Connect 4 检查获胜算法 [英] Connect 4 check for a win algorithm

问题描述

我知道有很多关于 connect 4 check for a win 的问题.问题是大多数其他算法使我的程序出现运行时错误，因为它们试图访问我的数组之外的索引.我的算法是这样的:

I know there is a lot of of questions regarding connect 4 check for a win. The issue is that most of other algorithms make my program have runtime errors, because they try to access an index outside of my array. My algorithm is like this:

private int checkWin(int[][] gridTable,int rowNum,int colNum, int maxRow, int maxCol) 
{
//  For checking whether any win or lose condition is reached. Returns 1 if win or lose is reached. else returns 0
//  gridTable[][] is the game matrix(can be any number of rows and columns between 4 and 40)
//  colNum is the column number where the last token was placed
//  rowNum is the row number where the last token was placed
//  maxRow is the number of rows in my grid
//  maxCol is the number of columns in my grid

int player = gridTable[rowNum][colNum]; //player ID
int count=0;

// Horizontal check
for (int i=0;i<maxCol;i++)
{
    if (gridTable[rowNum][i]==player)
        count++;
    else
        count=0;

    if (count>=4)
        return 1;
}
//Vertical check
for (int i=0;i<maxRow;i++)
{
    if (gridTable[i][colNum]==player)
        count++;
    else
        count=0;

    if (count>=4)
        return 1;
} 
count=0;
// 4 in a row diagonally
for(int i=colNum+1,j=rowNum+1;i<maxRow && j<maxCol;i++,j++) 
{ 
    if(gridTable[j][i]!=player)
    {
        count=1;
        break;        
    }
    count++;
}
// 4 in a row diagonally
for(int i=colNum-1,j=rowNum-1;i>=0 && j>=0;i--,j--) 
{ 
    if(gridTable[j][i]!=player)
    {
        count=1;
        break;        
    }
    count++;
}
// 4 in a row diagonally
for(int i=colNum+1,j=rowNum-1;i<maxRow && j>=0;i++,j--) 
{ 
    if(gridTable[j][i]!=player)
    {
        count=1;
        break;        
    }
    count++;
}

for(int i=colNum-1,j=rowNum+1;i>=0 && j<maxCol;i--,j++) 
{ // 4 in a row diagonally
    if(gridTable[j][i]!=player)
    {
        count=1;
        break;        
    }
    count++;
}

if(count>=4)
    return 1;

return 0;
}

count 是检查获胜的变量，如果计数等于或大于 4 意味着它们应该是同一玩家的 4 个或更多连续令牌.

count is the variable that checks for a win if count is equal or more than 4 means they should be 4 or more consecutive tokens of the same player.

问题:有时该方法在没有按顺序排列 4 个令牌的情况下检查获胜，而其他时候在顺序排列 4 个令牌时不检查获胜.

THE PROBLEM: sometimes the method checks for a win without being 4 tokens in order and other times does not check for a win when 4 tokens are in order.

推荐答案

看起来您的代码对于水平和垂直情况都是正确的.棘手的部分是对角线情况.

Looks like your code is correct for the horizontal and vertical cases. The tricky part is the diagonal case.

让我们尝试一张图片:

对于绿线，您的起始行位置为 0 ... maxRow - 4.该列将为 0 ...startingRow -

For the green lines, your starting row position is 0 ... maxRow - 4. The column would be 0 ... startingRow -

伪代码:

// top-left to bottom-right - green diagonals
for( rowStart = 0; rowStart < rowMax - 4; rowStart++){
    count = 0;
    int row, col;
    for( row = rowStart, col = 0; row < rowMax && col < colMax; row++, col++ ){
        if(gridTable[row][col] == player){
            count++;
            if(count >= 4) return 1;
        }
        else {
            count = 0;
        }
    }
}

// top-left to bottom-right - red diagonals
for( colStart = 1; colStart < colMax - 4; colStart++){
    count = 0;
    int row, col;
    for( row = 0, col = colStart; row < rowMax && col < colMax; row++, col++ ){
        if(gridTable[row][col] == player){
            count++;
            if(count >= 4) return 1;
        }
        else {
            count = 0;
        }
    }
}

您可以对相反方向的对角线(从左下角到右上角)执行类似的操作.

You could do something similar for diagonals going the other way (from bottom-left to top-right).

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