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Triangles and 3-coloring of planar graphs Bernard Lidick Iowa State - PowerPoint PPT Presentation

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Triangles and 3-coloring of planar graphs Bernard Lidick Iowa State University February 4, 2015 Outline what is graph coloring? Grtzschs theorem: triangle-free planar graphs are 3-colorable extensions of GT preserving

  1. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face (use minimality to 3-color H ) v 0 v 1 v 3 v 2 12

  2. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face (use minimality to 3-color H ) v 0 v 1 v 3 v 2 12

  3. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face (use minimality to 3-color H ) v 0 v 3 = v 1 v 2 12

  4. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face (use minimality to 3-color H ) x 2 v 0 x 1 v 3 = v 1 v 2 12

  5. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face (use minimality to 3-color H ) x 2 v 0 v 1 x 1 v 3 v 2 12

  6. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face (use minimality to 3-color H ) x 2 v 0 v 1 x 1 v 3 v 2 12

  7. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face (use minimality to 3-color H ) x 2 y 2 v 0 v 1 x 1 y 1 v 3 v 2 12

  8. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face (use minimality to 3-color H ) x 2 y 2 v 0 v 1 x 1 y 1 v 3 v 2 12

  9. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face Case 2: G contains no 4-faces | E ( G ) | = e , | V ( G ) | = v , | F ( G ) | = f F ( G ) is the set of faces of G • v − 2 + f = e by Euler’s formula 12

  10. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face Case 2: G contains no 4-faces | E ( G ) | = e , | V ( G ) | = v , | F ( G ) | = f F ( G ) is the set of faces of G • v − 2 + f = e by Euler’s formula • 2 e ≥ 5 f since each face has length ≥ 5 (no triangles) 12

  11. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face Case 2: G contains no 4-faces | E ( G ) | = e , | V ( G ) | = v , | F ( G ) | = f F ( G ) is the set of faces of G • v − 2 + f = e by Euler’s formula • 2 e ≥ 5 f since each face has length ≥ 5 (no triangles) • 5 v − 10 + 5 f = 5 e 12

  12. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face Case 2: G contains no 4-faces | E ( G ) | = e , | V ( G ) | = v , | F ( G ) | = f F ( G ) is the set of faces of G • v − 2 + f = e by Euler’s formula • 2 e ≥ 5 f since each face has length ≥ 5 (no triangles) • 5 v − 10 + 5 f = 5 e • 5 v − 10 + 2 e ≥ 5 e 12

  13. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face Case 2: G contains no 4-faces | E ( G ) | = e , | V ( G ) | = v , | F ( G ) | = f F ( G ) is the set of faces of G • v − 2 + f = e by Euler’s formula • 2 e ≥ 5 f since each face has length ≥ 5 (no triangles) • 5 v − 10 + 5 f = 5 e • 5 v − 10 + 2 e ≥ 5 e • 5 v − 10 ≥ 3 e (our G ) 12

  14. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face Case 2: G contains no 4-faces | E ( G ) | = e , | V ( G ) | = v , | F ( G ) | = f F ( G ) is the set of faces of G • v − 2 + f = e by Euler’s formula • 2 e ≥ 5 f since each face has length ≥ 5 (no triangles) • 5 v − 10 + 5 f = 5 e • 5 v − 10 + 2 e ≥ 5 e • 5 v − 10 ≥ 3 e (our G ) • 3 ( e + 4 ) ≥ 5 ( v + 1 ) − 2 ( H is 4-critical graph) 12

  15. Planar triangle-free graph and a 4-vertex v H is 4-critical, minimal counterexample H G plane, triangle-free, G = H − v G Case 1: G contains a 4-face Case 2: G contains no 4-faces | E ( G ) | = e , | V ( G ) | = v , | F ( G ) | = f F ( G ) is the set of faces of G • v − 2 + f = e by Euler’s formula • 2 e ≥ 5 f since each face has length ≥ 5 (no triangles) • 5 v − 10 + 5 f = 5 e • 5 v − 10 + 2 e ≥ 5 e • 5 v − 10 ≥ 3 e (our G ) • 3 ( e + 4 ) ≥ 5 ( v + 1 ) − 2 ( H is 4-critical graph) • 5 v − 10 ≥ 3 e ≥ 5 v − 9, contradiction 12

  16. Precoloring Theorem (Grötzsch ’59) Every precoloring of a face of length at most 5 in any triangle-free plane graph G can be extended to a (proper) 3 -coloring of G. G G 13

