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

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/

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

假设表的大小(或模数)是T = (B*C)。如果你输入的散列是(N*A*B) N可以是任何整数，那么你的输出就不会很好地分布。因为每次n变成C、2C、3C等，你的输出就会开始重复。也就是说，你的输出只会分布在C位。注意这里的C是(T / HCF(表大小，哈希))。

这个问题可以通过制造hcf1来消除。质数是很好的选择。

另一个有趣的现象是当T = 2^N时。这些将给出与所有输入哈希的低N位完全相同的输出。由于每个数都可以表示为2的幂，当我们对任意数取T的模时，我们将减去所有2的幂形式的数，即>= N，因此总能得到特定模式的数，取决于输入。这也是一个糟糕的选择。

类似地，T作为10^N也是不好的，因为类似的原因(模式是十进制数而不是二进制数)。

因此，质数往往会给出更好的分布结果，因此是表大小的好选择。

我想说，这个链接的第一个答案是我找到的关于这个问题的最清晰的答案。

考虑键K ={0,1，…，100}和一个哈希表，其中桶数为m = 12。因为3是12的因数，所以是3倍数的键将被散列到是3倍数的存储桶中:

键{0,12、24、36…}将被散列到bucket 0。 键{3,15日,27日,39岁,…}将被散列到桶3。 键{42 6日,18日,30日,…}将被散列到桶6。 键{9日,21日,33岁,45岁,…}将被散列到桶9。

如果K是均匀分布的(即K中的每个键出现的可能性都是相等的)，那么m的选择就不是那么关键了。但是，如果K不是均匀分布的呢?想象最有可能出现的键是3的倍数。在这种情况下，所有不是3倍数的桶都很可能是空的(这在哈希表性能方面非常糟糕)。

这种情况比看起来更常见。例如，想象一下，您正在根据对象在内存中的存储位置来跟踪它们。如果您的计算机的字大小是4个字节，那么您将哈希键是4的倍数。不用说，选择m是4的倍数将是一个糟糕的选择:你将有3m/4个桶完全空了，所有的键都在剩下的m/4个桶中碰撞。

一般来说:

K中每一个与桶数m有公因数的键都将被哈希为这个因数的倍数。

因此，为了尽量减少碰撞，减少m和k的元素之间的公因数的数量是很重要的，这是如何实现的呢?通过选择m是一个因数很少的数，一个质数。

来自马里奥的回答。