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### Re: The statistics of hashing.

by tobyink (Canon)
 on Mar 31, 2012 at 22:58 UTC ( #962804=note: print w/replies, xml ) Need Help??

in reply to [OT] The statistics of hashing.

What is the probability of getting false positives?

Seems almost certain to me, but my reasoning may well be faulty. Stats were never my forte.

You're effectively running four different 32-bit hash functions on each string. According to Wikipedia for a 32-bit hash you only need about 110,000 hashes to get a 75% chance of a collision.

Thus with 110,000 strings, getting a collision in all four stands at 75%**4, about 30% chance. That's with 110,000 strings. You have "several billion".

With several billion strings you are going to have a very close to 100% chance of a collision on a 32-bit hash function. With four 32-bit hash functions, that's still close to 100% chance of a collision.

Extending this to having 10 effective hash functions, this cannot increase the chance of collisions. Worst case scenario it has no effect. Here I think it does decrease the chance (difficult , but not enough to be useful.

perl -E'sub Monkey::do{say\$_,for@_,do{(\$monkey=[caller(0)]->[3])=~s{::}{ }and\$monkey}}"Monkey say"->Monkey::do'

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Re^2: The statistics of hashing.
by BrowserUk (Pope) on Mar 31, 2012 at 23:26 UTC
Seems almost certain to me,

That's a given. As with bloom filters, false positives are a fact of life. The question is how likely, or more to the point, how frequently, they will occur.

According to Wikipedia for a 32-bit hash you only need about 110,000 hashes to get a 75% chance of a collision.

Unfortunately, a good deal of what you read on wikipedia is less than reliable.

The following is empirical evidence from an on-going experiment.

After 3/4 of a billion trials, there were zero (possible) dups. After 1.2 billion, there are only 61.

These are the positions at which those possible dups were detected:

779967210 782382025 830840395 882115420 883031614 897571477 923155269 948108745 954455113 967783959 988381482 992691311 995935158 1011301141 1013742872 1022258193 1031874023 1034026887 1036254774 1037064360 1037193945 1037309710 1037661519 1041973643 1042577179 1045603197 1046414487 1047056233 1048652112 1048960948 1052042118 1057413610 1059736720 1067108427 1068042962 1069550972 1070612878 1075526924 1079515116 1079995733 1080999379 1086256410 1098797420 1110115220 1121074450 1121078210 1121376561 1132692265 1133452843 1138397053 1141581831 1143980294 1148289231 1149508590 1149743067 1151539362 1155227820 1156087258 1156164711 1158795083 1163191879 1165623466 1167648959 1168308302 1169005605 1173266361 1173729947 1174309756 1176812463 1178780066 1181065368 1183536832 1183744519 1183964959 1185195302 1185303999 1186231527 1186435292 1187445584 1189170990 1192726357 1195889596 1198540465 1198563209 1198727961 1201384004 1203571325 1204470932 1205045006 1205186694 1206609472 1209383402 1211297256 1212504859 1214473828 1217800153 1218236491 1219282533 1219801192 1220616524 1222841081 1226558804 1227645351 1229411437 1230159513 1230637415 1232098200 1232312863 1232491027 1233958398 1234452741 1237085887 1237298288 1237708765 1238281678 1238455038 1239452137 1240723226 1240980106 1241466028 1242695443 1242936270 1243921116 1245408660 1245914539 1247643815 1248306653 1249238533 1250144683 1250278843 1253111263 1254066285 1254097515 1254117382 1254371720 1255160872 1255282569 1255309049 1255779636 1257210934 1257614617 1259207077 1259923475 1261856200 1262605109 1262859049

As you can see, having filled 1/4 quarter of the total space available, false positives are just beginning to occur. And as the vector space fills, they are (as expected) becoming more frequent. (Of course, some of those may be actual duplicates. It will take a second run to determine that.)

But the question remains, how to calculate the probabilities of the mechanism.

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?

Unfortunately, a good deal of what you read on wikipedia is less than reliable.

Indeed, but the errors on Wikipedia are not evenly distributed. On a subject such as the birthday attack I'd expect Wikipedia's article to be on par with any other authority (short-lived events of vandalism notwithstanding).

But the question remains, how to calculate the probabilities of the mechanism.

The exact calculation involves some big numbers. But assuming that c(\$x, \$y) is the "pick \$y from \$x" function used in combinatorics, then the probability of a collision for \$n strings and an evenly distributed 32-bit hash function should be:

\$p = 1 - ( factorial(\$n) * c(2**32, \$n) / 365**\$n )

Big numbers. Horrible to calculate. Can be approximated though...

sub e () { 2.718281828 } my \$t = (\$n**2) / (2**33); \$p = 1 - ( e ** -\$t );

Calculating \$p is still horrible, but calculating \$t is easier. If \$t is above 20 then \$p is 1.00000 when rounded to 6 significant figures.

Thus you can effectively be sure to have a collision with a 32-bit hash function once \$t is above 20. You can figure out an \$n which triggers \$t to be 20 using:

\$n = sqrt(20 * (2 ** 33));

It's about 414,000. So with 414,000 strings, you are effectively certain to get collision on a 32-bit hash function.

Where I think my reasoning and tye's differ (and tye is almost certainly correct here - blame it on me answering late at night) is that I was then looking at the probabilities that you will have had collisions in all four (or ten) hash functions at the end of the entire run. With even half a million strings, that is a given.

What you're actually doing is looking at events where a single string triggers a simultaneous collision in all the hash functions. I defer to tye's calculations for that.

perl -E'sub Monkey::do{say\$_,for@_,do{(\$monkey=[caller(0)]->[3])=~s{::}{ }and\$monkey}}"Monkey say"->Monkey::do'
Indeed, but the errors on Wikipedia are not evenly distributed. On a subject such as the birthday attack I'd expect Wikipedia's article to be on par with any other authority

I think the first mistake that you (and Tye) are making here, is assuming that "the birthday paradox" (as applied to the each of the 32-bit hashes individually) is the controlling factor. It isn't.

The controlling factor is the combinatorial affect of the 4 (or, in my case 10) separate hashes & vectors.

I think that applying the birthday paradox to the 128-bit hash may prove to be an accurate upper bound. But it is so large relative to the limits of the 32-bit vectors as to never actually come into play.

I had tried applying various formulae I found on wikipedia and elsewhere, but found they do not match with reality. Hence, this post.

You might also look at the math described for Bloom filters. It also contradicts these naive applications of the birthday paradox, but is sufficiently well researched and documented (sources outside of wikipedia) to show that naive application of the birthday paradox is not applicable when combining multiple hashes in this way.

The problem with the Bloom filter math is that it is based upon setting the bits derived from all the different hashes into the same vector. Ie. Each Bloom filter insert sets N bits in a single vector, where N is the number of different hashes being used. This obviously causes the vector to fill up much more quickly than my mechanism which uses a separate vector for each hash.

My gut feel, is that as my vectors fill up N times more slowly, the probabilities of collisions for my mechanism are 2**N times lower than with a Bloom filter (at the expense of N*the memory requirement). This is the notion I am trying to verify here.

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?

Re^2: The statistics of hashing. (**4)
by tye (Sage) on Apr 01, 2012 at 07:54 UTC

Wikipedia is correct (well, agrees quite closely with some of my calculations). But you have extrapolated from that quite incorrectly.

0.75**4 is the odds of, after inserting 1.1e5 strings, that there has been at least one collision in each of the four hashes. It is not even close to the odds of there being a single string that produced a simultaneous collision in all 4 hashes.

Those odds would be more like 0.75*(0.75/1.1e5)**3 or about 2e-16. My calculations came up with an upper bound of 9.5e-15.

I believe the 9.5e-15 number is much, much more accurate. It only discounts the possibility of there being fewer than 1.1e5 bits set. I suspect the ratio of "expected bits set"/1.1e5 to be very close to 1.

While the 2e-16 number is based on there being a 75% chance of exactly one collision. But there is a 25% chance of 0 collisions and a 75% chance of 1 or more collisions. And the possibility of 2 collisions approximately doubles the odds for some subset of the possible outcomes and so has a much bigger impact on the final total than does the possibility of there being 1.1e5-1 bits set instead of 1.1e5.

For example, consider tossing a coin twice. There is a 75% chance of getting at least one "head". Do 4 such trials. For each head, assign it a random position in 1.1e5 slots (but only one head per slot per trial). The odds of the 4 trials producing 4 heads in one of those slots (one from each trial) one might estimate as:

.75* (.75/x) * (.75/x) * (.75/x)

Where x==1.1e5. Or, 2.3e-16. But to find the accurate odds you have to realize that there is a 50% chance of one "head" (per trial) and a 25% change two "heads" (per trial). So the actual odds are found via:

.5 * (.5/x) * (.5/x) * (.5/x) + .5 * (.25*2/x) * (.5/x) * (.5/x) + .5 * (.5/x) * (.25*2/x) * (.5/x) + .5 * (.25*2/x) * (.25*2/x) * (.5/x) + .5 * (.5/x) * (.5/x) * (.25*2/x) + .5 * (.25*2/x) * (.5/x) * (.25*2/x) + .5 * (.5/x) * (.25*2/x) * (.25*2/x) + .5 * (.25*2/x) * (.25*2/x) * (.25*2/x) + .25* (.5/x1) * (.5/x1) * (.5/x1) + .25* (.25*x12) * (.5/x1) * (.5/x1) + .25* (.5/x1) * (.25*x12) * (.5/x1) + .25* (.25*x12) * (.25*x12) * (.5/x1) + .25* (.5/x1) * (.5/x1) * (.25*x12) + .25* (.25*x12) * (.5/x1) * (.25*x12) + .25* (.5/x1) * (.25*x12) * (.25*x12) + .25* (.25*x12) * (.25*x12) * (.25*x12)

Where x1==1.1e5-1 and x12==1/(1.1e5-1)+1/(1.1e5-2). That raises the odds by a factor of 2.37 to 5.6e-16.

That type of magnification of the odds will be much more pronounced in the case originally discussed and I bet would result in nearly the 48-fold increase required to get to the odds of my upper bound calculation.

- tye

Re^2: The statistics of hashing.
by JavaFan (Canon) on Apr 01, 2012 at 11:17 UTC
Thus with 110,000 strings, getting a collision in all four stands at 75%**4, about 30% chance. That's with 110,000 strings. You have "several billion".
If, after 110,000 strings you get a 75% on a collision, 75%4 gives you the chance 4 sets of 110,000 each give you a collision. But that's nowhere near the probability the same pair gives you a collision. Which is what BrowserUK needs.

It would take me too much time to dig up the math required to do the calculation. But what BrowserUK may try to do: generate a billion strings that are all unique. Run the proposed algorithm. See how many collisions are reported: they're all false positives. Repeat several times (with different strings). You may even want to calculate a confidence interval. Of course, I've no idea how feasible this is. If it takes a week to process a data set this size, you may not want to do so.

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