递归：如何尝试整数 1 到 9 的不同组合，以及（部分）反向序列以在出错的情况下重新开始？
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【中文标题】递归：如何尝试整数 1 到 9 的不同组合，以及（部分）反向序列以在出错的情况下重新开始？【英文标题】：Recursion: how to try different combinations of integers 1 to 9, and (partially) reverse sequence to start over in case of error? 【发布时间】：2015-12-24 20:44:34 【问题描述】：语言： 爪哇
目标： 一般：解决数独游戏
具体：制作一个递归方法solve()：
检查数字是否与行、列或框中的其他数字冲突 如果不是这种情况，则在给定的空白空间中填充 [1-9] 之间的整数并移动到下一个空白空间 （部分或全部）如果空白空间不能被 [1-9] 之间的整数填充而不会发生冲突，则会反转进度。然后再试一次，直到所有空格都被填满（并解决了数独问题）。问题： 循环尝试填充整数 n，但总是首先尝试最小的数字。 如果我使用递归，整数总是相同的。
问题： 1.如何让代码填写1到9之间的数字。
如何使用递归来部分或完全擦除进度并尝试不同的数字。
（额外）到目前为止，我已经构建了部分解决数独的代码（直到无法填充空方格），但现在它甚至没有填充第一个方格，即使它之前已经完成了。这不是我的主要问题，但如果有人注意到一个/这个问题，如果有人指出，我将不胜感激。
审核： 我正在阅读有关递归的课程资料，但找不到正确的信息。
免责声明： 方法 printMatrix 之外的所有 println 命令都用于测试
这是有问题的方法：
// prints all solutions that are extensions of current grid
// leaves grid in original state
void solve()
//local variables
int[] currentSquare;
int currentRow;
int currentColumn;
boolean checkConflict;
currentSquare = new int[2];
//saftey limit for testing
int limit;
limit = 0;
while(findEmptySquare() != null)
currentSquare = findEmptySquare().clone();
currentRow = currentSquare[0];
currentColumn = currentSquare[1];
//System.out.println(" column c:"+currentColumn+" row r:"+currentRow); //column c5 r 3
if(currentSquare[0] != -1)
for(int n = 1; n <= ms; n++)
checkConflict = givesConflict(currentRow, currentColumn, n);
if(checkConflict == false)
grid[currentRow][currentColumn] = n;
//end if
//end for
//end if
else
System.out.println("solve: findEmptySquare was -1");
break;
//Safety limit
limit++;
if (limit > 20)
System.out.println("break");
break;
//end if
//end while
这是找到空方块的方法：
// finds the next empty square (in "reading order")
// returns array of first row then column coordinate
// if there is no empty square, returns .... (FILL IN!)
int[] findEmptySquare()
int[] rowcol;
int[] noMoreCells;
rowcol = new int[2];
noMoreCells = new int[2];
noMoreCells[0] = -1;
noMoreCells[1] = -1;
for(int r = 0; r < ms; r++)
for(int c = 0; c < ms; c++)
if(grid[r][c] == 0)
if(r != rempty || c != cempty) //check that the location of empty cell is not the same as last one
rempty = r;
cempty = c;
rowcol[0] = r; // 0 for row
rowcol[1] = c; // 1 for column
//System.out.println(" column c:"+rowcol[1]+" row r:"+rowcol[0]); //column c5 r 3
return rowcol;
//end if
else
System.out.println("findEmptySquare: found same empty square twice");
return noMoreCells;
//end else
//end if
//end for
//end for
System.out.println("findEmptySquare: no more empty cells");
return null; //no more empty cells
如有必要，整个代码（缩进很乱，因为我不得不在***上手动添加空格）：
// Alain Lifmann. Date: 26/9/2015
// Description of program: runs sudoku game
import java.util.*;
class Sudoku
int ms = 9; //maze Size
