The 3-Color Problem: A relaxation of Havel s problem

Transcription

1 The 3-Color Problem: A relaxation of Havel s problem Mickaël Montassier, André Raspaud ( ) LaBRI, Bordeaux, France Weifan Wang, Yingqian Wang Normal Zhejiang University, Jinhua, PRC Coloring Seminar January

2 Definition-Coloring A k-proper coloring of the vertices of a graph G is a mapping c : V (G) {1,,k} such that uv E(G),c(u) c(v).

3 Definition-Coloring A k-proper coloring of the vertices of a graph G is a mapping c : V (G) {1,,k} such that uv E(G),c(u) c(v).

4 The Four Color Theorem - [Appel, Haken 76] Every planar graph is 4-colorable.

5 The Four Color Theorem - [Appel, Haken 76] Every planar graph is 4-colorable. Problem asked by Francis Guthrie in 1852.

6 The Three Color Problem What about 3-colorable planar graphs.

7 The Three Color Problem What about 3-colorable planar graphs. Theorem - [Garey et al. 76] Decide whether a planar graph is 3-colorable is a NP-complet problem.

8 The Three Color Problem What about 3-colorable planar graphs. Theorem - [Garey et al. 76] Decide whether a planar graph is 3-colorable is a NP-complet problem. Question What are the conditions which imply that a planar graph is 3-colorable?

9 Some conditions Theorem - [Grötzsch 59] Every planar graph without triangle is 3-colorable.

10 Some conditions Theorem - [Grötzsch 59] Every planar graph without triangle is 3-colorable. Theorem - [Grünbaum 63] Every planar with at most three triangles is 3-colorable

11 Steinberg s Conjecture Conjecture - [Steinberg 76] Every planar graph without 4- and 5-cycles is 3-colorable.

12 Steinberg s Conjecture Conjecture - [Steinberg 76] Every planar graph without 4- and 5-cycles is 3-colorable. Question - [Erdös 91] Determine the smallest value of C, if it exists, such that every planar graph without any cycles of length 4 to C is 3-colorable?

13 Abbott,Zhou 91 C 11 Sanders, Zhao 95 C 9 Borodin 96 C 9 Borodin et al. 05 C 7

14 Abbott,Zhou 91 C 11 Sanders, Zhao 95 C 9 Borodin 96 C 9 Borodin et al. 05 C 7 Theorem - [Xu 06] Every planar graph without 5- and 7-cycle and without adjacent triangle is 3-colorable. C 7

15 Planar graphs without cycles of given length Reference X Grötzsch 59 X X X X X X X X Abbott, Zhou 91 X X X X X X X Borodin 96 X X X X X X Sanders, Zhao, Borodin 95 X X X X Borodin et al. 05 X X X Xu 06 X X X X Chen, Wang 07 X X X Chen, Wang 07 X X X X Zhang, Wu 05 (χ l ) X X X X Zhang, Wu 04 (χ l ) X X X X Wang et al. 07 (χ l ) X X X X Chen, Lu, Wang 07 (χ l ) X X X X Shen, Wang 07 (χ l ) X X X X Chen, Shen, Wang 07 (χ l )

16 Havel s problem Problem - [Havel 70] Does there exist a constant C such that every planar graph with the minimal distance between triangles at least C is 3-colorable?

17 If C exists, C 2.

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19 If C exists, C 3.

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21 If C exists, C 4.

22 Examples given by Aksionov, Mel nikov 80 Conjecture - [Aksionov, Mel nikov 80] C exists and C = 5.

23 At the crossroad of Havel s and Steinberg s problems Theorem - [Borodin, R. 03] every planar graph without 3-cycles at distance less than four and without 5-cycles is 3-colorable.

24 At the crossroad of Havel s and Steinberg s problems Theorem - [Borodin, R. 03] every planar graph without 3-cycles at distance less than four and without 5-cycles is 3-colorable. Bordeaux 3-color Conjecture - [Borodin, R. 03] Every planar graph without 5-cycle and without intersecting triangle is 3 colorable.

25 At the crossroad of Havel s and Steinberg s problems Theorem - [Borodin, R. 03] every planar graph without 3-cycles at distance less than four and without 5-cycles is 3-colorable. Bordeaux 3-color Conjecture - [Borodin, R. 03] Every planar graph without 5-cycle and without intersecting triangle is 3 colorable. Strong Bordeaux Conjecture - [Borodin, R. 03] Every planar graph without 5-cycle and without adjacent triangle is 3 colorable Steinberg s Conjecture.

26 Not too far from the Havel s problem Theorem - [Borodin et al. 06] Every planar graphs without triangles adjacent to cycles of length from 3 to 9 is 3-colorable.

27 Not too far from the Havel s problem Theorem - [Borodin et al. 06] Every planar graphs without triangles adjacent to cycles of length from 3 to 9 is 3-colorable. Novosibirsk 3-Color Conjecture - [Borodin et al. 06] Every planar graph without triangle adjacent to cycles of length 3 or 5 is 3-colorable. Strong Bordeaux Conjecture Conjecture de Steinberg.

