Planar graphs have exponentially many 3-arboricities
PPlanar graphs have exponentially many3-arboricities
Ararat Harutyunyan ∗ Department of MathematicsSimon Fraser UniversityBurnaby, B.C. V5A 1S6email: [email protected]
Bojan Mohar †‡ Department of MathematicsSimon Fraser UniversityBurnaby, B.C. V5A 1S6email: [email protected]
December 4, 2018
Abstract
It is well-known that every planar or projective planar graph canbe 3-colored so that each color class induces a forest. This bound issharp. In this paper, we show that there are in fact exponentially many3-colorings of this kind for any (projective) planar graph. The sameresult holds in the setting of 3-list-colorings.
Keywords:
Planar graph, vertex-arboricity, digraph chromatic number.
Motivation for this paper comes from two directions. One is related to thearboricity of undirected planar graphs, the other one to colorings of planardigraphs. Let us recall that a partition of vertices of a graph G into classes V ∪ · · · ∪ V k is an arboreal partition if each V i (1 ≤ i ≤ k ) induces a forestin G . A function f : V ( G ) → { , . . . , k } is called an arboreal k -coloring if V i = f − ( i ), i = 1 , . . . , k , form an arboreal partition. The vertex-arboricity a ( G ) of the graph G is the minimum k such that G admits an arboreal ∗ Research supported by FQRNT (Le Fonds qu´eb´ecois de la recherche sur la nature etles technologies) doctoral scholarship. † Supported in part by an NSERC Discovery Grant (Canada), by the Canada ResearchChair program, and by the Research Grant P1–0297 of ARRS (Slovenia). ‡ On leave from: IMFM & FMF, Department of Mathematics, University of Ljubljana,Ljubljana, Slovenia. a r X i v : . [ m a t h . C O ] O c t -coloring. Note that a ( G ) ≤ χ ( G ) ≤ a ( G ), where χ ( G ) is the chromaticnumber of G . Long ago, people asked if every planar graph has arboricity 2since this would imply the Four Color Theorem. However, planar graphs ofvertex-arboricty 3 have been found (see Chartrand et al. [1]).Let D be a digraph without cycles of length ≤
2, and let G be theunderlying undirected graph of D . A function f : V ( D ) → { , . . . , k } is a k-coloring of the digraph D if V i = f − ( i ) is acyclic in D for every i = 1 , . . . , k .Here we treat the vertex set V i acyclic if the induced subdigraph D [ V i ]contains no directed cycles (but G [ V i ] may contain cycles). The minimum k for which D admits a k -coloring is called the chromatic number of D , andis denoted by χ ( D ) (see Neumann-Lara [6]). Clearly, χ ( D ) ≤ a ( G ) . While planar graphs with arboricity 3 are known, no planar digraph(without cycles of length ≤
2) with χ ( D ) > Conjecture 1.1.
Every planar digraph D with no directed cycles of lengthat most has χ ( D ) ≤ . It is an easy consequence of 5-degeneracy of planar graphs that everyplanar digraph D without cycles of length at most 2 and its associatedunderlying planar graph G satisfy χ ( D ) ≤ a ( G ) ≤ . (1)The main result of this paper is a relaxation of Conjecture 1.1 and astrengthening of the above stated inequality (1). In doing so, we also extendthe result from planar graphs to graphs embedded in the projective plane.In particular, we prove the following. Theorem 1.2.
Every planar or projective planar graph of order n has atleast n/ arboreal -colorings. Corollary 1.3.
Every planar or projective planar digraph of order n withoutcycles of length at most has at least n/ -colorings. Let us observe that Theorem 1.2 cannot be extended to graphs embeddedin the torus since a ( K ) = 4 and K admits an embedding in the torus.However, for every orientation D of K , we have χ ( D ) ≤ χ ( D ) = 3); and it follows from the main result in [3] that every orientation2f a (simple) graph embeddable in the torus satisfies χ ( D ) ≤
3. So it ispossible that Corollary 1.3 extends to the torus. Graphs on the Klein Bottlebehave nicer since K can not be embedded in the Klein Bottle. ˇSkrekovski[8] and Kronk and Mitchem [4] have shown that these graphs have arboricityat most 3.It can be shown that a graph on the torus has arboricity at most 3unless it contains K as a subgraph. This can be used to prove that forevery graph G embeddable in the torus, there exists an edge e ∈ E ( G ) suchthat a ( G − e ) ≤
3. In this vein, we conjecture the following.
Conjecture 1.4.
