更新: 我已经意识到,由于涉及大量数据(15k项以上),因此无法以当前形式回答以下问题。 我刚刚发现,我要帮助的小组只允许其运行一个月,然后终止该小组以使用结果(这就是为什么他们希望在更快的时间内获得更多结果)。这对我来说似乎很疯狂,因为它们仅使用前几组数据(大列表中的最后一项永远不会使用)。因此,我正在修订此问题,以获取预期输出的样本(解决方案的近似值而不是完整的解决方案)。在更短的时间内完成此操作的最佳方法是什么?他们似乎想要多样化的结果样本,是遗传算法起作用还是某种抽样技术?其余问题保持不变(相同的输入/输出),但我
我的问题不完全是背包问题,而是非常接近的问题。基本上,我正在尝试查找等于特定值的X个项目的每种组合。我从一个我的朋友那里听说过这个问题,他在一个小型学校研究实验室工作,该过程运行了大约25天才能完成。看起来真的很可怕,所以我提供了帮助(对我来说是好处,是我可以学习和帮助一些非常好的人),所以我想出了如何通过多线程对其进行扩展(我将在下面的代码中提供),减少了几天的处理时间,但我仍然不满意,所以我只是将代码移植到GPU上工作,但我感到不满意(尽管他们很高兴,因为它速度更快,我捐赠了我的旧显卡),因为我只是在利用硬件,而不是在利用任何算法。马上,
因此,在这种背景下,我可以做些什么来通过算法加快速度吗?我的直觉告诉我没有,因为既然他们需要每种组合,从逻辑上看,计算机必须处理每种组合(数十亿),但是我之前在这里已经看到了令人惊奇的事情,即使很小的提速也可以在几天的处理中有所作为。
我喜欢超过10个版本的代码,但这是一个使用多线程的Java版本(但此代码与gpu之间的逻辑几乎相同)。
基本逻辑:
for (int c = 100; c >= 0; c--) { if (c * x_k == current.sum) { //if result is correct then save solutions.add(new Context(0, 0, newcoeff)); continue; } else if (current.k > 0) { //if result is not equal but not end of list then send to queue contexts.add(new Context(current.k - 1, current.sum - c * x_k, newcoeff)); } }
完整代码:
import java.util.Arrays; import java.util.ArrayDeque; import java.util.Queue; import java.util.concurrent.ConcurrentLinkedQueue; import java.util.concurrent.LinkedBlockingDeque; import java.util.concurrent.Callable; import java.util.concurrent.ExecutorService; import java.util.concurrent.ThreadPoolExecutor; import java.util.concurrent.TimeUnit; public class MixedParallel { // pre-requisite: sorted values !! private static final int[] data = new int[] { -5,10,20,30,35 }; // Context to store intermediate computation or a solution static class Context { int k; int sum; int[] coeff; Context(int k, int sum, int[] coeff) { this.k = k; this.sum = sum; this.coeff = coeff; } } // Thread pool for parallel execution private static ExecutorService executor; // Queue to collect solutions private static Queue<Context> solutions; static { final int numberOfThreads = 2; executor = new ThreadPoolExecutor(numberOfThreads, numberOfThreads, 1000, TimeUnit.SECONDS, new LinkedBlockingDeque<Runnable>()); // concurrent because of multi-threaded insertions solutions = new ConcurrentLinkedQueue<Context>(); } public static void main(String[] args) { System.out.println("starting.."); int target_sum = 100; // result vector, init to 0 int[] coeff = new int[data.length]; Arrays.fill(coeff, 0); mixedPartialSum(data.length - 1, target_sum, coeff); executor.shutdown(); // System.out.println("Over. Dumping results"); while(!solutions.isEmpty()) { Context s = solutions.poll(); printResult(s.coeff); } } private static void printResult(int[] coeff) { StringBuffer sb = new StringBuffer(); for (int i = coeff.length - 1; i >= 0; i--) { if (coeff[i] > 0) { sb.append(data[i]).append(" * ").append(coeff[i]).append(" + "); } } System.out.println(sb); } private static void mixedPartialSum(int k, int sum, int[] coeff) { int x_k = data[k]; for (int c = 0; c <= 100; c++) { coeff[k] = c; int[] newcoeff = Arrays.copyOf(coeff, coeff.length); if (c * x_k == sum) { //printResult(newcoeff); solutions.add(new Context(0, 0, newcoeff)); continue; } else if (k > 0) { if (data.length - k < 2) { mixedPartialSum(k - 1, sum - c * x_k, newcoeff); // for loop on "c" goes on with previous coeff content } else { // no longer recursive. delegate to thread pool executor.submit(new ComputePartialSum(new Context(k - 1, sum - c * x_k, newcoeff))); } } } } static class ComputePartialSum implements Callable<Void> { // queue with contexts to process private Queue<Context> contexts; ComputePartialSum(Context request) { contexts = new ArrayDeque<Context>(); contexts.add(request); } public Void call() { while(!contexts.isEmpty()) { Context current = contexts.poll(); int x_k = data[current.k]; for (int c = 0; c <= 100; c++) { current.coeff[current.k] = c; int[] newcoeff = Arrays.copyOf(current.coeff, current.coeff.length); if (c * x_k == current.sum) { //printResult(newcoeff); solutions.add(new Context(0, 0, newcoeff)); continue; } else if (current.k > 0) { contexts.add(new Context(current.k - 1, current.sum - c * x_k, newcoeff)); } } } return null; } } }
