knuth multiplicative hash

Question

Is this a correct implementation of the Knuth multiplicative hash.

int hash(int v)
{
    v *= 2654435761;
    return v >> 32;
}

Does over

Answer 1

Ok, I looked it up in TAOCP volume 3 (2nd edition), section 6.4, page 516.

This implementation is not correct, though as I mentioned in the comments it may give the correct result anyway.

A correct way (I think - feel free to read the relevant chapter of TAOCP and verify this) is something like this: (important: yes, you must shift the result right to reduce it, not use bitwise AND. However, that is not the responsibility of this function - range reduction is not properly part of hashing itself)

uint32_t hash(uint32_t v)
{
    return v * UINT32_C(2654435761);
    // do not comment about the lack of right shift. I'm not ignoring it. read on.
}

Note the uint32_t's (as opposed to int's) - they make sure the multiplication overflows modulo 2^32, as it is supposed to do if you choose 32 as the word size. There is also no right shift by k here, because there is no reason to give responsibility for range-reduction to the basic hashing function and it is actually more useful to get the full result. The constant 2654435761 is from the question, the actual suggested constant is 2654435769, but that's a small difference that as far as I know does not affect the quality of the hash.

Other valid implementations shift the result right by some amount (not the full word size though, that doesn't make sense and C++ doesn't like it), depending on how many bits of hash you need. Or they may use an other constant (subject to certain conditions) or an other word size. Reducing the hash modulo something is not a valid implementation, but a common mistake, likely it is a de-facto standard way to do range-reduction on a hash. The bottom bits of a multiplicative hash are the worst-quality bits (they depend on less of the input), you only want to use them if you really need more bits, while reducing the hash modulo a power of two would return only the worst bits. Indeed that is equivalent to throwing away most of the input bits too. Reducing modulo a non-power-of-two is not so bad since it does mix in the higher bits, but it's not how the multiplicative hash was defined.

So to be clear, yes there is a right shift, but that is range reduction not hashing and can only be the responsibility of the hash table, since it depends on its internal size.

The type should be unsigned, otherwise the overflow is unspecified (thus possibly wrong, not just on non-2's-complement architectures but also on overly clever compilers) and the optional right shift would be a signed shift (wrong).

On the page I mention at the top, there is this formula:

Here we have A = 2654435761 (or 2654435769), w = 2³² and M = 2³². Calculating AK/w gives a fixed-point result with the format Q32.32, the mod 1 step takes only the 32 fraction bits. But that's just the same thing as doing a modular multiplication and then saying that the result is the fraction bits. Of course when multiplied by M, all the fraction bits become integer bits because of how M was chosen, and so it simplifies to just a plain old modular multiplication. When M is a lower power of two, that just right-shifts the result, as mentioned.