很久以前,我花1.25美元在便宜货桌上买了一本数据结构的书。在这篇文章中,哈希函数的解释说,由于“数学的本质”,它最终应该被一个质数mod。

你对一本1.25美元的书有什么期待?

不管怎么说,我花了很多年思考数学的本质,但还是没弄明白。

当有质数个桶时,数字的分布真的更均匀吗?

或者这是一个老程序员的故事,每个人都接受,因为其他人都接受?


当前回答

只是把从答案中得到的一些想法写下来。

Hashing uses modulus so any value can fit into a given range We want to randomize collisions Randomize collision meaning there are no patterns as how collisions would happen, or, changing a small part in input would result a completely different hash value To randomize collision, avoid using the base (10 in decimal, 16 in hex) as modulus, because 11 % 10 -> 1, 21 % 10 -> 1, 31 % 10 -> 1, it shows a clear pattern of hash value distribution: value with same last digits will collide Avoid using powers of base (10^2, 10^3, 10^n) as modulus because it also creates a pattern: value with same last n digits matters will collide Actually, avoid using any thing that has factors other than itself and 1, because it creates a pattern: multiples of a factor will be hashed into selected values For example, 9 has 3 as factor, thus 3, 6, 9, ...999213 will always be hashed into 0, 3, 6 12 has 3 and 2 as factor, thus 2n will always be hashed into 0, 2, 4, 6, 8, 10, and 3n will always be hashed into 0, 3, 6, 9 This will be a problem if input is not evenly distributed, e.g. if many values are of 3n, then we only get 1/3 of all possible hash values and collision is high So by using a prime as a modulus, the only pattern is that multiple of the modulus will always hash into 0, otherwise hash values distributions are evenly spread

其他回答

关于素数幂模的“数学的本质”是它们是有限域的一个组成部分。另外两个构建块是加法运算和乘法运算。素模的特殊性质是,它们用“常规”的加法和乘法运算形成一个有限域,只是取到模。这意味着每一个乘法都映射到一个不同的整数对质数求模,每一个加法也是如此。

质模的优势在于:

它们在二次哈希中选择次乘数时给予了最大的自由,除了0之外的所有乘数最终都将访问所有元素一次 如果所有哈希值都小于模量,则根本不会发生碰撞 随机质数比两个模的幂更好地混合,并压缩所有比特的信息,而不仅仅是一个子集

然而,它们有一个很大的缺点,它们需要整数除法,这需要很多(~ 15-40)个周期,即使在现代CPU上也是如此。用大约一半的计算就可以确保散列混合得很好。两次乘法和异移运算比一个质数模更容易混合。然后,我们可以使用任何哈希表的大小,哈希约简是最快的,对于2个表大小的幂,总共给出7个操作,对于任意大小的表,大约9个操作。

我最近研究了许多最快的哈希表实现,其中大多数都不使用质数模块。

哈希表索引的分布主要依赖于所使用的哈希函数。质数模量不能修复一个坏的哈希函数,一个好的哈希函数也不能从质数模量中受益。然而,在某些情况下,它们可能是有利的。例如,它可以修复半坏的哈希函数。

我想为Steve Jessop的回答补充一些东西(我不能评论,因为我没有足够的声誉)。但我找到了一些有用的材料。他的回答很有帮助,但他犯了一个错误:桶的大小不应该是2的幂。我引用Thomas Cormen, Charles Leisersen等人写的《算法导论》263页

When using the division method, we usually avoid certain values of m. For example, m should not be a power of 2, since if m = 2^p, then h(k) is just the p lowest-order bits of k. Unless we know that all low-order p-bit patterns are equally likely, we are better off designing the hash function to depend on all the bits of the key. As Exercise 11.3-3 asks you to show, choosing m = 2^p-1 when k is a character string interpreted in radix 2^p may be a poor choice, because permuting the characters of k does not change its hash value.

希望能有所帮助。

插入/从哈希表中检索时要做的第一件事是计算给定键的hashCode,然后通过执行hashCode % table_length将hashCode修剪为哈希表的大小来找到正确的bucket。这里有两个“陈述”,你很可能在某处读到过

如果对table_length使用2的幂,那么查找(hashCode(key) % 2^n)就像查找(hashCode(key) & (2^n -1))一样简单快捷。但是如果你为一个给定的键计算hashCode的函数不是很好,你肯定会在几个散列桶中聚集许多键。 但是,如果table_length使用质数,即使使用稍微愚蠢的hashCode函数,计算出来的hashCode也可以映射到不同的散列桶中。

这就是证明。

如果假设你的hashCode函数的结果是以下hashCode {x, 2x, 3x, 4x, 5x, 6x…},那么所有这些都将聚集在m个桶中,其中m = table_length/GreatestCommonFactor(table_length, x)。(验证/推导这个很简单)。现在可以执行以下操作之一来避免集群

确保你不会生成太多的hashCode,这些hashCode是另一个hashCode的倍数,比如{x, 2x, 3x, 4x, 5x, 6x…}。但如果你的hashTable应该有数百万个条目,这可能有点困难。 或者通过使GreatestCommonFactor(table_length, x)等于1使m等于table_length,即使table_length与x为coprime。如果x可以是任何数字,则确保table_length是质数。

来自- http://srinvis.blogspot.com/2006/07/hash-table-lengths-and-prime-numbers.html

http://computinglife.wordpress.com/2008/11/20/why-do-hash-functions-use-prime-numbers/

解释得很清楚,还有图片。

编辑:作为一个总结,使用质数是因为当数值乘以所选质数并将它们全部相加时,获得唯一值的可能性最大。例如,给定一个字符串,将每个字母的值与质数相乘,然后将它们全部相加,就会得到它的哈希值。

一个更好的问题是,为什么是数字31?

Primes are unique numbers. They are unique in that, the product of a prime with any other number has the best chance of being unique (not as unique as the prime itself of-course) due to the fact that a prime is used to compose it. This property is used in hashing functions. Given a string “Samuel”, you can generate a unique hash by multiply each of the constituent digits or letters with a prime number and adding them up. This is why primes are used. However using primes is an old technique. The key here to understand that as long as you can generate a sufficiently unique key you can move to other hashing techniques too. Go here for more on this topic about http://www.azillionmonkeys.com/qed/hash.html

http://computinglife.wordpress.com/2008/11/20/why-do-hash-functions-use-prime-numbers/