问题描述
我开始了解递归任务和递归操作的实现。根据我的理解和一些Java文档,我想到了以下代码,将所有数字加到一个数组中。
在纠正此问题时,我需要帮助,请帮助我指出我哪里出了问题。
import java.util.concurrent.ForkJoinPool;
import java.util.concurrent.ForkJoinTask;
import java.util.concurrent.RecursiveTask;
public class ForkJoinPoolTest {
public static void main(String[] args) {
ForkJoinPool pool = new ForkJoinPool(4);
long[] numbers = {1,2,3,4,5,6,7,8,9};
AdditionTask newTask = new AdditionTask(numbers,numbers.length -1 );
ForkJoinTask<Long> submit = pool.submit(newTask);
System.out.println(submit.join());
}
}
class AdditionTask extends RecursiveTask<Long> {
long[] numbers;
int start;
int end;
public AdditionTask(long[] numbers,int start,int end) {
this.numbers = numbers;
this.start = start;
this.end = end;
}
@Override
protected Long compute() {
if ((end - start) > 2) {
int length = numbers.length;
int mid = (length % 2 == 0) ? length / 2 : (length - 1) / 2;
AdditionTask leftSide = new AdditionTask(numbers,mid);
leftSide.fork();
AdditionTask rightSide = new AdditionTask(numbers,mid+1,length-1);
return rightSide.compute() + leftSide.join();
} else {
return numbers[0] + numbers[1];
}
}
}
新代码[固定] 这是我修复的代码,似乎仅适用于小型数组。在下面的示例中,数组大小为10000,总和错误。为什么会计算错误的总和?
import java.util.Random;
import java.util.concurrent.ForkJoinPool;
import java.util.concurrent.ForkJoinTask;
import java.util.concurrent.RecursiveTask;
public class ForkJoinPoolTest {
public static void main(String[] args) {
Random r = new Random();
int low = 10000;
int high = 100000;
int size = 100000;
long[] numbers = new long[size];
int sum = 0;
for (int i = 0; i < size; i++) {
int n = r.nextInt(high - low) + low;
numbers[i] = n;
sum += numbers[i];
}
long s = System.currentTimeMillis();
ForkJoinPool pool = new ForkJoinPool(1);
AdditionTask newTask = new AdditionTask(numbers,numbers.length-1);
ForkJoinTask<Long> submit = pool.submit(newTask);
System.out.println("Expected Answer: " + sum + ",Actual: " + submit.join());
long e = System.currentTimeMillis();
System.out.println("Total time taken: " + (e - s) + " ms in parallel Operation");
long s2 = System.currentTimeMillis();
System.out.println("Started: " + s2);
int manualSum = 0;
for (long number : numbers) {
manualSum += number;
}
System.out.println("Expected Answer: " + sum + ",Actual: " + manualSum);
long e2 = System.currentTimeMillis();
System.out.println("Ended: " + e2);
System.out.println("Total time taken: " + (e2 - s2) + " ms in sequential Operation");
}
}
class AdditionTask extends RecursiveTask<Long> {
long[] numbers;
int start;
int end;
public AdditionTask(long[] numbers,int end) {
this.numbers = numbers;
this.start = start;
this.end = end;
}
@Override
protected Long compute() {
int length = (start == 0) ? end +1 : (end - (start - 1));
if (length > 2) {
int mid = (length % 2 == 0) ? length / 2 : (length - 1) / 2;
AdditionTask leftSide = new AdditionTask(numbers,start,(start+mid));
leftSide.fork();
AdditionTask rightSide = new AdditionTask(numbers,(start+mid)+1,end);
Long rightSideLong = rightSide.compute();
Long leftSideLong = leftSide.join();
Long total = rightSideLong + leftSideLong;
return total;
} else {
if (start == end) {
return numbers[start];
}
return numbers[start] + numbers[end];
}
}
}
解决方法
您的并行计算的第二个版本是正确的。但是代码中的两个非并行计算都被破坏了,因为它们使用int
作为它们的和,这对于大型数组将溢出。修复它们后,如果还使用long
,它们将产生与并行计算相同的结果。
不过,还有一些地方需要改进。首先,您应该摆脱这些条件:
int length = (start == 0) ? end +1 : (end - (start - 1));
和
int mid = (length % 2 == 0) ? length / 2 : (length - 1) / 2;
与简单的方法相比,它们没有任何好处
int length = end - (start - 1); // or end - start + 1
和
int mid = length / 2;
然后,有效的并行处理不应尽可能地分解,而应包含实际可实现的并行性。您可以为此使用getSurplusQueuedTaskCount()
@Override
protected Long compute() {
int length = end - (start - 1);
// only split when benefit from parallel processing is likely
if (length > 2 && getSurplusQueuedTaskCount() < 2) {
int mid = length / 2;
AdditionTask leftSide = new AdditionTask(numbers,start,(start+mid));
leftSide.fork();
AdditionTask rightSide = new AdditionTask(numbers,(start+mid)+1,end);
Long rightSideLong = rightSide.compute();
// do in this thread if no worker thread has picked it up yet
Long leftSideLong = leftSide.tryUnfork()? leftSide.compute(): leftSide.join();
Long total = rightSideLong + leftSideLong;
return total;
} else { // do sequential
long sum = 0;
for(int ix = start; ix <= end; ix++) sum += numbers[ix];
return sum;
}
}