For every graph G embeddable in the torus, there exists anedge e ∈ E ( G ) such that G − e has exponentially many 3-arboreal colorings. The proof of Theorem 1.2 is deferred until Section 4. Actually, we shallprove an extended version in the setting of list-colorings which we definenext.Let C be a finite set of colors. Given a graph G , let L : v (cid:55)→ L ( v ) ⊆ C be a list-assignment for G , which assigns to each vertex v ∈ V ( G ) a set ofcolors. The set L ( v ) is called the list (or the set of admissible colors ) for v . We say G is L -colorable if there is an L -coloring of G , i.e., each vertex v is assigned a color from L ( v ) such that every color class induces a forest in G . A k -list-assignment for G is a list-assignment L such that | L ( v ) | = k forevery v ∈ V ( G ). Theorem 1.5.
Let L be a -list-assignment for a planar or projective planargraph G of order n . Then G has at least n/ L -colorings. Similarly, we define list colorings for digraphs, where we insist that colorclasses induce acyclic subdigraphs. Corollary 1.3 then extends, as a corollaryto Theorem 1.5 to the list coloring setting as well.
We define a configuration as a plane graph C together with a function δ : V ( C ) → N such that δ ( v ) ≥ deg C ( v ) for every v ∈ V ( C ). A planegraph G contains the configuration ( C, δ ) if there is an injective mapping h : V ( C ) → V ( G ) such that the following statements hold:(i) For every edge ab ∈ E ( C ), h ( a ) h ( b ) is an edge of G .(ii) For every facial walk a . . . a k in C , except for the unbounded face, theimage h ( a ) . . . h ( a k ) is a facial walk in G .3iii) For every a ∈ V ( C ), the degree of h ( a ) in G is equal to δ ( a ).If v is a vertex of degree k in G , then we call it a k -vertex , and a vertexof degree at least k (at most k ) will also be referred to as a k + -vertex ( k − -vertex ). A neighbor of v whose degree is k is a k -neighbor (similarly k + - and k − -neighbor ).The goal of this section is to prove the following theorem. Theorem 2.1.
Every planar or projective planar triangulation contains oneof the configurations listed in Fig. 1.Proof.
The proof uses the discharging method. Assume, for a contradiction,that there is a (projective) planar triangulation G that contains none of theconfigurations shown in Fig. 1. We shall refer to these configurations as Q , Q , . . . , Q .Let G be a counterexample of minimum order. To each vertex v of G , weassign a charge of c ( v ) = deg( v ) −
6. A well-known consequence of Euler’sformula is that the total charge is always negative, (cid:80) v ∈ V ( G ) c ( v ) = −
12 inthe plane and (cid:80) v ∈ V ( G ) c ( v ) = − discharging rules :R1: A 7-vertex sends charge of 1 / / + -vertex sends charge of 1 / + -vertex sends charge of 2 / + , 8 + , 8 + , 6.R5: An 8 + -vertex sends charge of 3 / + , 8 + , 7, 6.R6: An 8 + -vertex sends charge of 1 / + , + , + , + .R7: An 8 + -vertex sends charge of 1 to each adjacent 4-vertex whose neigh-bors have degrees 8 + , + , , + , , , + -vertex sends charge of 3 / + , , , − − −
643 6 4 6 6 5 − − − − − − − − Q Q Q Q Q Q Q Q
75 55 5745 − −
85 55 5 5 Q Q Q
84 46 − Q
84 5 − Q − Q Q − − − Q
44 674 5 − − Q − − − − Q
10 7 664766 4 Q Q Q Q
84 44 769 445 − Q − Q Figure 1: Unavoidable configurations. The listed numbers refer to the degreefunction δ , and the notation d − at a vertex v means all such configurationswhere the value δ ( v ) is either d or d − c ∗ ( v ) be the final charge obtained after applying rules R1–R8 to allvertices in G . We will show that every vertex has non-negative final charge.This will yield a contradiction since the initial total charge of −
12 (or − bad if its neighbors have degrees 8 + , , , + -neighbor. Let us observe that theclockwise order of degrees of the neighbors of a bad vertex is 8 + , , , + , , ,
7) since Q is excluded.First, note that G has no 3 − -vertices since the configuration Q is ex-5luded and since a triangulation cannot have 2 − -vertices. We will also havein mind that Q is excluded, so every neighbor of a 4-vertex is a 6 + -vertex. Let v be a 4-vertex. Note that v has only 6 + -neighbor. Ifall neighbors have degree at most 7, then they all have degree exactly 7 since Q , Q and Q are excluded. Since the vertex v has initial charge of −
2, andeach 7-neighbor sends a charge of 1 / v is 0.Now, assume that v is adjacent to an 8 + -vertex. First, assume that theremaining three neighbors v , v , v of v are all 7 − -vertices. The vertices v , v , v cannot all have degree 6 since Q is excluded. If deg( v ) = 7 anddeg( v ) = deg( v ) = 6, then the rules R2 and R8 imply that v receives acharge of 2, resulting in the final charge of 0. If deg( v ) = deg( v ) = 7 anddeg( v ) = 6, then by rules R2 and R7, v again receives a charge of 2. Thecase where deg( v ) = deg( v ) = deg( v ) = 7 is similar through rules R2 andR6.Next, assume that v has exactly two 8 + -neighbors v , v . If the remainingtwo vertices v , v are both 7-vertices, then rules R2 and R6 imply that v receives a total charge of 2, giving it the final charge of 0. If the remainingtwo vertices are both 6-vertices, then rule R7 implies that v receives a totalcharge of 2, resulting in 0 final charge. Therefore, we may assume thatdeg( v ) = 7 and deg( v ) = 6. In this case, both v and v send a charge of3 / v by R5, and v sends a charge of 1 /
2, resulting in a final charge of0 for v .Finally, assume that v has at least three 8 + -neighbors. By rule R4 (if v has a 6-neighbor), or by rules R2 and R6 (if v has a 7-neighbor), or by ruleR6 (otherwise), we see that v receives total charge of 2, so c ∗ ( v ) = 0. Let v be a 5-vertex. Note that v is not adjacent to a 4-vertex. If all neighbors of v are 7 − -vertices, then exclusion of Q , Q and Q implies that v has at least three 7-neighbors. By R1, each such neighborsends a charge of 1 / v . Since v has initial charge of −
1, its final chargeis at least 0. Next, suppose that v has an 8 + -neighbor. If v has at leasttwo 8 + -neighbors, then by rule R3, v receives a charge of 1 / c ∗ ( v ) ≥
0. Therefore, we may suppose that v has exactlyone 8 + -neighbor. If v has at least two 7-neighbors, then by R1 and R3, v receives a total charge of at least 1 / / / >
1, resulting in a positivefinal charge for v . Finally, if v has at most one 7-neighbor, then we get theconfiguration Q , Q or Q . They have initial charge of 0, and by the discharging rules,they do not give or receive any charge, which implies that they have a final6harge of 0.
