[Haskell-cafe] Re: Very imperfect hash function

[Holger Siegel]

Am Donnerstag, den 28.01.2010, 19:37 +0000 schrieb Maciej Piechotka:
> On Thu, 2010-01-28 at 14:07 -0500, Steve Schafer wrote:
> > I'm looking for some algorithmic suggestions:
> > 
> > I have a set of several hundred key/value pairs. The keys are 32-bit
> > integers, and are all distinct. The values are also integers, but the
> > number of values is small (only six in my current problem). So,
> > obviously, several keys map to the same value.
> > 
> > For some subsets of keys, examining only a small portion of the key's
> > bits is enough to determine the associated value. For example, there may
> > be 250 keys that all have the same most-significant byte, and all 250
> > map to the same value. There are also keys at the other extreme, where
> > two keys that differ in only one bit position map to different values.
> > 
> > The data are currently in a large lookup table. To save space, I'd like
> > to convert that into a sort of hash function:
> > 
> >  hash :: key -> value
> > 
> > My question is this: Is there any kind of generic approach that can make
> > use of the knowledge about the internal redundancy of the keys to come
> > up with an efficient function?
> > 
> > Steve Schafer
> > Fenestra Technologies Corp.
> > http://www.fenestra.com/
> 
> Maybe:
> 
> data TTree a = TTree Int (TTree a) (TTree a)
>              | TNode a
> --              | THashNode <some hash table>
> 
> hash :: TTree a -> Int32 -> a
> hash (TNode v)     _ = v
> hash (TTree b l r) k = if testBit k b then hash r k else hash l k
> -- hash (THashNode h) k = lookupHashTable h k

This looks like you have re-invented Binary Decision Diagrams (BDDs). :)

> Of course you need to code efficiently the tree.

When you fix the order in which the bits are tested, you can take
advantage of sharing. This way you reach an efficient representation
called Reduced Ordered Binary Decision Diagram (ROBDD). Unfortunately, a
bad order may lead to exponential size (in the number of bits), and
finding a good order can be NP-hard.

Regards,

Holger