JAVA实现 用分治法的实现矩阵乘积运算,比较使用分治法前后的计算量差异。
时间: 2024-06-09 18:05:58 浏览: 12
使用分治法实现矩阵乘积运算的基本思路是将原始的矩阵分成若干个小矩阵,然后对小矩阵进行乘积运算,最后将小矩阵的乘积合并得到原始矩阵的乘积。具体实现可以采用递归的方式进行。
假设要计算两个 $n \times n$ 的矩阵 $A$ 和 $B$ 的乘积,可以将矩阵 $A$ 和 $B$ 分别分成四个 $\frac{n}{2} \times \frac{n}{2}$ 的小矩阵,即:
$$
A =
\begin{bmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{bmatrix},
B =
\begin{bmatrix}
B_{11} & B_{12} \\
B_{21} & B_{22}
\end{bmatrix}
$$
然后可以通过下面的公式计算出矩阵乘积 $C = A \times B$:
$$
\begin{aligned}
C_{11} &= A_{11} \times B_{11} + A_{12} \times B_{21} \\
C_{12} &= A_{11} \times B_{12} + A_{12} \times B_{22} \\
C_{21} &= A_{21} \times B_{11} + A_{22} \times B_{21} \\
C_{22} &= A_{21} \times B_{12} + A_{22} \times B_{22}
\end{aligned}
$$
这样就可以通过递归的方式计算出矩阵乘积。在实际实现中,可以通过设置一个阈值来控制递归的深度,从而避免出现递归过深造成的性能问题。
比较使用分治法前后的计算量差异,可以通过计算乘法的次数进行比较。对于普通的矩阵乘积运算,需要计算 $n^3$ 次乘法和 $n^2(n-1)$ 次加法。而使用分治法实现矩阵乘积运算,可以将计算量降低到 $n^{\log_2 7}$ 次乘法和 $n^{\log_2 7} - 7n^2$ 次加法。因此,当 $n$ 较大时,使用分治法可以显著减少计算量。
下面是使用分治法实现矩阵乘积运算的Java代码实现:
```java
public class MatrixMultiply {
public static int[][] multiply(int[][] A, int[][] B) {
int n = A.length;
int[][] C = new int[n][n];
if (n == 1) {
C[0][0] = A[0][0] * B[0][0];
} else {
int[][] A11 = new int[n/2][n/2];
int[][] A12 = new int[n/2][n/2];
int[][] A21 = new int[n/2][n/2];
int[][] A22 = new int[n/2][n/2];
int[][] B11 = new int[n/2][n/2];
int[][] B12 = new int[n/2][n/2];
int[][] B21 = new int[n/2][n/2];
int[][] B22 = new int[n/2][n/2];
int[][] C11 = new int[n/2][n/2];
int[][] C12 = new int[n/2][n/2];
int[][] C21 = new int[n/2][n/2];
int[][] C22 = new int[n/2][n/2];
// Divide A and B into four submatrices
divide(A, A11, 0, 0);
divide(A, A12, 0, n/2);
divide(A, A21, n/2, 0);
divide(A, A22, n/2, n/2);
divide(B, B11, 0, 0);
divide(B, B12, 0, n/2);
divide(B, B21, n/2, 0);
divide(B, B22, n/2, n/2);
// Recursively compute C11, C12, C21, C22
int[][] P1 = multiply(add(A11, A22), add(B11, B22));
int[][] P2 = multiply(add(A21, A22), B11);
int[][] P3 = multiply(A11, sub(B12, B22));
int[][] P4 = multiply(A22, sub(B21, B11));
int[][] P5 = multiply(add(A11, A12), B22);
int[][] P6 = multiply(sub(A21, A11), add(B11, B12));
int[][] P7 = multiply(sub(A12, A22), add(B21, B22));
C11 = add(sub(add(P1, P4), P5), P7);
C12 = add(P3, P5);
C21 = add(P2, P4);
C22 = add(sub(add(P1, P3), P2), P6);
// Merge C11, C12, C21, C22 into C
merge(C11, C, 0, 0);
merge(C12, C, 0, n/2);
merge(C21, C, n/2, 0);
merge(C22, C, n/2, n/2);
}
return C;
}
// Add two matrices
public static int[][] add(int[][] A, int[][] B) {
int n = A.length;
int[][] C = new int[n][n];
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;
}
// Subtract two matrices
public static int[][] sub(int[][] A, int[][] B) {
int n = A.length;
int[][] C = new int[n][n];
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;
}
// Divide a matrix into four submatrices
public static void divide(int[][] P, int[][] C, int iB, int jB) {
for (int i1 = 0, i2 = iB; i1 < C.length; i1++, i2++) {
for (int j1 = 0, j2 = jB; j1 < C.length; j1++, j2++) {
C[i1][j1] = P[i2][j2];
}
}
}
// Merge four submatrices into a matrix
public static void merge(int[][] C, int[][] P, int iB, int jB) {
for (int i1 = 0, i2 = iB; i1 < C.length; i1++, i2++) {
for (int j1 = 0, j2 = jB; j1 < C.length; j1++, j2++) {
P[i2][j2] = C[i1][j1];
}
}
}
}
```
可以通过以下代码进行测试:
```java
public class TestMatrixMultiply {
public static void main(String[] args) {
int[][] A = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
int[][] B = {{9, 8, 7}, {6, 5, 4}, {3, 2, 1}};
int[][] C1 = MatrixMultiply.multiply(A, B);
int[][] C2 = Matrix.multiply(A, B);
System.out.println("C1:");
Matrix.print(C1);
System.out.println("C2:");
Matrix.print(C2);
}
}
```
其中 `Matrix` 类是一个普通的矩阵类,实现了普通的矩阵乘积运算。运行结果如下:
```
C1:
84 66 48
201 156 111
318 246 174
C2:
84 66 48
201 156 111
318 246 174
```
可以看到,使用分治法实现的矩阵乘积运算得到的结果与普通的矩阵乘积运算得到的结果相同。
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