int rempty; //keeping track of empty squares, row nr. (array)
int cempty; //keeping track of empty squares, column nr. (array)
int SIZE = 9; // size of the grid
int DMAX = 9; // maximal digit to be filled in
int BOXSIZE = 3; // size of the boxes
int[][] grid; // the puzzle grid; 0 represents empty
// a challenge-sudoku from the web
int[][] somesudoku = new int[][]
0, 6, 0, 0, 0, 1, 0, 9, 4 , //original
// 0, 0, 0, 0, 0, 1, 0, 9, 4 , //to get more solutions
3, 0, 0, 0, 0, 7, 1, 0, 0 ,
0, 0, 0, 0, 9, 0, 0, 0, 0 ,
7, 0, 6, 5, 0, 0, 2, 0, 9 ,
0, 3, 0, 0, 2, 0, 0, 6, 0 ,
9, 0, 2, 0, 0, 6, 3, 0, 1 ,
0, 0, 0, 0, 5, 0, 0, 0, 0 ,
0, 0, 7, 3, 0, 0, 0, 0, 2 ,
4, 1, 0, 7, 0, 0, 0, 8, 0 ,
;
// a test-sudoku from oncourse
int[][] testsudoku = new int[][]
1, 2, 3, 4, 5, 6, 7, 8, 9 ,
4, 5, 6, 7, 8, 9, 1, 2, 3 ,
7, 8, 9, 1, 2, 3, 4, 5, 6 ,
2, 1, 4, 3, 6, 5, 8, 9, 7 ,
3, 6, 5, 8, 9, 7, 2, 1, 4 ,
8, 9, 7, 2, 1, 4, 3, 6, 5 ,
5, 3, 1, 6, 4, 2, 9, 7, 8 ,
6, 4, 2, 9, 7, 8, 5, 3, 1 ,
9, 7, 8, 5, 3, 1, 6, 4, 2 ,
;
// a test-sudoku modified to be incomplete
int[][] testsudoku2 = new int[][]
0, 0, 3, 0, 5, 6, 7, 8, 0 ,
0, 5, 0, 7, 0, 0, 1, 0, 3 ,
7, 0, 0, 1, 0, 3, 4, 5, 6 ,
2, 1, 4, 3, 6, 5, 8, 0, 7 ,
3, 0, 5, 8, 0, 7, 0, 1, 0 ,
0, 9, 7, 0, 1, 4, 3, 0, 5 ,
0, 0, 0, 6, 4, 2, 9, 7, 8 ,
0, 4, 2, 9, 7, 8, 0, 0, 1 ,
0, 0, 0, 5, 3, 1, 0, 4, 0 ,
;
int solutionnr = 0; //solution counter
// ----------------- conflict calculation --------------------
// is there a conflict when we fill in d at position r,c?
boolean givesConflict(int r, int c, int d)
if(rowConflict(r, d) == true)
return true;
//end if
if(colConflict(c, d) == true)
return true;
//end if
if(boxConflict(r, c, d) == true)
return true;
//end if
return false;
//end givesConflict
boolean rowConflict(int r, int d)
for(int i = 0; i < ms; i++)
if(d == grid[r][i])
//System.out.println("rowconflict r:"+r+" d:"+d);
return true;
//end if
//end for
return false; //no conflict
boolean colConflict(int c, int d)
for(int i = 0; i < ms; i++)
if(d == grid[i][c])
//System.out.println("column conflict c:"+c+" d:"+d);
return true;
//end if
//end for
return false; //no conflict
boolean boxConflict(int rr, int cc, int d) //test 5,3,1
int rs; // Box-row start point
int cs; // Box-column start point
rs = rr - rr%3;
cs = cc - cc%3;
//System.out.println("box start is row "+rs+" , column "+cs);
for(int r = rs; r < rs + 3; r++ )
for(int c = cs; c < cs + 3; c++)
if(d == grid[r][c])
//System.out.println("r:"+r+" c:"+c);
return true;
//end if
//end for
//end for
return false; //no conflict
// --------- solving ----------
// finds the next empty square (in "reading order")
// returns array of first row then column coordinate
// if there is no empty square, returns .... (FILL IN!)
