📅  最后修改于: 2023-12-03 15:11:14.791000             🧑  作者: Mango
Strassen 算法是一种用于矩阵乘法的算法,采用分治的思想,能有效降低矩阵乘法的时间复杂度。在一般情况下,矩阵乘法的时间复杂度是 $O(n^3)$,而 Strassen 算法的时间复杂度可以达到 $O(n^{log_2(7)}) \approx O(n^{2.81})$。
在本文中,我们将讨论如何使用Java实现 Strassen 算法。
首先,我们需要实现矩阵乘法的基本算法。矩阵乘法的基本算法可以使用三个循环实现。请参考以下代码:
public static int[][] matrixMultiplication(int[][] a, int[][] b) {
int m = a.length;
int n = a[0].length;
int p = b[0].length;
int[][] c = new int[m][p];
for (int i = 0; i < m; i++) {
for (int j = 0; j < p; j++) {
for (int k = 0; k < n; k++) {
c[i][j] += a[i][k] * b[k][j];
}
}
}
return c;
}
接下来,我们将实现 Strassen 算法。Strassen 算法的实现主要分为三个步骤:矩阵分解、递归求解、矩阵合并。
首先,我们需要将输入的两个 $n \times n$ 的矩阵 $A$ 和 $B$ 分解成 $2 \times 2$ 的小矩阵。这可以通过如下的方式实现:
public static int[][][] splitMatrix(int[][] a) {
int n = a.length / 2;
int[][][] result = new int[2][n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
result[0][i][j] = a[i][j];
result[1][i][j] = a[i][j + n];
result[2][i][j] = a[i + n][j];
result[3][i][j] = a[i + n][j + n];
}
}
return result;
}
然后,我们将分解后的小矩阵使用如下的公式计算出 $S_1$ 至 $S_{10}$:
$$ \begin{aligned} S_1 &= B_{12} - B_{22} \ S_2 &= A_{11} + A_{12} \ S_3 &= A_{21} + A_{22} \ S_4 &= B_{21} - B_{11} \ S_5 &= A_{11} + A_{22} \ S_6 &= B_{11} + B_{22} \ S_7 &= A_{12} - A_{22} \ S_8 &= B_{21} + B_{22} \ S_9 &= A_{11} - A_{21} \ S_{10} &= B_{11} + B_{12} \end{aligned} $$
接下来,我们将使用递归的方式计算出 $P_1$ 至 $P_7$:
$$ \begin{aligned} P_1 &= A_{11}S_1 \ P_2 &= S_2B_{22} \ P_3 &= S_3B_{11} \ P_4 &= A_{22}S_4 \ P_5 &= S_5S_6 \ P_6 &= S_7S_8 \ P_7 &= S_9S_{10} \end{aligned} $$
递归的方式为:
public static int[][] strassen(int[][] a, int[][] b) {
int n = a.length;
// 如果矩阵的大小为1,则直接计算矩阵乘积
if (n == 1) {
int[][] result = new int[1][1];
result[0][0] = a[0][0] * b[0][0];
return result;
}
// 将矩阵分解成2*2的小矩阵
int[][][] A = splitMatrix(a);
int[][][] B = splitMatrix(b);
// 计算S1~S10
int[][] S1 = subtract(B[0], B[3]);
int[][] S2 = add(A[0], A[1]);
int[][] S3 = add(A[2], A[3]);
int[][] S4 = subtract(B[2], B[0]);
int[][] S5 = add(A[0], A[3]);
int[][] S6 = add(B[0], B[3]);
int[][] S7 = subtract(A[1], A[3]);
int[][] S8 = add(B[2], B[3]);
int[][] S9 = subtract(A[0], A[2]);
int[][] S10 = add(B[0], B[1]);
// 计算P1~P7
int[][] P1 = strassen(A[0], S1);
int[][] P2 = strassen(S2, B[3]);
int[][] P3 = strassen(S3, B[0]);
int[][] P4 = strassen(A[3], S4);
int[][] P5 = strassen(S5, S6);
int[][] P6 = strassen(S7, S8);
int[][] P7 = strassen(S9, S10);
// 合并子矩阵
int[][][] C = new int[2][n/2][n/2];
C[0] = add(subtract(add(P5, P4), P2), P6);
C[1] = add(P1, P2);
C[2] = add(P3, P4);
C[3] = subtract(subtract(add(P5, P1), P3), P7);
// 合并成结果矩阵
int[][] result = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i < n/2 && j < n/2) {
result[i][j] = C[0][i][j];
} else if (i < n/2 && j >= n/2) {
result[i][j] = C[1][i][j-n/2];
} else if (i >= n/2 && j < n/2) {
result[i][j] = C[2][i-n/2][j];
} else {
result[i][j] = C[3][i-n/2][j-n/2];
}
}
}
return result;
}
最后,我们需要将递归求解出来的矩阵 $P_1$ 至 $P_7$ 合并成乘积矩阵。这可以通过如下的方式实现:
public static int[][] mergeMatrix(int[][] a, int[][] b) {
int n = a.length;
int[][] result = new int[n*2][n*2];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
result[i][j] = a[i][j];
result[i][j+n] = b[i][j];
result[i+n][j] = a[i+n][j];
result[i+n][j+n] = b[i+n][j];
}
}
return result;
}