  17. Precoloring Theorem (Grötzsch ’59) Every precoloring of a face of length at most 5 in any triangle-free plane graph G can be extended to a (proper) 3 -coloring of G. Theorem (Aksenov, Borodin, Glebov ’02) Every precoloring of two non-adjacent vertices in any triangle-free planar graph G can be extended to a (proper) 3 -coloring of G. G G G 13

  18. Precoloring Theorem (Grötzsch ’59; BKLY ’14) Every precoloring of a face of length at most 5 in any triangle-free plane graph G can be extended to a (proper) 3 -coloring of G. Theorem (Aksenov, Borodin, Glebov ’02; BKLY ’14) Every precoloring of two non-adjacent vertices in any triangle-free planar graph G can be extended to a (proper) 3 -coloring of G. (Proof similar to the previous one.) G G G 13

  19. Precoloring Theorem (Grötzsch ’59; BKLY ’14) Every precoloring of a face of length at most 5 in any triangle-free plane graph G can be extended to a (proper) 3 -coloring of G. Theorem (Aksenov, Borodin, Glebov ’02; BKLY ’14) Every precoloring of two non-adjacent vertices in any triangle-free planar graph G can be extended to a (proper) 3 -coloring of G. (Proof similar to the previous one.) G G G Both theorems are tight. 13

  20. Tightness for precoloring a 6-face 14

  21. Tightness for precoloring a 6-face 14

  22. Tightness for precoloring a 6-face 14

  23. Tightness for precoloring a 6-face 14

  24. Tightness for precoloring a 6-face 14

  25. Tightness for precoloring a 6-face Not every precoloring of a 6-face extends to a 3-coloring. 14

  26. Precoloring larger face Description of “critical” graphs with precolored • 6-face by Gimbel, Thomassen ’97; Aksenov, Borodin, Glebov ’03 15

  27. Precoloring larger face Description of “critical” graphs with precolored • 6-face by Gimbel, Thomassen ’97; Aksenov, Borodin, Glebov ’03 • 7-face by Aksenov, Borodin, Glebov ’04 (discharging); Dvoˇ rák, L. ’14 (network flows) 15

  28. Precoloring larger face Description of “critical” graphs with precolored • 6-face by Gimbel, Thomassen ’97; Aksenov, Borodin, Glebov ’03 • 7-face by Aksenov, Borodin, Glebov ’04 (discharging); Dvoˇ rák, L. ’14 (network flows) • 8-face by Dvoˇ rák, L. ’14 15

  29. Precoloring larger face Description of “critical” graphs with precolored • 6-face by Gimbel, Thomassen ’97; Aksenov, Borodin, Glebov ’03 • 7-face by Aksenov, Borodin, Glebov ’04 (discharging); Dvoˇ rák, L. ’14 (network flows) • 8-face by Dvoˇ rák, L. ’14 • 9-face by Choi, Ekstein, Holub, L. ’15+ 15

  30. New proof Theorem (Grötzsch ’59, BKLY ’14) Every precoloring of a face of length at most 5 in any triangle-free plane graph G can be extended to a (proper) 3 -coloring of G. G G Our proof is significantly easier. 16

  31. Proof If G is a triangle-free planar graph and F is a precolored 4-face or 5-face, then the precoloring of F extends. Case 1: F is a 4-face 2 1 G 1 2 Case 2: F is a 5-face 17

  32. Proof If G is a triangle-free planar graph and F is a precolored 4-face or 5-face, then the precoloring of F extends. Case 1: F is a 4-face H is 3-colorable 2 1 v G G 1 2 H Case 2: F is a 5-face 17

  33. Proof If G is a triangle-free planar graph and F is a precolored 4-face or 5-face, then the precoloring of F extends. Case 1: F is a 4-face H is 3-colorable 2 1 2 1 v G G G 3 1 2 1 2 H H Case 2: F is a 5-face 17

  34. Proof If G is a triangle-free planar graph and F is a precolored 4-face or 5-face, then the precoloring of F extends. Case 1: F is a 4-face H is 3-colorable 2 1 2 1 v G G G 3 1 2 1 2 H H 3 1 G 1 2 Case 2: F is a 5-face 17

  35. Proof If G is a triangle-free planar graph and F is a precolored 4-face or 5-face, then the precoloring of F extends. Case 1: F is a 4-face H is 3-colorable 2 1 2 1 v G G G 3 1 2 1 2 H H 3 1 G G H 1 2 Case 2: F is a 5-face 17

  36. Proof If G is a triangle-free planar graph and F is a precolored 4-face or 5-face, then the precoloring of F extends. Case 1: F is a 4-face H is 3-colorable 2 1 2 1 v G G G 3 1 2 1 2 H H 3 1 3 1 G G H G H 1 2 1 2 Case 2: F is a 5-face 17

