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java - 离散余弦变换实现

问题描述

我正在尝试实现此网页的离散余弦变换部分下给出的二维余弦变换方程：https ://en.wikipedia.org/wiki/JPEG 

    public void transform() {
        int u;
        int v;
        double alphau;
        double alphav;

        double[][] test = {
            {52, 55, 61, 66, 70, 61, 64, 73},   
            {63, 59, 55, 90, 109, 85, 69, 72},
            {62, 59, 68, 113, 144, 104, 66, 73},
            {63, 58, 71, 122, 154, 106, 70, 69}, 
            {67, 61, 68, 104, 126, 88, 68, 70},  
            {79, 65, 60, 70, 77, 68, 58, 75},
            {85, 71, 64, 59, 55, 61, 65, 83},
            {87, 79, 69, 68, 65, 76, 78, 94}
        };

        double summation = 0;
        double[][] coefficients = new double[8][8];

        for(int x = 0; x < test.length; x++) 
        {

            for(int y = 0; y < test[x].length; y++)
            {
                //Inner discrete transform.
                u = x % 8;
                v = y % 8;

                double cosu = Math.cos(Math.toRadians((((2 * x) + 1) * u * Math.PI) / 16));
                double cosv = Math.cos(Math.toRadians((((2 * y) + 1) * v * Math.PI) / 16));

                summation = ((test[y][x]) * cosu * cosv) + summation;
                System.out.print(test[y][x] - 128 + ", ");
            }
            System.out.println("");
        }

        System.out.println("");

        for(int i = 0; i < 8; i++)
        {
            for(int j = 0; j < 8; j++) 
            {
                //Outer discrete transform.
                alphau = 1.0;
                alphav = 1.0;

                if(i == 0) {
                    alphav = 1 / Math.sqrt(2);
                }
                if(j == 0) {
                    alphau = 1 / Math.sqrt(2);
                }

                coefficients[i][j] = .25 * alphau * alphav * summation;

                System.out.print(coefficients[i][j] + ", ");
            }
            System.out.println("");
        }

    }

可以在此处找到我尝试实现的等式的快速链接：1 第一个 for 循环旨在计算测试数组的余弦的双重求和，而第二个循环旨在将其与各自的 alpha 值相乘。我的系数输出返回：

589, 833, 833, 833, 833, 833, 833, 833, 833, 1179, 1179, 1179, 1179, 1179, 1179, 1179, 833, 1179, 1179, 1179, 1179, 1179, 83,9 1179, 1179, 1179, 1179, 1179, 1179, 1179, 833, 1179, 1179, 1179, 1179, 1179, 1179, 1179, 833, 1179, 1179, 1179, 1179, 1179, 83, 1179, 8179, 1179, 831 1179、1179、1179、1179、1179、1179、833、1179、1179、1179、1179、1179、1179、1179、

虽然可以在这里找到预期的输出：2

标签： javaalgorithmmathmultidimensional-arrayequation

解决方案

以下程序与在线 DCT 计算器相匹配 https://asecuritysite.com/comms/dct2 请检查并告诉我这是否适合您

 void dctTransform(double matrix[][])
    {
        int i, j, k, l;

        // dct will store the discrete cosine transform
        double[][] dct = new double[m][n];

        double ci, cj, dct1, sum;

        for (i = 0; i < m; i++)
        {
            for (j = 0; j < n; j++)
            {
                // ci and cj depends on frequency as well as
                // number of row and columns of specified matrix
                if (i == 0)
                    ci = 1 / Math.sqrt(m);
                else
                    ci = Math.sqrt(2) / Math.sqrt(m);

                if (j == 0)
                    cj = 1 / Math.sqrt(n);
                else
                    cj = Math.sqrt(2) / Math.sqrt(n);

                // sum will temporarily store the sum of
                // cosine signals
                sum = 0;
                for (k = 0; k < m; k++)
                {
                    for (l = 0; l < n; l++)
                    {
                        dct1 = matrix[k][l] *
                                Math.cos((2 * k + 1) * i * pi / (2 * m)) *
                                Math.cos((2 * l + 1) * j * pi / (2 * n));
                        sum = sum + dct1;
                    }
                }
                dct[i][j] = ci * cj * sum;
            }
        }

        for (i = 0; i < m; i++)
        {
            for (j = 0; j < n; j++)
                System.out.printf("%f\t", dct[i][j]);
            System.out.println();
        }
    }

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