the coloring problem in graphs with structural
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The coloring problem in graphs with structural restrictions Lucas - PowerPoint PPT Presentation

The coloring problem in graphs with structural restrictions Lucas Pastor Joint-work with Frdric Maffray G-SCOP March 4, 2016 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 1 / 36 The k -coloring problem k -coloring For


  1. Tools Lemma It is sufficient to produce a 4-coloring for any ( P 6 , bull)-free graph G that satisfies the following properties: 1 G and G are connected. 2 G is quasi-prime. 3 G is K 5 -free and double-wheel-free. Warning We do not claim that K 5 and the double-wheel are the only 5-vertex-critical graphs! Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 12 / 36

  2. Tools Proof of 1 If G is not connected we can examine each component of G separately. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 13 / 36

  3. Tools Proof of 1 If G is not connected we can examine each component of G separately. If G is not connected, V ( G ) can be partitioned into two non-empty sets V 1 and V 2 complete to each other. We have χ ( G ) = χ ( G [ V 1 ]) + χ ( G [ V 2 ]). Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 13 / 36

  4. Tools Proof of 1 If G is not connected we can examine each component of G separately. If G is not connected, V ( G ) can be partitioned into two non-empty sets V 1 and V 2 complete to each other. We have χ ( G ) = χ ( G [ V 1 ]) + χ ( G [ V 2 ]). ◮ We need G [ V 1 ] and G [ V 2 ] to be 3-colorable. We use the known algorithms for that. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 13 / 36

  5. Tools Proof of 1 If G is not connected we can examine each component of G separately. If G is not connected, V ( G ) can be partitioned into two non-empty sets V 1 and V 2 complete to each other. We have χ ( G ) = χ ( G [ V 1 ]) + χ ( G [ V 2 ]). ◮ We need G [ V 1 ] and G [ V 2 ] to be 3-colorable. We use the known algorithms for that. ◮ If they are 3-colorable, we can precisely determine their chromatic number by testing if they are bipartite or edgeless. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 13 / 36

  6. Tools Proof of 2 Assume that G is not quasi-prime. So it has a homogeneous set X that is not a clique. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 14 / 36

  7. Tools Proof of 2 Assume that G is not quasi-prime. So it has a homogeneous set X that is not a clique. A necessary condition for G to be 4-colorable is that χ ( G [ X ]) ≤ 3. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 14 / 36

  8. Tools Proof of 2 Assume that G is not quasi-prime. So it has a homogeneous set X that is not a clique. A necessary condition for G to be 4-colorable is that χ ( G [ X ]) ≤ 3. If it is 3-colorable, we determine χ ( G [ X ]) by testing if it is bipartite or edgeless. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 14 / 36

  9. Tools Proof of 2 Assume that G is not quasi-prime. So it has a homogeneous set X that is not a clique. A necessary condition for G to be 4-colorable is that χ ( G [ X ]) ≤ 3. If it is 3-colorable, we determine χ ( G [ X ]) by testing if it is bipartite or edgeless. Contract X into a clique of size χ ( G [ X ]). Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 14 / 36

  10. Tools Proof of 2 Assume that G is not quasi-prime. So it has a homogeneous set X that is not a clique. A necessary condition for G to be 4-colorable is that χ ( G [ X ]) ≤ 3. If it is 3-colorable, we determine χ ( G [ X ]) by testing if it is bipartite or edgeless. Contract X into a clique of size χ ( G [ X ]). Repeat this until we obtain a quasi-prime graph G ′ . Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 14 / 36

  11. Tools Proof of 2 Assume that G is not quasi-prime. So it has a homogeneous set X that is not a clique. A necessary condition for G to be 4-colorable is that χ ( G [ X ]) ≤ 3. If it is 3-colorable, we determine χ ( G [ X ]) by testing if it is bipartite or edgeless. Contract X into a clique of size χ ( G [ X ]). Repeat this until we obtain a quasi-prime graph G ′ . We can show that G is 4-colorable if and only if G ′ is 4-colorable. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 14 / 36

  12. Tools Quasi-prime Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 15 / 36

  13. Tools Quasi-prime Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 15 / 36

  14. Tools Quasi-prime Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 15 / 36

  15. Tools Proof of 3 It is easy to check that K 5 and the double-wheel are not 4-colorable. Hence, if G contains a K 5 or the double-wheel it is not 4-colorable. K 5 double-wheel Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 16 / 36

  16. Tools Structural Lemma Let G be a quasi-prime bull-free graph that contains no K 5 and no double-wheel. Then at least one of the following holds: Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 17 / 36

  17. Tools Structural Lemma Let G be a quasi-prime bull-free graph that contains no K 5 and no double-wheel. Then at least one of the following holds: G contains a magnet. special graph Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 17 / 36

  18. Tools Structural Lemma Let G be a quasi-prime bull-free graph that contains no K 5 and no double-wheel. Then at least one of the following holds: G contains a magnet. G contains a gem and a special graph. special graph Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 17 / 36

  19. Tools Structural Lemma Let G be a quasi-prime bull-free graph that contains no K 5 and no double-wheel. Then at least one of the following holds: G contains a magnet. G contains a gem and a special graph. G is gem-free. special graph Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 17 / 36

  20. Tools Magnet A subgraph F of G is a magnet if every vertex of G \ F has two neighbors u , v ∈ V ( F ) such that uv ∈ E ( F ). Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 18 / 36

  21. Tools Magnet A subgraph F of G is a magnet if every vertex of G \ F has two neighbors u , v ∈ V ( F ) such that uv ∈ E ( F ). How to use it? We can fix a coloring on F and use the 2-list-coloring algorithms to try extend it to the rest of the graph in polynomial time. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 18 / 36

  22. Tools List-coloring problem Every vertex v of the graph G has a list L ( v ) of admissible colors. We want to find a proper coloring such that c ( v ) ∈ L ( v ), for every vertex v ∈ V ( G ). Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 19 / 36

  23. Tools List-coloring problem Every vertex v of the graph G has a list L ( v ) of admissible colors. We want to find a proper coloring such that c ( v ) ∈ L ( v ), for every vertex v ∈ V ( G ). 2-list coloring List-coloring problem where all lists are of size 2. The 2-list coloring problem can be solved in polynomial time. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 19 / 36

  24. Tools Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 20 / 36

  25. Tools v 1 {• , •} Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 20 / 36

  26. Tools v 1 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 20 / 36

  27. Tools v 2 {• , •} v 1 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 20 / 36

  28. Tools v 2 v 1 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 20 / 36

  29. Tools v 2 v 1 v 3 {• , •} Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 20 / 36

  30. Tools v 2 v 1 v 3 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 20 / 36

  31. Tools v 2 v 1 v 3 v 4 {• , •} Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 20 / 36

  32. Tools {• , •} v 1 {• , •} {• , •} {• , •} Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 20 / 36

  33. Tools {• , •} v 1 {• , •} {• , •} Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 20 / 36

  34. Tools {• , •} {• , •} {• , •} Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 20 / 36

  35. Tools Magnet coloring Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 21 / 36

  36. Tools Magnet coloring Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 21 / 36

  37. Tools Magnet coloring Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 21 / 36

  38. Tools Magnet coloring Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 21 / 36

  39. Tools Magnet coloring 2-list-coloring problem Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 21 / 36

  40. Tools F 0 F 1 F 2 F 3 F 4 F 5 F 6 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 22 / 36

  41. Gem-free case Theorem [ Maffray, P. ] For any fixed k , there is a polynomial algorithm that determines if a ( P 6 , bull, gem)-free graph is k -colorable and if it is, produces a k -coloring. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 23 / 36

  42. Gem-free case Theorem [ Maffray, P. ] For any fixed k , there is a polynomial algorithm that determines if a ( P 6 , bull, gem)-free graph is k -colorable and if it is, produces a k -coloring. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 23 / 36

  43. Gem-free case Perfect graph A graph G is perfect if every induced subgraph H of G satisfies χ ( H ) = ω ( H ). Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 24 / 36

  44. Gem-free case Perfect graph A graph G is perfect if every induced subgraph H of G satisfies χ ( H ) = ω ( H ). Strong Perfect Graph Theorem [ Chudnovsky, Robertson, Seymour and Thomas 2006 ] A graph is perfect if and only if it contains no C ℓ and no C ℓ for any odd ℓ ≥ 5 . Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 24 / 36

  45. Gem-free case Coloring ( P 6 , bull, gem)-free graphs Since G is P 6 -free, it contains no C ℓ with ℓ ≥ 7, and since it is gem-free, it contains no C ℓ with ℓ ≥ 7. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 25 / 36

  46. Gem-free case Coloring ( P 6 , bull, gem)-free graphs Since G is P 6 -free, it contains no C ℓ with ℓ ≥ 7, and since it is gem-free, it contains no C ℓ with ℓ ≥ 7. If G contains no C 5 , then it is bull-free perfect. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 25 / 36

