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### Re: [OT] The statistics of hashing.

by moritz (Cardinal)
 on Apr 01, 2012 at 10:44 UTC ( #962873=note: print w/replies, xml ) Need Help??

in reply to [OT] The statistics of hashing.

I'm not sure I understood the question correctly, but I hope my answer is still helpful for you.

Further to that, if the number of sub hashes and vectors is increased to 10, by permuting each of the four sub hashes against the other 3 and XORing, does that substantially decrease the odds of false matches? (Or increase them? Or leave them the same?)

Statistically, the only thing that matters for the probability of collisions is the total entropy of the hash. If you create a 128bit hash over a string, you have 128 bits of entropy. If you create two 128bit hashes, one over the first part and one over the second part of the string, you have 256 bits of entropy. If you XOR the two hashes, and use the results, you have 128 bits again.

If you use one 128bit hash, the probabilities for getting no collision on inserting the nth number is

```P(no_collision, n) = (1 - 1/2**128) ** n

I'm pretty sure that the number of collisions follows (at least approximately) a Poisson distribution, though I haven't yet figured out how to calculate the expectation value.

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Re^2: [OT] The statistics of hashing.
by BrowserUk (Pope) on Apr 01, 2012 at 11:30 UTC

Hm. Attempting calculate 1 - 1/2**128 using my calculator just returns 1.

So then I tried Math::BigFloat, and still got just 1. What Am I doing wrong?

```use Math::BigFloat;

\$moritzFactor = Math::BigFloat->new( 1 )->bsub(
Math::BigFloat->new( 1 )->bdiv(
Math::BigFloat->new( 2 )->bpow( 128 )
)
);

print \$moritzFactor;
;;
1

[0] Perl> print \$moritzFactor->bstr;;
1

[0] Perl> print \$moritzFactor->bsstr;;
1e+0

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The start of some sanity?

You'd need a floating-point number with a mantissa of about 128 bits to hold the value of (1-1/2**128)

Luckily there are other ways to evaluate the expression. If we set x = 1/2**128, we can do a taylor series to evaluate (1-x)**n around x = 0 (and since x is very small, it's a good approximation:

```(1 - x)^n = 1 - n * x + 1 / 2 * n * (n - 1) * x ** 2 + ...

Where higher orders can be neglected. If you chose an n that is high enough, you can get non-1 values as the result.

Though I'm not sure if that's really helpful. What you'd really need is the expectation value of the number of collisions (the lambda symbol in Poisson Distribution, then you could easily calculate everything you wanted.

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