Question

据我所知，该算法使用8次乘法和4次加时间复杂度：

乘法在每个n/2 * n/2矩阵上完成。我对此几乎没有问题：

通过执行n * n，每个n=1矩阵最终会缩小到T(n/2)吗？如果是这样，那么返回a11*b11似乎毫无意义，就像为1*6返回以下矩阵的a11*b11一样：

然后基本情况应该是n==2执行else部分，因为下面的操作似乎是合法的。

为什么添加部分需要0(n^2)？我的意思是，我们完全没有处理矩阵加法，而仅仅是数字，因为每个矩阵都减少到2 * 2，如下所示：

所以加法部分应该只贡献4个？（为什么0(n^2)？）

Answer 1

如果我正确理解了这个问题，可以按如下方式回答这些问题。

事实上，所有矩阵最终都缩减为1*1矩阵;这不应该太令人惊讶，因为矩阵乘法的基本定义最终是根据底层环的乘法来定义的。
加法部分在递归的每个级别都有复杂度0(n^2)，因为对递归的递归求值结果执行加法。

Answer 2

1）矩阵最终会减少到1 * 1矩阵。但是没关系，你甚至可以为n == 2设置基本情况，并且它仍然是O（1）时间，因为乘以2 * 2矩阵仍然需要恒定时间并且复杂性仍将保持不变。

2）加法部分应该具有复杂度O（n ^ 2），因为每个子问题都有（n ^ 2）/ 4个元素，并且有4个这样的子问题，这意味着你实际上正在执行n ^ 2个运算，这导致O（ n ^ 2）复杂性。

Answer 3

仅从算法上看，可能不清楚加法步骤花费theta（n ^ 2）时间的原因。我也有同样的困惑，即加法需要花费固定的时间。如果执行更改，则在addMatrices（）方法中使用2 * 2矩阵

C[rowC][columnC] = A[0][0] + B[0][0];

然后它也给出相同的结果。

但是一旦我们采用4 * 4矩阵，就可以看到在调用堆栈中发生了addMatrices（）方法调用，该方法正在从矩阵A和B中添加多个元素。这就是为什么需要在内部运行加法的原因循环。

实施该程序后，更容易理解。我试图进行解释，有关详细信息，请参阅方法注释。

package matrix;

/***
 * Square Matrix multiplication(2^x) using divide and conquer technique
 * 
 * @author kmandal
 *
 */
public class MatrixMultiplication {

    public static void main(String[] args) {
        int[][] A = { { 1, 2 }, { 3, 4 } };
        int[][] B = { { 5, 6 }, { 7, 8 } };
        int C[][] = squareMatrixMultiplyRecursive(A, B);

        for (int i = 0; i < C.length; i++) {
            for (int j = 0; j < C.length; j++) {
                System.out.print(C[i][j] + "    ");
            }
            System.out.println();
        }
    }

    private static int[][] squareMatrixMultiplyRecursive(int[][] A, int[][] B) {
        return squareMatrixMultiplicationDNC(A, B, 0, 0, 0, 0, A.length);
    }

    /**
     * <pre>
     * Let A and B are 2 square matrices with dimension 2^x
     * A = [
     *      A00     A01
     *      A10     A11
     *      ]
     * ,
     * B = [
     *      B00     B01
     *      B10     B11
     *      ]
     * 
     * C be another matrix stores the result of multiplication of A and B.
     * 
     *  C = A.B;
     *  
     *  C = [
     *      C00     C01
     *      C10     C11
     *      ]
     *  
     *  where
     *  for C00 calculation, elements in 0th row of A and 0th column of B considered
     *  C00 = A00*B00+A01*B10;  
     *  
     *  for C01 calculation, elements in 0th row of A and 1st column of B considered
     *  C01 = A00*B01+A01*B11; 
     *  
     *  for C10 calculation, elements in 1st row of A and 0th column of B considered
     *  C10 = A10*B00+A11*B10; 
     *  
     *  for C11 calculation, elements in 1st row of A and 1st column of B considered
     *  C11 = A10*B01+A11*B11;
     * 
     * Here we are using index based calculation, 
     * hence time complexity for index calculation is Theta(1). 
     * 
     * We have divided the problem into 8 sub-problems with size n/2.
     * Hence the recurrence for this divide part is: 8T(n/2).
     * 
     * Additionally we need to consider the cost of matrix addition step, 
     * which is Theta(n^2). For more details refer addMatrices() method.
     * 
     * Hence the recurrence relation become 
     * T(n) = Theta(1) + 8T(n/2)+ Theta(n^2);
     * 
     * Applying Master theorem, 
     * the time complexity of this algorithm become O(n^3)
     * </pre>
     * 
     * @param A
     * @param B
     * @param rowA
     * @param columnA
     * @param rowB
     * @param columnB
     * @param size
     * @return
     */
    private static int[][] squareMatrixMultiplicationDNC(int[][] A, int[][] B,
            int rowA, int columnA, int rowB, int columnB, int size) {
        int[][] C = new int[size][size];
        if (size == 1) {
            C[0][0] = A[rowA][columnA] * B[rowB][columnB];
        } else {
            int newSize = size / 2;
            // calculate C00 = A00*B00+A01*B10;
            addMatrices(
                    C,
                    squareMatrixMultiplicationDNC(A, B, rowA, columnA, rowB,
                            columnB, newSize),
                    squareMatrixMultiplicationDNC(A, B, rowA,
                            columnA + newSize, rowB + newSize, columnB, newSize),
                    0, 0);
            // calculate C01 = A00*B01+A01*B11;
            addMatrices(
                    C,
                    squareMatrixMultiplicationDNC(A, B, rowA, columnA, rowB,
                            columnB + newSize, newSize),
                    squareMatrixMultiplicationDNC(A, B, rowA,
                            columnA + newSize, rowB + newSize, columnB
                                    + newSize, newSize), 0, newSize);
            // calculate C10 = A10*B00+A11*B10;
            addMatrices(
                    C,
                    squareMatrixMultiplicationDNC(A, B, rowA + newSize,
                            columnA, rowB, columnB, newSize),
                    squareMatrixMultiplicationDNC(A, B, rowA + newSize, columnA
                            + newSize, rowB + newSize, columnB, newSize),
                    newSize, 0);
            // calculate C11 = A10*B01+A11*B11;
            addMatrices(
                    C,
                    squareMatrixMultiplicationDNC(A, B, rowA + newSize,
                            columnA, rowB, columnB + newSize, newSize),
                    squareMatrixMultiplicationDNC(A, B, rowA + newSize, columnA
                            + newSize, rowB + newSize, columnB + newSize,
                            newSize), newSize, newSize);

        }
        return C;
    }

    /**
     * Matrix I represented by 2 dimensional array hence for addition of 2
     * matrices, need to fetch same element from both the matrices and then
     * add them. Traversing 2D array mean need to access elements by row and
     * column index thus need to loop inside loop. Hence time complexity of
     * addition is Theta(n^2)
     * 
     * @param C
     * @param A
     * @param B
     * @param rowC
     * @param columnC
     */
    private static void addMatrices(int[][] C, int[][] A, int[][] B, int rowC,
            int columnC) {
        int n = A.length;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                C[i + rowC][j + columnC] = A[i][j] + B[i][j];
            }
        }
    }
}

Output:
19    22    
43    50

使用分而治之的矩阵乘法，时间复杂度

3 个答案: