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### Re^2: [OT] The statistics of hashing.

by syphilis (Bishop)
 on Mar 31, 2012 at 23:55 UTC ( #962810=note: print w/replies, xml ) Need Help??

in reply to Re: [OT] The statistics of hashing.
in thread [OT] The statistics of hashing.

IIUC it's just (N/4294967296)**4

I believe the figure I should have provided is:
(1 - ((4294967295/4294967296) ** N)) ** 4

And if that's not right, I give up.

Update: The logic is simple. (Danger, Will Robinson ;-)
If you're picking numbers at random in the range (1 .. 4294967296), then the probability that the N+1th pick has already come up is:
1 minus the probability that it has *not* already come up
and the probability that it has *not* already come up is (4294967295/4294967296) ** N

So that gives us the
1 - ((4294967295/4294967296) ** N)

But because we're looking for the case where all *4* numbers have already come up, that probability needs to be raised to the 4th power ... which yields the final figure.

Cheers,
Rob

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Re^3: [OT] The statistics of hashing.
by BrowserUk (Pope) on Apr 01, 2012 at 09:04 UTC

Thanks syphilis. Your calculations make sense to me. But I'm not sure that it gels with the actual data?

Assuming I've coded your formula correctly (maybe not!), then using 10 hashes & vectors, I get the odds of having seen a dup after 1e9 inserts as (1 - ((4294967295/4294967296)**1e9) ) **10 := 0.00000014949378123.

By that point I had actually seen 13 collisions:

```printf "After %5.5s inserts the odds of no dups: %.17f\n",
\$_, ( 1 - ( 0.99999999976716935634613037109375 ** \$_ ) ) )**10
for map "\${_}e8", 1..40;;
After   1e8 inserts the odds of no dups: 0.00000000000000004
After   2e8 inserts the odds of no dups: 0.00000000000003802
After   3e8 inserts the odds of no dups: 0.00000000000195354
After   4e8 inserts the odds of no dups: 0.00000000003092706
After   5e8 inserts the odds of no dups: 0.00000000025689926
After   6e8 inserts the odds of no dups: 0.00000000141936970
After   7e8 inserts the odds of no dups: 0.00000000591942653
After   8e8 inserts the odds of no dups: 0.00000002009607516
After   9e8 inserts the odds of no dups: 0.00000005831019113
After  10e8 inserts the odds of no dups: 0.00000014949378123 **
After  11e8 inserts the odds of no dups: 0.00000034677629037
After  12e8 inserts the odds of no dups: 0.00000074067890139
After  13e8 inserts the odds of no dups: 0.00000147618719432
After  14e8 inserts the odds of no dups: 0.00000277379244752
After  15e8 inserts the odds of no dups: 0.00000495435410008
After  16e8 inserts the odds of no dups: 0.00000846740103378
After  17e8 inserts the odds of no dups: 0.00001392227490012
After  18e8 inserts the odds of no dups: 0.00002212133818548
After  19e8 inserts the odds of no dups: 0.00003409433217130
After  20e8 inserts the odds of no dups: 0.00005113288027133
After  21e8 inserts the odds of no dups: 0.00007482409152212
After  22e8 inserts the odds of no dups: 0.00010708222525783
After  23e8 inserts the odds of no dups: 0.00015017742682200
After  24e8 inserts the odds of no dups: 0.00020676062948530
After  25e8 inserts the odds of no dups: 0.00027988383247111
After  26e8 inserts the odds of no dups: 0.00037301510162471
After  27e8 inserts the odds of no dups: 0.00049004779031911
After  28e8 inserts the odds of no dups: 0.00063530363659898
After  29e8 inserts the odds of no dups: 0.00081352955192231
After  30e8 inserts the odds of no dups: 0.00102988807161069
After  31e8 inserts the odds of no dups: 0.00128994158265037
After  32e8 inserts the odds of no dups: 0.00159963057715543
After  33e8 inserts the odds of no dups: 0.00196524629692540
After  34e8 inserts the odds of no dups: 0.00239339823431121
After  35e8 inserts the odds of no dups: 0.00289097703606607
After  36e8 inserts the odds of no dups: 0.00346511341973626
After  37e8 inserts the odds of no dups: 0.00412313375677405
After  38e8 inserts the odds of no dups: 0.00487251300375936
After  39e8 inserts the odds of no dups: 0.00572082567410912
After  40e8 inserts the odds of no dups: 0.00667569553892502

And looking at the figure for 4e9 := 0.00667569553892502, by which time the 10 vectors will be almost fully populated, it looks way too low to me?

I would have expected that calculation (for N=4e9) to have yielded odds of almost 1?

With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.

The start of some sanity?

I get the odds of having seen a dup after 1e9 inserts as (1 - ((4294967295/4294967296)**1e9) ) **10 := 0.00000014949378123

That's not the probability of "having seen a dup", but the probability that the 1000000001st random selection of 10 numbers would be reported as a dup (ie the probability that each of the relevant bits in all 10 bit vectors was already set for that 1000000001st random selection of the 10 numbers).

If I get a chance I'll try to work out the probability of "having seen a dup" in the first 1e9 iterations. (But, judging by some of the figures being bandied about, it probably has little bearing on this actual case where we're looking at MD5 hashes instead of random selections.)

Cheers,
Rob
If I get a chance I'll try to work out the probability of "having seen a dup" in the first 1e9 iterations.

Thank you if you do find that time at some point. If you could also show your workings, I might be able to wrap my muddled brain around it and stand a chance of re-applying your derivation.

it probably has little bearing on this actual case where we're looking at MD5 hashes instead of random selections.

Whilst the MD5 hash is known to be imperfect, it has been well analysed and has been demonstrated to produce a close to perfect random distribution of bits from the input data by several practical measures.

Eg. If you take any single input text, and produce its MD5; and then vary a single bit in the input and produce a new MD5, then -- on average -- half of the bits in the new MD5 will have changed relative to the original.

And if you repeat that process -- varying a single bit in the input and then compare the original and new MD5s -- the average number of bits changed in the outputs will tend towards 1/2. That is about as good a measure of randomness as you can hope for from a deterministic process.

I am aware of the limitations on the distribution of the hashes when derived from a non-full spectrum of inputs; but given that 2**32 (the maximum capacity of the vectors), represent such a minuscule proportion of the 1e44 possible inputs, I'd have to be extremely unlucky in my random selection from the total inputs for the hashing bias to actually have a measurable affect upon the probabilities of false positives.

With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.

The start of some sanity?

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