为什么使用2的幂作为哈希​​大小会使哈希表比使用素数差很多？

Question

I'm implementing a hash table that is supposed to store pairs of 32-bit values. Considering my elements are fixed size, I'm using a very simple hashing function:

hash(a,b) = asUint64(a) + (asUint64(b) << 32)

With that, I get the index of an element in a hash table (that is, its corresponding bucket) with:

index(a,b) = hash(a,b) % hash_size

Where hash_size is the number of entries/buckets on my table. I've realized, though, that I could speed up this implementation a little bit if I replaced the "modulus" operator by a bitwise mod of 2 , by fixing hash_size as a power of 2. Except, when I do that, most of my pairs end up on the first bucket! Why is that happening?

Answer 1

My guess is that your data is not evenly distributed in a . Consider the concatenation of a and b as your hash code:

b31b30...b1b0a31a30...a1a0, where ai, bi is the ith bit of a,b

Suppose you have a table with a million entries, your hash index is then

a9a8...a1a0 (as an integer)

Worse, suppose a only ever ranges from 1 to 100. Then you have even less dependence on the higher order bits of a .

As you can see, if your hash table doesn't have at least 4 billion entries, your hashcode has no dependence on b at all, and hash(x, a) will collide for all x .