Bordeaux 3-color conjecture and 3-choosability
Discrete Mathematics 306 (2006) 573 – 579 www.elsevier.com/locate/disc
Bordeaux 3-color conjecture and 3-choosability夡 Mickaël Montassier, André Raspaud, Weifan Wang∗ LaBRI UMR CNRS 5800, Université Bordeaux I, 33405 Talence Cedex, France Received 22 September 2005; received in revised form 1 February 2006; accepted 2 February 2006
Abstract A graph G = (V , E) is list L-colorable if for a given list assignment L = {L(v) : v ∈ V }, there exists a proper coloring c of G such that c(v) ∈ L(v) for all v ∈ V . If G is list L-colorable for every list assignment with |L(v)| k for all v ∈ V , then G is said to be k-choosable. In this paper, we prove that (1) every planar graph either without 4- and 5-cycles, and without triangles at distance less than 4, or without 4-, 5- and 6-cycles, and without triangles at distance less than 3 is 3-choosable; (2) there exists a non-3-choosable planar graph without 4-cycles, 5-cycles, and intersecting triangles. These results have some consequences on the Bordeaux 3-color conjecture by Borodin and Raspaud [A sufficient condition for planar graphs to be 3-colorable. J. Combin. Theory Ser. B 88 (2003) 17–27]. © 2006 Elsevier B.V. All rights reserved. Keywords: Bordeaux 3-color conjecture; Choosability
1. Introduction Let G be a graph. Let V (G) be its set of vertices and E(G) be its set of edges. A triangle in G is synonymous with a 3-cycle. We say that two cycles (or faces) of a graph are adjacent if they share at least one common edge. Two cycles (or faces) are intersecting if they share at least one common vertex. The distance, denoted by dist(x, y), between two vertices x and y is the length of a shortest path connecting them in G. The distance between two triangles T and T is defined as the value min{dist(x, y)|x ∈ V (T ) and y ∈ V (T )}. A proper vertex coloring of a graph G is an assignment c of integers (or labels) to the vertices of G such that c(u) = c(v) if the vertices u and v are adjacent in G. A graph G is list L-colorable if for a given list assignment L = {L(v) : v ∈ V (G)} there exists a proper coloring c of the vertices such that ∀v ∈ V (G), c(v) ∈ L(v). We say that c is an L-coloring of G. If G is list L-colorable for every list assignment with |L(v)| k for all v ∈ V (G), then G is k-choosable. All 2-choosable graphs were characterized completely in [3]. Thomassen [10] proved that every planar graph is 5-choosable, whereas Voigt [12] presented an example of a planar graph which is not 4-choosable. Thus, it remains to determine whether a given planar graph is 3- or 4-choosable. In [5], Gutner proved that these problems are NP-hard. Therefore, many authors tried to find sufficient conditions for a planar graph to be 3- or 4-choosable. Alon and Tarsi 夡 This
work was supported by the French CNRS.
∗ On leave of absence from the Department of Mathematics, Zhejiang Normal University, Jinhua 321004, PR China.
E-mail addresses: [email protected] (M. Montassier), [email protected] (A. Raspaud), [email protected] (W. Wang). 0012-365X/$ - see front matter © 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2006.02.001
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[1] proved that every planar bipartite graph is 3-choosable. Thomassen [11] proved that every planar graph of girth 5 is 3-choosable. More recently, some new sufficient conditions for a planar graph to be 3-choosable have been given. These conditions include: having no 3-, 5- and 6-cycles [8]; having no 4-, 5-, 6- and 9-cycles [18]; or having no 4-, 5-, 7- and 9-cycles [17]. The reader is referred to [7,9,15,16] for results about the 4-choosability of planar graphs. In this paper, we investigate the 3-choosability of planar graphs with spare triangles and without cycles of special length. We prove that: (1) Every planar graph without 4- and 5-cycles, and without two triangles at distance less than 4 is 3-choosable. (2) Every planar graph without 4-, 5- and 6-cycles, and without two triangles at distance less than 3 is 3-choosable. Moreover we show that: (3) There exists a non-3-choosable planar graph without 4-cycles, 5-cycles and intersecting triangles. The paper is organized as follows. In the remaining of this section, we define some notation used throughout the paper. In Section 2, we establish the two sufficient conditions for planar graphs to be 3-choosable. An example of a planar graph concerning the Bordeaux 3-coloring conjecture [2] will be provided in Section 3. Section 4 includes two open problems on 3-choosability of planar graphs. Let G be a plane graph. Let F (G) denote the set of faces of G. For a face f ∈ F (G), we use b(f ) to represent the boundary walk of f and write f = [u1 u2 . . . un ] if u1 , u2 , . . . , un are the boundary vertices in the cyclic order. Moreover, we write simply that V (f ) = V (b(f )). The degree of a face is the number of edge-steps in its boundary walk. Note that each cut-edge is counted twice. If b(f ) forms a cycle, we call f a simple face. Obviously, each face of degree at most 5 is a simple face when G contains no vertices of degree less than 2. For x ∈ V (G) ∪ F (G), let dG (x), or simply d(x), denote the degree of x in G. A vertex (or face) of degree k is called a k-vertex (or k-face). When v is a k-vertex, we say that there are k-faces incident to v. However, these faces are not required to be distinct, i.e., v may have repeated occurrences on the boundary walk of some of its incident faces. 2. Two sufficient conditions on 3-choosability A vertex v of G is called minor if d(v)5. A minor vertex is said to be light if it is incident to a 3-face. We say that a face f of G is good if either d(f )7, or d(f ) = 6 and f is not incident to a light 3-vertex. Theorem 1. Every plane graph without 4-cycles, 5-cycles, and without triangles at distance less than 4 is 3-choosable. Proof. The proof is proceeded by contradiction. Suppose that G is a minimal counterexample to the theorem (i.e., with the smallest vertex number). Let L be a list assignment of G with |L(v)| = 3 for all v ∈ V (G) such that G is not list L-colorable. Thus, G is connected, without 4-faces and 5-faces, and satisfies the following claims: Claim 1. There are no two triangles at distance less than 4. Proof. By hypothesis. Claim 2. G does not contain any vertices of degree less than 3. Proof. Suppose that G contains a vertex u of degree less than 3. By the minimality of G, G − u admits an L-coloring c. In G, we can color u with a color in L(u) different from the colors of its neighbors to extend c to an L-coloring of G. This is a contradiction. Claim 3. G does not contain any 6-cycle C such that each vertex of C is of degree 3. Proof. Suppose that G contains a 6-cycle C with d(x) = 3 for all vertices x ∈ V (C). For each x ∈ V (C), we use x to represent the neighbor of x belonging to V (G)\V (C). By the minimality of G, G − V (C) has an L-coloring c. We
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define a list assignment L (x) = L(x)\{c(x )} for every x ∈ V (C). It is easy to see that |L (x)| |L(x)| − 1 = 3 − 1 = 2. Since every even cycle is 2-choosable, C is list L -colorable and, therefore, G is list L-colorable, producing a contradiction. By Euler’s formula |V (G)| − |E(G)| + |F (G)| = 2 and the relation d(v) = d(f ) = 2|E(G)|, v∈V (G)
f ∈F (G)
we can derive the following identity: (d(v) − 6) + (2d(f ) − 6) = −12. v∈V (G)
f ∈F (G)
Define a weight function w by w(x) = d(x) − 6 if x ∈ V (G) and w(x) = 2d(x) − 6 if x ∈ F (G). Thus, the total sum of weights is equal to −12. Now we are going to describe a discharging process that will redistribute the weights w(x) of faces to its incident vertices while the total sum of weights is kept fixed. In precise, we define the following discharging rules: (R1) Every face of degree at least seven gives 23 to each incident light 3-vertex, and 1 to each other incident minor vertex (i.e., 4-vertices, 5-vertices, and 3-vertices which are not light). (R2) Every good 6-face gives 1 to each incident minor vertex. (R3) Every 6-face which is not good gives 23 to each incident light 3-vertex, 21 to each incident light 4-vertex, and 1 to each other incident 3-vertex. We introduce a notation (f → v) that denotes the amount of weight transferred from a face f to its incident vertex v according to (R1)–(R3). Let w (x) denote the final weight function when the discharging is complete. It suffices to show that w (x)0 for all x ∈ V (G) ∪ F (G) to obtain the following contradiction: w (x) = w(x) = −12. 0 x∈V (G)∪F (G)
x∈V (G)∪F (G)
Let v ∈ V (G). Then d(v) 3 by Claim 2 and v is incident to at most one 3-face by Claim 1. • If d(v) = 3, then w(v) = −3. When v is incident to a 3-face, that is, v is a light 3-vertex, each of the other faces incident to v is of degree at least six and gives 23 to v by (R1) and (R3). It follows that w (v) = −3 + 2 × 23 = 0. When v is not incident to a 3-face, we have w (v) − 3 + 3 × 1 = 0 by (R1)–(R3). • Suppose that d(v) = 4. We see that w(v) = −2. Let v1 , v2 , v3 , v4 denote the neighbors of v in G arranged around v in the clockwise direction. For i = 1, 2, 3, 4, let fi denote the face of G incident to v such that b(fi ) contains the edges vv i and vv i+1 , where indices are taken modulo 4. First, assume that v is light. Without loss of generality, suppose that f1 is a 3-face, that is f1 =[vv 1 v2 ]. It follows from Claim 1 that d(fi )6 for i = 2, 3, 4. If f2 is a good face, then (f2 → v) = 1 by (R1) and (R2). If f2 is not a good face, then (f2 → v)= 21 by (R3). In a word, we get that (f2 → v) 21 . With the same reasoning, (f4 → v) 21 . Moreover, if d(f3 )7, then (f3 → v) = 1 by (R1). So suppose that f3 is a 6-face. Since dist(v, t) 3 for any vertex t ∈ V (f3 ), f3 cannot be adjacent to a 3-face by Claim 1 and furthermore cannot be incident to a light 3-vertex. This implies that f3 is a good 6-face and henceforth (f3 → v) = 1 by (R2). Consequently, we have w (v) − 2 + 1 + 2 × 21 = 0. Next assume that v is not light. (R1), (R2) and the following Claim 4 imply clearly that w (v) − 2 + 1 + 1 = 0. Claim 4. The vertex v is incident to at least two good faces. Proof. In order to prove Claim 4, we suppose, without loss of generality, that at most f4 is a good face. Thus f1 , f2 , f3 are 6-faces, each of which must be adjacent to some 3-face. We set f1 = [vv 1 x1 x2 x3 v2 ],
f2 = [vv 2 y1 y2 y3 v3 ],
and
f3 = [vv 3 z1 z2 z3 v4 ].