Let v be a 7-vertex, and note that v has initial chargeof 1. If v has no 4-neighbors then it has at most three 5-neighbors since Q is excluded. Therefore, it sends a charge of 1 / v has at leastone 4-neighbor. Since Q is excluded, v has at most one other 5 − -neighbor.Therefore, v sends a charge of at most 1 / / v . An 8-vertex v has initial charge of +2. Since Q is excluded, v has at most three 4-neighbors. First, suppose that v has exactly three 4-neighbors. Let u be one of them. We claim that v sends charge of atmost 2 / u . Since Q and Q are excluded, we have that N ( u ) \{ v } contains vertices of degrees either 7 + , + , + or 8 + , + ,
6. In the first case, v sends charge 1 / u , and in the second case charge 2 /
3. Since v has no5-neighbors (again, by exclusion of Q ), c ∗ ( v ) ≥ − × / v has exactly two 4-neighbors, say v and v . Weconsider two subcases. First, assume that v has a 5-neighbor. Excluding Q and Q , no vertex in N ( v ) ∩ N ( v ) and N ( v ) ∩ N ( v ) has degree at most6. If the two vertices in N ( v ) ∩ N ( v ) are both 7-vertices, then v has no6 − -neighbor ( Q and Q being excluded). This implies that v sends chargeof 1 / v . Otherwise, the two vertices in N ( v ) ∩ N ( v ) are an 8 + and a7 + -vertex, respectively. This implies that by rules R4, R5 and R6, v sendscharge of 3 /
4, 2 / / v . Therefore, in all cases, v sends no more than3 / v . An identical argument shows that v sends a charge of atmost 3 / v . Since v sends a charge of 1 / v sends a total charge of at most 3 / / / v has no 5-neighbors. Consider v . Excluding Q and Q , v is nota bad 4-vertex. Therefore, v sends charge of at most 1 to v . An identicalargument shows that v sends charge of at most 1 to v . Therefore, the finalcharge of v is non-negative.Next, suppose that v has exactly one 4-neighbor, say v . First, supposethat v is a bad 4-vertex. Excluding Q and Q , v has at most one 5-neighbor. Since v sends a charge of at most 3 / v and charge 1 / v isnot a bad 4-vertex. Then v sends at most charge of 1 to v . Because Q isexcluded, v has at most two 5-neighbors, to each of which it sends a chargeof 1 /
2. Therefore, v sends a total charge of at most 1 + 1 / / c ∗ ( v ) ≥ v has no 4-neighbors. Excluding Q and Q , v has7t most four 5-neighbors, to each of which it sends charge of 1 /
2. Therefore,the final charge of v is non-negative. A 9-vertex v has a charge of +3. Since Q is excluded, v has at most four 4-neighbors. First, suppose that v has exactly four 4-neighbors or three 4-neighbors and at least one 5-neighbor; let u be one ofthe 4-neighbors. We claim that v sends charge of at most 2 / u . Since Q and Q are excluded, we have that N ( u ) \{ v } contains vertices of degrees7 + , + , + or 8 + , + ,
6. In the first case, v sends charge 1 / u , and in thesecond case charge 2 /
3. Since v has only one 5-neighbor (again, by exclusionof Q ), c ∗ ( v ) ≥ − × / > v has exactly three 4-neighbors and no 5-neighbors.Since Q and Q are excluded, none of the 4-neighbors are bad. Therefore,in this case v sends charge of at most 1 to each 4-neighbor, resulting in anon-negative final charge.If v has exactly two 4-neighbors, we consider two subcases. For the firstsubcase, suppose that none of the 4-neighbors are bad. Now, v has at mosttwo 5-neighbors since Q is excluded. This implies that v sends total chargeof at most 1 + 1 + 1 / / v . For the second subcase, assume that v has at least onebad 4-neighbor. Now, the exclusion of Q implies that v has no 5-neighbors.Thus, v sends total charge of at most 3 / / c ∗ ( v ) ≥ v has exactly one 4-neighbor. The exclusion of Q implies that v has at most three 5-neighbors, and hence it sends out a totalcharge of at most 3 / / / / c ∗ ( v ) ≥
0. Lastly,assume that v has no 4-neighbors. Excluding Q we see that v has at most six5-neighbors. This implies that v sends a total charge of at most 6 × / c ∗ ( v ) ≥ A 10-vertex v has a charge of +4. Let v , . . . , v be theneighbors of v in the cyclic order around v . If v i is a bad 4-neighbor of v and deg( v i − ) = 7, deg( v i +1 ) = 6, then the absence of Q and Q impliesthat deg( v i +2 ) ≥ v i − ) ≥
5. The absence of Q also implies thatif v i +3 is another bad 4-neighbor, then deg( v i +2 ) = 7, thus deg( v i +4 ) = 6and deg( v i +5 ) ≥ Q and Q , weconclude that if v has two bad 4-neighbors, then it has no other 4-neighborand has at most two 5-neighbors. This implies that c ∗ ( v ) ≥
0. Suppose nowthat v has precisely one bad 4-neighbor, say v . We may assume deg( v ) = 7,deg( v ) = 6 and by the arguments given above, deg( v ) ≥
5, deg( v ) ≥
6. Excluding Q , v can have at most four 5-neighbors. Thus, the onlypossibility that c ∗ ( v ) < v has three more 4-neighbors (and the only8ay to have this is that the 4-neighbors are v , v , v ) or that v has two more4-neighbors and two 5-neighbors (in which case 4-neighbors are v , v and5-neighbors are v , v ). In each of these cases, we see, by excluding Q and Q , that deg( v ) ≥
