我看到了一些数独求解器的实现,但我无法弄清楚我的代码中的问题。我有一个功能sudokusolver成为sudoku董事会,必须返回解决sudoku板。用python跟回溯的数独求解器
def sudokutest(s,i,j,z):
# z is the number
isiValid = np.logical_or((i+1<1),(i+1>9));
isjValid = np.logical_or((j+1<1),(j+1>9));
iszValid = np.logical_or((z<1),(z>9));
if s.shape!=(9,9):
raise(Exception("Sudokumatrix not valid"));
if isiValid:
raise(Exception("i not valid"));
if isjValid:
raise(Exception("j not valid"));
if iszValid:
raise(Exception("z not valid"));
if(s[i,j]!=0):
return False;
for ii in range(0,9):
if(s[ii,j]==z):
return False;
for jj in range(0,9):
if(s[i,jj]==z):
return False;
row = int(i/3) * 3;
col = int(j/3) * 3;
for ii in range(0,3):
for jj in range(0,3):
if(s[ii+row,jj+col]==z):
return False;
return True;
def possibleNums(s , i ,j):
l = [];
ind = 0;
for k in range(1,10):
if sudokutest(s,i,j,k):
l.insert(ind,k);
ind+=1;
return l;
def sudokusolver(S):
zeroFound = 0;
for i in range(0,9):
for j in range(0,9):
if(S[i,j]==0):
zeroFound=1;
break;
if(zeroFound==1):
break;
if(zeroFound==0):
return S;
x = possibleNums(S,i,j);
for k in range(len(x)):
S[i,j]=x[k];
sudokusolver(S);
S[i,j] = 0;
return S;
sudokutest和possibleNums是正确的,只是sudokusolver给予RecursionError
为什么在sudokusolver中使用S [i,j] = 0;?以及你如何用Numpy构建矩阵S [i,j]或? –
是的,有numpy,我用s [i,j] = 0来回溯,当num不是正确的时候 –
好的...然后检查我要安装numpy :-)让我们来看看。要启动你的软件,那么我只需要s = numpy.zeros(shape =(9,9)) sudokusolver(s),对吗? –