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Let f be a 3-face adjacent to f2 . Denote by e an edge shared by f2 and f . It is immediate to see that e ∈ {v2 y1 , y1 y2 , y2 y3 , y3 v3 }. If e = v2 y1 , then it is easy to check that dist(v2 , t) 3 for all t ∈ V (f3 )\{z2 }, implying that f3 cannot be adjacent to a 3-face by Claim 1, which is a contradiction. If e = y1 y2 , then dist(y1 , t) 3 for all t ∈ V (f1 )\{x1 }. This means that x1 is the only vertex in V (f1 ) that is incident to a 3-face possibly. Thus, no edge in b(f1 ) can be incident to a 3-face, i.e., f1 is not adjacent to any 3-face, also a contradiction. For e = y2 y3 , or e = y3 v3 , we can derive similarly that f1 or f3 cannot be adjacent to any 3-face. Claim 4 is proved. • Suppose that d(v) = 5. Then w(v) = −1. Similar to the previous argument, we can show that v is incident to at least one good face and thus w (v) − 1 + 1 = 0 by (R1) and (R2). • If d(v) 6, then it holds trivially that w (v) = w(v) 0. Let f ∈ F (G). Then d(f ) = 4, 5. • If d(f ) = 3, then w (f ) = w(f ) = 0. • Suppose that d(f ) = 6. Then w(f ) = 6. If f is good, then w (f )6 − 6 = 0 by (R2). If f is not good, then f is adjacent to a 3-face f = [xyz] such that xy ∈ b(f ) ∩ b(f ) with d(x) = 3. If d(y) = 3, then at least one vertex in V (f )\{x, y} is of degree greater than 3 by Claim 3. Thus w (f )6 − 2 × 23 − 3 × 1 = 0 by (R3). If d(y) = 4, then f gives 21 to y and w (f )6 − 23 − 21 − 4 × 1 = 0. If d(y)5, then f gives nothing to y and hence w (f )6 − 23 − 4 × 1 = 21 . • If d(f ) = 7, then w(f ) = 8. Since f is adjacent to at most one 3-face and hence is incident to at most two light 3-vertices, we derive that w (f )8 − 2 × 23 − 5 × 1 = 0 by (R1). • Suppose that d(f )8. Let l(f ) denote the number of light 3-vertices incident to f. By Claim 1, it is not difficult to conclude that l(f ) 25 d(f ). Henceforth, w (f )(2d(f )−6)− 23 l(f )−(d(f )−l(f ))=d(f )−6− 21 l(f ) d(f )− 6 − 21 · 25 d(f ) = 45 d(f ) − 6 > 0 by (R1). This completes the proof of the theorem.
Theorem 2. Every plane graph without cycles of length from 4 to 6 and without triangles at distance less than 3 is 3-choosable. Proof. Suppose that the theorem is false. Let G be a minimal counterexample, and let L be a list assignment of G with |L(v)| = 3 for all v ∈ V (G) such that G is not list L-colorable. Then G is connected, having neither vertices of degree less than 3 nor two triangles at distance less than 3. Since G contains no 4-cycles and 5-cycles, G contains no 4-faces and 5-faces. If G contains a 6-face f, then f is not a simple face because of the absence of 6-cycles. It is easy to notice that the boundary of f consists of two 3-cycles C and C which intersect at the only one common vertex. It follows that the distance between C and C is equal to 0, contradicting the assumption on G. This shows that G cannot contain a 6-face. To complete the proof, we rewrite Euler formula into the following new form:
(d(v) − 4) +
(d(f ) − 4) = −8,
f ∈F (G)
v∈V (G)
and define the initial weight function w(x) = d(x) − 4 for all x ∈ V (G) ∪ F (G). This time, the only one discharging rule is designed below: (R) Every face of degree at least 7 gives 21 to each incident light 3-vertex, each adjacent 3-face across every common boundary edge.
1 3
to each other incident 3-vertex, and
1 3
to
Analogously to the proof of Theorem 1, if we can verify that the resulting weight function w satisfies w (x) 0 for all x ∈ V (G) ∪ F (G), then the following contradictory inequality will finish our proof: 0
x∈V (G)∪F (G)
w (x) =
x∈V (G)∪F (G)
w(x) = −8.
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Suppose that v is a vertex. Then d(v) 3. • If d(v) 4, then it is evident that w (v) = w(v) = d(v) − 4 0. • Assume that d(v) = 3 and so w(v) = −1. Since G contains no two triangles at distance less than 3, which implies that G has no two adjacent 3-faces, so v is adjacent to at most one 3-face. If v is not adjacent to any 3-face, then w (v) = −1 + 3 × 13 = 0 by (R). Otherwise, it follows that w (v) = −1 + 2 × 21 = 0. Suppose that f is a face. Then d(f ) = 4, 5, 6 by the foregoing argument. • Assume that d(f ) = 3, then w(f ) = −1. Noting that each of the faces adjacent to f is of degree at least 7, we derive that w (f ) = −1 + 3 × 13 = 0. • Assume that d(f ) = 7. We see that w(f ) = 3. Since there do not exist two triangles at distance less than 3, f is adjacent to at most one 3-face, and furthermore is incident to at most two light 3-vertices. It turns out that w (f )3 − 2 × 21 − 5 × 13 − 13 = 0 by (R). • Assume that d(f ) 8. Let t (f ) denote the number of 3-faces adjacent to f, and again let l(f ) denote the number of light 3-vertices incident to f. Since G contains no triangles at distance less than 3, it is easy to deduce that t (f ) ≤ d(f )/4 . Moreover, we note that a 3-face adjacent to f leads to increase at most two light 3-vertices for f, thus l(f ) 2t (f ). Applying these facts, we estimate the total sum, denoted by S(f ), of weights transferred from f to its adjacent 3-faces and incident 3-vertices according to the rule (R): S(f ) 21 l(f ) + =
1 3 13 13 21
1 1 3 (d(f ) − l(f )) + 3 t (f ) d(f ) + 16 l(f ) + 13 t (f ) d(f ) + 16 · 2t (f ) + 13 t (f ) = 13 d(f ) + 23 d(f ) + 23 · 41 d(f )
t (f )
d(f ) d(f ) − 4 = w(f ).
