1. Finding Conjugate Orthogonal Diagonal Latin Squares Using Finite Model Generators Hantao Zhang Jian Zhang 2012-10-26

2. Outline Introduction (Latin squares, SAT, finite models) Concepts and Definitions -- Conjugate Orthogonal Diagonal Latin Squares Solution of an Open Case -- formalization in first-order logic Conclusion

3. Introduction

4. Latin square problems Latin squares (quasigroups) 1 2 3 0 1 2 3 1 2 2 0 1 2 3 1 1 2 0 Challenging problems in mathematics –Counting problem –Existence of Latin squares with certain properties, e.g., Mutually orthogonal Latin squares (MOLS)

5. Orthogonal Latin squares A pair of latin squares A=(a ij ) and B=(b ij ) are orthogonal ( A ┴ B) iff the ordered pairs (a ij,b ij ) are distinct for all i and j A ┴ A T The pairs (a ij,a t ij ) are (0,0) (2,3) (3,1) (1,2) (3,2) (1,1) (0,3) (2,0) (1,3) (3,0) (2,2) (0,1) (2,1) (0,2) (1,0) (3,3)

6. Solving Latin square problems with computers Computer-aided research in combinatorics –existence of small Latin squares – crucial ! Clement Lam and other mathematicians AI researchers (using CSP / SAT / …) –Jian Zhang (1990/1991) –M. Fujita, J. Slaney, M. Stickel (IJCAI-1993, 1995) –Hantao Zhang, W. McCune, … –C. Gomes, O. Dubois and G. Dequen, …

7. SAT Given a formula in the propositional logic, decide whether it is satisfiable or not. (p ∨ ┐q) ∧ (q ∨ ┐p) satisfiable: p = TRUE, q = TRUE. Many problems can be transformed into SAT. Tools (SAT solvers): SATO (1997), zChaff (2001), minisat (2005), …

8. Finite model generation/searching Given a formula in the first-order logic, decide whether it is satisfiable or not in a fixed finite domain ([0, 1, …, n-1]). forall x,y,z: y≠z  f(x,y) ≠f(x,z) forall x,y,z: x≠z  f(x,y) ≠f(z,y) Tools (finite model generators): Finder (~1994), SEM (1995), Mace4 (~2003), …

9. SEM input: 3. y = z | f(x,y) != f(x,z). x = z | f(x,y) != f(z,y). SEM output: f | 0 1 2 | 0 | 0 1 2 1 | 1 2 0 2 | 2 0 1

10. Mace4 Input for qg5.8 assign(domain_size, 8). % clear(negprop). set(verbose). clauses(theory). x * z != y * z | x = y. x * y != x * z | y = z. x * x = x. (((y * x) * y) * y) = x. end_of_list.

11. Mace4 Output for qg5.8 interpretation( 8, [number=1, seconds=0], [ function(*(_,_), [ 0, 2, 6, 5, 1, 4, 7, 3, 7, 1, 5, 6, 2, 3, 0, 4, 3, 6, 2, 1, 5, 7, 4, 0, 1, 7, 4, 3, 0, 6, 2, 5, 6, 3, 0, 7, 4, 1, 5, 2, 2, 0, 3, 4, 7, 5, 1, 6, 5, 4, 7, 0, 3, 2, 6, 1, 4, 5, 1, 2, 6, 0, 3, 7 ]) ]).

12. Concepts and Definitions

13. transversal in a Latin square: a set of positions, one per row and one per column, among which the elements {0, 1, …, n-1} occur exactly once each. diagonal Latin square -- main diagonal is a transversal; back diagonal is also a transversal. The following is a DLS: 0 2 3 1 3 1 0 2 1 3 2 0 2 0 1 3

14. Conjugates Given a Latin square defined by f(x,y), we may obtain its conjugates in the following way: f123(x,y) = z; f132(x,z) = y; f213(y,x) = z; f231(y,z) = x; f312(z,x) = y; f321(z,y) = x.

15. LS (2;1;3)-cjg (3;2;1)-cjg 1 4 2 3 1 2 4 3 1 3 2 4 2 3 1 4 4 3 1 2 2 4 1 3 4 1 3 2 2 1 3 4 4 2 3 1 3 2 4 1 3 4 2 1 3 1 4 2 (2;3;1)-cjg (1;3;2)-cjg (3;1;2)-cjg 1 2 4 3 1 3 4 2 1 3 2 4 3 4 2 1 3 1 2 4 3 1 4 2 2 1 3 4 2 4 3 1 4 2 3 1 4 3 1 2 4 2 1 3 2 4 1 3

16. Latin square with a hole Given a symbol set S and a subset H of S. If a Latin square L satisfies the following equations forall x,y,z: y≠z  f(x,y) ≠f(x,z) forall x,y,z: y≠z  f(y,x) ≠f(z,x) other than (x,y), (x,z), (y,x), (z,x) in H 2 we say L is a Latin square with the hole H. Example: S = { 0, 1, …, 7 } H = { 3, 4} 0 7 5 6 2 3 4 1 7 1 6 5 0 4 2 3 5 6 2 7 1 0 3 4 2 0 1 6 7 5 1 2 0 7 5 6 4 3 7 0 6 5 1 2 3 5 4 2 7 1 6 0 6 4 3 1 5 2 0 7

