java - 如何计算成功和失败的平均探测长度 - 线性探测(哈希表)

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我正在为我的数据结构类(class)做作业。我们被要求研究负载因子为 0.1、.2、.3、.... 和 0.9 的线性探测。测试公式为:

使用线性探测的平均探针长度大致为

成功--> ( 1 + 1/(1-L)**2)/2
或者
失败--> (1+1(1-L))/2。

我们需要使用我所做的上面的公式找到理论值(只需在公式中插入负载因子)，然后我们必须计算经验值(我不太确定该怎么做)。这是其余的要求

**For each load factor, 10,000 randomly generated positive ints between 1 and 50000 (inclusive) will be inserted into a table of the "right" size, where "right" is strictly based upon the load factor you are testing. Repeats are allowed. Be sure that your formula for randomly generated ints is correct. There is a class called Random in java.util. USE it! After a table of the right (based upon L) size is loaded with 10,000 ints, do 100 searches of newly generated random ints from the range of 1 to 50000. Compute the average probe length for each of the two formulas and indicate the denominators used in each calculationSo, for example, each test for a .5 load would have a table of > > size approximately 20,000 (adjusted to be prime) and similarly each test for a .9 load would have a table of approximate size 10,000/.9 (again adjusted to be prime).

The program should run displaying the various load factors tested, the average probe for each search (the two denominators used to compute the averages will add to 100), and the theoretical answers using the formula above. .**

我如何计算经验成功？

到目前为止，这是我的代码:

import java.util.Random;
/**
 *
 * @author Johnny
 */
class DataItem
{
    private int iData;
    public DataItem(int it)
    {iData = it;}
    public int getKey()
    {
        return iData;
    }
}

class HashTable
{
private DataItem[] hashArray;
private int arraySize;
public HashTable(int size)
{
    arraySize = size;
    hashArray = new DataItem[arraySize];
}
public void displayTable()
{
    int sp=0;
    System.out.print("Table: ");
    for(int j=0; j<arraySize; j++)
{
    if(sp>50){System.out.println("");sp=0;}

    if(hashArray[j] != null){
        System.out.print(hashArray[j].getKey() + " ");sp++;}
    else
    {System.out.print("** "); sp++;}
}
    System.out.println("");
}

public int hashFunc(int key)
{
    return key %arraySize;
}

public void insert(DataItem item)
{
    int key = item.getKey();
    int hashVal = hashFunc(key);

    while(hashArray[hashVal] != null &&
                    hashArray[hashVal].getKey() != -1)
    {
        ++hashVal;
        hashVal %= arraySize;
    }
    hashArray[hashVal]=item;
}
public int hashFunc1(int key)
{
    return key % arraySize;
}

public int hashFunc2(int key)
{
// non-zero, less than array size, different from hF1
// array size must be relatively prime to 5, 4, 3, and 2
    return 5 - key % 5;
}


public DataItem find(int key) // find item with key
// (assumes table not full)
    {
    int hashVal = hashFunc1(key); // hash the key
    int stepSize = hashFunc2(key); // get step size
    while(hashArray[hashVal] != null) // until empty cell,
    { // is correct hashVal?
        if(hashArray[hashVal].getKey() == key)
            return hashArray[hashVal]; // yes, return item
        hashVal += stepSize; // add the step
        hashVal %= arraySize; // for wraparound
    }
    return null; // can’t find item
    }
}
public class n00645805 {
/**
 * @param args the command line arguments
 */
public static void main(String[] args) {
    double b=1;
    double L;
    double[] tf = new double[9];
    double[] ts = new double[9];
    double d=0.1;
    DataItem aDataItem;
    int aKey;
    HashTable h1Table = new HashTable(100003); //L=.1
    HashTable h2Table = new HashTable(50051);  //L=.2
    HashTable h3Table = new HashTable(33343);  //L=.3
    HashTable h4Table = new HashTable(25013);  //L=.4
    HashTable h5Table = new HashTable(20011);  //L=.5
    HashTable h6Table = new HashTable(16673);  //L=.6
    HashTable h7Table = new HashTable(14243);  //L=.7
    HashTable h8Table = new HashTable(12503);  //L=.8
    HashTable h9Table = new HashTable(11113);  //L=.9

    fillht(h1Table);
    fillht(h2Table);
    fillht(h3Table);
    fillht(h4Table);
    fillht(h5Table);
    fillht(h6Table);
    fillht(h7Table);
    fillht(h8Table);
    fillht(h9Table);
    pm(h1Table);
    pm(h2Table);
    pm(h3Table);
    pm(h4Table);
    pm(h5Table);
    pm(h6Table);
    pm(h7Table);
    pm(h8Table);
    pm(h9Table);

    for (int j=1;j<10;j++)
    {
        //System.out.println(j);
        L=Math.round((b-d)*100.0)/100.0;
        System.out.println(L);
        System.out.println("ts "+(1+(1/(1-L)))/2);
        System.out.println("tf "+(1+(1/((1-L)*(1-L))))/2);
        tf[j-1]=(1+(1/(1-L)))/2;
        ts[j-1]=(1+(1/((1-L)*(1-L))))/2;
        d=d+.1;
    }
    display(ts,tf);
}
public static void fillht(HashTable a)
{
    Random r = new Random();
    for(int j=0; j<10000; j++)
    {
        int aKey;
        DataItem y;
        aKey =1+Math.round(r.nextInt(50000));
        y = new DataItem(aKey);
        a.insert(y);

    }
}
public static void pm(HashTable a)
{
    DataItem X;
    int numsuc=0;
    int numfail=0;
    int aKey;
    Random r = new Random();
    for(int j=0; j<100;j++)
    {
        aKey =1+Math.round(r.nextInt(50000));
        X = a.find(aKey);
        if(X != null)
        {
            //System.out.println("Found " + aKey);
            numsuc++;
        }
        else
        {
            //System.out.println("Could not find " + aKey);
            numfail++;
        }

    }
    System.out.println("# of succ is "+ numsuc+" # of failures is "+ numfail);
}
public static void display(double[] s, double[] f)
{

}

}

Answer 1

您应该考虑到 Java 的 HashTable使用封闭寻址(无探测)实现，因此您有单独的存储桶，可以在其中放置许多项目。这不是您在基准测试中寻找的内容。我不确定 HashMap实现，但我认为它也使用开放寻址。

所以忘记 JDK 类......因为你想计算经验值，你应该编写自己的哈希表版本，它使用带有线性探测的开放寻址实现，但是当你试图从哈希图..

例如，您可以编写哈希图，然后处理

class YourHashMap
{
   int empiricalGet(K key)
   {
     // search for the key but store the probe length of this get operation

     return probeLength;
   }
}

然后您可以通过搜索您想要的键数并计算平均探针长度来轻松地对其进行基准测试。

否则，您可以只为 hasmap 提供存储总探测长度和请求的获取计数的能力，并在基准运行后检索它们以计算平均值。

这种练习必须证明经验值与理论值一致。因此还要考虑到您可能需要许多基准的事实，然后对所有基准进行平均，确保方差不会太高。