Therefore, w (f )w(f ) − S(f )0.
3. An example on Bordeaux 3-color conjecture In studying the colorability and choosability of graphs, a naturally logical question is to ask if a sufficient condition for k-colorability is also a sufficient condition for k-choosability? Unfortunately, the answer is usually negative, even for planar graphs. The well-known Grötzsch’s Theorem [4] states that every triangle-free planar graph is 3-colorable. However, Voigt [13] gave an example of a triangle-free planar graph that is not 3-choosable. In 1976, Steinberg conjectured that every planar graph without 4- and 5-cycles is 3-colorable. Recently, Voigt [14] proved that there exist planar graphs without 4- and 5-cycles which are not 3-choosable. So, what are the “good” sufficient conditions? In 1969, Havel asked if there existed a constant C such that every planar graph with the minimal distance between triangles at least C is 3-colorable [6]. This problem remains widely open. Borodin and Raspaud [2] proved that every planar graph with neither 3-cycles at distance less than 4 nor 5-cycles is 3-colorable. Moreover, they made the following conjecture: Conjecture 1 (Bordeaux 3-color conjecture). Every planar graph without intersecting 3-cycles and without 5-cycles is 3-colorable. In this section, we shall prove that, like Grötzsch’s Theorem and Steinberg’s Conjecture, the Bordeaux 3-color conjecture cannot be extended to the list coloring situation. In precise, we have the following: Theorem 3. There exists a non-3-choosable planar graph without 4-cycles, 5-cycles and intersecting triangles. Proof. Let H (x, t) be the graph depicted in Fig. 1 and La,b its list assignment. Given the colors a and b of the vertices x and t, we cannot properly color the vertices of H (x, t) with a color from their lists. To see this, observe first that if
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Fig. 1. The graph H (x, t) and its list assignment.
the colors of the vertices x and t (a and b with a = b) are given then we have only two choices to color the vertices y and z: either we color y with 1 and z with 2, or we color y with 2 and z with 1. In the first case (resp., second case), we consider the top part of the graph (resp., the bottom part). Assume that we color y with 1 and z with 2. The coloring of x, y, z, t implies that u1 must be colored with 3, v1 with 6 and w1 with 7. Now there are two cases for the coloring of the vertices r1 and s1 : (1) We color r1 with 8 and s1 with 9. And so, the vertex o1 must be colored with 3, p1 with 4, and q1 with 5. Finally, we cannot color the triangle T1 : for each incident vertex, we have only the two available colors 1 and 2. (2) We color r1 with 9 and s1 with 8. Consequently, the adjacent vertices m1 and n1 must be colored with 5, which is impossible. Now, if we color y with 2 and z with 1, by the same way on the bottom part of H (x, t), we cannot color properly H (x, t). We construct the graph H ∗ as follows: for each pair (a, b) ∈ {10, 11, 12} × {13, 14, 15}, let H (xi , ti ) (1 i 9) be a copy of H (x, t) with the list assignment La,b . So, we have nine copies of H (x, t): H (x1 , t1 ) with the list assignment L10,13 , H (x2 , t2 ) with the list assignment L10,14 , H (x3 , t3 ) with the list assignment L10,15 , and so on. We then identify all the vertices xi (resp., ti ), 1 i 9, to a vertex x ∗ (resp., t ∗ ). We assign the list {10, 11, 12} to the vertex x ∗ and the list {13, 14, 15} to the vertex t ∗ . Now, for any coloring of the vertices x ∗ and t ∗ with the colors c1 and c2 , there exists a copy H (xj , tj ) with the list assignment Lc1 ,c2 for which we cannot color properly. The graph H ∗ contains 56 × 9 + 2 = 506 vertices and does not contain any 4-, 5-cycles and intersecting triangles. This completes the proof of the theorem.
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4. Concluding remarks It is unknown if Theorem 1 is best possible in the sense that there exists a planar graph without 4- and 5-cycles and without triangles at distance less than 3 which is not 3-choosable. A similar question may be made for Theorem 2. Let d1 denote the least integer k such that every planar graph without 4- and 5-cycles, and without triangles at distance less than k is 3-choosable. Theorems 1 and 3 assert that 2 ≤ d1 4. Problem 1. What is the exact value of d1 ? Though the Bordeaux 3-coloring conjecture cannot be strengthened to the list coloring case, we like to propose the following problem: Problem 2. Does it exist a constant d2 such that every planar graph without 5-cycles and without triangles at distance less than d2 is 3-choosable? And if it exists, what is the minimum value of d2 ? It follows from Theorem 3 that d2 2, if it exists. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18]
N. Alon, M. Tarsi, Colorings and orientations of graphs, Combinatorica 12 (2) (1992) 125–134. O.V. Borodin, A. Raspaud, A sufficient condition for planar graphs to be 3-colorable, J. Combin. Theory Ser. B 88 (2003) 17–27. P. Erdös, A.L. Rubin, H. Taylor, Choosability in graphs, Congr. Numer. 26 (1979) 125–157. H. Grötzsch, Ein dreifarbensatz für dreikreisfreie Netze auf der Kugel, Math.-Nat. Reihe 8 (1959) 109–120. S. Gutner, The complexity of planar graph choosability, Discrete Math. 159 (1996) 119–130. I. Havel, On a conjecture of B. Grünbaum, J. Combin. Theory 7 (1969) 184–186. B.C.P. Lam, W.C. Shiu, B. Xu, On structure of some plane graphs with applications to choosability, J. Combin. Theory Ser. B 82 (2001) 285–296. P. Lam, W.C. Shiu, Z.M. Song, The 3-choosability of plane graphs of girth 4, Disrete Math. 294 (2005) 297–301. P. Lam, B. Xu, J. Liu, The 4-choosability of planar graphs without 4-cycles, J. Combin. Theory Ser. B 76 (1999) 117–126. C. Thomassen, Every planar graph is 5-choosable, J. Combin. Theory Ser. B 62 (1994) 180–181. C. Thomassen, 3-List coloring planar graphs of girth 5, J. Combin. Theory Ser. B 64 (1995) 101–107. M. Voigt, List colourings of planar graphs, Discrete Math. 120 (1993) 215–219. M. Voigt, A non-3-choosable planar graph without 3-cycles, Discrete Math. 146 (1995) 325–328. M. Voigt, A non-3-choosable planar graph without cycles of length 4 and 5, 2003, Manuscript. W. Wang, K.W. Lih, Choosability and edge choosability of planar graphs without five cycles, Appl. Math. Lett. 15 (2002) 561–565. W. Wang, K.W. Lih, Choosability and edge choosability of planar graphs without intersecting triangles, SIAM J. Discrete Math. 15 (2002) 538–545. L. Zhang, B. Wu, Three-choosable planar graphs without certain small cycles, Graph Theory Notes New York 46 (2004) 27–30. L. Zhang, B. Wu, A note on 3-choosability of planar graphs without certain cycles, Discrete Math. 297 (2005) 206–209.