7, deg( v ) ≥ v ) ≥
7. Thus, excluding Q , v sends charge of at most 3 / v and v and at most 1 together toboth v and v . Hence, c ∗ ( v ) ≥ − / − × / − v has no bad 4-neighbors. If v has five 4-neighbors,then they are (without loss of generality) v , v , v , v , v , and excluding Q we see that deg( v j ) ≥ j = 2 , , , ,
10. This implies (by the argumentas used above) that v sends charge of at most 3 / c ∗ ( v ) ≥ − × / >
0. Similarly, if v has one 5-neighbor v and four 4-neighbors v , v , v , v , then we see as above that v sends charge of at most3 / c ∗ ( v ) ≥ − × / − / >
0. If v hasthree 4-neighbors, then the exclusion of Q implies that it has at most two5-neighbors. Similarly, if v has two 4-neighbors, then it has at most four5-neighbors. If v has one 4-neighbor, then it has at most five 5-neighbors. If v has no 4-neighbors, it has at most six 5-neighbors. In each case, c ∗ ( v ) ≥ + -vertices: Let v be a d -vertex, with d ≥
11. Let v , . . . , v d be theneighbors of v in cyclic clockwise order, indices modulo d . Suppose that v i isa bad 4-vertex. Then we may assume that deg( v i − ) = 7 and deg( v i +1 ) = 6(or vice versa), since Q is excluded. By noting that the fourth neighborof v i has degree 6, we see that deg( v i +2 ) ≥ Q is excluded) anddeg( v i − ) ≥ Q is excluded). If v i is a good 4-vertex, then itsneighbors are 6 + -vertices. Now, we redistribute the charge sent from v toits neighbors so that from each bad 4-vertex v i we give 1 / v i − and1 / v i +1 , and from each good 4-vertex v i we give 1 / v i − and 1 / v i +1 . We claim that after the redistribution, each neighbor of v receivesfrom v at most 1 / v . A5-neighbor of v is not adjacent to a 4-vertex, so it gets charge of at most1 / v since it is adjacentto at most one 4-vertex ( Q is excluded). If a 7-neighbor v j of v satisfiesdeg( v j +1 ) = deg( v j − ) = 4, the exclusion of Q implies that both v j − and v j +1 are good 4-vertices. Thus, the claim holds for 7-neighbors of v . An8 + -neighbor of v cannot be adjacent to a bad 4-neighbor of v , and thereforeit receives charge of at most 1 / v after the redistribution. This impliesthat if d ≥
12, then the final charge at v is c ∗ ( v ) ≥ c ( v ) − d ≥ d = 11. In this case the sameconclusion as above can be made if we show that either the redistributedcharge at one of the vertices v i is 0, or that there are two vertices whose re-9istributed charge is at most 1 /
4. If there exists a good 4-vertex, then thereexists a good 4-vertex v i , one of whose neighbors, say v i − , gets 1 / d = 11 is odd and Q and Q are excluded. Let t ≥ v i , v i +2 , . . . , v i +2 t areall good 4-neighbors of v . Then it is clear that v i +2 t +1 has total redistributedcharge 1 / v i − (cid:54) = v i +2 t +1 (by parity). This shows that the totalcharge sent from v is at most 5, thus the final charge c ∗ ( v ) is non-negative.Thus, we may assume that v has no good 4-neighbors. If v has a bad 4-neighbor v i , then we may assume that deg( v i − ) = 7 and deg( v i +1 ) = 6.As mentioned above, we conclude that deg( v i +2 ) ≥
6. We are done if thisvertex has 0 redistributed charge. Otherwise, v i +2 is adjacent to anotherbad 4-neighbor v i +3 of v . Since v i , v i +1 , v i +2 , v i +3 do not correspond to theexcluded configuration Q , we conclude that deg( v i +2 ) = 7. Now we canrepeat the argument with v i +3 to conclude that v i +6 , v i +9 are also bad 4-vertices and deg( v i +8 ) = 7. However, since deg( v i − ) = 7, we concludethat v i +9 cannot be a bad 4-vertex and hence there is a neighbor of v withredistributed charge 0.Thus, v has no 4-neighbors. Now the only way to send charge 1 / v is that all neighbors of v are 5-vertices. However, in thiscase we have the configuration Q .To summarize, we have shown that the final charge of each vertex isnon-negative and this completes the proof. This section is devoted to the reducibility part of the proof of our mainresult (Theorem 1.5) using the unavoidable configurations in Fig. 1. Let G be a (projective) planar graph and L a 3-list-assignment. It is sufficient toprove the theorem when G is a triangulation. Otherwise, we triangulate G and any L -coloring of the triangulation is an L -coloring of G . Of course,we only consider arboreal L -colorings, and we omit the adverb “arboreal”in the sequel.A configuration C contained in G is called reducible if | C | ≤ L -coloring of G − V ( C ) can be extended to an L -coloring of G in at least twoways. Showing that every triangulation G contains a reducible configurationwill imply that G has at least 2 | V ( G ) | / arboreal L -colorings. While this argument is standard for planar graphs, it is much less clear (and onlyconditionally true) for the case of projective plane. The details about this case are providedin the next section.
Lemma 3.1.
Let G be a planar graph, L a -list-assignment for G , and v , . . . , v k ∈ V ( G ) . Let G i = G − { v i +1 , . . . , v k } for i = 0 , . . . , k andconsider the following properties: (1) For every i = 1 , . . . , k , deg G i ( v i ) ≤ . (2) There exists an i such that deg G i ( v i ) ≤ .If (1) holds, then every arboreal L -coloring of G can be extended to G . Ifboth (1) and (2) hold, then every arboreal L -coloring of G can be extendedto G in at least two ways.Proof. Let f be an L -coloring of G . Since v has degree at most 5 in G ,there is a color c ∈ L ( v ) such that c appears at most once on N G ( v ).Therefore, coloring v with c gives an L -coloring of G . Repeating thisargument, we see that the L -coloring of G can be extended to an L -coloringof G by consecutively L -coloring v , v , . . . , v k . If (2) holds for i , then thereare actually two possible colors that can be used to color v i . Therefore,every L -coloring of G can be extended to G in at least two ways. Lemma 3.2.
Configurations Q , . . . , Q , Q , . . . , Q , Q , . . . , Q listedin Fig. 1 are reducible. The configuration Q (cid:48) that is obtained from Q bydeleting the pendant vertex with δ ( v ) = 4 is also reducible.Proof. For these configurations Q i and Q (cid:48) we simply apply Lemma 3.1.The corresponding enumeration v , . . . , v k ( k = | V ( Q i ) | or k = | V ( Q (cid:48) ) | ) isshown in Figure 2. The vertex for which condition (2) of Lemma 3.1 appliesis always v ; it is shown by a larger circle. Lemma 3.3.
Configuration Q in Fig. 1 is reducible.Proof. Let u be the 4-vertex and let u , u , u , u be its neighbors in cyclicorder and let C be the cycle u u u u . Suppose that deg( u ) = deg( u ) = 7,deg( u ) ≤ u ) = 6. Let f be an L -coloring of G −{ u, u , u , u , u } .Now, consider u . If there are at least two ways to extend the coloring f to u , then we can obtain at least two different colorings for G by sequen-tially coloring u , u , u , u using Lemma 3.1. Therefore, we may assumethat L ( u ) = { , , } and that colors 1 and 2 each appear exactly twice on11 − − − − − − − − − − Q Q Q Q Q Q
75 55 5 745 − −
85 55 5 5 Q Q Q − Q Q − − − Q
44 674 5 − − Q − − − − Q Q Q Q v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v Q
84 44 769 445 − Q − Q v v v v v v v v v v v v v v v v v v v Figure 2: Lemma 3.1 applies to several configurations. N ( u ). Now, let us color u with color 3. We now consider coloring u and u . We claim that at least one of u and u must be forced to be colored3. Otherwise, we color u and u without using color 3, then we color u arbitrarily (this is possible since u is yet uncolored). Now, if 3 ∈ L ( u ), thenwe can color u with 3 since u has no neighbor of color 3 and hence it isnot possible to make a cycle colored 3. Moreover, there is at most one color(other than color 3) that can appear on the neighborhood of u twice. There-fore, u has another available color in its list and so there are two ways tocolor u . Similarly, we get two different colorings of u when 3 / ∈ L ( u ). Thisproves the claim, and we may assume that L ( u ) = { a, b, } , u is forced tobe colored 3, and that the four colored neighbors of u not on C have colors12 , a, b, b . Now, we color u arbitrarily with a color c . We may assume that c (cid:54) = 3, for otherwise we color u arbitrarily and we will have two availablecolors for u . To complete the proof it is sufficient to show that u can becolored with a color that is not c , for then we could color u with at leasttwo different colors. If u is forced to be colored c , then for every color x ∈ L ( u ), x (cid:54) = c , the color x must appear at least twice on N ( u ). Thisimplies that the three colored neighbors of u not on the cycle have colors3 , y, y , for some color y and that 3 , y ∈ L ( u ). But recall that u and u haveno neighbors outside C having color 3. Therefore, coloring u with color 3gives a proper coloring of G − u . Now, u can be colored with at least twocolors to obtain a coloring of G . Lemma 3.4.
