CODEWITHSUNDEEP
mappingintegerrange
David Rutten
David Rutten

Reputation: 4806

Mapping integers onto the entire range

I'm using a hash table (DotNET Dictionary object) as part of a sparse two-dimensional data set. Most of the entries in the hash table will be close together. I'll probably end up with 100 ~ 10,000 entries, all of them clustered near zero. I've read that a Hash table performs better when the hashes are spread across the entire integer(32 bit) range.

Is there a cheap way to map consecutive integers onto wildly different values in a 1:1 fashion? I don't have to map them back, it's purely a one-way thing.

Upvotes: 0

Views: 303

Answers (3)

dan-gph
dan-gph

Reputation: 16909

Maybe I'm misunderstanding what you are saying, but Dictionary will already hash your integers. There shouldn't be any need to pre-hash them. Why not try out the default implementation and see how it goes instead of attempting a pre-optimizion that will in all likelihood be pointless.

Upvotes: 3

Mitch Wheat
Mitch Wheat

Reputation: 300559

If you know the maximum value of your keyset (kmax), you could expand by a constant factor (multiplier), say multiply by a fixed prime number that keeps the product below the max integer size (2^31 - 1):

i.e. the nearest prime number to (2^30) / kmax

Note: make sure the prime used is not the same as the number of buckets in the Hash table.

Here is another solution: Since the .NET Random class will generate the same value for the same seed, you could use that to distribute the incoming keys.

Upvotes: 1

Erich
Erich

Reputation: 655

Instead of using Integer, write a class that Inherits from Integer, and override the GetHashCode function. This way you don't have to do anything but create this function!

The easiest way I can think of to spread out the values evenly is to do something like:

public class MyInteger:Integer
{
    public override int GetHashCode()
    {
       unchecked
       {
           return (int)Math.Pow(this,this);
       }
    }
}

Nice and evenly split up, while keeping the effort to a minimum.

Upvotes: 1

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