具有最小优先级队列的复杂性问题

时间:2016-11-16 17:25:35

标签: java algorithm queue complexity-theory timing

我是数据结构课程中的学生,我目前的任务是制作一个最小优先级队列,我已经完成并测试过,据我所知,它可行。我的问题是我们还需要为我们的课程计时并分析他们的Big-O复杂性;我设计了一个计时课程(我已经为过去的作业完成了),收集了数据并绘制了它。我的deleteMin和add方法应该有复杂度O(logn),而findMin应该有O(c),但是我的add方法由于某种原因返回O(c)。由于我反复测试了minPQ,我怀疑这个问题与我的计时方式有关,但是我很难过,希望有人可以接受我所做的事情。忽视。

TLD;我的add方法运行得比它应该更快和/或我的方法测试add方法存在问题。

编辑:有关计时器如何工作的一些其他信息;基本上它只是将随机数添加到队列中以使其大小合适,然后计算添加一个多长时间所需的时间。它从大小2 ^ startPow到2 ^ stopPow,它重复每个大小iterCount次的时间并输出平均值。

这是我的队列类:

package assignment11;

import java.io.IOException;
import java.io.PrintWriter;
import java.util.Comparator;
import java.util.NoSuchElementException;

/**
 * Represents a priority queue of generically-typed items. 
 * The queue is implemented as a min heap. 
 * The min heap is implemented implicitly as an array.
 * 
 * @author Christopher Nielson
 * @uid u0777607
 */
@SuppressWarnings("unchecked")
public class PriorityQueue<AnyType> {

    private int currentSize;

    private AnyType[] array;

    private Comparator<? super AnyType> cmp;

    /**
     * Constructs an empty priority queue. Orders elements according
     * to their natural ordering (i.e., AnyType is expected to be Comparable)
     * AnyType is not forced to be Comparable.
     */

    public PriorityQueue() {
        currentSize = 0;
        cmp = null;
        array = (AnyType[]) new Object[10]; // safe to ignore warning
    }

    /**
     * Construct an empty priority queue with a specified comparator.
     * Orders elements according to the input Comparator (i.e., AnyType need not
     * be Comparable).
     */
    public PriorityQueue(Comparator<? super AnyType> c) {
        currentSize = 0;
        cmp = c;
        array = (AnyType[]) new Object[10]; // safe to ignore warning
    }

    /**
     * @return the number of items in this priority queue.
     */
    public int size() {
        return currentSize;
    }

    /**
     * Makes this priority queue empty.
     */
    public void clear() {
        currentSize = 0;
    }

    /**
     * @return the minimum item in this priority queue.
     * @throws NoSuchElementException if this priority queue is empty.
     * 
     * (Runs in constant time.)
     */
    public AnyType findMin() throws NoSuchElementException {
        if (currentSize == 0) {
            throw new NoSuchElementException();
        }
        return array[0];
    }


    /**
     * Removes and returns the minimum item in this priority queue.
     * 
     * @throws NoSuchElementException if this priority queue is empty.
     * 
     * (Runs in logarithmic time.)
     */
    public AnyType deleteMin() throws NoSuchElementException {
        if (currentSize == 0) {
            throw new NoSuchElementException();
        }
        AnyType tmp = array[0];

        array[0] = array[currentSize - 1];
        array[currentSize - 1] = null;
        --currentSize;

        downHeap(0);

        return tmp;
    }


    /**
     * Adds an item to this priority queue.
     * 
     * (Runs in logarithmic time.) Can sometimes terminate early.
     * 
     * @param x -- the item to be inserted
     */
    public void add(AnyType x) {
        if (currentSize == array.length) {
            AnyType[] tmp = array;
            array = (AnyType[]) new Object[array.length * 2];
            for (int currentIndex = 0; currentIndex < tmp.length; currentIndex++) {
                array[currentIndex] = tmp[currentIndex];
            }
        }
        array[currentSize] = x;
        ++currentSize;

        upHeap(currentSize - 1);
    }

    /**
     * Generates a DOT file for visualizing the binary heap.
     */
    public void generateDotFile(String filename) {
        try(PrintWriter out = new PrintWriter(filename)) {
            out.println("digraph Heap {\n\tnode [shape=record]\n");

            for(int i = 0; i < currentSize; i++) {
                out.println("\tnode" + i + " [label = \"<f0> |<f1> " + array[i] + "|<f2> \"]");
                if(((i*2) + 1) < currentSize)
                    out.println("\tnode" + i + ":f0 -> node" + ((i*2) + 1) + ":f1");
                if(((i*2) + 2) < currentSize)
                    out.println("\tnode" + i + ":f2 -> node" + ((i*2) + 2) + ":f1");
            }
            out.println("}");
        } catch (IOException e) {
            System.out.println(e);
        }
    }

