问题描述
我很难解决创建可以解决超级数据的算法的问题。我的数独游戏是21x9，由三个9x9 Sudokus组成（第1-9行是第一行，7-15秒和12-21行）。我试着写一个算法，但它没有正确解决数独。我认为在函数中检查是否可以插入数字（）是错误的。也许有人可以注意到我犯了什么错误。
我尝试了什么：
这是我试图做的事情：
I struggle with a problem of creating an algorithm which can solve a Super-sudoku. My Sudoku is 21x9 and consists of three 9x9 Sudokus (rows 1-9 it's first, 7-15 second and 12-21 thrid). I tried to write an algorithm but it doesn't solve sudoku properly. I think something is wrong in function that checks whether it's possible to insert the number(). Maybe someone can notice where I made a mistake.
What I have tried:
Here's what I tried to do:
#include<stdio.h>
#include<algorithm>
bool checkCol(int workArray[21][9], short begin, short end, short checkingValue, short y){
for (int i = begin; i < end; ++i){
if (workArray[i][y] == checkingValue){
return false;
}
}
return true;
}
bool notValid(int workArray[21][9], short value, short pointX, short pointY){
short rows = (pointX / 3) * 3;
short columns = (pointY / 3) * 3;
for (int i = 0; i < 9; ++i){
if (workArray[rows + (i % 3)][columns + (i / 3)] == value){
return false;
}
}
for (int i = 0; i < 9; ++i){
if (workArray[pointX][i] == value){
return false;
}
}
if (pointX >= 0 && pointX <= 5){
if (!checkCol(workArray, 0, 9, value, pointY) && !checkCol(workArray, 6, 15, value, pointY)/* &&
!checkCol(workArray, 12, 21, value, pointY)*/){
return false;
}
}
else if (pointX >= 6 && pointX <= 8){
if (!checkCol(workArray, 0, 9, value, pointY) && !checkCol(workArray, 6, 15, value, pointY) /*&&
!checkCol(workArray, 12, 21, value, pointY)*/){
return false;
}
}
else if (pointX >= 9 && pointX <= 11){
if (!checkCol(workArray, 6, 15, value, pointY) && !checkCol(workArray, 0, 9, value, pointY) &&
!checkCol(workArray, 12, 21, value, pointY)){
return false;
}
}
else if (pointX >= 12 && pointX <= 14){
if (!checkCol(workArray, 12, 21, value, pointY) && !checkCol(workArray, 6, 15, value, pointY)/* &&
!checkCol(workArray, 0, 9, value, pointY)*/){
return false;
}
}
else if (pointX >= 15 && pointX <= 20){
if (/*!checkCol(workArray, 6, 15, value, pointY) &&*/ !checkCol(workArray, 12, 21, value, pointY)){
return false;
}
}
return true;
}
bool findEmptySpace(int workArray[21][9], short &pointX, short &pointY);
bool solveTheSudoku(int workArray[21][9]){
short pointX, pointY;
if (!findEmptySpace(workArray, pointX, pointY)){
return true;
}
for (int i=1; i<=9; ++i){
if (notValid(workArray, i, pointX, pointY)){
workArray[pointX][pointY] = i;
if (solveTheSudoku(workArray)){
return true;
}
workArray[pointX][pointY] = 0;
}
}
return false;
}
bool findEmptySpace(int workArray[21][9], short &pointX, short &pointY){
for (pointX = 20; pointX >= 0; --pointX){
for (pointY = 8; pointY >= 0; --pointY){
if (workArray[pointX][pointY] == 0){
return true;
}
}
}
return false;
}
void printSudoku(int sudoku[21][9]){
for (int i = 0; i < 21; ++i){
for (int j = 0; j < 9; ++j){
printf("%d ",sudoku[i][j]);
}printf("\n");
}
}
int main(){
int sudoku[21][9] = {{0, 0, 0, 0, 0, 0, 0, 0, 9},
{4, 7, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 5, 6, 2, 0, 0, 3},
{0, 6, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 4, 0, 0, 3, 0, 6, 0},
{0, 5, 9, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 2, 0, 0, 0, 0, 0},
{6, 0, 0, 4, 0, 0, 0, 0, 0},
{0, 4, 8, 0, 0, 0, 6, 0, 0},
{0, 0, 4, 0, 0, 0, 0, 0, 0},
{0, 2, 0, 0, 0, 0, 1, 0, 0},
{0, 9, 1, 0, 0, 4, 0, 0, 5},
{0, 0, 0, 0, 0, 0, 0, 0, 0},
{4, 0, 0, 6, 0, 0, 0, 0, 5},
{5, 0, 0, 0, 0, 0, 8, 6, 0},
{0, 0, 0, 0, 8, 7, 0, 2, 1},
{0, 1, 0, 0, 0, 3, 0, 8, 4},
{0, 2, 0, 0, 0, 0, 3, 0, 0},
{8, 0, 0, 3, 0, 0, 6, 4, 2},
{0, 4, 0, 7, 0, 0, 5, 0, 8},
{3, 0, 0, 4, 2, 0, 0, 7, 0}};
//int arrayTmp[21][9];
//std::copy(sudoku, sudoku+21, arrayTmp);
if (solveTheSudoku(sudoku)){
printSudoku(sudoku);
}
else{
printf("No solution find\n");
printSudoku(sudoku);
}
return 0;
}推荐答案
Input Expected output Actual output
1 2 1
2 4 4
3 6 9
4 8 16然后很明显问题出在将它加倍的位 - 它不会将自身加到自身上，或者将它乘以2，它会将它自身相乘并返回输入的平方。
所以，你可以查看代码和很明显，它在某处：
Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:
private int Double(int value)
{
return value * value;
}
一旦你知道可能出现的问题，就开始使用teh调试器找出原因。在你的行上设一个断点：
Once you have an idea what might be going wrong, start using teh debugger to find out why. Put a breakpoint on your line:
myaverage.DisplayAverage();
并运行你的应用程序。在执行代码之前，请考虑代码中的每一行应该做什么，并将其与使用Step over按钮依次执行每一行时实际执行的操作进行比较。它符合您的期望吗？如果是这样，请转到下一行。
如果没有，为什么不呢？它有何不同？
这是一项非常值得开发的技能，因为它可以帮助你在现实世界和发展中。和所有技能一样，它只能通过使用来改善！
所以试试看：在面对之前，学会在这样一个相对简单的例子中使用调试器在他离开公司的那天，别人写了200,000行代码中的一个错误！
and run your app. Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?
This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!
So give it a try: learn to use the debugger on a relatively trivial example like this before you are faced with a bug in a 200,000 line lump of code someone else wrote the day he left the company!
这些较小的Sudokus重叠：第1-9行是第一个Sudoku，但第7-9行也是第二个Sudoku的一部分（整个是7-15行）。类比是在第三个数独的情况下，最后三行（12-15）是第三个数独（第12-21行）的一部分。
These smaller Sudokus overlap: rows 1-9 are first Sudoku, but rows 7-9 are also part of second Sudoku (the whole is 7-15 rows). The analogy is in case of the third Sudoku, the three last rows (12-15) are parts of the third Sudoku (rows 12-21).
据我所知，你有 3数独在0-8行， 6-14， 12-20。
如果你不明白你的代码在做什么或为什么它做了什么，答案是调试器。
使用调试器查看代码正在执行的操作。只需设置断点并查看代码执行情况，调试器允许您逐行执行第1行并在执行时检查变量，这是一个令人难以置信的学习工具。
[]
[]
[]
调试器在这里显示你的代码在做什么你的任务是与它应该做的事情进行比较。
调试器中没有魔法，它没有发现错误，它只是帮助你。当代码没有达到预期效果时，你就接近一个错误。
[更新]
As I understand, you have 3 Sudoku that are at rows 0-8, 6-14, 12-20.
When you don't understand what your code is doing or why it does what it does, the answer is debugger.
Use the debugger to see what your code is doing. Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute, it is an incredible learning tool.
Debugger - Wikipedia, the free encyclopedia[^]
Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't find bugs, it just help you to. When the code don't do what is expected, you are close to a bug.
[Update]
它正在使用回溯。从底部开始，它试图解决数独。如果有一个空位我检查数字是否合适并输入它，如果不合适我会返回上一个输入的数字。
It's using backtracking. Starting from a bottom it tries to solve the Sudoku. If there is a empty place I check whether the number is fit and enter it, if not fit I return to the previous entered number.
这是一次暴力攻击，它是高度的效率低下。人类喜欢求解算法更有效率。
This is a brut force attack, it is highly inefficient. Human like solving algorithm is way more efficient.
