我在 SciLab 中编写了一个解决数独问题的程序。 但它只能解决总是有一个可能值为 1 的正方形的数独。 就像 brainbashers.com 上非常简单和简单的数独游戏。
中等数独总是会到达一个点，即它们没有具有 1 个可能值的正方形。 我如何修改我的代码来解决这些更难的数独问题？
///////////////////////////////////////////////////////////////////////////
////////////////////////// Check Sudoku ///////////////////////////////
///////////////////////////////////////////////////////////////////////////
function r=OneToNine(V) // function checks if the given vector V contains 1 to 9
r = %T // this works
u = %F
index = 1
while r == %T & index < 10
for i=1 : length(V)
if V(i)==index then
u = %T
end
end
index=index+1
if u == %F then r = %F
else u = %F
end
end
if length(V) > 9 then r = %F
end
endfunction
function y=check(M) // Checks if the given matrix M is a solved sudoku
y = %T // this works too
if size(M,1)<>9 | size(M,2)<>9 then // if it has more or less than 9 rows and columns
y = %F // we return false
end
for i=1 : size(M,1) // if not all rows have 1-9 we return false
if OneToNine(M(i,:)) == %F then
y = %F
end
end
endfunction
function P=PossibilitiesPosition(board, x, y)
// this one works
// we fill the vector possibilites with 9 zeros
// 0 means empty, 1 means it already has a value, so we don't need to change it
possibilities = [] // a vector that stores the possible values for position(x,y)
for t=1 : 9 // sudoku has 9 values
possibilities(t)=0
end
// Check row f the value (x,y) for possibilities
// we fill the possibilities further by puttin '1' where the value is not possible
for i=1 : 9 // sudoku has 9 values
if board(x,i) > 0 then
possibilities(board(x,i))=1
end
end
// Check column of the value (x,y) for possibilities
// we fill the possibilities further by puttin '1' where the value is not possible
for j=1 : 9 // sudoku has 9 values
if board(j, y) > 0 then
possibilities(board(j, y))=1
end
end
// Check the 3x3 matrix of the value (x,y) for possibilities
// first we see which 3x3 matrix we need
k=0
m=0
if x >= 1 & x <=3 then
k=1
else if x >= 4 & x <= 6 then
k = 4
else k = 7
end
end
if y >= 1 & y <=3 then
m=1
else if y >= 4 & y <= 6 then
m = 4
else m = 7
end
end
// then we fill the possibilities further by puttin '1' where the value is not possible
for i=k : k+2
for j=m : m+2
if board(i,j) > 0 then
possibilities(board(i,j))=1
end
end
end
P = possibilities
// we want to see the real values of the possibilities. not just 1 and 0
for i=1 : 9 // sudoku has 9 values
if P(i)==0 then
P(i) = i
else P(i) = 0
end
end
endfunction
function [x,y]=firstEmptyValue(board) // Checks the first empty square of the sudoku
R=%T // and returns the position (x,y)
for i=1 : 9
for j=1 : 9
if board(i,j) == 0 & R = %T then
x=i
y=j
R=%F
end
end
end
endfunction
function A=numberOfPossibilities(V) // this checks the number of possible values for a position
A=0 // so basically it returns the number of elements different from 0 in the vector V
for i=1 : 9
if V(i)>0 then
A=A+1
end
end
endfunction
function u=getUniquePossibility(M,x,y) // this returns the first possible value for that square
pos = [] // in function fillInValue we only use it
pos = PossibilitiesPosition(M,x,y) // when we know that this square (x,y) has only one possible value
for n=1 : 9
if pos(n)>0 then
u=pos(n)
end
end
endfunction
///////////////////////////////////////////////////////////////////////////
////////////////////////// Solve Sudoku ///////////////////////////////
///////////////////////////////////////////////////////////////////////////
function G=fillInValue(M) // fills in a square that has only 1 possibile value
x=0
y=0
pos = []
for i=1 : 9
for j=1 : 9
if M(i,j)==0 then
if numberOfPossibilities(PossibilitiesPosition(M,i,j)) == 1 then
x=i
y=j
break
end
end
end
if x>0 then
break
end
end
M(x,y)=getUniquePossibility(M,x,y)
G=M
endfunction
function H=solve(M) // repeats the fillInValue until it is a fully solved sudoku
P=[]
P=M
if check(M)=%F then
P=fillInValue(M)
H=solve(P)
else
H=M
end
endfunction
//////////////////////////////////////////////////////////////////////////////
所以它解决了第一个
// Very easy and easy sudokus from brainbashers.com get solved completely
// Very Easy sudoku from brainbashers.com
M = [0 2 0 0 0 0 0 4 0
7 0 4 0 0 0 8 0 2
0 5 8 4 0 7 1 3 0
0 0 1 2 8 4 9 0 0
0 0 0 7 0 5 0 0 0
0 0 7 9 3 6 5 0 0
0 8 9 5 0 2 4 6 0
4 0 2 0 0 0 3 0 9
0 1 0 0 0 0 0 8 0]
但它并没有解决这个媒介:
M2= [0 0 6 8 7 1 2 0 0
0 0 0 0 0 0 0 0 0
5 0 1 3 0 9 7 0 8
1 0 7 0 0 0 6 0 9
2 0 0 0 0 0 0 0 7
9 0 3 0 0 0 8 0 1
3 0 5 9 0 7 4 0 2
0 0 0 0 0 0 0 0 0
0 0 2 4 3 5 1 0 0]
尝试解决中等数独时的错误代码:
-->solve(M2)
!--error 21
Invalid index.
at line 14 of function PossibilitiesPosition called by :
at line 3 of function getUniquePossibility called by :
at line 20 of function fillInValue called by :
at line 182 of function solve called by :
at line 183 of function solve called by :
at line 183 of function solve called by :
at line 183 of function solve called by :
at line 183 of function solve called by :
solve(M2)
at line 208 of exec file called by :
_SCILAB-6548660277741359031.sce', 1
while executing a callback
最佳答案
好吧，编写数独求解器的最简单方法之一(不是最有效的)可能是使用所有可能的选项递归地求解每个单元格(这可能类似于“回溯”算法)，直到得到完整答案找到了。
另一种选择(我会说它更好)是遍历所有方 block 解决所有“简单”方 block 并将可能的答案存储在其他方 block 中，然后重复(现在你已经解决了更多)，重复这个过程直到解出数独或不能直接解出更多的方 block 。然后你可以尝试剩下的用暴力或回溯(也许数独已经解出一半或更多，所以它可能相对有效)
无论如何，通过快速搜索我找到了this Wikipedia page其中一些数独求解算法用伪代码示例进行了解释，希望这些对您有用
关于algorithm - 数独求解器 - Scilab，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/30288165/