LGI2P Research Center Coloration des graphes de reines 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  n 2 vertices and  E   n 3 edges

6 The Queen Graph 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-7 2 is 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  E. Y. Gik, 1983 : Shakhmaty i matematika, Bibliotechka Kvant, vol. 24, Nauka, Moscow. The chromatic number of the Queen Graph is equal to n if and only if n is prime with 2 and 3

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 vertices of the Queen-n 2 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-n 2 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) Find n 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-10 2

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 1 second (maximum depth of backtrack in the search tree : 5 rather than 10)   (12)  solutions 6963 seconds (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, R 2 et R 3 Search tree depth: (n/4) - 1  (21)  21

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

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

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

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

48 Geometric operator (3) n = ( 4  p ) + 1   /2 rotations: R, R 2 et R 3 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

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

52 Results synthesis  New results for the graphs counting more than ………………… 81 vertices  (10)  11 1 sec.  (20)  20 1 sec.  (12)  12 1 sec.  (21)  sec.  (14)  14 5 sec.  (22)  sec.  (15)  sec.  (24)  sec.  (16)  16 1 sec.  (28)  sec.  (18)  sec.  (32)  sec. … 1024 vertices

53 Results synthesis  Some improvements :  fixing the first stable set (and its symmetric set) according to other certificates,  branching heuristic,  …  (26)  26 1,400,000 seconds

54 The 26 letters of the alphabet are enough for coloring the 26 X 26 chessboard

55 26 = ( 3 x 2 ) + …

56 26 = ( 5 x 4 ) + ( 3 x 2 )

57  (26) = sec. (better than 1,400,000)

58 30 = ( 6 x 4 ) + ( 3 x 2 )  “Too long”  still more heuristic  Evaluate the nodes by counting the number of edges in the no colored sub graph  Partial branching : at each level of the search tree take only the best node among 10

59 X(30)=30 cpu 2965 sec.

60 Results synthesis Nous avons 13 contre exemples qui prouvent que n n’a pas besoin d’être premier avec 6 pour que  (n)  n

61 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

62 Remark An independent set with 12 vertices for n=12

63 … remains an independent set for n = Remark

64 Idea … replacing 1 square that uses 1 color by 5 2 squares that use only 5 colors...

65 Idea …  (5  12) =  (60) = 60...

66 Complementary Diagonals

67 Polynomial formula for n no multiple of 2 or 3 r(i,j)  ( 2.i + j ) modulo p  complementary diagonals

68 Characteristic of Complementary Diagonals

69 Characteristic of Complementary Diagonals

70 Coloring Extension Formula  If  (n)  n and if p is prime with 2 and 3 then  (np)  np  given a coloring C(i,j) for the chessboard n  n  r(i,j)  ( 2i + j ) modulo p R(i,j) = r(i,j) + p  C(i/p,j/p) (elementary algebra in the ring  /p  )

71 Coloring Extension Formula R(i,j) = r(i,j) + p.C(i/p,j/p) If R(i,j) = R(i’,j)  r(i,j) + p.C(i/p,j/p) = r(i,j) + p.C(i’/p,j/p)  r(i,j)=r(i’,j) and C(i/p,j/p)= C(i’/p,j/p) because r < p  i=i’ modulo p because 2 is prime with p ( recall : r(i,j)  2i+j ) and i/p=i’/p because C is a coloration and we are on the same column then i = i’

72 Coloring Extension Formula R(i,j) = r(i,j) + p.C(i/p,j/p) Exercise Prove the same kind of results for lines and diagonals

73  (60)   (5  12)  5  12  60,  (70)   (5  14)  5  14  70,  (75)   (5  15)  5  15  75,  (84)   (7  12)  7  12  84, …  (2010)   (67  30)  67  30  2010 …

74 Conclusion We know that  (n)  n for n  12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 30 and 32 We know that there is an infinity of integers n that are multiples of 2 or 3 such that  (n)  n  Is  (n)  n  n  11 ? But we don’t know, today, the value of  (27)

75 Publications Comptes Rendus Mathématiques de l’Académie des Sciences, Paris, Elsevier, Ser. I 342 (2006) p “Coloration des graphes de reines”. Journal of Heuristics (2004), 10 : p : “New Results on Queen Graph Coloring Problem”. ECAI'04, 16 th European Conference on Artificial Intelligence, Valencia, Spain, August 22-27, 2004 : “Complete and Incomplete Algorithms for the Queen Graph Coloring Problem”. Programmation en logique avec contraintes, JFPLC 2004, Hermes Science, ISBN , p : “Algorithmes complet et incomplet pour la coloration des graphes de reines”.