The SAT+CAS method for combinatorial search with applications to best matrices

Abstract

In this paper, we provide an overview of the SAT+CAS method that combines satisfiability checkers (SAT solvers) and computer algebra systems (CAS) to resolve combinatorial conjectures, and present new results vis-à-vis best matrices. The SAT+CAS method is a variant of the Davis–Putnam–Logemann–Loveland DPLL(T) architecture, where the T solver is replaced by a CAS. We describe how the SAT+CAS method has been previously used to resolve many open problems from graph theory, combinatorial design theory, and number theory, showing that the method has broad applications across a variety of fields. Additionally, we apply the method to construct the largest best matrices yet known and present new skew Hadamard matrices constructed from best matrices. We show the best matrix conjecture (that best matrices exist in all orders of the form r² + r + 1) which was previously known to hold for r ≤ 6 also holds for r = 7. We also confirmed the results of the exhaustive searches that have been previously completed for r ≤ 6.

Appendix

Let A, B, C, D be a set of circulant best matrices of order n = r² + r + 1 (note that all numbers of this form are odd). As described in Section ?? the rowsums of the first rows of A, B, C are 1 and the rowsum of the first row of D is ± (2r + 1) where the sign of sum(D) is positive when r ≡ 0, 1 (mod 4) and negative otherwise.

Proof

Since the matrix A is skew we have a_i + a_n−i = 0 for i≠ 0. Thus

$$ \text{sum}(A)=a_{0}+{\sum}_{i=1}^{(n-1)/2}(a_{i}+a_{n-i}) = 1 $$

and similarly for B and C. Taking the relationship AA^T + BB^T + CC^T + DD^T = 4nI and multiplying by the row vector of ones (on the left) and the column vector of ones (on the right) we obtain

$$ \text{sum}(A)^{2} + \text{sum}(B)^{2} + \text{sum}(C)^{2} + \text{sum}(D)^{2} = 4n $$

and therefore sum(D)² = 4n − 3 = (2r + 1)² and sum(D) = s(2r + 1) where s = ± 1.

Since D is symmetric and 2d_i ≡ 2 (mod 4)

$$ \text{sum}(D) = 1 + 2{\sum}_{i=1}^{(n-1)/2}d_{i} \equiv n \pmod{4} . $$

Therefore r² + r + 1 ≡ s(2r + 1) (mod 4). Since

$$ r^{2}+r+1 \equiv (-1)^{\lfloor{(r+1)/2}\rfloor} \pmod{4} \qquad\text{and}\qquad 2r+1 \equiv (-1)^{r} \pmod{4} $$

we have s = 1 when r ≡ 0, 1 (mod 4) and s = − 1 otherwise. □

As described in Section 5.4 we have that the entries of these matrices satisfy the relationship

$$ a_{k} b_{k} c_{k} d_{k} a_{2k} b_{2k} c_{2k} = -1 $$

for k≠ 0 with indices reduced mod n.

Proof

We can equivalently consider circulant best matrices to be polynomials given by the generating function of the entries of their first rows. In this formulation A, B, C, D are polynomials with ± 1 coefficients and of degree n − 1 that satisfy

$$ A(x)A(x^{-1}) + B(x)B(x^{-1}) + C(x)C(x^{-1}) + D(x)D(x^{-1}) = 4n $$

(1)

modulo the ideal generated by xⁿ − 1 (all computations will take place modulo this ideal).

Let A₊ denote the polynomial containing the terms of A with positive coefficients and let $\lvert {A_{+}}\rvert$ denote the number of terms in A₊. Then A = 2A₊ − T where $T(x):={\sum }_{i=0}^{n-1}x^{i}$. Since xⁱT = T we have $A_{+}T=\lvert {A_{+}}\rvert T$ and T² = nT.

Since A is anti-symmetric (i.e., A(x) + A(x^− 1) = 2) we have A(1) = 1 and $\lvert {A_{+}}\rvert =(T(1)+A(1))/2=(n+1)/2$. Furthermore,

$$ \begin{array}{@{}rcl@{}} A(x)A(x^{-1}) &=& 2A-A^{2} \\ &=& 2(2A_{+}-T)-(2A_{+}-T)^{2} \\ &=& 4A_{+} - 4A_{+}^{2} - (2T - 4\lvert{A_{+}}\rvert T + nT) \\ &=& 4A_{+} - 4A_{+}^{2} + nT \end{array} $$

(2)

and similarly for B and C.

Since D is symmetric (i.e., D(x) = D(x^− 1)) we have

$$ D(x)D(x^{-1}) = (2D_{+}-T)^{2} = 4D_{+}^{2} + (n-4\lvert{D_{+}}\rvert)T . $$

(3)

By the symmetry of D we have $D(x)=1+D^{\prime }(x)+D^{\prime }(x^{-1})$ where $D^{\prime }(x):={\sum }_{i=1}^{(n-1)/2}d_{i}x^{i}$. Then $\lvert {D_{+}}\rvert =1+2\lvert {{D}_{+}^{\prime }}\rvert$ and thus $\lvert {D_{+}}\rvert$ is odd.

Equating (1)–(3) and dividing by four we have

$$ A_{+}-A_{+}^{2} + B_{+}-B_{+}^{2} + C_{+}-C_{+}^{2} + D_{+}^{2} + (n-\lvert{D_{+}}\rvert)T = n . $$

(4)

Since $A_{+}={\sum }_{a_{i}=1}x^{i}$ we have $A_{+}^{2}\equiv {\sum }_{a_{i}=1}x^{2i}\pmod {2}$ and (4) reduces to

$$ \sum\limits_{a_{i}=1}(x^{2i}+x^{i}) + \sum\limits_{b_{i}=1}(x^{2i}+x^{i}) + \sum\limits_{c_{i}=1}(x^{2i}+x^{i}) + \sum\limits_{d_{i}=1}x^{2i} \equiv 1 \pmod{2} $$

since both n and $\lvert {D_{+}}\rvert$ are odd.

Since n is odd the congruence i ≡ 2y (mod n) has exactly one solution 0 ≤ y < n for each 0 ≤ i < n. Denoting this solution by i/2 we have

$$ \sum\limits_{a_{i/2}=1}x^{i} + \sum\limits_{a_{i}=1}x^{i} + \sum\limits_{b_{i/2}=1}x^{i} + \sum\limits_{b_{i}=1}x^{i} + \sum\limits_{c_{i/2}=1}x^{i} + \sum\limits_{c_{i}=1}x^{i} + \sum\limits_{d_{i}=1}x^{2i} \equiv 1 \pmod{2} . $$

In other words, we have that the number of entries in {a_i/2,a_i,b_i/2,b_i,c_i/2,c_i,d_i/2} that are positive is 1 (mod 2) for i = 0 and 0 (mod 2) for i≠ 0. Letting k = i/2 for i≠ 0 this means a_ka_2kb_kb_2kc_kc_2kd_k = − 1 as required. □

One new skew Hadamard matrix that we constructed was given in Fig. 3 and the other two new skew Hadamard matrices are given in Fig. 4. The first rows of the best matrices used to construct these skew Hadamard matrices are given here:

+--++------+-++++-+-+--+--++-+--++-++-+-+----+-++++++--++ +-+-++++-++--++-+--+---+-+-+-+-+-+-+++-++-+--++--+----+-+ ++-++-+--+---++--+++----+---+-+++-++++---++--+++-++-+--+- +---+------++-----+++----++++++++----+++-----++------+--- +--++------+-++++-+-+--+--++-+--++-++-+-+----+-++++++--++ +-+-++++-++--++-+--+---+-+-+-+-+-+-+++-++-+--++--+----+-+ ++-++-+--+---++--+++----+---+-+++-++++---++--+++-++-+--+- +---+------++-----+++----++++++++----+++-----++------+--- ++----+------++--++-+--+---+-+-+++-++-+--++--++++++-++++- +--+-+-+--+-++----++--++++++-+------++--++++--+-++-+-+-++ ++-+--++++++---++-++----+-+-+-+-+-++++--+--+++------++-+- +---++++--+-+---+--++------+--+------++--+---+-+--++++---

Fig. 4

Two new skew Hadamard matrices of order 4 ⋅ 57 constructed using best matrices