28 A relaxation of Havel s problem Definition - Distance between cycles Let C and C be to cycles. d(c,c ) = min{d G (x,y) : x V (C),y V (C )} Definition - (k, l)-property A planar graph G satisfies the distance-(k,l)-property if every pair of cycles {C,C } of G of length at most k verifies d(c,c ) l.

29 A relaxation of Havel s problem Montassier s Problem Given an integer i 3, does there exist d i such that any planar graph G with the distance-(i,d i )-property is 3-colorable? The case i = 3 corresponds to the Havel.

30 Results Corollary - [Borodin et al. 06] Every planar graph with the distance-(9,1)-property is 3-colorable. Theorem Every planar graph satisfying one of the following properties: 1. the distance-(7, 2)-property 2. the distance-(6, 3)-property 3. the distance-(5, 4)-property is 3-colorable.

31 Proof of the Theorem, (1) et (2) Lemma Every planar graph with the distance-(7,2)-property (resp. the distance-(6, 3)-property) is 2-degenerate. Theorem, (1) et (2)

32 We will prove that if a graph satisfies the distance-(7,2)-property δ(g) 2 Let G be a counterexample with the minimum order. G is connected and δ(g) 3. Euler s formula : n m + f = 2 (1) 6n + 6m 6f = 12 (2) 6n + 2 d(v) + d(f ) 6f = 12 (3) v V(G) f F(G) (2 d(v) 6) + (d(f ) 6) = 12 (4) v V(G) f F(G)

33 Discharging procedure : Initial charges v V (G),ω(v) = 2 d(v) 6 f F(G),ω(f ) = d(f ) 6 By Euler formula, 12 = ω(v) v V(G) F(G)

34 The discharging rules (R) Each 8-face gives 1 to each adjacent 5-face. 1 1

35 After the discharging procedure : ω x V (G) F(G),ω (x) 0 or, ω(v) = ω (v) v V(G) F(G) v V(G) F(G) 12 = ω(v) = ω (v) 0 v V(G) F(G) v V(G) F(G)

36 We show : x V (G) F(G),ω (x) 0 x V (G),d(x) 3 et ω(x) = ω (x) = 2 k 6 0 let x be a k-face. Cas 3 k 5. By (R), x receives 1 from each adjacent faces and so, ω (x) = k 6 + k 0. Cas k = 6,7. ω (x) = ω(x) = k 6 0 Cas k 8. By (R), x gives 1 to at most k/3 adjacent faces and so, ω (x) = k 6 k/3 0. This completes the proof.

37 Theorem, (2) :For graphs satisfying the distance-(6, 3)-property, the proof is the same as the previous one with the following discharging rule. (R) Each 7-face gives 1 to each adjacent 5-face.

38 Lemma Every planar graph with the distance-(7,2)-property (resp. the distance-(6, 3)-property) is 2-degenerate. This lemma is optimal in term of distance: Observation There exist planar graphs with the distance-(9,1)-property (resp. distance-(6,2)-property) which are not 2-degenerate.

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40 Proof of Theorem, (3) Let G be a counterexample satisfying the distance-(5,4)-property with the minimum order. Observation 1 The counterexample G does not contain : 1. A 2-vertex. 2. An even cycle composed of 3-vertices. We call a light 6-face a 6-face adjacent to a 5-face. Observation 2 In the counterexample G, 1. a 4-vertex is incident to at most two light 6-faces ; 2. a 5-vertex is incident to at most three light 6-faces.

41 Discharging procedure : Initial charges v V (G),ω(v) = 2 d(v) 6 f F(G),ω(f ) = d(f ) 6 By Euler s formula, 12 = ω(v) v V(G) F(G)

42 Discharging procedure (R1) Each 4-vertex gives 1 to each incident light 6-face. (R2) Each 6-face gives 1 to each adjacent 5-face. Let x be a k-vertex, we have k 3 (Obs 1.1). k = 3. ω (x) = ω(x) = 0. k = 4. x gives at most 2 (Obs 2.1); so, ω (x) k = 5. x gives at most 3 ; so, ω (x) k 6. x gives at most k times 1 ; so, ω (x) 2 k 6 k 1 0. Let x be k-face. 3 k 5.x receives 1 from each adjacent face ; so, ω (x) = k 6 + k 0. k = 6.x is incident to a 4-vertex (Obs 1.2). If x is light, then x receives 1 from at least one of their incident vertices. Hence, by (R2), if x is light or not, then ω (x) 0. k 7. By (R2), x gives 1 to at most k/5 adjacent faces and so, ω (x) = k 6 k/5 0.

43 x V (G) F(G),ω (x) 0 or, ω(v) = ω (v) v V(G) F(G) v V(G) F(G) 12 = ω(v) = ω (v) 0 This completes the proof. v V(G) F(G) v V(G) F(G)

44 Conclusion and problems i d i Question 1 For a given i, exhibit a graph which gives lower bounds on d i. Question 2 For a given i, decrease d i.