最终的完整代码如下:
public class StrassenMatrixMultiplication {
public static void main(String[] args) {
int[][] A = { {1, 3, 5, 7},
{2, 4, 6, 8},
{9, 7, 5, 3},
{8, 6, 4, 2} };
int[][] B = { {8, 6, 4, 2},
{9, 7, 5, 3},
{2, 4, 6, 8},
{1, 3, 5, 7} };
int[][] C = strassen(A, B);
printMatrix(C);
}
public static int[][] matrixMultiplication(int[][] a, int[][] b) {
int m = a.length;
int n = a[0].length;
int p = b[0].length;
int[][] c = new int[m][p];
for (int i = 0; i < m; i++) {
for (int j = 0; j < p; j++) {
for (int k = 0; k < n; k++) {
c[i][j] += a[i][k] * b[k][j];
}
}
}
return c;
}
public static int[][][] splitMatrix(int[][] a) {
int n = a.length / 2;
int[][][] result = new int[4][n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
result[0][i][j] = a[i][j];
result[1][i][j] = a[i][j + n];
result[2][i][j] = a[i + n][j];
result[3][i][j] = a[i + n][j + n];
}
}
return result;
}
public static int[][] add(int[][] a, int[][] b) {
int n = a.length;
int[][] result = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
result[i][j] = a[i][j] + b[i][j];
}
}
return result;
}
public static int[][] subtract(int[][] a, int[][] b) {
int n = a.length;
int[][] result = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
result[i][j] = a[i][j] - b[i][j];
}
}
return result;
}
public static int[][] strassen(int[][] a, int[][] b) {
int n = a.length;
// 如果矩阵的大小为1,则直接计算矩阵乘积
if (n == 1) {
int[][] result = new int[1][1];
result[0][0] = a[0][0] * b[0][0];
return result;
}
// 将矩阵分解成2*2的小矩阵
int[][][] A = splitMatrix(a);
int[][][] B = splitMatrix(b);
// 计算S1~S10
int[][] S1 = subtract(B[1], B[3]);
int[][] S2 = add(A[0], A[1]);
int[][] S3 = add(A[2], A[3]);
int[][] S4 = subtract(B[2], B[0]);
int[][] S5 = add(A[0], A[3]);
int[][] S6 = add(B[0], B[3]);
int[][] S7 = subtract(A[1], A[3]);
int[][] S8 = add(B[2], B[3]);
int[][] S9 = subtract(A[0], A[2]);
int[][] S10 = add(B[0], B[1]);
// 计算P1~P7
int[][] P1 = strassen(A[0], S1);
int[][] P2 = strassen(S2, B[3]);
int[][] P3 = strassen(S3, B[0]);
int[][] P4 = strassen(A[3], S4);
int[][] P5 = strassen(S5, S6);
int[][] P6 = strassen(S7, S8);
int[][] P7 = strassen(S9, S10);
// 合并子矩阵
int[][][] C = new int[4][n/2][n/2];
C[0] = subtract(add(P5, P4), subtract(P2, P6));
C[1] = add(P1, P2);
C[2] = add(P3, P4);
C[3] = subtract(subtract(P5, P3), subtract(P1, P7));
// 合并成结果矩阵
int[][] result = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i < n/2 && j < n/2) {
result[i][j] = C[0][i][j];
} else if (i < n/2 && j >= n/2) {
result[i][j] = C[1][i][j-n/2];
} else if (i >= n/2 && j < n/2) {
result[i][j] = C[2][i-n/2][j];
} else {
result[i][j] = C[3][i-n/2][j-n/2];
}
}
}
return result;
}
public static int[][] mergeMatrix(int[][] a, int[][] b) {
int n = a.length;
int[][] result = new int[n*2][n*2];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
result[i][j] = a[i][j];
result[i][j+n] = b[i][j];
result[i+n][j] = a[i+n][j];
result[i+n][j+n] = b[i+n][j];
}
}
return result;
}
public static void printMatrix(int[][] matrix) {
int m = matrix.length;
int n = matrix[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
System.out.print(matrix[i][j] + " ");
}
System.out.println();
}
}
}
本文介绍了如何使用Java实现 Strassen 矩阵乘法算法。这个算法虽然时间复杂度很优秀,但是在实际应用中,由于其常数较大,因此不一定能够比普通的矩阵乘法算法更快速。但是本文的重点在于介绍一个高效算法的实现过程,希望可以对读者有所启示。