【算法】数独解题——用python代码
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【算法】数独解题——用python代码
小編覺得挺不錯的,現在分享給大家,幫大家做個參考.
最近陪小孩做數獨游戲,一時興起,寫了一段代碼在解題,僅供大家娛樂!
思路:
1、獲取題目,存放于二維列表a中,無數字的地方為0
2、準備一個三維數組b,存放每個格子中可能的數字列表,剩下一個可能數字時填寫到a,并將b中此處為[0]
每填入一個數字到a,就將b中不可能的數字移除。
3、無限循環,直到所有a中沒有0,或者填寫完成了81個數字。
好的,廢話少說,直接上代碼!
def main():a = [[2,0,9,0,0,0,0,6,0],[0,1,0,4,6,3,0,2,0],[0,0,4,0,0,5,7,0,0],[0,0,0,0,0,8,0,3,9],[0,0,0,5,7,9,0,0,0],[9,8,0,3,0,0,0,0,0],[0,0,2,6,0,0,4,0,0],[0,7,6,9,8,4,0,5,0],[0,9,0,0,0,0,8,0,6]]# a = []# for i in range(0, 9):# a0 = input("please input requestion line"+str(i))# a[i].appand(a0.split(' '))for iii in range(0, 9):print(a[iii])b = []for i in range(0,9):b.append([])for j in range(0,9):b[i].append([1,2,3,4,5,6,7,8,9])print(b[i])#初始化num = 0for i in range(0,9):for j in range(0,9):if a[i][j]!=0:num += 1a,b = movenot(i,j,a,b)print(num)#解題while num<81:for i in range(0, 9):for j in range(0, 9):if len(b[i][j]) == 1 and b[i][j] != [0] and a[i][j] == 0:a[i][j] = b[i][j][0]b[i][j] = [0]num += 1print(num)a, b = movenot(i, j, a, b)print("解題完成!")for iii in range(0,9):print(a[iii])def movenot(i,j,a,b):#移除不可能的數字for jj in range(0,9):if a[i][j] in b[i][jj]:b[i][jj].remove(a[i][j])if a[i][j] in b[jj][j]:b[jj][j].remove(a[i][j])if i<3:m = 0elif i<6:m = 3else:m=6if j<3:n = 0elif j<6:n = 3else:n=6if a[i][j] in b[0+m][0+n]:b[0+m][0+n].remove(a[i][j])if a[i][j] in b[0+m][1+n]:b[0+m][1+n].remove(a[i][j])if a[i][j] in b[0+m][2+n]:b[0+m][2+n].remove(a[i][j])if a[i][j] in b[1+m][0+n]:b[1+m][0+n].remove(a[i][j])if a[i][j] in b[1+m][1+n]:b[1+m][1+n].remove(a[i][j])if a[i][j] in b[1+m][2+n]:b[1+m][2+n].remove(a[i][j])if a[i][j] in b[2+m][0+n]:b[2+m][0+n].remove(a[i][j])if a[i][j] in b[2+m][1+n]:b[2+m][1+n].remove(a[i][j])if a[i][j] in b[2+m][2+n]:b[2+m][2+n].remove(a[i][j])return a,bif __name__ == '__main__':main()以上!請各位大佬指正!謝謝!
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