17. An Open Problem Given integers v and n, let ICODLS(v, n) denote a pair of orthogonal Latin squares (A, B) of order v with a hole of size n, where B is the (3,2,1)-conjugate of A; both A and B are diagonal. We knew the existence of ICODLS(v, n) for all possible feasible values, except possibly (n, v) = (11, 3). 0 7 5 6 2 3 4 1 7 1 6 5 0 4 2 3 5 6 2 7 1 0 3 4 2 0 1 6 7 5 1 2 0 7 5 6 4 3 7 0 6 5 1 2 3 5 4 2 7 1 6 0 6 4 3 1 5 2 0 7 Example: ICODLS(8, 2) A 0 3 4 5 1 2 7 6 4 1 3 7 2 6 5 0 3 4 2 6 0 7 1 5 6 5 7 0 2 1 5 7 6 1 0 2 2 6 0 1 7 5 4 3 7 2 1 0 5 3 6 4 1 0 5 2 6 4 3 7 B

18. Solution of an Open Case

19. First Order Logic Formulas Latin Square Constraints: A(x,y) = A(x,z) → y = z A(x,y) = A(z,y) → x = z A┴B: A(x 1,y 1 )=A(x 2,y 2 ) ∧ B(x 1,y 1 )=B(x 2,y 2 ) → x 1 =x 2 ∧ y 1 =y 2 Latin Square Constraints with a hole: (A(x,y) = A(x,z) → y = z) \/ h(x,y) \/ h(x,z) (A(x,y) = A(z,y) → x = z) \/ h(x,y) \/ h(z,y) A┴B: (A(x 1,y 1 )=A(x 2,y 2 ) ∧ B(x 1,y 1 )=B(x 2,y 2 ) → x 1 =x 2 ∧ y 1 =y 2 ) \/ h(x 1,y 1 ) \/ h(x 2,y 2 )

20. Finding the ICODLS with SAT solvers Generating propositional formulas: Introduce boolean variables V ijk (i,j,k ∈ [0, n-1]) V ijk = 1 iff A(i,j) = k V ijk = 0 iff A(i,j) ≠ k Using SAT solvers (minisat, Glucose)  not completed (> two weeks)

21. Mace4 input set(arithmetic). assign(domain_size, 11). assign(selection_order, 2). op(400, infix, [@,g,f]). formulas(theory). % define hole h(4,4). h(4,5). h(4,6). h(5,5). h(5,6). h(6,6). -h(y,x) | h(x,y). -h(x,y) | x = 4 | x = 5 | x = 6. -h(y,x) | x = 4 | x = 5 | x = 6. x @ y != z | -h(x,z). x @ y != z | -h(y,z).

22. Mace4 input (cont.) % define main and back diagonals x = y | (x @ x) != (y @ y) | h(x,x) | h(y,y). x = y | (x @ SUB(10, x)) != (y @ SUB(10, y)) | h(x,SUB(10,x)) | h(y, SUB(10, y)). x = y | z @ x != x | z @ y != y | h(z,x) | h(z,y). x = y | z @ SUB(10, x) != x | z @ SUB(10, y) != y | h(x,SUB(10,x)) | h(y, SUB(10, y)). % quasigroup axioms (equational) x @ (x g y) = y | h(x,y). x g (x @ y) = y | h(x,y). (x f y) @ y = x | h(x,y). (x @ y) f y = x | h(x,y). % Orthogonal to its (3,2,1)-conjugate (u @ y) @ y != (u @ w) @ w | y = w | h(u,y) | h(u,w) | h(w,y). end_of_list.

23. Mace4 Output ======== DOMAIN SIZE 11============ ======== MODEL =================== interpretation( 11, [number=1, seconds=1154566], [ function(@(_,_), [ 3, 9, 2, 5, 7, 8, 0,10, 6, 4, 1, 0, 2, 9, 6, 8, 7, 1, 3, 4,10, 5, 8, 4, 7, 3,10, 1, 2, 5, 9, 6, 0, 4, 6, 5,10, 9, 0, 3, 2, 8, 1, 7, 9, 0,10, 8, 0, 0, 0, 1, 7, 3, 2, 1,10, 0, 2, 0, 0, 0, 9, 3, 7, 8, 2, 7, 1, 9, 0, 0, 0, 0,10, 8, 3, 5, 3, 4, 0, 1,10, 7, 8, 2, 9, 6, 6, 5, 3, 7, 0, 9, 8, 4, 1, 2,10, 10, 8, 6, 1, 2, 3, 9, 7, 5, 0, 4, 7, 1, 8, 4, 3, 2,10, 6, 0, 5, 9 ]), …

24. Conclusion  We solved an open case: existence of (3, 2, 1)-ICODLS(11, 3).  We presented the formalization of the problem in FOL.  Other cases remain open: e.g., the existence of (3, 2, 1)-CODLS(10), …  Challenging benchmarks for constraint solving, SAT solving and finite model generation.

25. Thank you!