sat-solver.py

# Author: Nicholas Pilkington, 2015
# Blog Post: https://nickp.svbtle.com/sudoku-satsolver

import pycosat

N = 9
M = 3

def exactly_one(variables):
    cnf = [ variables ]
    n = len(variables)

    for i in range(n):
        for j in range(i+1, n):
            v1 = variables[i]
            v2 = variables[j]
            cnf.append([-v1, -v2])

    return cnf

def transform(i, j, k):
    return i*N*N + j*N + k + 1

def inverse_transform(v):
    v, k = divmod(v-1, N)
    v, j = divmod(v, N)
    v, i = divmod(v, N)
    return i, j, k

if __name__ == '__main__':
    cnf = []

    # Cell, row and column constraints
    for i in range(N):
        for s in range(N):
            cnf = cnf + exactly_one([ transform(i, j, s) for j in range(N) ])
            cnf = cnf + exactly_one([ transform(j, i, s) for j in range(N) ])

        for j in range(N):
            cnf = cnf + exactly_one([ transform(i, j, k) for k in range(N) ])

    # Sub-matrix constraints
    for k in range(N):
        for x in range(M):
            for y in range(M):
                v = [ transform(y*M + i, x*M + j, k) for i in range(M) for j in range(M)]
                cnf = cnf + exactly_one(v)

    # See contribution from @GregoryMorse below
    cnf = { frozenset(x) for x in cnf }
    cnf = list(cnf)

    # A 16-constraint Sudoku
    constraints = [
        (0, 3, 7),
        (1, 0, 1),
        (2, 3, 4),
        (2, 4, 3),
        (2, 6, 2),
        (3, 8, 6),
        (4, 3, 5),
        (4, 5, 9),
        (5, 6, 4),
        (5, 7, 1),
        (5, 8, 8),
        (6, 4, 8),
        (6, 5, 1),
        (7, 2, 2),
        (7, 7, 5),
        (8, 1, 4),
        (8, 6, 3),

    ]

    cnf = cnf + [[transform(z[0], z[1], z[2])-1] for z in constraints]

    for solution in pycosat.itersolve(cnf):
        X = [ inverse_transform(v) for v in solution if v > 0]
        for i, cell in enumerate(sorted(X, key=lambda h: h[0] * N*N + h[1] * N)):
            print(cell[2]+1, end=" ")
            if (i+1) % N == 0: print("")

Injected at line 49:

    print(len(cnf))
    print(len({frozenset(x) for x in cnf}))

11988
10530

So there are a lot of duplicated constraints in here and makes the formula above a worst case not an optimal one - there is cell visibility reflexivity as well to take into account. The minimum number of inclusive and negation constraints should be computable though.

I can confirm 10530 is the correct number and it can be mathematically proven:
inclusive - must have each value in each mutally exclusive region 27*9=243 or 3n^2
inclusive - must have a value of 1-9 in each cell 81 or n^2
exclusive - must not have 2 values in any cell 81 * (2 choose 9)=2916 or n^2*n(n-1)/2
exclusive - pairs of cells which see each other 81*(9-1+9-1+9+1-sqrt(9)-sqrt(9))/2*9=7290 or n^2*(n-1+n-1+n+1-sqrt(n)-sqrt(n))/2*n

so we have: 3n^2+n^2+n^2*n(n-1)/2+n^2*(n-1+n-1+n+1-sqrt(n)-sqrt(n))/2*n
which simplifies to 2n^4-n^3+4n^2-n^(7/2)

indeed 9^(7/2) is 2187 and 9^3=729 and 11988-2187+729=10530.

So 2187-729=1458 extraneous constraints are in the current solution and is not ideal. I hope this proof suffices.

Its also highly questionable that:
exclusive - must not have 2 values in any cell 81 * (2 choose 9)=2916 or n^2*n(n-1)/2
conditions are necessary. As the not having more than 2 values would break the other constraints on not having the same value in cells which see each other. I have already tested and it works fine without this constraint. It only needs to know that some value 1-9 is there the inclusive rule. Which excludes 2916 rules and really reduces this dramatically to a smaller amount of boolean constraints.

In fact it is even better. You can use the 2916 rules to avoid duplicates -OR- you can exclude based on the pairs of cells which is 7290 constraints.

So in fact eliminating the 7290 constraints since the inclusive constraints already imply them is possible. The minimum constraints is thus:
10530-7290=3240 constraints.

Cool! Thanks for the observation @GregoryMorse, I've updated the gist.

Cool! Thanks for the observation @GregoryMorse, I've updated the gist.

Hi Nick, you forgot to put {} instead of [] for a set instead of a list to do the elimination.

Here is a better fix to get down to 3240 constraints:

37: cnf = cnf + exactly_one([ transform(i, j, s) for j in xrange(N) ])
38: cnf = cnf + exactly_one([ transform(j, i, s) for j in xrange(N) ])
48: cnf = cnf + exactly_one(v)

just change them from exactly_one(…) to [] e.g.

cnf = cnf + [[ transform(i, j, s) for j in xrange(N) ]]
cnf = cnf + [[ transform(j, i, s) for j in xrange(N) ]]
cnf = cnf + [v]

Now you don't need any sets and frozensets and should have 3240 minimized constraints and it will work faster and more efficiently. exactly_one is only needed for preventing duplicate cell values. Since the rest are inclusive restraints - they are not redundant. Not sure if even further conditions can be eliminated.

Ah was using Python 2 at that point, updated to Python 3.

I think Python 2 used the set([…]) notation maybe did not have the {} shorthand syntax which can be a dictionary if empty, or a set if individual values instead of key:value values.