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I have a simple struct consists of a fixed size string and an integer. I need to use this struct as the key for a hash table. I have a hash function for sting, Hs(string), and a hash function for integer, Hi(int), I'm wondering if the hash function for this simple struct would just be H(struct) = Hs(string) + Hi(int)? Alternatively, I could encode the integer into a string and append it to the string, then just use the string hash function. Any suggestions? Thanks.

atsu
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In order to figure the "how" we need to answer a few questions first:
a) how many items you should be able to accommodate ?
b) what's the size of the hash table ?
c) how many combinations of <string X int> are there (since the string has a fixed size it's easy to calculate) ?

Once you figure that out - it will be easier to find a hashing function that will minimize collisions, for example, there might be a case where using Hs(string) is good enough!

Nir Alfasi
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  • the string portion is something like uuid, evenly distributed, the integer is something like the elapsed seconds since an early starting time. I would say the number of entries in the table could reach a few millions and more. – atsu Sep 26 '13 at 03:23
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    According to what I read, a loading factor of 75% is considered good. If the UUID provides a good distribution then I don't see any advantage of using the int in the hashing algorithm. – Nir Alfasi Sep 26 '13 at 03:31
  • There are entries with the same uuid but different integer values. – atsu Sep 26 '13 at 03:35
  • @atsu right, and you can use it to identify in which `bucket` the `` is (when there is a collision). – Nir Alfasi Sep 26 '13 at 03:37
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Either of those would work. My default would be Hs(string) XOR Hs(int) but plus is fine too. It just needs to probably not collide and both of those will probably not collide although Hs(string) XOR Hs(int) or Hs(string) + Hs(int) will be faster.

Frankie Robertson
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  • `int` = 32 byte (at least in Java it is) while `String` is (theoretically) limited only by the JVM - how do you XOR two items of different sizes ? – Nir Alfasi Sep 26 '13 at 03:14
  • You hash them both first. You now have something fixed size of the correct length. – Frankie Robertson Sep 26 '13 at 03:25
  • @FrankieRobertson without knowing some details of the platform/hash implementations, you might introduce a lot of bias (collisions) by simply XORing the two values. – David-SkyMesh Sep 26 '13 at 04:16