and { v } . The coloring can be extended as anedge with lists of size 1 and 2 is clearly L -colorable.(FORB): Let I ⊂ V ( H ) \ B where | I | ≤
3. If | I | ≥
2, then I is not { C + } -forbidding since connecting a newvertex to any pair of vertices in H always creates a C + . It remains to consider the cases where | I | = (cid:3) Lemma 26.
A 4-cycle v v v v with consecutive vertices of degrees 4 , , v has no degree restriction,(RC3), is ( F , { v } . Proof.
Let v v v v be a 4-cycle with deg( v ) = deg( v ) = v ) =
5. Let H be the graph on v , v , v , v and set the boundary B = { v } .(FIX): If we fix the color of one vertex, then the coloring can be extended as the other two vertices haveat least two available colors.(FORB): Let I ⊂ V ( H ) \ B where | I | ≤
3. If | I | ≥
2, then I is not F -forbidding, since connecting a newvertex to any pair of vertices in H creates either a C + or a K , . It remains to consider the cases where | I | = (cid:3) Lemma 27. (RC4) is weakly ( F , -boundary-reducible with empty boundary.Proof. Let v v v v be a 4-cycle with deg( v ) = v ) = deg( v ) = deg( v ) =
5. Let H be the graphon v , v , v , v and set the boundary B = ∅ .(FIX): Since our aim is to show weak reducibility it is su ffi cient to show (FIX) only for one vertex. If wefix v , then we obtain a 4-cycle where every vertex has two available colors. If we fix v i where i (cid:44)
1, then weobtain a 4-cycle where the list sizes are 1 , , ,
3. In both cases, the coloring can be extended.(FORB): Let I ⊂ V ( H ) \ B where | I | ≤
3. If | I | ≥
2, then I is not F -forbidding, since connecting a newvertex to any pair of vertices in H creates either a C + or a K , . It remains to consider the cases where | I | = (cid:3) Lemma 28. (RC5) is ( ∅ , -boundary-reducible with empty boundary.Proof. Let v , v , v , v be consecutive 4-vertices on a path. Let H be the graph on v , v , v , v and set theboundary B = ∅ .(FIX): If we fix the color of one vertex v , then the coloring can be extended to the remaining vertices onthe path as the remaining vertices have at least two available colors prior to fixing the color of v .(FORB): Let I ⊂ V ( H ) \ B where | I | ≤
3. Up to symmetry it su ffi ces to check the cases I = { v , v , v } and I = { v , v , v } . In the first and second case, we are left with a path with lists of size 1 , , , , , , (cid:3) For each vertex v and each face f , let ch ( v ) = deg( v ) − ch ( f ) = −
2. The discharging rules are asfollows:(D1) Every 6 + -vertex sends 2 / / / + -face.(D3) Every 4-vertex sends 2 / / + -face.14e now check that each vertex and each face has non-negative final charge. Lemma 29.
Each vertex has non-negative final charge.Proof.
There are no 3 − -vertices by (RC1). If v is a 6 + -vertex then ch ∗ ( v ) = deg( v ) − − deg( v ) · ≥
0. If v is a5-vertex, then it is adjacent to at most three 3-faces, since there is no C + , so ch ∗ ( v ) ≥ − · − · =
0. If v isa 4-vertex, then it is adjacent to at most two 3-faces, since there is no C + , so ch ∗ ( v ) ≥ − · − · = (cid:3) Lemma 30.
Each face has non-negative final charge.Proof. If f is a 3-face, then in all situations each vertex on f gives 2 / f , so ch ∗ ( f ) = − + · =
0. If f is a 4-face, then f is incident with at most two 4-vertices by (RC2). We distinguish the following cases basedon the number of 4-vertices incident with f . • If there are exactly two 4-vertices, then by (RC2) the 4-vertices cannot be consecutive on f . By (RC4)the other two vertices are 6 + -vertices. Therefore, ch ∗ ( f ) = − + · + · = • If there is exactly one 4-vertex, then f is incident with at least one other 6 + -vertex by (RC3). Therefore, ch ∗ ( f ) ≥ − + + · + = • If there is no 4-vertex, then all other vertices have degree at least 5. Therefore, ch ∗ ( f ) ≥ − + · = f is a 5 + -face, then f is incident with five 4 + -vertices, two of which are 5 + -vertices, by (RC5), so ch ∗ ( f ) ≥ − + · + · = (cid:3) Besides the open questions given in the introduction, we propose some open areas of inquiry. In addition tofurther exploring the notion of weak flexibility, we propose two possible directions that align with the generale ff ort to distinguish between flexibility and choosability in the class of planar graphs.First, Cohen-Addad, Hebdige, Král’, Li, and Salgado [8] constructed a planar graph with neither C nor C that is not even 3-colorable, refuting Steinberg’s Conjecture (see [21]). In this vein, we feel it would beinteresting to determine whether it is possible to strengthen Theorem 6 to graphs without C and C with listsof size 4.As pointed out in [11] it would be nice to narrow the gap between d -degenerate graphs and (weighted) ε -flexible graphs with lists of size k . Theorem 1 shows that d -degenerate graphs are ε -flexible when the sizeof the lists is at least d +
2, but the bound conjectured in [11] is d +
1. We propose a study of outer-planargraphs, which are 2-degenerate, but it is not known whether lists of size 3 su ffi ce to achieve ε -flexibility. Acknowledgement
This work was completed in part at the 2019 Graduate Research Workshop in Combinatorics, which wassupported in part by National Science Foundation grants
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