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第四章 分治策略 4.2 矩阵乘法的Strassen算法

2014-06-08 16:48 393 查看
package chap04_Divide_And_Conquer;

import static org.junit.Assert.*;

import java.util.Arrays;

import org.junit.Test;

/**
* 矩阵相乘的算法
*
* @author xiaojintao
*
*/
public class MatrixOperation {
/**
* 普通的矩阵相乘算法,c=a*b。其中,a、b都是n*n的方阵
*
* @param a
* @param b
* @return c
*/
static int[][] matrixMultiplicationByCommonMethod(int[][] a, int[][] b) {
int n = a.length;
int[][] c = new int

;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
c[i][j] = 0;
for (int k = 0; k < n; k++) {
c[i][j] += a[i][k] * b[k][j];
}
}
}
return c;
}

/**
* strassen 算法求矩阵乘法 n为2的幂
*
* @param a
* @param b
* @return
*/
static int[][] matrixMultiplicationByStrassen(int[][] a, int[][] b) {
int n = a.length;
if (n == 1) {
int[][] c = new int[1][1];
c[0][0] = a[0][0] * b[0][0];
return c;
}
int m = n / 2;
int[][] a11, a12, a21, a22, b11, b12, b21, b22;
int[][] c = new int

;
a11 = new int[m][m];
a12 = new int[m][m];
a21 = new int[m][m];
a22 = new int[m][m];
b11 = new int[m][m];
b12 = new int[m][m];
b21 = new int[m][m];
b22 = new int[m][m];

for (int i = 0; i < m; i++) {
for (int j = 0; j < m; j++) {
a11[i][j] = a[i][j];
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < m; j++) {
b11[i][j] = b[i][j];
}
}
for (int i = 0; i < m; i++) {
for (int j = m; j < n; j++) {
a12[i][j - m] = a[i][j];
}
}
for (int i = 0; i < m; i++) {
for (int j = m; j < n; j++) {
b12[i][j - m] = b[i][j];
}
}
for (int i = m; i < n; i++) {
for (int j = 0; j < m; j++) {
a21[i - m][j] = a[i][j];
}
}
for (int i = m; i < n; i++) {
for (int j = 0; j < m; j++) {
b21[i - m][j] = b[i][j];
}
}
for (int i = m; i < n; i++) {
for (int j = m; j < n; j++) {
a22[i - m][j - m] = a[i][j];
}
}
for (int i = m; i < n; i++) {
for (int j = m; j < n; j++) {
b22[i - m][j - m] = b[i][j];
}
}
int[][] s1 = matrixMinus(b12, b22);
int[][] s2 = matrixAdd(a11, a12);
int[][] s3 = matrixAdd(a21, a22);
int[][] s4 = matrixMinus(b21, b11);
int[][] s5 = matrixAdd(a11, a22);
int[][] s6 = matrixAdd(b11, b22);
int[][] s7 = matrixMinus(a12, a22);
int[][] s8 = matrixAdd(b21, b22);
int[][] s9 = matrixMinus(a11, a21);
int[][] s10 = matrixAdd(b11, b12);

int[][] p1 = matrixMultiplicationByStrassen(a11, s1);
int[][] p2 = matrixMultiplicationByStrassen(s2, b22);
int[][] p3 = matrixMultiplicationByStrassen(s3, b11);
int[][] p4 = matrixMultiplicationByStrassen(a22, s4);
int[][] p5 = matrixMultiplicationByStrassen(s5, s6);
int[][] p6 = matrixMultiplicationByStrassen(s7, s8);
int[][] p7 = matrixMultiplicationByStrassen(s9, s10);

int[][] t1, t2, t3;
t1 = matrixAdd(p5, p4);
t2 = matrixMinus(t1, p2);
int[][] c11 = matrixAdd(t2, p6);
int[][] c12 = matrixAdd(p1, p2);
int[][] c21 = matrixAdd(p3, p4);
t1 = matrixAdd(p5, p1);
t2 = matrixMinus(t1, p3);
int[][] c22 = matrixMinus(t2, p7);
c = matrixConbine(c11, c12, c21, c22);
return c;
}

/**
* 矩阵加法 c=a+b
*
* @param a
* @param b
* @return
*/
static int[][] matrixAdd(int[][] a, int[][] b) {
int n = a.length;
int[][] c = new int

;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
c[i][j] = a[i][j] + b[i][j];
}
}
return c;
}

/**
* 矩阵减法 c=a-b
*
* @param a
* @param b
* @return
*/
static int[][] matrixMinus(int[][] a, int[][] b) {
int n = a.length;
int[][] c = new int

;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
c[i][j] = a[i][j] - b[i][j];
}
}
return c;
}

/**
* 将矩阵的四个部分组合
*
* @param t11
* @param t12
* @param t21
* @param t22
* @return
*/
protected static int[][] matrixConbine(int[][] t11, int[][] t12,
int[][] t21, int[][] t22) {
int n = t11.length;
int m = 2 * n;
int[][] c = new int[m][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
c[i][j] = t11[i][j];
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
c[i][j + n] = t12[i][j];
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
c[i + n][j] = t21[i][j];
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
c[i + n][j + n] = t22[i][j];
}
}
return c;
}

@Test
public void testName() throws Exception {
// int[][] a = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
// int[][] b = { { 1, 3, 5 }, { 2, 4, 6 }, { 9, 8, 7 } };
// int[][] c = commonMatrixMultiplication(a, b);
// int[][] c = matrixAdd(a, b);

int[][] m = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
int[][] n = { { 1, 3, 5, 7 }, { 2, 4, 6, 8 }, { 4, 3, 2, 1 },
{ 9, 8, 7, 6 } };

int[][] c = matrixMultiplicationByStrassen(m, n);
System.out.println(Arrays.deepToString(c));
int[][] d = matrixMultiplicationByCommonMethod(m, n);
System.out.println(Arrays.deepToString(d));
}
}


暴力求解复杂度为O(n3),Strassen算法为O(n log7)
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