Let u be a 4-vertex, and suppose u , u , u , u are the neighborsof u in cyclic order. Suppose that deg( u ) ≤ , deg( u ) ≤ and deg( u ) ≤ .This configuration is reducible. In particular, the configuration Q in Fig. 1is reducible.Proof. Let f be an L -coloring of G (cid:48) = G − { u, u , u , u } . Suppose that f ( u ) = 3. Now, consider u . Note that we can extend the coloring of G (cid:48) to u , u , u , u (in this order) by Lemma 3.1. Suppose, for a contradiction,that f has only one extension to an L -coloring of G . Then colors of each of u , u , u , u are uniquely determined in each step and two colors from eachvertex list are forbidden. Now, consider u . Since only four of its neighborsare colored and f ( u ) = 3, we can color u with a color other than 3, say2, and we may further assume that its colored neighbors use colors 1 and3 twice, where L ( u ) = { , , } . Now, consider coloring u . The color 2at u cannot create a monochromatic cycle containing u . Thus, the onlyway for a color of u to be forced is that L ( u ) = { a, b, x } and colors a and b each appear twice on N ( u ) \{ u } . In this case, we color u withthe color x . Similarly, x does not give any restriction for a color at u , so u satisfies L ( u ) = { , c, y } and the three neighbors of u distinct from u are colored with colors 3 , c, c . Now, if u does not have two colors on N ( u ), each appearing twice, we have two different available colors in L ( u ).Therefore, we may assume that { x, y } = { , } and that 2 , ∈ L ( u ). Since L ( u ) = { , c, y } , it follows that y = 2 and x = 3. Now we see that coloring u with color 3 does not create a monochromatic cycle, so u has two availablecolors: color 3 and z ∈ L ( u ) \{ , } . Lemma 3.5.
The configuration Q is reducible.Proof. Let u be an 8-vertex and assume its neighbors (in the clockwise cyclicorder) are u , . . . , u and let C be the 8-cycle u u . . . u u . Suppose that13eg( u i ) = deg( u j ) = 4, deg( u k ) ≤ u l ) = 6, where i, j, k, l ∈{ , . . . , } and i (cid:54) = j . Assume that u l and u j are adjacent on C . We mayassume that u l = u j +1 . If u i = u j +2 , then we can use Lemma 3.1 (with v = u i , v = u, v = u j +1 , v = u j , v = u k ), where property (2) applies for v . Therefore, we may assume that u i (cid:54) = u j +2 . Let L ( u ) = { , , } andconsider an L -coloring f of G − { u, u i , u j , u k , u l } . Without loss of generality,we may assume that colors 1 and 2 each appear exactly twice on N ( u ) inthe coloring f . Otherwise, there are two ways to extend the coloring f of G − { u, u i , u j , u k , u l } to a coloring of G − { u i , u j , u k , u l } , and applyingLemma 3.1 we can extend each of these to a coloring of G . Therefore, color3 does not appear in the neighborhood of u in the coloring f . We color u with color 3 to obtain a coloring g of G − { u i , u j , u k , u l } . Now, considerthe 6-vertex u j +1 . Since u j +1 has at most five colored neighbors so far,we have at least one available color for it from its list. If 3 / ∈ L ( u j +1 )we color u j +1 arbitrarily with an available color. If 3 ∈ L ( u j +1 ), we color u j +1 with 3 if color 3 does not appear on N ( u j +1 ) \{ u } . If color 3 appearson N ( u j +1 ) \{ u } , we color u j +1 with any other available color from its listexcept 3 (this is possible since the remaining three colored neighbors of u j +1 can forbid only one additional color from L ( u j +1 )). Now, consider u i . Weknow that u i (cid:54) = u j +2 . First, assume that 3 / ∈ L ( u i ). Since u i has onlythree colored neighbors and u is colored 3, there are at least two availablecolors in L ( u i ) that can be used to color u i . Each coloring then can beextended to a coloring of G by Lemma 3.1. Therefore, we may assume