获得与此哈希码函数的HashCode冲突的可能性有多大？

问题描述

1. 对于键[0]的随机int值，它在多大程度上获得与下面的函数的HashCode冲突]，键[1]，键[2]，键[3]


2. 带有以下约束的随机键值
   
   
   
   • 键[0]< 1,000,000
   
   
   • 键[1]< 10,000
   
   
   • 键[2]< 1,000
   
   
   • 键[3]< 1,000
   
   

假设我们有1000万

  int [] key = new int [4]; 
 public override int GetHashCode（）
 {
 //使用大的素数来创建唯一的哈希键
 //使用偶数幂减去2创建散列偏移量方法，这给了
 //大部分时间。 
 int hashKey = 0; 
 hashKey + = 2047 * key [0]; 
 hashKey + = 8191 * key [1]; 
 hashKey + = 32767 * key [2]; 
 hashKey + = 131071 * key [3]; 
 return hashKey; 
} 
  

解决方案

我写了一个快速脚本

  import random 
 
 def hash（key）：
 hashKey = 0 
 hashKey + = 2047 * key [0] 
 hashKey + = 8191 * key [1] 
 hashKey + = 32767 * key [2] 
 hashKey + = 131071 * key [ 3] 
 return hashKey 
 
 seen = set（）
 collisions = 0 
 for i in range（0,10000000）：
x = hash（[ random.randint（0,1000000），random.randint（0,10000），random.randint（0,1000），random.randint（0,1000）]）
如果看到x：
碰撞+ = 1 
 else：
 seen.add（x）
 
打印冲突
  

当我运行它告诉我我有23735次碰撞。
我也尝试了一百万个元素，我有247个碰撞。这两个数字是4次以上的平均值。

How likely is it to get a HashCode collision with the function below in following scenarios.

1. With random int values for key[0],key[1], key[2], key[3]
2. With random key values with the following constraints
   • key[0] <1,000,000
   • key[1] <10,000
   • key[2] <1,000
   • key[3] <1,000

Assume we have 10 Million objects.

int[] key=new int[4];
public override int GetHashCode()
{
    // Use large prime multiples to create a unique hash key
    // Create the hash offsets using a "even powers of 2 minus 1" method, which gives
    // primes most of the time.
    int hashKey = 0;
    hashKey += 2047 * key[0];
    hashKey += 8191 * key[1];
    hashKey += 32767 * key[2];
    hashKey += 131071 * key[3];
    return hashKey;
}

解决方案

I wrote a quick script to test this.

import random

def hash(key):
    hashKey = 0
    hashKey += 2047 * key[0]
    hashKey += 8191 * key[1]
    hashKey += 32767 * key[2]
    hashKey += 131071 * key[3]
    return hashKey

seen = set()
collisions = 0
for i in range(0,10000000):
    x = hash([random.randint(0,1000000), random.randint(0,10000), random.randint(0,1000), random.randint(0,1000)])
    if x in seen:
        collisions += 1
    else:
        seen.add(x)

print collisions

When I ran it, it told me I got 23735 collisions.I also tried it on one million elements, and I got 247 collisions. Both numbers are averages over 4 runs.

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