Finding Conjugate Orthogonal Diagonal Latin Squares Using Finite Model Generators Hantao Zhang Jian Zhang

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

Introduction

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

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)

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, …

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), …

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), …

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

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.

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 ]) ]).

Concepts and Definitions

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:

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.

LS (2;1;3)-cjg (3;2;1)-cjg (2;3;1)-cjg (1;3;2)-cjg (3;1;2)-cjg

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}

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) Example: ICODLS(8, 2) A B

Solution of an Open Case

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 )

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)

Mace4 input set(arithmetic). assign(domain_size, 11). assign(selection_order, 2). op(400, infix, 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. y != z | -h(x,z). y != z | -h(y,z).

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

Mace4 Output ======== DOMAIN SIZE 11============ ======== MODEL =================== interpretation( 11, [number=1, seconds= ], [ [ 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 ]), …

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.

Thank you!