  47. Gem-free case Coloring ( P 6 , bull, gem)-free graphs Since G is P 6 -free, it contains no C ℓ with ℓ ≥ 7, and since it is gem-free, it contains no C ℓ with ℓ ≥ 7. If G contains no C 5 , then it is bull-free perfect. If G contains a C 5 , we prove that it is triangle-free. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 25 / 36

  48. Gem-free case U 1 U 5 U 2 U 4 U 3 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 26 / 36

  49. Gem-free case U 1 U i is anticomplete to U i − 2 ∪ U i +2 U 5 U 2 U 4 U 3 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 26 / 36

  50. Gem-free case U 1 U i is anticomplete to U i − 2 ∪ U i +2 U i contains a vertex that is complete to U i − 1 ∪ U i +1 U 5 U 2 U 4 U 3 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 26 / 36

  51. Gem-free case U 1 U i is anticomplete to U i − 2 ∪ U i +2 U i contains a vertex that is complete to U i − 1 ∪ U i +1 Each of U 1 , . . . , U 5 is a stable set U 5 U 2 U 4 U 3 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 26 / 36

  52. Gem-free case U 1 U i is anticomplete to U i − 2 ∪ U i +2 U i contains a vertex that is complete to U i − 1 ∪ U i +1 Each of U 1 , . . . , U 5 is a stable set There is no blue triangle U 5 U 2 U 4 U 3 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 26 / 36

  53. Gem-free case U 1 U i is anticomplete to U i − 2 ∪ U i +2 U i contains a vertex that is complete to U i − 1 ∪ U i +1 Each of U 1 , . . . , U 5 is a stable set There is no blue triangle U 5 U 2 There is no red triangle U 4 U 3 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 26 / 36

  54. Gem-free case U 1 U i is anticomplete to U i − 2 ∪ U i +2 U i contains a vertex that is complete to U i − 1 ∪ U i +1 Each of U 1 , . . . , U 5 is a stable set There is no blue triangle U 5 U 2 There is no red triangle There is no green triangle U 4 U 3 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 26 / 36

  55. Gem-free case U 1 U i is anticomplete to U i − 2 ∪ U i +2 U i contains a vertex that is complete to U i − 1 ∪ U i +1 Each of U 1 , . . . , U 5 is a stable set There is no blue triangle U 5 U 2 There is no red triangle There is no green triangle Hence, there is no triangle U 4 U 3 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 26 / 36

  56. Gem-free case Theorem [ Brandstädt et al. 2006 ] For any fixed k , k -coloring a ( P 6 , K 3 )-free graph is polynomial solvable. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 27 / 36

  57. Gem-free case Theorem [ Brandstädt et al. 2006 ] For any fixed k , k -coloring a ( P 6 , K 3 )-free graph is polynomial solvable. Lemma Let G be a prime ( P 6 , bull, gem)-free graph that contains a 5-hole. Then G is triangle-free. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 27 / 36

  58. Gem case [ Maffray, P. ] There is a polynomial time algorithm that determines whether a ( P 6 , bull)-free graphs is 4-colorable, and if it is, produces a 4-coloring. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 28 / 36

  59. Gem case [ Maffray, P. ] There is a polynomial time algorithm that determines whether a ( P 6 , bull)-free graphs is 4-colorable, and if it is, produces a 4-coloring. Structural Lemma Let G be a quasi-prime bull-free graph that contains no K 5 and no double-wheel. Then at least one of the following holds: G contains a magnet. G contains a gem and a special graph. G is gem-free. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 28 / 36

  60. Gem case [ Maffray, P. ] There is a polynomial time algorithm that determines whether a ( P 6 , bull)-free graphs is 4-colorable, and if it is, produces a 4-coloring. Structural Lemma Let G be a quasi-prime bull-free graph that contains no K 5 and no double-wheel. Then at least one of the following holds: G contains a magnet. G contains a gem and a special graph. G is gem-free. Structure By the structural Lemma and further examining the structure we can partition the graph into sets as in the next drawing. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 28 / 36

  61. Gem case How to use this structure? We will see that we can use this structure to color all the graph. In fact we color a fixed number of vertices and try to extend it. Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 29 / 36

  62. Gem case Z 1 Z 0 X W V 1 V 5 V 4 V 2 V 3 Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 30 / 36

  63. Gem case Z 1 Z 0 X W V 1 V 5 V 4 V 2 V 3 induced gem Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 30 / 36

  64. Gem case Z 1 Z 0 X W V 1 V 5 V 4 V 2 V 3 induced special graph Lucas Pastor (G-SCOP) Coloring with restrictions March 4, 2016 30 / 36

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