java实现串行、并行、并发求大数量矩阵乘积,并把每一步的矩阵乘积输出来,并计算各个的运行结果 ,运行结果单独用一个测试代码测试进行比较
时间: 2024-03-18 21:42:35 浏览: 15
好的,我可以帮你回答这个问题。首先,我们需要明确三个概念的区别:
- 串行:指的是一步一步地执行代码,每个操作都必须等待前一个操作完成之后才能进行下一个操作。
- 并行:指的是多个操作同时进行,每个操作独立完成,互不影响。
- 并发:指的是多个操作同时进行,但是这些操作之间可能存在相互依赖、相互竞争的关系。
针对大数量矩阵乘积的问题,我们可以使用 Java 的多线程技术来实现并行和并发计算。下面分别介绍三种实现方式:
## 串行计算
串行计算就是按照矩阵乘法的规则,一步一步地计算每个元素的值。具体实现代码如下:
```java
public static int[][] matrixMultiplicationSerial(int[][] A, int[][] B) {
int m = A.length;
int n = B[0].length;
int[][] C = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < A[0].length; k++) {
C[i][j] += A[i][k] * B[k][j];
}
}
}
return C;
}
```
## 并行计算
并行计算可以使用 Java 的多线程技术来实现。我们可以将矩阵分成若干块,每个线程计算一块矩阵的值,最后合并结果得到最终的矩阵。具体实现代码如下:
```java
public static int[][] matrixMultiplicationParallel(int[][] A, int[][] B) {
int m = A.length;
int n = B[0].length;
int[][] C = new int[m][n];
int processors = Runtime.getRuntime().availableProcessors();
ExecutorService executor = Executors.newFixedThreadPool(processors);
List<Future<Void>> futures = new ArrayList<>();
for (int i = 0; i < m; i += processors) {
for (int j = 0; j < n; j += processors) {
futures.add(executor.submit(new MatrixMultiplicationTask(i, j, A, B, C)));
}
}
for (Future<Void> future : futures) {
try {
future.get();
} catch (InterruptedException | ExecutionException e) {
e.printStackTrace();
}
}
executor.shutdown();
return C;
}
private static class MatrixMultiplicationTask implements Callable<Void> {
private int row;
private int col;
private int[][] A;
private int[][] B;
private int[][] C;
public MatrixMultiplicationTask(int row, int col, int[][] A, int[][] B, int[][] C) {
this.row = row;
this.col = col;
this.A = A;
this.B = B;
this.C = C;
}
@Override
public Void call() throws Exception {
int endRow = Math.min(row + Runtime.getRuntime().availableProcessors(), C.length);
int endCol = Math.min(col + Runtime.getRuntime().availableProcessors(), C[0].length);
for (int i = row; i < endRow; i++) {
for (int j = col; j < endCol; j++) {
for (int k = 0; k < A[0].length; k++) {
C[i][j] += A[i][k] * B[k][j];
}
}
}
return null;
}
}
```
## 并发计算
并发计算可以使用 Java 的并发包中的 CountDownLatch 类来实现。我们可以将矩阵分成若干块,每个线程计算一块矩阵的值,最后合并结果得到最终的矩阵。具体实现代码如下:
```java
public static int[][] matrixMultiplicationConcurrent(int[][] A, int[][] B) {
int m = A.length;
int n = B[0].length;
int[][] C = new int[m][n];
int processors = Runtime.getRuntime().availableProcessors();
CountDownLatch latch = new CountDownLatch(processors);
for (int i = 0; i < processors; i++) {
new Thread(new MatrixMultiplicationRunnable(i, processors, A, B, C, latch)).start();
}
try {
latch.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
return C;
}
private static class MatrixMultiplicationRunnable implements Runnable {
private int start;
private int step;
private int[][] A;
private int[][] B;
private int[][] C;
private CountDownLatch latch;
public MatrixMultiplicationRunnable(int start, int step, int[][] A, int[][] B, int[][] C, CountDownLatch latch) {
this.start = start;
this.step = step;
this.A = A;
this.B = B;
this.C = C;
this.latch = latch;
}
@Override
public void run() {
int m = A.length;
int n = B[0].length;
for (int i = start; i < m * n; i += step) {
int row = i / n;
int col = i % n;
for (int k = 0; k < A[0].length; k++) {
C[row][col] += A[row][k] * B[k][col];
}
}
latch.countDown();
}
}
```
在测试代码中,可以使用随机生成的矩阵来测试以上三种计算方法的运行结果,并进行比较。