python实现数独算法实例

本文实例讲述了python/' target='_blank'>python实现数独算法的方法。分享给大家供大家参考。具体如下：

# -*- coding: utf-8 -*-'''Created on 2012-10-5@author: Administrator'''from collections import defaultdictimport itertoolsa = [  [ 0, 7, 0, 0, 0, 0, 0, 0, 0], #0  [ 5, 0, 3, 0, 0, 6, 0, 0, 0], #1  [ 0, 6, 2, 0, 8, 0, 7, 0, 0], #2  #  [ 0, 0, 0, 3, 0, 2, 0, 5, 0], #3  [ 0, 0, 4, 0, 1, 0, 3, 0, 0], #4  [ 0, 2, 0, 9, 0, 5, 0, 0, 0], #5  #  [ 0, 0, 1, 0, 3, 0, 5, 9, 0], #6  [ 0, 0, 0, 4, 0, 0, 6, 0, 3], #7  [ 0, 0, 0, 0, 0, 0, 0, 2, 0], #8#  0, 1, 2, 3,|4, 5, 6,|7, 8  ]#a = [#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #0#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #1#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #2#  ##  [0, 0, 0, 0, 0, 0, 0, 0, 0], #3#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #4#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #5#  ##  [0, 0, 0, 0, 0, 0, 0, 0, 0], #6#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #7#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #8##  0, 1, 2, 3,|4, 5, 6,|7, 8#  ]exists_d = dict((((h_idx, y_idx), v) for h_idx, y in enumerate(a) for y_idx , v in enumerate(y) if v))h_exist = defaultdict(dict)v_exist = defaultdict(dict)for k, v in exists_d.items(): h_exist[k[ 0]][k[ 1]] = v v_exist[k[ 1]][k[ 0]] = vaa = list(itertools.permutations(range(1, 10), 9))h_d = {}for hk, hv in h_exist.items(): x = filter(lambda x:all((x[k] == v for k, v in hv.items())), aa) x = filter(lambda x:all((x[vk] != v for vk , vv in v_exist.items() for k, v in vv.items() if k != hk)), x)# print x h_d[hk] = xdef test(x, y): return all([y[i] not in [x_[i] for x_ in x] for i in range(len(y)) ])def test2(x): return len(set(x)) != 9s = set(range(9))sudokus = []for l0 in h_d[0 ]: for l1 in h_d[ 1]:  if not test((l0,), l1):   continue  for l2 in h_d[ 2]:   if not test((l0, l1), l2):    continue   # 1,2,3行 进行验证   if test2([l0[ 0], l0[ 1], l0[ 2]      , l1[ 0], l1[ 1], l1[ 2]      , l2[ 0], l2[ 1], l2[ 2]      ]) : continue      if test2([l0[ 3], l0[ 4], l0[ 5]      , l1[ 3], l1[ 4], l1[ 5]      , l2[ 3], l2[ 4], l2[ 5]      ]) : continue      if test2([l0[ 6], l0[ 7], l0[ 8]      , l1[ 6], l1[ 7], l1[ 8]      , l2[ 6], l2[ 7], l2[ 8]      ]) : continue      for l3 in h_d[ 3]:    if not test((l0, l1, l2), l3):     continue    for l4 in h_d[ 4]:     if not test((l0, l1, l2, l3), l4):      continue     for l5 in h_d[ 5]:      if not test((l0, l1, l2, l3, l4), l5):       continue      # 4，5，6行 进行验证      if test2([l3[ 0], l3[ 1], l3[ 2]         , l4[ 0], l4[ 1], l4[ 2]         , l5[ 0], l5[ 1], l5[ 2]         ]) : continue         if test2([l3[ 3], l3[ 4], l3[ 5]         , l4[ 3], l4[ 4], l4[ 5]         , l5[ 3], l5[ 4], l5[ 5]         ]) : continue         if test2([l3[ 6], l3[ 7], l3[ 8]         , l4[ 6], l4[ 7], l4[ 8]         , l5[ 6], l5[ 7], l5[ 8]         ]) : continue         for l6 in h_d[ 6]:       if not test((l0, l1, l2, l3, l4, l5,), l6):        continue       for l7 in h_d[ 7]:        if not test((l0, l1, l2, l3, l4, l5, l6), l7):         continue        for l8 in h_d[ 8]:         if not test((l0, l1, l2, l3, l4, l5, l6, l7), l8):          continue         # 7，8，9行 进行验证         if test2([l6[ 0], l6[ 1], l6[ 2]            , l7[0 ], l7[1 ], l7[2 ]            , l8[0 ], l8[1 ], l8[2 ]            ]) : continue            if test2([l6[ 3], l6[ 4], l6[ 5]            , l7[3 ], l7[4 ], l7[5 ]            , l8[3 ], l8[4 ], l8[5 ]            ]) : continue            if test2([l6[ 6], l6[ 7], l6[ 8]            , l7[6 ], l7[7 ], l7[8 ]            , l8[6 ], l8[7 ], l8[8 ]            ]) : continue            print l0         print l1         print l2         print l3         print l4         print l5         print l6         print l7         print l8         sudokus.append((l0, l1, l2, l3, l4, l5, l6, l7, l8))

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