递归FFT Java算法返回null吗？

我目前正在尝试在Java中实现FFT算法,但遇到了一些麻烦！我已经对该算法的所有其他部分进行了很好的测试,它们似乎运行良好.

我遇到的麻烦是,在基本情况下,它会在基本情况A [0]内返回一个复数数组.执行完基本情况后,将执行for循环,其中y0 [0]和y1 [0]被发现为空,尽管将它们分配给基本情况,对此感到很困惑.这显示在System.out.println行中

谁能告诉我我的方式的错误？

    //This method implements the recursive FFT algorithm, it assumes the input length
//N is some power of two
private static Complex[] FFT(Complex[] A, int N) { 
    double real, imag;
    Complex A0[] = new Complex[((int) Math.ceil(N/2))]; 
    Complex A1[] = new Complex[((int) Math.ceil(N/2))];
    Complex[] omega = new Complex[N];
    Complex[] y = new Complex[N];
    Complex[] y0 = new Complex[((int) Math.ceil(N/2))];
    Complex[] y1 = new Complex[((int) Math.ceil(N/2))]; 

    //base case
    if (N == 1) {
        return A;
    }
    else {
        real = Math.cos((2*Math.PI)/N); if (real < 1E-10 && real > 0) real = 0;
        imag = Math.sin((2*Math.PI)/N); if (imag < 1E-10 && imag > 0) imag = 0;
        omega[N-1] = new Complex(real, imag);
        omega[0] = new Complex(1, 0);
        A0 = splitInput(A, 1);
        A1 = splitInput(A, 0);
        //recursive calls
        y0 = FFT(A0, N/2);
        y1 = FFT(A1, N/2);
        for (int k = 0; k < ((N/2)-1); k++) {
            System.out.print("k: " + k + ", y0: " + y0[k]); System.out.println(", y1: " + y1[k]);
            y[k] = y0[k].plus(omega[k].times(y1[k]));
            y[k+(N/2)] = y0[k].minus(omega[k].times(y1[k]));
            omega[0] = omega[0].times(omega[N]);
        }
        return y;
    }
}

这是请求的我的splitInput方法的代码

//This method takes a double array as an argument and returns every even or odd
//element according to the second int argument being 1 or 0
private static Complex[] splitInput(Complex[] input, int even) {
    Complex[] newArray = new Complex[(input.length/2)];

    //Return all even elements of double array, including 0
    if (even == 1) {
        for (int i = 0; i < (input.length/2); i++) {
            newArray[i] = new Complex(input[i*2].re, 0.0);
        }
        return newArray;
    }
    //Return all odd elements of double array
    else {
        for (int i = 0; i < (input.length/2); i++) {
            newArray[i] = new Complex (input[(i*2) + 1].re, 0.0);
        }
    return newArray;
    }
}

编辑：我根据您的建议更新了代码,仍然从行y [k] = y0 [k] .plus(omega [k] .times(y1 [k]))中获取空指针异常；为y0&在基本情况之后y1仍然为空:(还有其他想法吗？这是更新的算法

//This method implements the recursive FFT algorithm, it assumes the input length
//N is some power of two
private static Complex[] FFT(Complex[] A, int N) { 
    double real, imag;
    Complex[] omega = new Complex[N];
    Complex[] y = new Complex[N];
    Complex[] A0;
    Complex[] A1;
    Complex[] y0;
    Complex[] y1;

    //base case
    if (N == 1) {
        return A;
    }
    else {
        real = Math.cos((2*Math.PI)/N); if (real < 1E-10 && real > 0) real = 0;
        imag = Math.sin((2*Math.PI)/N); if (imag < 1E-10 && imag > 0) imag = 0;;
        omega[N-1] = new Complex(real, imag);
        omega[0] = new Complex(1, 0);
        A0 = splitInput(A, 1);
        A1 = splitInput(A, 0);
        //recursive calls
        y0 = FFT(A0, N/2);
        y1 = FFT(A1, N/2);
        for (int k = 0; k < ((N/2)-1); k++) {
            y[k] = y0[k].plus(omega[k].times(y1[k]));
            y[k+(N/2)] = y0[k].minus(omega[k].times(y1[k]));
            omega[0] = omega[0].times(omega[N-1]);
        }
        return y;
    }
}

解决方法:

跳出来的问题：

>(int)Math.ceil(N / 2)您仍在进行int除法,因此Math.ceil()无效,并且您的拆分数组对于奇数n可能不正确
>您只会填充omega [0]和omega [N-1],当您尝试访问omega [1]时会出现NullPointerException,这会在N> = 6时发生.
> omega [N],也如sarnold所述
>您分配A0和A1,然后将它们分配给splitInput结果

标签：complex-numbers,fft,java,algorithm,recursion
来源： https://codeday.me/bug/20191201/2084616.html