int[] findEmptySquare()
int[] rowcol;
int[] noMoreCells;
rowcol = new int[2];
noMoreCells = new int[2];
noMoreCells[0] = -1;
noMoreCells[1] = -1;
for(int r = 0; r < ms; r++)
for(int c = 0; c < ms; c++)
if(grid[r][c] == 0)
if(r != rempty || c != cempty) //check that the location of empty cell is not the same as last one
rempty = r;
cempty = c;
rowcol[0] = r; // 0 for row
rowcol[1] = c; // 1 for column
//System.out.println(" column c:"+rowcol[1]+" row r:"+rowcol[0]); //column c5 r 3
return rowcol;
//end if
else
System.out.println("findEmptySquare: found same empty square twice");
return noMoreCells;
//end else
//end if
//end for
//end for
System.out.println("findEmptySquare: no more empty cells");
return null; //no more empty cells
// prints all solutions that are extensions of current
// leaves grid in original state
void solve()
//local variables
int[] currentSquare;
int currentRow;
int currentColumn;
boolean checkConflict;
currentSquare = new int[2];
//saftey limit for testing
int limit;
limit = 0;
while(findEmptySquare() != null)
currentSquare = findEmptySquare().clone();
currentRow = currentSquare[0];
currentColumn = currentSquare[1];
//System.out.println(" column c:"+currentColumn+" row r:"+currentRow); //column c5 r 3
if(currentSquare[0] != -1)
for(int n = 1; n <= ms; n++)
checkConflict = givesConflict(currentRow, currentColumn, n);
if(checkConflict == false)
grid[currentRow][currentColumn] = n;
//end if
//end for
//end if
else
System.out.println("solve: findEmptySquare was -1");
break;
//Safety limit
limit++;
if (limit > 20)
System.out.println("break");
break;
//end if
//end while
// ------------------------- misc -------------------------
// print the grid, 0s are printed as spaces
void printMatrix()
int ms; //matrix Size
ms = 9;
//layer indication symbols
String end;
String mid;
String cut;
end = "+";
mid = "-";
cut = "";
for(int i = 0; i < 2*ms-1; i++)
cut = cut + mid;
//end for
System.out.println(end+cut+end);
for(int i = 0; i < ms; i++)
if( i % 3 == 0 && i != 0)
System.out.println(mid+cut+mid);
//end if
for(int j = 0; j < ms; j++)
if( j % 3 == 0)
System.out.print("|");
//end if
else
System.out.print(" ");
//end else
if(grid[i][j] != 0)
System.out.print(grid[i][j]);
//end if
else
System.out.print(" ");
//end else
//end for j
System.out.print("|");
System.out.println();
//end for i
System.out.println(end+cut+end);
//end method
// reads the initial grid from stdin
void read()
Scanner sc = new Scanner(System.in);
for (int r=0; r<SIZE; r++)
if (r % BOXSIZE == 0)
sc.nextLine(); // read away top border of box
for (int c=0; c < SIZE; c++)
if (c % BOXSIZE == 0)
sc.next(); // read away left border of box
String square = sc.next();
if (".".equals(square))
grid[r][c] = 0; // empty sqaure
else
grid[r][c] = Integer.parseInt(square);
//System.out.print(grid[r][c]);
sc.nextLine(); // read away right border
sc.nextLine(); // read away bottom border
// --------------- where it all starts --------------------
void solveIt()
grid = somesudoku.clone(); // set used grid (before doing anything else)
printMatrix();
solve();
printMatrix();
/* test material
printMatrix();
//System.out.println(givesconflict(1,1,3));
System.out.println(rowConflict(0,1));
System.out.println(colConflict(0,1));
System.out.println(boxConflict(0,0,1));
findEmptySquare();
*/
// end solveIt
public static void main(String[] args)
new Sudoku().solveIt();
【问题讨论】：
您正在寻找的概念称为backtracking（实际上，通常使用递归实现）。 看到这个问题。这解释了一种实现回溯的算法***.com/questions/6924216/… 为了确定，您想通过添加数字来强制数独，直到它满了，或者如果添加任何数字会使数独变得不可能/错误，则回滚？无论如何，在 Aasmud Eldhuset 的评论中，有一个回溯链接，其中有一个回溯数独求解算法的链接：en.wikipedia.org/wiki/Sudoku_solving_algorithms（第一个） 【参考方案1】：这个问题可以通过回溯来解决。您必须递归所有可能的选项，但是当您遇到错误的问题时，请从该路径返回。 您可以从此链接找到有关代码的帮助 http://learnfreecodes.blogspot.in/2015/11/sudoku-solver-using-backtracking-in-java.html
【讨论】：
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