Question

所以我试图使用递归和回溯来解决4x4数独。当我打电话给SolveSmallSudoku（L）; ＆＃34;现在解决......＆＃34; 它给了我这个＆＃34;错误，（在SolveSmallSudoku中）矩阵索引超出范围＆＃34; 但我无法发现任何与我的矩阵L，指数相关的错误。似乎我的程序没有正确地执行我的回溯部分。我认为我的findPossibleEntries程序运行正常。它确实找到了该特定单元格的所有可能值。有人有任何暗示吗？

> L := Matrix(4,4,[ [0,4,0,0],[2,0,0,3],[4,0,0,1],[0,0,3,0] ]);
 > isFull := proc(L)
    local x, y;    
        for x from 1 to 4 do
        for y from 1 to 4 do
         if L[x,y]=0 then
                 return false;
              end if;
           end do;
        end do;

        return true;
        end proc;

>findPossibleEntries := proc(L, i, j)
    local x, y, possible:=[0,0,0,0];
    local r:=1, c:=1;


#Checking possible entries in ith row
for y from 1 to 4 do
    if not L[i,y] = 0 then
       possible[L[i,y]] := 1;     
    end if;
end do;

#Checking possible entries in jth col
for x from 1 to 4 do
    if not L[x,j] = 0 then
       possible[L[x,j]] := 1;

    end if;
end do;

#Checking possible entries block by block
if i >= 1 and i <= 2 then
   r := 1;
elif i >= 3 and i <= 4 then
   r := 3;
end if;

if j >= 1 and j <= 2 then
   c := 1;
elif j >= 3 and j <= 4 then
   c := 3;
end if;

#Using for-loop to find possible entries in the block
for x in range(r, r+1) do
      for y in range(c, c+1) do

            if not L[x,y] = 0 then
               possible[L[x,y]] := 1;
            end if;
      end do;
end do;


#Now the list, possible, only holds the possible entries
for x from 1 to 4 do
    if possible[x] = 0 then
       possible[x] := x;
    else
       possible[x] := 0;
    end if;
end do;

return possible;

end proc;

>SolveSmallSudoku := proc(L)
local x, y, i:=0, j:=0, possibleVal:=[0,0,0,0];

if isFull(L) then
   print("Solved!");
   print(L);
   return;
else
   print("Solving now...");

   for x from 1 to 4 do
       for y from 1 to 4 do
          if L[x,y] = 0 then
               i:=x;
               j:=y;
               break;
          end if
       end do;
       #Breaks the outer loop as well
       if L[x,y] = 0 then
          break;
       end if

   end do;    

#Finds all the possibilities for i,j
possibleVal := findPossibleEntries(L,i,j);

#Traverses the list, possibleVal to find the correct entries and finishes the sudoku recursively 
for x from 1 to 4 do
    if not possibleVal[x] = 0 then
       L[i,j]:= possibleVal[x];
       SolveSmallSudoku(L);
    end if;
end do;
#Backtracking
 L[i,j]:= 0;

end if;


end proc;

Answer 1

摆脱，

#Breaks the outer loop as well
   if L[x,y] = 0 then
      break;
   end if

正如你原来的那样，外部检查试图为你给出的例子L访问L [1,5]。

而是用，

替换内循环中的break

x:=4; break;

这将导致外部循环也在下一次迭代时完成（这恰好发生在内部循环结束或中断之后。因此，您将获得所需的完整中断。

然后代码似乎按照您的意图工作，并为您的输入示例打印解决方案。

在枫树中解决4X4数独游戏

1 个答案: