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Re^5: [OT] The statistics of hashing. (formula)by tye (Sage)
|on Apr 01, 2012 at 21:32 UTC||Need Help??|
If you could distill your write-up to a formula, that would be helpful.
I already did that for the most useful and most accurate calculation (as well as for other parts):
It is just $prod += $odds-$prod*$odds (and that last term doesn't matter for a very long time). Just to be clear, $odds is just $b1*$b2*$b3*$b4/2**(32*4) where $b1 is the number of bits set in the first vector.
Looking at what is not explicit there, the only thing I find moderately difficult to determine is that you can start with $prod initialized to 0. The value being calculated is the probability (a value between 0 and 1) that there has been at least one 4-hash collision (due to random chance) so far.
The upper bound on this value that I pre-calculated by discounting the impact of single-bit collisions is achieved by simply assuming $b1 through $b4 are all equal to the number of insertions done so far.
This is substantially accurate for quite a while. Making it more accurate is seriously difficult other than by computing run-specific odds as is done by the two formulae I quoted above.
I find it difficult to predict how tightly this upper bound is likely to fit against some run-specific odds when the number of inserts is a significant fraction of the number of bits per hash, as you are trying to do.
Your 10 hash-values via XOR is not something that I can predict the results of, other than that I expect it to lie somewhere between the two very remote extremes of "no better than 4 independent hash values" and "just as good as 10 fully independent hash values". Your results so far seem to be somewhat close to what I would expect if you have 10 fully independent hash values per string.
So my calculations completely ignored the "10 hashes"
If I wanted to gain additional buckets, then I would compute the additional hash values using a (moderately) cryptographically sound process. For example, you could compute md5($string) and md5('BrowserUk'.$string) to get 8 hashes that should be substantially independent (based on prior difficult analysis of the MD5 algorithm). Though, your experiment may prove that your strategy is sufficient.