Python implements a Sudoku algorithm instance

In this paper, we describe the method of Python's implementation of Sudoku algorithm. Share to everyone for your reference. Specific as follows:

#-*-Coding:utf-8-*-"Created on 2012-10-5@author:administrator" from collections import Defaultdictimport Itertool SA = [[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], #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_id x), V) for H_idx, y in Enumerate (a) for Y_idx, V in enumerate (y) if v) h_exist = defaultdict (dict) v_exist = Defaultdict (d ICT) 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, ten), 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 al L ([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 # A-line verification if Test2 ([l0[0], l0[1], l0[2], l1[0], l1[1], l1[2], l2[0], l2[1], L       2[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 forL3 in h_d[3]: "If not" test ((L0, L1, L2), L3): Continue for L4 on 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 row for validation i         f 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]]): cont  Inue 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 te             St ((l0, L1, L2, L3, L4, L5, L6, L7), L8): Continue # 7,8,9 line to verify 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 PR int L7 Print L8 sudokus.append ((l0, L1, L2, L3, L4, L5, L6, L7, L8))

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