1 / 75

Coloration des graphes de reines

Coloration des graphes de reines . michel.vasquez@mines-ales.fr LGI2P Ecole des Mines d’Alès. Outline. About the Queen Graph Coloring Problem Definition Conjecture ? A Complete Algorithm Reformulation of the coloring problem Efficient filtering A Geometric Based Heuristic

brit
Download Presentation

Coloration des graphes de reines

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Coloration des graphes de reines michel.vasquez@mines-ales.fr LGI2P Ecole des Mines d’Alès

  2. Outline • About the Queen Graph Coloring Problem • Definition • Conjecture ? • A Complete Algorithm • Reformulation of the coloring problem • Efficient filtering • A Geometric Based Heuristic • Geometric Operators • Results synthesis • Coloring Extension

  3. Rule for moving the queen on the chessboard • Each queen controls: • 1 column • 1 row • 2 diagonals

  4. Graph definition • 1 square of the chessboard  vertex • 2 squares controlled by the same queenedge

  5. Graph definition: from chessboard to queen graph a queen graph instance  G(V,E) with :V n2 vertices and E  n3 edges

  6. The QueenGraph Coloring Problem: definition Given a chessboard, what is the minimum number of colors required to cover it without clash between two queens of the same color ?

  7. The Queen Graph Coloring Problem: what we know The chromatic number of Queen-72is 7 :  (7)  7 (and (n)  n if n is prime with 2 and 3)

  8. Conjecture ? The chromatic number of the Queen Graph is equal to n if and only if n is prime with 2 and 3 • M. Gardner,1969 : The Unexpected Hanging and Other Mathematical Diversions, Simon and Schuster, New York.

  9. Conjecture ? The chromatic number of the Queen Graph is equal to n if and only if n is prime with 2 and 3 The chromatic number of the Queen Graph is equal to n if and only if n is prime with 2 and 3 • E. Y. Gik,1983 : Shakhmaty i matematika, Bibliotechka Kvant, vol. 24, Nauka, Moscow.

  10. Intox…

  11. Intox…

  12. Until 2003 no result are available for the queen graph chromatic number when n is greater than 9 and n is multiple of 2 or 3

  13. Outline • About the Queen Graph Coloring Problem • A Complete Algorithm • Reformulation of the coloring problem • Efficient filtering • A Geometric Based Heuristic • Geometric Operators • Results synthesis • Coloring Extension

  14. Property (1) • The n rows, the n columns and the 2 main diagonals are cliques with n verticesof the Queen-n2 graph •  (n)  n

  15. Question (1) For a given n, is  (n) equal to n ? saying it differently Is there a partition of the Queen-n2 graph in n independent sets ?

  16. Property (2) • A stable set cannot contain more than n vertices To answer yes to question (1) and cover nn squares : each independent set must contain at least n vertices

  17. Question (2) • Are there n independent sets with exactly n vertices which do not cover themselves ?

  18. General Algorithm Step 1) Enumerate the independent sets with n vertices (n queens that do not attack themselves) Step 2) Findn among them which do not intersect (solve the CSP)

  19. Avoiding many equivalent coloring permutations n squares belonging to a same clique are colored once for all:

  20. Computing IS by backtracking • Enumeration : backtracking

  21. A CSP with n variables (corresponding to a n squares) • Spreading of the independent sets for Queen-102

  22. Branching on the smallest domain variable • Non overlapping constraints propagation • The search space size is decreasing geometrically

  23. First result n = 10 : no solution 7000 seconds   (10) = 11

  24. Filtering (principle) • Consider the cliques of the graph constituted by the uncolored vertices • If such a clique contains k vertices then you need at least k colors (i.e. k independent sets) to complete the process

  25. Efficient Filtering (computationally) • Diagonals constitute cliques (and are easy to handle): • for a given diagonal there is at most one vertex that can come from a specific stable set, • at level k of the search tree, diagonals must contain less than n-k empty squares Delete all the independent sets that do not verify this condition

  26. Efficient Filtering (experimentally) • At the root of the search tree this independent set is excluded from the search space

  27. Efficient Filtering (experimentally) • Search space reduction

  28. Efficient Filtering (experimentally) • At each level : 4 more constraints

  29. First Results : complete method •  (10) no solution 1second (maximum depth of backtrack in the search tree : 5 rather than 10) •  (12) 12 454 solutions 6963seconds (exhaustive search) •  (14) 14 1 solution en 142 hours (search aborted after one week)

  30. Interest of filtering • Comparative results on n=12

  31. Outline • About the Queen Graph Coloring Problem • Definition • Intox/Conjecture ? • A Complete Algorithm • Reformulation of the coloring problem • Efficient filtering • A Geometric Based Heuristic • Geometric Operators • Results synthesis • Coloring Extension

  32. Certificate for n = 12

  33. Certificate for n = 12

  34. Certificate for n = 12

  35. Certificate for n = 12

  36. Exact but incomplete method • Assumption on the distribution of the colors on the chessboard • Enumerate several independent sets at the same time

  37. Geometric operator (1) n = 2  p symmetry H Search tree depth: n/2  (22)  22

  38. Geometric operator (1) n = 2  p symmetry H Search tree depth: n/2  (22)  22

  39. Geometric operator (1) n = 2  p symmetry H Search tree depth: n/2  (22)  22

  40. Geometric operator (2) n = 3  p  central symmetry Search tree depth: (n/2) - 1  (15)  15

  41. Geometric operator (2) n = 3  p  central symmetry Search tree depth: (n/2) - 1  (15)  15

  42. Geometric operator (2) n = 3  p  central symmetry Search tree depth: (n/2) - 1  (15)  15

  43. Geometric operator (3) n = ( 4  p ) + 1 /2 rotations: R, R2 et R3 Search tree depth: (n/4) - 1  (21)  21

  44. Geometric operator (3) n = ( 4  p ) + 1 /2 rotations: R, R2 et R3 Search tree depth: (n/4) - 1  (21)  21

  45. Geometric operator (3) n = ( 4  p ) + 1 /2 rotations: R, R2 et R3 Search tree depth: (n/4) - 1  (21)  21

  46. Geometric operator (3) n = ( 4  p ) + 1 /2 rotations: R, R2 et R3 Search tree depth: (n/4) - 1  (21)  21

  47. Geometric operator (3) n = ( 4  p ) + 1 /2 rotations: R, R2 et R3 Search tree depth: (n/4) - 1  (21)  21

  48. Geometric operator (3) n = ( 4  p ) + 1 /2 rotations: R, R2 et R3 Search tree depth: (n/4) - 1  (21)  21

  49. Geometric operator (4) n = ( 4  p )  symmetries H & V Search tree depth: (n/4)  (32)  32

  50. Geometric operator (4) n = ( 4  p )  symmetries H & V Search tree depth: (n/4)  (32)  32

More Related