perfect hash function

Question

I'm attempting to hash the values

10, 100, 32, 45, 58, 126, 3, 29, 200, 400, 0

I need a function that will map them to an array that has a size of 13 without causing any collisions.

I've spent several hours thinking this over and googling and can't figure this out. I haven't come close to a viable solution.

How would I go about finding a hash function of this sort? I've played with gperf, but I don't really understand it and I couldn't get the results I was looking for.

Answer 1

Try the following which maps your n values to unique indices between 0 and 12 (1369%(n+1))%13

Answer 2

On some platforms (e.g. embedded), modulo operation is expensive, so % 13 is better avoided. But AND operation of low-order bits is cheap, and equivalent to modulo of a power-of-2.

I tried writing a simple program (in Python) to search for a perfect hash of your 11 data points, using simple forms such as ((x << a) ^ (x << b)) & 0xF (where & 0xF is equivalent to % 16, giving a result in the range 0..15, for example). I was able to find the following collision-free hash which gives an index in the range 0..15 (expressed as a C macro):

#define HASH(x)    ((((x) << 2) ^ ((x) >> 2)) & 0xF)

Here is the Python program I used:

data = [ 10, 100, 32, 45, 58, 126, 3, 29, 200, 400, 0 ]

def shift_right(value, shift_value):
    """Shift right that allows for negative values, which shift left
    (Python shift operator doesn't allow negative shift values)"""
    if shift_value == None:
        return 0
    if shift_value < 0:
        return value << (-shift_value)
    else:
        return value >> shift_value

def find_hash():
    def hashf(val, i, j = None, k = None):
        return (shift_right(val, i) ^ shift_right(val, j) ^ shift_right(val, k)) & 0xF

    for i in xrange(-7, 8):
        for j in xrange(i, 8):
            #for k in xrange(j, 8):
                #j = None
                k = None
                outputs = set()
                for val in data:
                    hash_val = hashf(val, i, j, k)
                    if hash_val >= 13:
                        pass
                        #break
                    if hash_val in outputs:
                        break
                    else:
                        outputs.add(hash_val)
                else:
                    print i, j, k, outputs

if __name__ == '__main__':
    find_hash()

Answer 3

Found One

I tried a few things and found one semi-manually:

(n ^ 28) % 13

The semi-manual part was the following ruby script that I used to test candidate functions with a range of parameters:

t = [10, 100, 32, 45, 58, 126, 3, 29, 200, 400, 0]
(1..200).each do |i|
  t2 = t.map { |e| (e ^ i) % 13 }
  puts i if t2.uniq.length == t.length
end

Answer 4

Just some quasi-analytical ramblings:

In your set of numbers, eleven in all, three are odd and eight are even. Looking at the simplest forms of hashing - %13 - will give you the following hash values: 10 - 3, 100 - 9, 32 - 6, 45 - 6, 58 - 6, 126 - 9, 3 - 3, 29 - 3, 200 - 5, 400 - 10, 0 - 0

Which, of course, is unusable due to the number of collisions. Something more elaborate is needed.

Why state the obvious? Considering that the numbers are so few any elaborate - or rather, "less simple" - algorithm will likely be slower than either the switch statement or (which I prefer) simply searching through an unsigned short/long vector of size eleven positions and using the index of the match.

Why use a vector search?

1. You can fine-tune it by placing the most often occuring values towards the beginning of the vector.
2. I assume the purpose is to plug in the hash index into a switch with nice, sequential numbering. In that light it seems wasteful to first use a switch to find the index and then plug it into another switch. Maybe you should consider not using hashing at all and go directly to the final switch?
3. The switch version of hashing cannot be fine-tuned and, due to the widely differing values, will cause the compiler to generate a binary search tree which will result in a lot of comparisons and conditional/other jumps (especially costly) which take time (I've assumed you've turned to hashing for its speed) and require space.
4. If you want to speed up the vector search additionally and are using an x86-system you can implement a vector search based on the assembler instructions repne scasw (short)/repne scasd (long) which will be much faster. After a setup time of a few instructions you will find the first entry in one instruction and the last in eleven followed by a few instructions cleanup. This means 5-10 instructions best case and 15-20 worst. This should beat the switch-based hashing in all but maybe one or two cases.

Answer 5

Bob Jenkins has a program for this too: http://burtleburtle.net/bob/hash/perfect.html

Unless you're very lucky, there's no "nice" perfect hash function for a given dataset. Perfect hashing algorithms usually use a simple hashing function on the keys (using enough bits so it's collision-free) then use a table to finish it off.

Answer 6

I did a quick check and using the SHA256 hash function and then doing modular division by 13 worked when I tried it in Mathematica. For c++ this function should be in the openssl library. See this post.

If you were doing a lot of hashing and lookup though, modular division is a pretty expensive operation to do repeatedly. There is another way of mapping an n-bit hash function into a i-bit indices. See this post by Michael Mitzenmacher about how to do it with a bit shift operation in C. Hope that helps.

Answer 7

if you know the exact keys then it is trivial to produce a perfect hash function -

int hash (int n) {
  switch (n) {
    case 10:   return 0;
    case 100:  return 1;
    case 32:   return 2;
    // ...
    default:   return -1;
  }
}