Trivial backtracking implementation to solve http://puzzling.stackexchange.com/questions/25098/filling-an-11-by-11-square

11x11.py

from collections import Counter


size = 11
tab = [None] * size*size

def split(t):
    return [t[size*i:size*i+size] for i in range(size)]

def splitV(t):
    return [t[i::size] for i in range(size)]

not_none = lambda x: x is not None

calls_counter = 0
inconsistencies_counter = 0

def solve(t):
    global calls_counter
    global inconsistencies_counter
    calls_counter += 1
    cnt = Counter(sum(filter(not_none, row))
        for row in split(t)
            if len(list(filter(not_none, row))) == size)
    cnt += Counter(sum(filter(not_none, col))
        for col in splitV(t)
            if len(list(filter(not_none, col))) == size)
    if len(cnt) == size * 2:
        # OK
        return t
    if len(cnt) > 0 and cnt.most_common(1)[0][1] > 1:
        # Inconsistency
        inconsistencies_counter += 1
        return None
    for val in [-1, 0, 1]:
        tt = t[:]
        if None not in t:
            print t
        tt[t.index(None)] = val
        res = solve(tt)
        if res:
            return res
    return None

print(solve(tab))
print(calls_counter, inconsistencies_counter)