    /**
     * Internal method for comparing lhs and rhs using Comparator if provided by the
     * user at construction time, or Comparable, if no Comparator was provided.
     */
    private int compare(AnyType lhs, AnyType rhs) {
        if (cmp == null) {
            return ((Comparable<? super AnyType>) lhs).compareTo(rhs); // safe to ignore warning
        }
        // We won't test your code on non-Comparable types if we didn't supply a Comparator

        return cmp.compare(lhs, rhs);
    }

    /**
     * Internal method to reheapify upward.
     * 
     * @param root  Item where reheapifying will begin
     */
    private void upHeap(int root) {
        // check if root is less than parent
        if (root >= 0 && compare(array[root], array[(root - 1) / 2]) < 0) {
            AnyType tmp = array[(root - 1) / 2];
            array[(root - 1) / 2] = array[root];
            array[root] = tmp;
            upHeap((root - 1) / 2);
        }
    }

    /**
     * Internal method to reheapify downward.
     * 
     * @param root  Item where reheapifying will begin.
     */
    private void downHeap(int root) {
        // check if left child is less than root
        if ((root * 2) + 1 < currentSize && array[(root * 2) + 1] != null && compare(array[(root * 2) + 1], array[root]) < 0) {
            // check if right child is less than left child
            if ((root * 2) + 2 < currentSize && array[(root * 2) + 2] != null && compare(array[(root * 2) + 2], array[(root * 2) + 1]) < 0) {
                // swap with right child
                AnyType tmp = array[root];
                array[root] = array[(root * 2) + 2];
                array[(root * 2) + 2] = tmp;
                downHeap((root * 2) + 2);
            } else {
                // swap with left child
                AnyType tmp = array[root];
                array[root] = array[(root * 2) + 1];
                array[(root * 2) + 1] = tmp;
                downHeap((root * 2) + 1);
            }
        } else if ((root * 2) + 2 < currentSize && array[(root * 2) + 2] != null && compare(array[(root * 2) + 2], array[root]) < 0) {
            // swap with right child
            AnyType tmp = array[root];
            array[root] = array[(root * 2) + 2];
            array[(root * 2) + 2] = tmp;
            downHeap((root * 2) + 2);
        }
    }

    // LEAVE IN for grading purposes
    public Object[] toArray() {    
        Object[] ret = new Object[currentSize];
        for(int i = 0; i < currentSize; i++) {
            ret[i] = array[i];
        }
        return ret;
    }
}

这是我的计时课:

package assignment11;

import java.io.File;
import java.io.FileOutputStream;
import java.io.OutputStreamWriter;
import java.util.Random;

/**
 * @author Christopher Nielson
 * @uid u0777607
 */
public class PriorityQueueTimer {

    private static int startPow = 10;
    private static int stopPow = 24;
    private static int iterCount = 10000;
    private static Random rand;
    private static OutputStreamWriter fileOut;

    public static void main(String[] args) {
        timeAdd();
//      timeDeleteMin();
//      timeFindMin();
        System.out.println("Finished timing!");
    }

    /**
     * Times add method of PriorityQueue for different data sizes and outputs results to a file.
     */
    private static void timeAdd() {
        try {
            fileOut = new OutputStreamWriter(new FileOutputStream(new File("assignment11-addTimer.csv")));
            PriorityQueue<Integer> addTimer;
            for (int currentPow = startPow; currentPow <= stopPow; currentPow++) {
                int dataSize = (int) Math.pow(2,  currentPow);
                System.out.print("Timing add on datasize " + dataSize + " (Power: " + currentPow + ")...");
                long totalTime = 0;
                long stopTime;

                addTimer = new PriorityQueue<Integer>();
                rand = new Random(13);

                for (int currentRand = 0; currentRand < dataSize; currentRand++) {
                    addTimer.add(rand.nextInt());
                }

                long startTime = System.nanoTime();
                while (System.nanoTime() - startTime < 1000000){}

                for (int currentIter = 0; currentIter < iterCount; currentIter++) {
                    startTime = System.nanoTime();
                    addTimer.add(rand.nextInt());
                    stopTime = System.nanoTime();
                    addTimer.deleteMin();
                    totalTime += stopTime - startTime;
                }
                System.out.println("Finished!");
                fileOut.write(dataSize + "\t" + (totalTime / iterCount) + "\n");
                fileOut.flush();
            }
            fileOut.close();
        } catch(Exception e) {
            System.err.println(e.getMessage());
            e.printStackTrace();
        }
    }