  37. Proof If G is a triangle-free planar graph and F is a precolored 4-face or 5-face, then the precoloring of F extends. Case 1: F is a 4-face H is 3-colorable 2 1 2 1 v G G G 3 1 2 1 2 H H 3 1 3 1 G G H G H 1 2 1 2 Case 2: F is a 5-face 1 2 3 G 2 3 17

  38. Proof If G is a triangle-free planar graph and F is a precolored 4-face or 5-face, then the precoloring of F extends. Case 1: F is a 4-face H is 3-colorable 2 1 2 1 v G G G 3 1 2 1 2 H H 3 1 3 1 G G H G H 1 2 1 2 Case 2: F is a 5-face 1 2 3 v G G 2 3 H 17

  39. Proof If G is a triangle-free planar graph and F is a precolored 4-face or 5-face, then the precoloring of F extends. Case 1: F is a 4-face H is 3-colorable 2 1 2 1 v G G G 3 1 2 1 2 H H 3 1 3 1 G G H G H 1 2 1 2 Case 2: F is a 5-face 1 1 2 2 3 3 v 1 G G G 2 3 H 2 3 H 17

  40. Allowing some triangles Theorem (Grötzsch ’59) Every planar triangle-free graph is 3 -colorable. 18

  41. Allowing some triangles Theorem (Grötzsch ’59) Every planar triangle-free graph is 3 -colorable. One triangle is easy! G Removing one edge of the triangle results in triangle-free G . 18

  42. Allowing some triangles Theorem (Grötzsch ’59) Every planar triangle-free graph is 3 -colorable. One triangle is easy! G Removing one edge of the triangle results in triangle-free G . Theorem ( Grünbaum ’63; Aksenov ’74; Borodin ’97; BKLY ’14) Every planar graph containing at most three triangles is 3 -colorable. G 18

  43. Theorem ( Grünbaum ’63; Aksenov ’74; Borodin ’97; BKLY ’14) Every planar graph containing at most three triangles is 3 -colorable. Proof • G is 4-critical (minimal counterexample) • Reductions: • every 3-cycle is a face • every 4-cycle is a face or has a triangle inside and outside • every 5-cycle is a face or has a triangle inside and outside Case 1: G has no 4-faces Case 2: G has a 4-face with a triangle (no identification) Case 3: G has a 4-face where identification is possible 19

  44. Three triangles - Proof sketch Case 2: G has a 4-face F with a triangle (no identification) v 0 v 1 F v 3 v 2 Both v 0 , v 1 , v 2 and v 0 , v 2 , v 3 are faces. G has 4 vertices! 20

  45. Three triangles - Proof sketch Case 3: G has a 4-face where identification is possible x 2 y 2 v 0 v 1 y 1 x 1 v 3 v 2 Since G is plane, some of these vertices are the same. 21

  46. Three triangles - Proof sketch Case 3: G has a 4-face where identification is possible x 2 y 2 v 0 v 1 x 1 y 1 v 3 v 2 Since G is plane, some of these vertices are the same. Only two cases left . . . z z y v 0 v 1 x v 0 v 1 v 3 v 2 v 3 v 2 x = y 21

  47. Problem (Erd˝ os ’90) Describe 4 -critical planar graphs containing 4 triangles. 22

  48. Havel ’69; Aksenov ’70s 23

  49. Havel ’69; Aksenov ’70s Problem (Sachs ’72) Can the triangles be partitioned into two pairs so that in each pair the distance between the triangles is less than two? 23

  50. Havel ’69; Aksenov ’70s; Aksenov, Melnikov ’78,’80 Problem (Sachs ’72) Can the triangles be partitioned into two pairs so that in each pair the distance between the triangles is less than two? 23

  51. Havel ’69; Aksenov ’70s; Aksenov, Melnikov ’78,’80; Borodin ’97 Problem (Sachs ’72) Can the triangles be partitioned into two pairs so that in each pair the distance between the triangles is less than two? 23

  52. Havel ’69; Aksenov ’70s; Aksenov, Melnikov ’78,’80; Borodin ’97 Thomas and Walls ’04 ... Problem (Sachs ’72) Can the triangles be partitioned into two pairs so that in each pair the distance between the triangles is less than two? 23

  53. Havel ’69; Aksenov ’70s; Aksenov, Melnikov ’78,’80; Borodin ’97 Thomas and Walls ’04 ... Problem (Sachs ’72) Can the triangles be partitioned into two pairs so that in each pair the distance between the triangles is less than two? 23

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