joana Posted October 25, 2008 Report Share Posted October 25, 2008 Foi feito para uma aula ... Uso de função recursiva m=[ [0,0,0,0,0,6,0,8,0], [0,8,0,0,5,0,6,7,0], [0,0,2,0,0,3,0,5,1], [1,0,0,2,7,0,0,0,0], [0,0,7,6,0,8,2,0,0], [0,0,0,0,1,4,0,0,7], [5,2,0,4,0,0,7,0,0], [0,4,3,0,8,0,0,6,0], [0,9,0,1,0,0,0,0,0] ] def doit(): i,j=pz() if i==-1:return 1 for n in range(1,10): if linok(i,j,n) and colok(i,j,n) and quadok(i,j,n): m[i][j]=n if doit():return 1 m[i][j]=0 return 0 def linok(i,p,n): for j in range(9): if m[i][j]==n:return 0 return 1 def colok(p,j,n): for i in range(9): if m[i][j]==n: return 0 return 1 def quadok(i,j,n): li=(i/3)*3 lj=(j/3)*3 for x in range(li,li+3): for y in range(lj,lj+3): if m[x][y]==n: return 0 return 1 def pz(): for i in range(9): for j in range(9): if m[i][j]==0:return i,j return -1,0 for p in m: print p doit() print for p in m: print p Acho que dá para todas as "grelhas" ... Link to comment Share on other sites More sharing options...
pedrotuga Posted October 25, 2008 Report Share Posted October 25, 2008 Yay! Está bem mais elegante que o meu, tenhoque afixar o meu código tambem, só ainda não o fiz porque há lá partes que são embaraçosas 🙂 Acabei de experimentar e funcionou sem problemas. Boa onda. Link to comment Share on other sites More sharing options...
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