java - 生成一个接近中心的值较小的矩阵

Question

我希望生成一个整数矩阵，其中整数在边缘处较大，如下所示:

[23] [23] [23] [23] [23] 
[15] [15] [15] [15] [15] 
[10] [10] [10] [10] [10] 
[10] [7] [7] [7] [10] 
[10] [4] [4] [4] [10] 
[10] [3] [3] [3] [10] 
[10] [3] [2] [3] [10] 
[10] [3] [1] [3] [10] 
[10] [3] [1] [3] [10] 
[10] [3] [1] [3] [10] 
[10] [3] [2] [3] [10] 
[10] [3] [3] [3] [10] 
[10] [4] [4] [4] [10] 
[10] [7] [7] [7] [10] 
[10] [10] [10] [10] [10] 
[15] [15] [15] [15] [15] 
[23] [23] [23] [23] [23] `

到目前为止一切顺利!但是，我遇到的问题是保持顶部和底部的最大边缘值与左右边缘相同，如下所示:

[23] [23] [23] [23] [23] 
[23] [15] [15] [15] [23] 
[23] [10] [10] [10] [23] 
[23] [7] [7] [7] [23] 
[23] [7] [4] [7] [23] 
[23] [7] [3] [7] [23] 
[23] [7] [2] [7] [23] 
[23] [7] [2] [7] [23] 
[23] [7] [2] [7] [23] 
[23] [7] [2] [7] [23] 
[23] [7] [2] [7] [23] 
[23] [7] [3] [7] [23] 
[23] [7] [4] [7] [23] 
[23] [7] [7] [7] [23] 
[23] [10] [10] [10] [23] 
[23] [15] [15] [15] [23] 
[23] [23] [23] [23] [23] 

注意边都等于相同的最大数字 23。让它们都等于相同的数字是我遇到问题的地方(这是由于不同的长度和宽度)。以下是我生成矩阵的方式:

public void generateChanceMatrix(){
    double ratio = numPointsX/numPointsY; //The difference in length of x and y

    for (int i = 0; i < numPointsX; i++){ //for each point
        for (int j = 0; j < numPointsY; j++){

            double max = Math.abs((numPointsX /2) - i); // these are responsible for generating larger numbers as you approach the edge
            double max2 = Math.abs((numPointsY /2) - j)* ratio; //this ratio is how i try to even out the difference in length of x and y

            if (max2 > max){ //see if the y axis is larger than the x, choose the larger
                max = max2;
            }

            chanceMatrix[i][j] = (int) Math.pow(2.2, max/2); //make it exponential instead of linear
        }
    }
}

我如何调整它以确保所有边缘的值接近相同的最大值？对于那些好奇的人，我正在使用它来生成一个 map 边界被封锁的游戏世界，见下文:

提前致谢!

Answer 1

我会根据到边缘的相对距离使用一些插值。 IE。相对距离 t 在矩阵中间为零，在边缘为 1。然后你可以根据这个值做简单的插值:

double relativeDiffX = 2 * abs(i - numPointsX / 2) / (numPointsX - 1);
double relativeDiffY = 2 * abs(j - numPointsY / 2) / (numPointsY - 1);
double t = max(relativeDiffX, relativeDiffY);
//now any interpolation
chanceMatrix[i][j] = (int) Math.pow(2.2, t); //this will produce values between 1 and 2.2