that3 ∈ L ( u i ). Recall that no neighbor of u , except possibly u j +1 , is colored 3,and if so, then u j +1 has no neighbor besides u of color 3. Therefore, u i canbe colored with color 3 without creating a monochromatic cycle of color 3.Consequently, the four colored neighbors of u i can forbid at most one colorfrom L ( u i ), which implies that we can color u i with two different colors.Now, applying Lemma 3.1 to G − { u k , u j } , we see that each of these twocolorings can be extended to a coloring of G . It is easy to see that every plane graph is a spanning subgraph of a tri-angulation; we can always add edges joining distinct nonadjacent verticesuntil we obtain a triangulation. However, graphs in the projective plane nolonger satisfy this property. The following extension will be sufficient forour purpose. 14 roposition 4.1.
Let G be a graph embeddable in the projective plane.Then one of the following holds: (a) G is a spanning subgraph of a triangulation of the plane or the projec-tive plane. (b) G contains vertices u, v of degree at most such that the graph G − u − v is planar. (c) G contains adjacent vertices u, v of degree at most such that thegraph G − u − v is planar.Proof. If G is a planar graph, then we have (a); so we may assume that G is not planar. The proof proceeds by induction on the number k =3 | V ( G ) | − | E ( G ) | −
3. If k = 0, then G triangulates the projective plane (cf.[5, Proposition 4.4.4]), and we have (a). If G is not 2-connected, then wecan add an edge joining two vertices in distinct blocks of G and keep theembeddability in the projective plane, and we win by induction. Thus wemay assume that G is 2-connected and non-planar. This assures that facialwalks of every embedding of G are cycles of G (cf. [5, Proposition 5.5.11]).If G is not a triangulation, then there is a facial cycle C = v v . . . v r v ,where r ≥
4. If two vertices of C are nonadjacent in G , we can add the edgejoining them and win by induction. Thus, the subgraph K of G induced on V ( C ) is the complete graph of order r . Since this subgraph has a facial walkof length r >
3, we conclude that r ∈ { , } and the induced embedding of K is as shown in Figure 3. v v v v v v v v v Figure 3: K and K embedded in the projective planeLet us consider the vertex v and the edges v v and v v (if r = 4), and v v , v v and v v (if r = 5). These edges are embedded as shown in Figure3. Suppose that v has two neighbors a, b / ∈ V ( C ) such that the cyclic order15round v is v v , v a, v v , v b when r = 4 and v v , v a, v v s , v b (where s = 3 or s = 4) when r = 5. Then we can re-embed the edge v v (if r = 4)or re-embed the edges v v and v v (if r = 5) into the face bounded by C and then add an edge joining two nonadjacent neighbors of v . Again, weare done by applying the induction hypothesis.Thus we may assume henceforth that all neighbors of each vertex v i thatare not on C are contained in a single face of K . If a face F of K containsat least one vertex that is not on C , then each vertex of C on the boundaryof F has a neighbor inside F . If not, we would be able to add an edge andwould be done by applying induction. Since any two faces of K have a vertexin common, the aforementioned property implies that at most one face of K contains any vertices of G . If r = 5, this implies that (c) is satisfied. Thus r = 4 and since G is non-planar, there is a face F of K that contains verticesof G in its interior. We may assume that F contains the edge v v on itsboundary. Now, if we re-embed the edge v v into the face of K distinctfrom F and C , we obtain a new face containing the former face bounded by C that is of length at least 5. Thus we get into one of the above cases, andwe are done. Proof of Theorem 1.5.