1
所以我試圖用遞歸和回溯來解決4x4數獨。 當我打電話SolveSmallSudoku(L); 「現在解決...」 它給了我這個「錯誤,(在SolveSmallSudoku中)矩陣索引超出範圍」 但我無法發現任何與我的矩陣L有關的索引。這似乎是我的程序沒有正確執行我的回溯部分。我認爲我的findPossibleEntries過程正常工作。它確實找到了該特定單元格的所有可能值。任何人有任何提示?解決楓4X4數獨
> 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;
請問您指出y在哪裏設置爲5? – harre
你可以通過在調試器中運行,或者在外部之前運行'print(x,y)',如果L [x,y] = 0,然後我建議你移除'',你就可以自己看到它。請注意以下常見行爲,當結束此循環時,會得到5作爲'y'的最終值:'對於y,從1到4做;結束了:y;'。 – acer
感謝您的澄清。你知道這個「通常行爲」背後的想法,for循環中的條件是「y <= 4」而不是「y <4」嗎? (請注意,我不是OP,爲什麼我沒有對發佈的代碼進行任何嚴格的調試。) – harre