Question

我一直在尝试编码鸿沟并征服矩阵乘法算法

有问题，当我尝试将矩阵划分为四个四分之一时，它会给我一个错误ArrayOutOfIndexBound

我不确定我的基本情况是否合适，所以你能帮助我吗？

我得到的问题是 double [] [] a21

    public static double[][] divideAndConquer(double[][] a , double[][] b, int dimension){

if (a.length == 1){
    double[][] result = new double[1][1];
    result[0][0]= a[0][0]*b[0][0];
    return result;
}
else {
    int m = dimension/2;
    double[][] a11 = new double[m][m];
    for(int i = 0; i < m ; i++){
        for (int j = 0 ; j< m ; j++)
            a11[i][j]= a[i][j];
    }

              double[][] a21 = new double[m][m];
            for(int i = m; i < dimension; i++){
        for (int j = 0 ; j< m ; j++)
            a21[i][j]= a[i][j];
    }
     double[][] a12 = new double[m][m];
            for(int i = 0; i < m ; i++){
        for (int j = m ; j< dimension ; j++)
            a12[i][j]= a[i][j];
    }



    double[][] a22 = new double[m][m];
            for(int i = m; i < dimension; i++){
        for (int j =  m; j < dimension; j++)
            a21[i][j]= a[i][j];
    }


    double[][] b11 = new double[m][m];
    for(int i = 0; i < m ; i++){
        for (int j = 0 ; j< m ; j++)
            b11[i][j]= b[i][j];
    }

     double[][] b12 = new double[m][m];
            for(int i = 0; i < m ; i++){
        for (int j = m ; j< dimension ; j++)
            b12[i][j]= b[i][j];
    }

      double[][] b21 = new double[m][m];
            for(int i = m; i < dimension; i++){
        for (int j = 0 ; j< m ; j++)
            b21[i][j]= b[i][j];
    }

    double[][] b22 = new double[m][m];
            for(int i = m; i < dimension; i++){
        for (int j =  m; j < dimension; j++)
            b21[i][j]= b[i][j];
    }

            double[][] x1 = divideAndConquer(a11,b11,m);
            double[][] x2 = divideAndConquer(a12,b21,m);
            double[][] x3 = divideAndConquer(a11,b12,m);
            double[][] x4 = divideAndConquer(a12,b22,m);
            double[][] x5 = divideAndConquer(a21,b11,m);
            double[][] x6 = divideAndConquer(a22,b21,m);
            double[][] x7 = divideAndConquer(a21,b12,m);
            double[][] x8 = divideAndConquer(a22,b22,m);
        ..........................etc

Answer 1

如上所述，您的问题是您需要减去数组偏移量;例如，

a12[i][j]= a[i][j];

应该是

a12[i][j-dimension]= a[i][j];

你更大的问题是你正在创建4个新的子矩阵，这将产生吨的垃圾。一旦你完成了这项工作，我会强烈想到通过操纵数组索引来“就地”这样做的方法。

例如，你的新api看起来像

public static double[][] divideAndConquer(double[][] a , double[][] b, int aMinIndex, int aMaxIndex, int bMinIndex, bMaxIndex){

并且你的分而治之将构建min＆amp;的子集。最大指数。

Answer 2

如何使用它进行分区？请考虑它在C中，N / 2是每个分区的块大小

for (int i = 0; i < N / 2; i++) {
    for (int j = 0; j < N / 2; j++) {

        ha11[i][j] = hA[i][j]; // top left
        ha12[i][j] = hA[i][j + N / 2]; // top right
        ha21[i][j] = hA[i + N / 2][j]; // bottom left
        ha22[i][j] = hA[i + N / 2][j + N / 2]; // bottom right

        hb11[i][j] = hB[i][j]; // top left
        hb12[i][j] = hB[i][j + N / 2]; // top right
        hb21[i][j] = hB[i + N / 2][j]; // bottom left
        hb22[i][j] = hB[i + N / 2][j + N / 2]; // bottom right
    }
}

划分和征服矩阵乘法基础案例+如何将矩阵分成4个季度

2 个答案: