关联非对易散列函数 [英] Associative noncommutative hash function
问题描述
- 是关联的
- 不可交换
- 可以在32位整数上轻松实现:
int32 hash(int32,int32)
如果我是正确的,这样的函数可以实现以下目标:
$ b $ ul
迄今为止我发现的最好的方法是乘以4x4矩阵的比特,但这很难实现,并将空间减少到16比特。
我很感激任何帮助。
这就是我想出的在Java中)。
基本思路是将32bit-int分成2个数字。包含乘法效应的旧比特和。年轻的位跟踪乘法效应。
它的工作原理。它具有良好的分布 - 也可以与(0,1),(1,0)等常见参数相比。
public class AssociativelyMergedIntegers {
/ **最大无符号16位素数* /
private static final int PRIME = 65521;
$ b $ / **联合式,不可交换散列函数* /
public static int merged(int first,int second){
int firstFactor = remainingOf(first& 0x0000FFFF);
int secondFactor = remainingOf(second& 0x0000FFFF);
int firstSum = remainingOf(first>>> 16& 0x0000FFFF);
int secondSum = remainingOf(second>>> 16& 0x0000FFFF);
int resultSum = remainingOf(firstSum +(long)firstFactor * secondSum);
int resultFactor = remainingOf((long)firstFactor * secondFactor);
返回resultSum<< 16 ^ resultFactor;
private static int remainingOf(long number){
int rest =(int)(number%PRIME);
return rest == 0
? PRIME - 2
:休息;
}
}
Is there a hash function with following properties?
- is associative
- is not commutative
- easily implementable on 32 bit integers:
int32 hash(int32, int32)
If I am correct, such function allows achieving following goals
- calculate hash of concatenated string from hashes of substrings
- calculate hash concurrently
- calculate hash of list implemented on binary tree - including order, but excluding how tree is balanced
The best I found so far is multiplication of 4x4 matrix of bits, but thats awkward to implement and reduces space to 16bits.
I am grateful for any help.
This is what I came up with (written in Java). Basic idea is to split 32bit-int into 2 numbers. Older bits sums including multiplication effect. Younger bits tracks that multiplication effect. It works. It has good distribution - also against common arguments like (0, 1), (1, 0).
public class AssociativelyMergedIntegers {
/** biggest unsigned 16-bit prime */
private static final int PRIME = 65521;
/** associative, not commutative hash function */
public static int merged(int first, int second) {
int firstFactor = remainderOf(first & 0x0000FFFF);
int secondFactor = remainderOf(second & 0x0000FFFF);
int firstSum = remainderOf(first >>> 16 & 0x0000FFFF);
int secondSum = remainderOf(second >>> 16 & 0x0000FFFF);
int resultSum = remainderOf(firstSum + (long) firstFactor * secondSum);
int resultFactor = remainderOf((long) firstFactor * secondFactor);
return resultSum << 16 ^ resultFactor;
}
private static int remainderOf(long number) {
int rest = (int) (number % PRIME);
return rest == 0
? PRIME - 2
: rest;
}
}
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