#include<stdio.h>
#include<algorithm>
bool checkCol(int workArray[21][9], short begin, short end, short checkingValue, short y){
for (int i = begin; i < end; ++i){
if (workArray[i][y] == checkingValue){
return false;
}
}
return true;
}
bool notValid(int workArray[21][9], short value, short pointX, short pointY){
short rows = (pointX / 3) * 3;
short columns = (pointY / 3) * 3;
for (int i = 0; i < 9; ++i){
if (workArray[rows + (i % 3)][columns + (i / 3)] == value){
return false;
}
}
for (int i = 0; i < 9; ++i){
if (workArray[pointX][i] == value){
return false;
}
}
if (pointX >= 0 && pointX <= 5){
if (!checkCol(workArray, 0, 9, value, pointY) && !checkCol(workArray, 6, 15, value, pointY)/* &&
!checkCol(workArray, 12, 21, value, pointY)*/){
return false;
}
}
else if (pointX >= 6 && pointX <= 8){
if (!checkCol(workArray, 0, 9, value, pointY) && !checkCol(workArray, 6, 15, value, pointY) /*&&
!checkCol(workArray, 12, 21, value, pointY)*/){
return false;
}
}
else if (pointX >= 9 && pointX <= 11){
if (!checkCol(workArray, 6, 15, value, pointY) && !checkCol(workArray, 0, 9, value, pointY) &&
!checkCol(workArray, 12, 21, value, pointY)){
return false;
}
}
else if (pointX >= 12 && pointX <= 14){
if (!checkCol(workArray, 12, 21, value, pointY) && !checkCol(workArray, 6, 15, value, pointY)/* &&
!checkCol(workArray, 0, 9, value, pointY)*/){
return false;
}
}
else if (pointX >= 15 && pointX <= 20){
if (/*!checkCol(workArray, 6, 15, value, pointY) &&*/ !checkCol(workArray, 12, 21, value, pointY)){
return false;
}
}
return true;
}
bool findEmptySpace(int workArray[21][9], short &pointX, short &pointY);
bool solveTheSudoku(int workArray[21][9]){
short pointX, pointY;
if (!findEmptySpace(workArray, pointX, pointY)){
return true;
}
for (int i=1; i<=9; ++i){
if (notValid(workArray, i, pointX, pointY)){
workArray[pointX][pointY] = i;
if (solveTheSudoku(workArray)){
return true;
}
workArray[pointX][pointY] = 0;
}
}
return false;
}
bool findEmptySpace(int workArray[21][9], short &pointX, short &pointY){
for (pointX = 20; pointX >= 0; --pointX){
for (pointY = 8; pointY >= 0; --pointY){
if (workArray[pointX][pointY] == 0){
return true;
}
}
}
return false;
}
void printSudoku(int sudoku[21][9]){
for (int i = 0; i < 21; ++i){
for (int j = 0; j < 9; ++j){
printf("%d ",sudoku[i][j]);
}printf("\n");
}
}
int main(){
int sudoku[21][9] = {{0, 0, 0, 0, 0, 0, 0, 0, 9},
{4, 7, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 5, 6, 2, 0, 0, 3},
{0, 6, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 4, 0, 0, 3, 0, 6, 0},
{0, 5, 9, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 2, 0, 0, 0, 0, 0},
{6, 0, 0, 4, 0, 0, 0, 0, 0},
{0, 4, 8, 0, 0, 0, 6, 0, 0},
{0, 0, 4, 0, 0, 0, 0, 0, 0},
{0, 2, 0, 0, 0, 0, 1, 0, 0},
{0, 9, 1, 0, 0, 4, 0, 0, 5},
{0, 0, 0, 0, 0, 0, 0, 0, 0},
{4, 0, 0, 6, 0, 0, 0, 0, 5},
{5, 0, 0, 0, 0, 0, 8, 6, 0},
{0, 0, 0, 0, 8, 7, 0, 2, 1},
{0, 1, 0, 0, 0, 3, 0, 8, 4},
{0, 2, 0, 0, 0, 0, 3, 0, 0},
{8, 0, 0, 3, 0, 0, 6, 4, 2},
{0, 4, 0, 7, 0, 0, 5, 0, 8},
{3, 0, 0, 4, 2, 0, 0, 7, 0}};
//int arrayTmp[21][9];
//std::copy(sudoku, sudoku+21, arrayTmp);
if (solveTheSudoku(sudoku)){
printSudoku(sudoku);
}
else{
printf("No solution find\n");
printSudoku(sudoku);
}
return 0;
}
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