    /**
     * Times deleteMin method of PriorityQueue for different data sizes and outputs results to a file.
     */
    private static void timeDeleteMin() {
        try {
            fileOut = new OutputStreamWriter(new FileOutputStream(new File("assignment11-deleteMinTimer.csv")));
            PriorityQueue<Integer> deleteTimer;
            for (int currentPow = startPow; currentPow <= stopPow; currentPow++) {
                int dataSize = (int) Math.pow(2,  currentPow);
                System.out.print("Timing deleteMin on datasize " + dataSize + " (Power: " + currentPow + ")...");
                long totalTime = 0;
                long stopTime;

                deleteTimer = new PriorityQueue<Integer>();
                rand = new Random(13);

                for (int currentRand = 0; currentRand < dataSize; currentRand++) {
                    deleteTimer.add(rand.nextInt());
                }

                long startTime = System.nanoTime();
                while (System.nanoTime() - startTime < 1000000){}

                for (int currentIter = 0; currentIter < iterCount; currentIter++) {
                    startTime = System.nanoTime();
                    deleteTimer.deleteMin();
                    stopTime = System.nanoTime();
                    deleteTimer.add(rand.nextInt());
                    totalTime += stopTime - startTime;
                }
                System.out.println("Finished!");
                fileOut.write(dataSize + "\t" + (totalTime / iterCount) + "\n");
                fileOut.flush();
            }
            fileOut.close();
        } catch(Exception e) {
            System.err.println(e.getMessage());
            e.printStackTrace();
        }
    }

    /**
     * Times findMin method of PriorityQueue for different data sizes and outputs results to a file.
     */
    private static void timeFindMin() {
        try {
            fileOut = new OutputStreamWriter(new FileOutputStream(new File("assignment11-findMinTimer.csv")));
            PriorityQueue<Integer> findTimer;
            for (int currentPow = startPow; currentPow <= stopPow; currentPow++) {
                findTimer = new PriorityQueue<Integer>();
                int dataSize = (int) Math.pow(2,  currentPow);
                System.out.print("Timing findMin on datasize " + dataSize + " (Power: " + currentPow + ")...");
                long totalTime = 0;
                long stopTime;

                rand = new Random(13);

                for (int currentRand = 0; currentRand < dataSize; currentRand++) {
                    findTimer.add(rand.nextInt());
                }

                long startTime = System.nanoTime();
                while (System.nanoTime() - startTime < 1000000){}

                for (int currentIter = 0; currentIter < iterCount; currentIter++) {
                    startTime = System.nanoTime();
                    findTimer.findMin();
                    stopTime = System.nanoTime();
                    totalTime += stopTime - startTime;
                }
                System.out.println("Finished!");
                fileOut.write(dataSize + "\t" + (totalTime / iterCount) + "\n");
                fileOut.flush();
            }
            fileOut.close();
        } catch(Exception e) {
            System.err.println(e.getMessage());
            e.printStackTrace();
        }
    }
}

这是我目前的图表结果: Timing Results

提前致谢!

2 个答案:

答案 0 :(得分:1)

以下是关于插入O(1)平均值(当然是O(log n)最差情况)效果的手工波形论证。

通过将最小值分配为根,将其余元素划分为随机子集,并在每个子集上构造子树作为根的子节点,在随机元素集上构造随机二进制堆。 (通过随机插入构建的堆的实际分布可能与此不同,但我猜测不是那么重要。)

在这个堆中,我们插入一个随机元素x。扩展数组是O(1)摊销的,因此主要成本是上堆,这与移位的元素数成正比。让我们计算一个期望值。

鉴于堆具有n个现有元素,新元素小于根的概率为1/(n+1),导致最多log (n+1)个替换元素。否则,位移被限制在(n-1)/2个元素的子树之一。 x和该子树的元素都以大于根的条件为条件,因此我们可以归纳地推断出预期成本是

       log (n+1)
T(n) = --------- + T((n - 1)/2)
         n + 1

T(0) = 0,

于是我们发现,对于n = 2^k - 1

T(2^k - 1) = k/2^k + T(2^(k-1) - 1)
           = sum_{j=0}^k j/2^j
           = O(1) by standard analysis techniques.

(所有对数均为基数2。)

答案 1 :(得分:0)

您是否尝试使用非随机输入运行add方法(例如,在增加/减少的数组上)?它可以使每次插入时实际执行~log n次操作。随机输入可能不是这种情况,因为新添加的元素不太可能是最大的元素,并且在添加随机元素时一直到根目录(可能是每个交换数量的预期值的情况)在这种情况下,插入是一个常量。虽然我没有试图实际计算它。