数独回溯算法

Posted 2023-02-22

【中文标题】数独回溯算法

【英文标题】：Sodoku Backtracking Algorithm

【发布时间】：2014-03-02 07:36:57

【问题描述】：

我编写了代码，使用回溯法从 Java 中的空白网格生成数独。但是我在运行程序时没有得到任何输出

 public class SodokuGenerator 

int[][] puzzle=new int[9][9];
int num_givens=0;

public static int get_random_value(int high, int low)

    //Returns a random value between the given maximum and minimum values(both     inclusive)
    Random r=new Random();
    return (r.nextInt(high+1-low) + low);


     public boolean check_column(int x, int y, int curr_value)

    for (int i=0;i<9;i++)
    
        if (this.puzzle[i][y]==curr_value)
        
            return false;
        
    
    return true;


public boolean check_row(int x, int y, int curr_value)

    for (int j=0;j<9;j++)
    
        if (this.puzzle[x][j]==curr_value)
        
            return false;
        
    
    return true;


public boolean check_block(int x, int y, int curr_value)

    int block_row_start=0, block_row_end=0,block_col_start=0, block_col_end=0;

    if (x==0 || x==3 || x==6)
    
        block_row_start=x;
        block_row_end=x+3-1;
    
    else if (x==2 || x==5 || x==8)//At the end of a block
    
        block_row_start=x-3+1; //both bounds are inclusive
        block_row_end=x;
    
    else if (x==1 || x==4 || x==7) //Neither multiples of 2 nor 3.
    
        block_row_start=x-1;
        block_row_end=x+1;
    

    if (y==0 || y==3 || y==6)
    
        block_col_start=y;
        block_col_end=y+3-1;
    
    else if (y==2 || y==5 || y==8)//At the end of a block
    
        block_col_start=y-3+1; //both bounds are inclusive
        block_col_end=y;
    
    else if (y==1 || y==4 || y==7) //Neither multiples of 2 nor 3.
    
        block_col_start=y-1;
        block_col_end=y+1;
    
    //Established the bounds of the block based on the current position
    //System.out.println("block_row_start="+block_row_start);
    //System.out.println("block_row_end= "+block_row_end);
    for (int i=block_row_start;i<=block_row_end;i++)
    
        for (int j=block_col_start;j<=block_col_end;j++)
        
            //System.out.println("i="+i);
            //System.out.println("j="+j);
            if (this.puzzle[i][j]==curr_value)
            
                return false;
            
        
    
    return true;




public void create_puzzle()

    int curr_value=0;
    int index=0;
    int[] possible_values=1,2,3,4,5,6,7,8,9;

    this.puzzle[0][0]=get_random_value(9,1);
    int x=0,y=1; //Holds the coordinates of the current position in the puzzle


    while (x<=8 && y<=8)
    
        this.puzzle[x][y]=0;
        curr_value= get_random_value(9,1);
        if (this.check_block(x,y, curr_value) &&  this.check_row(x,y,curr_value) 
        && this.check_column(x,y,curr_value))
        
            this.puzzle[x][y]=curr_value;
        
        else //If there is a conflict with another element
        
            index=-1;
            //Checks for a conflict using all possible values
            do //Using a do-while loop prevents a repeated computation
            
                index++;
                curr_value=possible_values[index];
            
            while (index<8 && (this.check_block(x,y, curr_value)==false || 
                    this.check_row(x,y,curr_value)==false 
                    || this.check_column(x,y,curr_value)==false));


            if (index==8)//This means that no possible solution was found
            
                //BACKTRACKING
                if (y==0 && x!=0)
                
                    y=8;
                    x--;
                
                else
                
                    y--;
                

                continue;
            
            else //If a possible solution was found
            
                this.puzzle[x][y]=curr_value;
            
        

        //Advancing the current position coordinates to the next position
        if (y==8)
        
            y=0;
            x++;
        
        else
        
            y++;
        
        

我确实在这个论坛上看过类似的问题，但它们并没有真正帮助我。调试告诉我有一个无限循环在起作用。谁能指出我正确的方向？我会非常感激。谢谢。

【问题讨论】：

将其余代码添加到问题中可能会有所帮助。特别是关于谜题。

@Asher 您花了多少时间才找到解决方案并对其进行编码:)。只是好奇而已

@Dexters 老实说不会太长.. 调试需要更长的时间:)

【参考方案1】：

而 (x

那个循环是无限的。尝试在循环中将值打印到屏幕上。

System.out.println("X: "+x+" "+y);

我没有深入检查您的代码...但我怀疑问题出在此处：

        if (index==8)//This means that no possible solution was found
        
            //BACKTRACKING
            if (y==0 && x!=0)
            
                y=8;
                x--;
            
            else
            
                y--;
            

            continue;
        

和

if (y==8) //This if statement only runs once. Y always gets subtracted back to 7 in the code above.
    
         y=0;
        x++;
    
    else
    
        y++;
    
    

一旦 index 达到 8，该函数将进入 else，将 y 减一。然后 y 加一。这个过程会反弹回来第四次，导致 y 值在无限循环中变为 -1 和 +1。

【讨论】：

你是对的。不过，我很难想出一个合适的替代方案。