Bordeaux 3-color conjecture and 3-choosability夡 Mickaël Montassier, André Raspaud, Weifan Wang∗ LaBRI UMR CNRS 5800, Université Bordeaux I, 33405 Talence Cedex, France Received 22 September 2005; received in revised form 1 February 2006; accepted 2 February 2006
Abstract A graph G = (V , E) is list L-colorable if for a given list assignment L = {L(v) : v ∈ V }, there exists a proper coloring c of G such that c(v) ∈ L(v) for all v ∈ V . If G is list L-colorable for every list assignment with |L(v)| k for all v ∈ V , then G is said to be k-choosable. In this paper, we prove that (1) every planar graph either without 4- and 5-cycles, and without triangles at distance less than 4, or without 4-, 5- and 6-cycles, and without triangles at distance less than 3 is 3-choosable; (2) there exists a non-3-choosable planar graph without 4-cycles, 5-cycles, and intersecting triangles. These results have some consequences on the Bordeaux 3-color conjecture by Borodin and Raspaud [A sufficient condition for planar graphs to be 3-colorable. J. Combin. Theory Ser. B 88 (2003) 17–27]. © 2006 Elsevier B.V. All rights reserved. Keywords: Bordeaux 3-color conjecture; Choosability
1. Introduction Let G be a graph. Let V (G) be its set of vertices and E(G) be its set of edges. A triangle in G is synonymous with a 3-cycle. We say that two cycles (or faces) of a graph are adjacent if they share at least one common edge. Two cycles (or faces) are intersecting if they share at least one common vertex. The distance, denoted by dist(x, y), between two vertices x and y is the length of a shortest path connecting them in G. The distance between two triangles T and T is defined as the value min{dist(x, y)|x ∈ V (T ) and y ∈ V (T )}. A proper vertex coloring of a graph G is an assignment c of integers (or labels) to the vertices of G such that c(u) = c(v) if the vertices u and v are adjacent in G. A graph G is list L-colorable if for a given list assignment L = {L(v) : v ∈ V (G)} there exists a proper coloring c of the vertices such that ∀v ∈ V (G), c(v) ∈ L(v). We say that c is an L-coloring of G. If G is list L-colorable for every list assignment with |L(v)| k for all v ∈ V (G), then G is k-choosable. All 2-choosable graphs were characterized completely in [3]. Thomassen [10] proved that every planar graph is 5-choosable, whereas Voigt [12] presented an example of a planar graph which is not 4-choosable. Thus, it remains to determine whether a given planar graph is 3- or 4-choosable. In [5], Gutner proved that these problems are NP-hard. Therefore, many authors tried to find sufficient conditions for a planar graph to be 3- or 4-choosable. Alon and Tarsi 夡 This
work was supported by the French CNRS.
∗ On leave of absence from the Department of Mathematics, Zhejiang Normal University, Jinhua 321004, PR China.