以下是数据/方法的一些特征:
newcoeff
Update2:我认为问题本身很容易/很常见,但是问题正在大规模运行,所以这是我进行测试时获得的真实数据(虽然不大,但大约有3000项,因此如果要测试您不必自己生成它):
private static final int target_sum = 5 * 1000; private static final List<Integer> data = Arrays.asList( -193, -138, -92, -80, -77, -70, -63, -61, -60, -56, -56, -55, -54, -54, -51, -50, -50, -50, -49, -49, -48, -46, -45, -44, -43, -43, -42, -42, -42, -42, -41, -41, -40, -40, -39, -38, -38, -38, -37, -37, -37, -37, -37, -36, -36, -36, -35, -34, -34, -34, -34, -34, -34, -34, -33, -33, -33, -32, -32, -32, -32, -32, -32, -32, -32, -31, -31, -31, -31, -31, -31, -31, -30, -30, -30, -30, -30, -29, -29, -29, -29, -29, -29, -29, -29, -29, -28, -28, -28, -28, -27, -27, -27, -27, -26, -26, -26, -26, -26, -26, -25, -25, -25, -25, -25, -25, -25, -25, -24, -24, -24, -24, -24, -24, -24, -24, -24, -24, -23, -23, -23, -23, -23, -23, -23, -23, -22, -22, -22, -22, -22, -22, -22, -22, -22, -21, -21, -21, -21, -21, -21, -21, -20, -20, -20, -20, -20, -20, -20, -19, -19, -19, -19, -19, -19, -19, -19, -19, -19, -19, -19, -19, -19, -18, -18, -18, -18, -18, -18, -18, -18, -18, -18, -18, -18, -18, -18, -18, 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13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 20, 20, 20, 20, 20, 20, 20, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 28, 28, 28, 28, 28, 28, 29, 29, 29, 29, 30, 30, 30, 30, 30, 30, 30, 31, 31, 31, 31, 31, 32, 32, 32, 32, 33, 33, 33, 33, 34, 34, 34, 34, 34, 34, 34, 35, 35, 35, 35, 36, 36, 36, 36, 37, 37, 38, 39, 39, 39, 40, 41, 41, 41, 41, 41, 42, 42, 43, 43, 44, 45, 45, 46, 47, 47, 48, 48, 49, 49, 50, 54, 54, 54, 55, 55, 56, 56, 57, 57, 57, 57, 57, 58, 58, 58, 59, 60, 66, 67, 68, 70, 72, 73, 73, 84, 84, 86, 92, 98, 99, 105, 114, 118, 120, 121, 125, 156);
我正在学习编程和算法,因此,如果我错过了任何东西或没有意义的东西,请告诉我。请注意,我了解到这可能是不可能的,因为数据量很大,但事实并非如此(我已经看到它运行了,并且已经运行了很多年)。请不要让规模分散实际问题的注意力,如果您使用10个变量并将其速度提高10%,那么我的10个变量的蛮力将使我确信,在更大的数据上它将更快。我不是想熄灭灯火,而是在寻找可以提供甚至更快的结果的小改进(或更大的设计改进)(我将研究每个建议)。另外,如果有任何需要放松的假设,请告诉我。
谢谢!
它使用动态编程来解决您在示例中给出的相同问题。通过跟踪值的索引而不是值来更新重复值以处理重复值,并更正了忽略了一些解决方案的错误。
public class TurboAdder { private static final int[] data = new int[] { 5, 10, 20, 25, 40, 50 }; private static class Node { public final int index; public final int count; public final Node prevInList; public final int prevSum; public Node(int index, int count, Node prevInList, int prevSum) { this.index = index; this.count = count; this.prevInList = prevInList; this.prevSum = prevSum; } } private static int target = 100; private static Node sums[] = new Node[target+1]; // Only for use by printString. private static boolean forbiddenValues[] = new boolean[data.length]; public static void printString(String prev, Node n) { if (n == null) { System.out.println(prev); } else { while (n != null) { int idx = n.index; // We prevent recursion on a value already seen. if (!forbiddenValues[idx]) { forbiddenValues[idx] = true; printString((prev == null ? "" : (prev+" + "))+data[idx]+"*"+n.count, sums[n.prevSum]); forbiddenValues[idx] = false; } n = n.prevInList; } } } public static void main(String[] args) { for (int i = 0; i < data.length; i++) { int value = data[i]; for (int count = 1, sum = value; count <= 100 && sum <= target; count++, sum += value) { for (int newsum = sum+1; newsum <= target; newsum++) { if (sums[newsum - sum] != null) { sums[newsum] = new Node(i, count, sums[newsum], newsum - sum); } } } for (int count = 1, sum = value; count <= 100 && sum <= target; count++, sum += value) { sums[sum] = new Node(i, count, sums[sum], 0); } } printString(null, sums[target]); } }