Java中DCT和IDCT算法的问题

时间:2010-11-21 21:31:24

标签: java algorithm dct

这里我的DCT算法类包含“applyDCT”和“applyIDCT”方法。技术上,在0到255之间的2x2随机整数表上进行正向DCT(离散余弦变换)之后,然后立即对这些数字进行反向DCT,我们应该回到我们首先得到的原始整数。就我而言,情况并非如此。我在这里做错了什么?

public class DCT {
    private static final int N = 2;
    private double[] c = new double[N];

    public DCT() {
          this.initializeCoefficients();
    }

    private void initializeCoefficients() {
        for (int i=1;i<N;i++) {
            c[i]=1;
        }
        c[0]=1/Math.sqrt(2.0);
    }

    public double[][] applyDCT(double[][] f) {
        double[][] F = new double[N][N];
        for (int u=0;u<N;u++) {
          for (int v=0;v<N;v++) {
            double sum = 0.0;
            for (int i=0;i<N;i++) {
              for (int j=0;j<N;j++) {
                sum+=Math.cos(((2*i+1)/(2.0*N))*u*Math.PI)*Math.cos(((2*j+1)/(2.0*N))*v*Math.PI)*f[i][j];
              }
            }
            sum*=((c[u]*c[v])/4.0);
            F[u][v]=sum;
          }
        }
        return F;
    }

    public double[][] applyIDCT(double[][] F) {
        double[][] f = new double[N][N];
        for (int u=0;u<N;u++) {
          for (int v=0;v<N;v++) {
            double sum = 0.0;
            for (int i=0;i<N;i++) {
              for (int j=0;j<N;j++) {
                sum+=((c[u]*c[v]))*Math.cos(((2*i+1)/(2.0*N))*u*Math.PI)*Math.cos(((2*j+1)/(2.0*N))*v*Math.PI)*F[i][j];
              }
            }
            sum/=4.0;
            //sum*=((c[u]*c[v])/4.0);
            f[u][v]=sum;
          }
        }
        return f;
    }
}

以下是与之相关的主要课程:

public class Main {
    private static final int N = 2;
    private static double[][] f = new double[N][N];
    private static Random generator = new Random();

    public static void main(String[] args) {
        // Generate random integers between 0 and 255
        int value;
        for (int x=0;x<N;x++) {
            for (int y=0;y<N;y++) {
              value = generator.nextInt(255);
              f[x][y] = value;
              System.out.println(f[x][y]+" => f["+x+"]["+y+"]");
            }
        }

        DCT dctApplied = new DCT();
        double[][] F = dctApplied.applyDCT(f);
        System.out.println("From f to F");
        System.out.println("-----------");
        for (int x=0;x<N;x++) {
            for (int y=0;y<N;y++) {
             try {
                 System.out.println(F[x][y]+" => F["+x+"]["+y+"]");
                 } catch (Exception e) {
                    System.out.println(e);
                 }
            }
        }

        double f[][] = dctApplied.applyIDCT(F);
        System.out.println("Back to f");
        System.out.println("---------");
        for (int y=0;y<N;y++) {
            for (int z=0;z<N;z++) {
              System.out.println(f[y][z]+" => f["+y+"]["+z+"]");
            }
        }
    }
}

以下是结果示例:

149.0 => f[0][0]
237.0 => f[0][1]
122.0 => f[1][0]
147.0 => f[1][1] 

From f to F
-----------
81.87499999999999 => F[0][0]
-14.124999999999993 => F[0][1]
14.62500000000001 => F[1][0]
-7.875 => F[1][1] 

Back to f
---------
9.3125 => f[0][0]
14.812499999999998 => f[0][1]
7.624999999999999 => f[1][0]
9.187499999999998 => f[1][1]

如上所示,“返回f”最初没有显示f中包含的相同值...

1 个答案:

答案 0 :(得分:9)

我已经解决了这个问题,如果我的问题不清楚,我很抱歉,但这里有什么不对:IDCT方法必须在i和j for循环中有系数:

public double[][] applyIDCT(double[][] F) {
        double[][] f = new double[N][N];
        for (int i=0;i<N;i++) {
          for (int j=0;j<N;j++) {
            double sum = 0.0;
            for (int u=0;u<N;u++) {
              for (int v=0;v<N;v++) {
                sum+=(c[u]*c[v])/4.0*Math.cos(((2*i+1)/(2.0*N))*u*Math.PI)*Math.cos(((2*j+1)/(2.0*N))*v*Math.PI)*F[u][v];
              }
            }
            f[i][j]=Math.round(sum);
          }
        }
        return f;
    }

这仅适用于8x8数据块,否则您必须更改此内容:

(c[u]*c[v])/4.0)

这样的事情:

(2*c[u]*c[v])/Math.sqrt(M*N)

其中M和N是表格的维数......

以下是2x2数据块的结果:

Original values
---------------
54.0 => f[0][0]
35.0 => f[0][1]
128.0 => f[1][0]
185.0 => f[1][1]

From f to F
-----------
200.99999999999994 => F[0][0]
-18.99999999999997 => F[0][1]
-111.99999999999997 => F[1][0]
37.99999999999999 => F[1][1]

Back to f
---------
54.0 => f[0][0]
35.0 => f[0][1]
128.0 => f[1][0]
185.0 => f[1][1]