E-mail addresses: [email protected] (M. Montassier), [email protected] (A. Raspaud), [email protected] (W. Wang). 0012-365X/$ - see front matter © 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2006.02.001
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[1] proved that every planar bipartite graph is 3-choosable. Thomassen [11] proved that every planar graph of girth 5 is 3-choosable. More recently, some new sufficient conditions for a planar graph to be 3-choosable have been given. These conditions include: having no 3-, 5- and 6-cycles [8]; having no 4-, 5-, 6- and 9-cycles [18]; or having no 4-, 5-, 7- and 9-cycles [17]. The reader is referred to [7,9,15,16] for results about the 4-choosability of planar graphs. In this paper, we investigate the 3-choosability of planar graphs with spare triangles and without cycles of special length. We prove that: (1) Every planar graph without 4- and 5-cycles, and without two triangles at distance less than 4 is 3-choosable. (2) Every planar graph without 4-, 5- and 6-cycles, and without two triangles at distance less than 3 is 3-choosable. Moreover we show that: (3) There exists a non-3-choosable planar graph without 4-cycles, 5-cycles and intersecting triangles. The paper is organized as follows. In the remaining of this section, we define some notation used throughout the paper. In Section 2, we establish the two sufficient conditions for planar graphs to be 3-choosable. An example of a planar graph concerning the Bordeaux 3-coloring conjecture [2] will be provided in Section 3. Section 4 includes two open problems on 3-choosability of planar graphs. Let G be a plane graph. Let F (G) denote the set of faces of G. For a face f ∈ F (G), we use b(f ) to represent the boundary walk of f and write f = [u1 u2 . . . un ] if u1 , u2 , . . . , un are the boundary vertices in the cyclic order. Moreover, we write simply that V (f ) = V (b(f )). The degree of a face is the number of edge-steps in its boundary walk. Note that each cut-edge is counted twice. If b(f ) forms a cycle, we call f a simple face. Obviously, each face of degree at most 5 is a simple face when G contains no vertices of degree less than 2. For x ∈ V (G) ∪ F (G), let dG (x), or simply d(x), denote the degree of x in G. A vertex (or face) of degree k is called a k-vertex (or k-face). When v is a k-vertex, we say that there are k-faces incident to v. However, these faces are not required to be distinct, i.e., v may have repeated occurrences on the boundary walk of some of its incident faces. 2. Two sufficient conditions on 3-choosability A vertex v of G is called minor if d(v)5. A minor vertex is said to be light if it is incident to a 3-face. We say that a face f of G is good if either d(f )7, or d(f ) = 6 and f is not incident to a light 3-vertex. Theorem 1. Every plane graph without 4-cycles, 5-cycles, and without triangles at distance less than 4 is 3-choosable. Proof. The proof is proceeded by contradiction. Suppose that G is a minimal counterexample to the theorem (i.e., with the smallest vertex number). Let L be a list assignment of G with |L(v)| = 3 for all v ∈ V (G) such that G is not list L-colorable. Thus, G is connected, without 4-faces and 5-faces, and satisfies the following claims: Claim 1. There are no two triangles at distance less than 4. Proof. By hypothesis. Claim 2. G does not contain any vertices of degree less than 3. Proof. Suppose that G contains a vertex u of degree less than 3. By the minimality of G, G − u admits an L-coloring c. In G, we can color u with a color in L(u) different from the colors of its neighbors to extend c to an L-coloring of G. This is a contradiction. Claim 3. G does not contain any 6-cycle C such that each vertex of C is of degree 3. Proof. Suppose that G contains a 6-cycle C with d(x) = 3 for all vertices x ∈ V (C). For each x ∈ V (C), we use x to represent the neighbor of x belonging to V (G)\V (C). By the minimality of G, G − V (C) has an L-coloring c. We
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define a list assignment L (x) = L(x)\{c(x )} for every x ∈ V (C). It is easy to see that |L (x)| |L(x)| − 1 = 3 − 1 = 2. Since every even cycle is 2-choosable, C is list L -colorable and, therefore, G is list L-colorable, producing a contradiction. By Euler’s formula |V (G)| − |E(G)| + |F (G)| = 2 and the relation d(v) = d(f ) = 2|E(G)|, v∈V (G)
f ∈F (G)
we can derive the following identity: (d(v) − 6) + (2d(f ) − 6) = −12. v∈V (G)
f ∈F (G)
Define a weight function w by w(x) = d(x) − 6 if x ∈ V (G) and w(x) = 2d(x) − 6 if x ∈ F (G). Thus, the total sum of weights is equal to −12. Now we are going to describe a discharging process that will redistribute the weights w(x) of faces to its incident vertices while the total sum of weights is kept fixed. In precise, we define the following discharging rules: (R1) Every face of degree at least seven gives 23 to each incident light 3-vertex, and 1 to each other incident minor vertex (i.e., 4-vertices, 5-vertices, and 3-vertices which are not light). (R2) Every good 6-face gives 1 to each incident minor vertex. (R3) Every 6-face which is not good gives 23 to each incident light 3-vertex, 21 to each incident light 4-vertex, and 1 to each other incident 3-vertex. We introduce a notation (f → v) that denotes the amount of weight transferred from a face f to its incident vertex v according to (R1)–(R3). Let w (x) denote the final weight function when the discharging is complete. It suffices to show that w (x)0 for all x ∈ V (G) ∪ F (G) to obtain the following contradiction: w (x) = w(x) = −12. 0 x∈V (G)∪F (G)
x∈V (G)∪F (G)
Let v ∈ V (G). Then d(v) 3 by Claim 2 and v is incident to at most one 3-face by Claim 1. • If d(v) = 3, then w(v) = −3. When v is incident to a 3-face, that is, v is a light 3-vertex, each of the other faces incident to v is of degree at least six and gives 23 to v by (R1) and (R3). It follows that w (v) = −3 + 2 × 23 = 0. When v is not incident to a 3-face, we have w (v) − 3 + 3 × 1 = 0 by (R1)–(R3). • Suppose that d(v) = 4. We see that w(v) = −2. Let v1 , v2 , v3 , v4 denote the neighbors of v in G arranged around v in the clockwise direction. For i = 1, 2, 3, 4, let fi denote the face of G incident to v such that b(fi ) contains the edges vv i and vv i+1 , where indices are taken modulo 4. First, assume that v is light. Without loss of generality, suppose that f1 is a 3-face, that is f1 =[vv 1 v2 ]. It follows from Claim 1 that d(fi )6 for i = 2, 3, 4. If f2 is a good face, then (f2 → v) = 1 by (R1) and (R2). If f2 is not a good face, then (f2 → v)= 21 by (R3). In a word, we get that (f2 → v) 21 . With the same reasoning, (f4 → v) 21 . Moreover, if d(f3 )7, then (f3 → v) = 1 by (R1). So suppose that f3 is a 6-face. Since dist(v, t) 3 for any vertex t ∈ V (f3 ), f3 cannot be adjacent to a 3-face by Claim 1 and furthermore cannot be incident to a light 3-vertex. This implies that f3 is a good 6-face and henceforth (f3 → v) = 1 by (R2). Consequently, we have w (v) − 2 + 1 + 2 × 21 = 0. Next assume that v is not light. (R1), (R2) and the following Claim 4 imply clearly that w (v) − 2 + 1 + 1 = 0. Claim 4. The vertex v is incident to at least two good faces. Proof. In order to prove Claim 4, we suppose, without loss of generality, that at most f4 is a good face. Thus f1 , f2 , f3 are 6-faces, each of which must be adjacent to some 3-face. We set f1 = [vv 1 x1 x2 x3 v2 ],
f2 = [vv 2 y1 y2 y3 v3 ],
and
f3 = [vv 3 z1 z2 z3 v4 ].
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Let f be a 3-face adjacent to f2 . Denote by e an edge shared by f2 and f . It is immediate to see that e ∈ {v2 y1 , y1 y2 , y2 y3 , y3 v3 }. If e = v2 y1 , then it is easy to check that dist(v2 , t) 3 for all t ∈ V (f3 )\{z2 }, implying that f3 cannot be adjacent to a 3-face by Claim 1, which is a contradiction. If e = y1 y2 , then dist(y1 , t) 3 for all t ∈ V (f1 )\{x1 }. This means that x1 is the only vertex in V (f1 ) that is incident to a 3-face possibly. Thus, no edge in b(f1 ) can be incident to a 3-face, i.e., f1 is not adjacent to any 3-face, also a contradiction. For e = y2 y3 , or e = y3 v3 , we can derive similarly that f1 or f3 cannot be adjacent to any 3-face. Claim 4 is proved. • Suppose that d(v) = 5. Then w(v) = −1. Similar to the previous argument, we can show that v is incident to at least one good face and thus w (v) − 1 + 1 = 0 by (R1) and (R2). • If d(v) 6, then it holds trivially that w (v) = w(v) 0. Let f ∈ F (G). Then d(f ) = 4, 5. • If d(f ) = 3, then w (f ) = w(f ) = 0. • Suppose that d(f ) = 6. Then w(f ) = 6. If f is good, then w (f )6 − 6 = 0 by (R2). If f is not good, then f is adjacent to a 3-face f = [xyz] such that xy ∈ b(f ) ∩ b(f ) with d(x) = 3. If d(y) = 3, then at least one vertex in V (f )\{x, y} is of degree greater than 3 by Claim 3. Thus w (f )6 − 2 × 23 − 3 × 1 = 0 by (R3). If d(y) = 4, then f gives 21 to y and w (f )6 − 23 − 21 − 4 × 1 = 0. If d(y)5, then f gives nothing to y and hence w (f )6 − 23 − 4 × 1 = 21 . • If d(f ) = 7, then w(f ) = 8. Since f is adjacent to at most one 3-face and hence is incident to at most two light 3-vertices, we derive that w (f )8 − 2 × 23 − 5 × 1 = 0 by (R1). • Suppose that d(f )8. Let l(f ) denote the number of light 3-vertices incident to f. By Claim 1, it is not difficult to conclude that l(f ) 25 d(f ). Henceforth, w (f )(2d(f )−6)− 23 l(f )−(d(f )−l(f ))=d(f )−6− 21 l(f ) d(f )− 6 − 21 · 25 d(f ) = 45 d(f ) − 6 > 0 by (R1). This completes the proof of the theorem.
Theorem 2. Every plane graph without cycles of length from 4 to 6 and without triangles at distance less than 3 is 3-choosable. Proof. Suppose that the theorem is false. Let G be a minimal counterexample, and let L be a list assignment of G with |L(v)| = 3 for all v ∈ V (G) such that G is not list L-colorable. Then G is connected, having neither vertices of degree less than 3 nor two triangles at distance less than 3. Since G contains no 4-cycles and 5-cycles, G contains no 4-faces and 5-faces. If G contains a 6-face f, then f is not a simple face because of the absence of 6-cycles. It is easy to notice that the boundary of f consists of two 3-cycles C and C which intersect at the only one common vertex. It follows that the distance between C and C is equal to 0, contradicting the assumption on G. This shows that G cannot contain a 6-face. To complete the proof, we rewrite Euler formula into the following new form:
(d(v) − 4) +
(d(f ) − 4) = −8,
f ∈F (G)
v∈V (G)
and define the initial weight function w(x) = d(x) − 4 for all x ∈ V (G) ∪ F (G). This time, the only one discharging rule is designed below: (R) Every face of degree at least 7 gives 21 to each incident light 3-vertex, each adjacent 3-face across every common boundary edge.
1 3
to each other incident 3-vertex, and
1 3
to
Analogously to the proof of Theorem 1, if we can verify that the resulting weight function w satisfies w (x) 0 for all x ∈ V (G) ∪ F (G), then the following contradictory inequality will finish our proof: 0
x∈V (G)∪F (G)
w (x) =
x∈V (G)∪F (G)
w(x) = −8.
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Suppose that v is a vertex. Then d(v) 3. • If d(v) 4, then it is evident that w (v) = w(v) = d(v) − 4 0. • Assume that d(v) = 3 and so w(v) = −1. Since G contains no two triangles at distance less than 3, which implies that G has no two adjacent 3-faces, so v is adjacent to at most one 3-face. If v is not adjacent to any 3-face, then w (v) = −1 + 3 × 13 = 0 by (R). Otherwise, it follows that w (v) = −1 + 2 × 21 = 0. Suppose that f is a face. Then d(f ) = 4, 5, 6 by the foregoing argument. • Assume that d(f ) = 3, then w(f ) = −1. Noting that each of the faces adjacent to f is of degree at least 7, we derive that w (f ) = −1 + 3 × 13 = 0. • Assume that d(f ) = 7. We see that w(f ) = 3. Since there do not exist two triangles at distance less than 3, f is adjacent to at most one 3-face, and furthermore is incident to at most two light 3-vertices. It turns out that w (f )3 − 2 × 21 − 5 × 13 − 13 = 0 by (R). • Assume that d(f ) 8. Let t (f ) denote the number of 3-faces adjacent to f, and again let l(f ) denote the number of light 3-vertices incident to f. Since G contains no triangles at distance less than 3, it is easy to deduce that t (f ) ≤ d(f )/4 . Moreover, we note that a 3-face adjacent to f leads to increase at most two light 3-vertices for f, thus l(f ) 2t (f ). Applying these facts, we estimate the total sum, denoted by S(f ), of weights transferred from f to its adjacent 3-faces and incident 3-vertices according to the rule (R): S(f ) 21 l(f ) + =
1 3 13 13 21
1 1 3 (d(f ) − l(f )) + 3 t (f ) d(f ) + 16 l(f ) + 13 t (f ) d(f ) + 16 · 2t (f ) + 13 t (f ) = 13 d(f ) + 23 d(f ) + 23 · 41 d(f )
t (f )
d(f ) d(f ) − 4 = w(f ).
Therefore, w (f )w(f ) − S(f )0.
3. An example on Bordeaux 3-color conjecture In studying the colorability and choosability of graphs, a naturally logical question is to ask if a sufficient condition for k-colorability is also a sufficient condition for k-choosability? Unfortunately, the answer is usually negative, even for planar graphs. The well-known Grötzsch’s Theorem [4] states that every triangle-free planar graph is 3-colorable. However, Voigt [13] gave an example of a triangle-free planar graph that is not 3-choosable. In 1976, Steinberg conjectured that every planar graph without 4- and 5-cycles is 3-colorable. Recently, Voigt [14] proved that there exist planar graphs without 4- and 5-cycles which are not 3-choosable. So, what are the “good” sufficient conditions? In 1969, Havel asked if there existed a constant C such that every planar graph with the minimal distance between triangles at least C is 3-colorable [6]. This problem remains widely open. Borodin and Raspaud [2] proved that every planar graph with neither 3-cycles at distance less than 4 nor 5-cycles is 3-colorable. Moreover, they made the following conjecture: Conjecture 1 (Bordeaux 3-color conjecture). Every planar graph without intersecting 3-cycles and without 5-cycles is 3-colorable. In this section, we shall prove that, like Grötzsch’s Theorem and Steinberg’s Conjecture, the Bordeaux 3-color conjecture cannot be extended to the list coloring situation. In precise, we have the following: Theorem 3. There exists a non-3-choosable planar graph without 4-cycles, 5-cycles and intersecting triangles. Proof. Let H (x, t) be the graph depicted in Fig. 1 and La,b its list assignment. Given the colors a and b of the vertices x and t, we cannot properly color the vertices of H (x, t) with a color from their lists. To see this, observe first that if
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Fig. 1. The graph H (x, t) and its list assignment.
the colors of the vertices x and t (a and b with a = b) are given then we have only two choices to color the vertices y and z: either we color y with 1 and z with 2, or we color y with 2 and z with 1. In the first case (resp., second case), we consider the top part of the graph (resp., the bottom part). Assume that we color y with 1 and z with 2. The coloring of x, y, z, t implies that u1 must be colored with 3, v1 with 6 and w1 with 7. Now there are two cases for the coloring of the vertices r1 and s1 : (1) We color r1 with 8 and s1 with 9. And so, the vertex o1 must be colored with 3, p1 with 4, and q1 with 5. Finally, we cannot color the triangle T1 : for each incident vertex, we have only the two available colors 1 and 2. (2) We color r1 with 9 and s1 with 8. Consequently, the adjacent vertices m1 and n1 must be colored with 5, which is impossible. Now, if we color y with 2 and z with 1, by the same way on the bottom part of H (x, t), we cannot color properly H (x, t). We construct the graph H ∗ as follows: for each pair (a, b) ∈ {10, 11, 12} × {13, 14, 15}, let H (xi , ti ) (1 i 9) be a copy of H (x, t) with the list assignment La,b . So, we have nine copies of H (x, t): H (x1 , t1 ) with the list assignment L10,13 , H (x2 , t2 ) with the list assignment L10,14 , H (x3 , t3 ) with the list assignment L10,15 , and so on. We then identify all the vertices xi (resp., ti ), 1 i 9, to a vertex x ∗ (resp., t ∗ ). We assign the list {10, 11, 12} to the vertex x ∗ and the list {13, 14, 15} to the vertex t ∗ . Now, for any coloring of the vertices x ∗ and t ∗ with the colors c1 and c2 , there exists a copy H (xj , tj ) with the list assignment Lc1 ,c2 for which we cannot color properly. The graph H ∗ contains 56 × 9 + 2 = 506 vertices and does not contain any 4-, 5-cycles and intersecting triangles. This completes the proof of the theorem.
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4. Concluding remarks It is unknown if Theorem 1 is best possible in the sense that there exists a planar graph without 4- and 5-cycles and without triangles at distance less than 3 which is not 3-choosable. A similar question may be made for Theorem 2. Let d1 denote the least integer k such that every planar graph without 4- and 5-cycles, and without triangles at distance less than k is 3-choosable. Theorems 1 and 3 assert that 2 ≤ d1 4. Problem 1. What is the exact value of d1 ? Though the Bordeaux 3-coloring conjecture cannot be strengthened to the list coloring case, we like to propose the following problem: Problem 2. Does it exist a constant d2 such that every planar graph without 5-cycles and without triangles at distance less than d2 is 3-choosable? And if it exists, what is the minimum value of d2 ? It follows from Theorem 3 that d2 2, if it exists. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18]
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