Re^2: [OT] The statistics of hashing.
by roboticus (Chancellor) on Apr 02, 2012 at 17:45 UTC

BrowserUk:
I'm done with the program now. I added a column to show the predicted number of false positives. The code is nearly unchanged. Basically, I moved:
my $expected = (1exp($cnt_rex/$VecSize))**$num_vecs;
outside of the reporting section, and then accumulate it:
$cum_collisions += $expected;
It's not very fast, so I only ran it a couple of times, but it seems like it gives reasonable agreement:
$ ./analyze_bloom.pl 14 3 17
NumVecs=3
avg avg calc  # collisions 
+ expected
BI Samples one bits pct 1s pct 1s all 1 2 3
+ p(coll) false pos
         
+ 
14 256: 253 1.55 1.55 7 7 0 0 0
+.0000037264 0.0002414791
14 512: 504 3.08 3.08 22 22 0 0 0
+.0000291236 0.0037774699
14 1024: 993 6.06 6.06 89 87 2 0 0
+.0002224011 0.0581208314
14 2048: 1926 11.76 11.75 336 309 26 1 0
+.0016223627 0.8630370129
14 4096: 3637 22.20 22.12 1185 1005 170 10 0
+.0108230772 11.9418752510
14 8192: 6414 39.15 39.35 3986 2802 1022 162 0
+.0609161842 144.4751474482
14 16384: 10300 62.87 63.21 11216 5615 4166 1435 0
+.2525804578 1374.7071068207
14 32768: 14151 86.37 86.47 27361 7789 10655 8917 0
+.6464623148 8946.4018915720
14 65536: 16105 98.30 98.17 60117 8201 15657 36259 0
+.9460533270 36391.1792023071
$ ./analyze_bloom.pl 16 4 18
NumVecs=4
avg avg calc  # collisions 
+ expected
BI Samples one bits pct 1s pct 1s all 1 2 3
+ 4 p(coll) false pos
         
+  
16 256: 255 0.39 0.39 3 3 0 0
+ 0 0.0000000002 0.0000000120
16 512: 510 0.78 0.78 5 5 0 0
+ 0 0.0000000037 0.0000003784
16 1024: 1018 1.55 1.55 21 21 0 0
+ 0 0.0000000578 0.0000119226
16 2048: 2021 3.08 3.08 107 106 1 0
+ 0 0.0000008960 0.0003713063
16 4096: 3979 6.07 6.06 444 423 20 1
+ 0 0.0000134746 0.0112771478
16 8192: 7749 11.82 11.75 1615 1466 141 8
+ 0 0.0001906326 0.3256727521
16 16384: 14569 22.23 22.12 5843 4608 1066 158
+ 11 0.0023940562 8.5228328538
16 32768: 25931 39.57 39.35 18293 11116 5475 1529
+ 173 0.0239686508 185.0883611550
16 65536: 41690 63.61 63.21 49093 18714 17246 10354
+ 2779 0.1596613002 2882.5123468408
16 131072: 56843 86.74 86.47 114278 21592 29488 36446
+ 26752 0.5589731543 26626.3779623356
The updated code:
Both runs are still underway, I'll update this when they complete. Done.
Update: Added rest of the tables. As you can see from the two sample runs, there's good agreement between the actual false positive count and the expected count.
...roboticus
When your only tool is a hammer, all problems look like your thumb.  [reply] [d/l] [select] 
Re^2: [OT] The statistics of hashing.
by roboticus (Chancellor) on Apr 02, 2012 at 22:50 UTC

OK, I think I'm done with this diversion. I've built a program that will compute the expected number of collisions with no looping. Example output:
$ ./ana_2.pl 16 4 65536
N=65536, V=4, X=65536 integral(65536)=139415.765849051, integral(0)=13
+6533.333333333
Expected collisions: 2882.43251571807
$ ./ana_2.pl 16 4 16384
N=65536, V=4, X=16384 integral(16384)=136541.854969116, integral(0)=13
+6533.333333333
Expected collisions: 8.52163578287582
$ ./ana_2.pl 14 3 16384
N=16384, V=3, X=16384 integral(16384)=31411.9141476821, integral(0)=30
+037.3333333333
Expected collisions: 1374.58081434877
$ ./ana_2.pl 16 10 32768
N=65536, V=10, X=32768 integral(32768)=191953.190301726, integral(0)=1
+91952.863492063
Expected collisions: 0.326809662627056
The code is straightforward:
Notes:
 The only testing I did is limited to the cases above. Be sure to run a few more before relying on it.
 I entered the ex_1 and ex_2 routines, but didn't test 'em.
 For the other numbers of vectors, just expand (1  exp(X/N))^V, and integrate it. The ex_1..n functions are coded straight from the integration.
...roboticus
When your only tool is a hammer, all problems look like your thumb.  [reply] [d/l] [select] 

#! perl slw
use strict;
sub ex_10 {
my ($N, $X) = @_;
return $N/10*exp(10*$X/$N)
+ 10*$N/9 *exp( 9*$X/$N)
 45*$N/8 *exp( 8*$X/$N)
+ 120*$N/7 *exp( 7*$X/$N)
 35*$N *exp( 6*$X/$N)
+ 252*$N/5 *exp( 5*$X/$N)
 105*$N/2 *exp( 4*$X/$N)
+ 40*$N *exp( 3*$X/$N)
 45*$N/2 *exp( 2*$X/$N)
+ 10*$N *exp( 1*$X/$N)
+ $X;
}
my $exp10_0 = ex_10( 2**32, 0 );
open I, '<', 'L25B32V10I32S1.posns' or die $!;
while( <I> ) {
my( $inserts, $collisions ) = split;
my $exp3 = ex_10( 2**32, $inserts )  $exp10_0;
printf "Insert %10d: Predicted %6d Actual: %6d Delta: %+.3f%%\n"
+,
$inserts, $exp3, $collisions, ( $collisions  $exp3 ) / $exp3
+* 100;
}
__END__
C:\test>rbtcsformverify
Insert 779967210: Predicted 1 Actual: 1 Delta: 18.051%
Insert 782382025: Predicted 1 Actual: 2 Delta: +58.813%
Insert 830840395: Predicted 2 Actual: 3 Delta: +29.292%
Insert 882115420: Predicted 4 Actual: 4 Delta: 5.961%
Insert 883031614: Predicted 4 Actual: 5 Delta: +16.325%
Insert 897571477: Predicted 5 Actual: 6 Delta: +18.390%
Insert 923155269: Predicted 6 Actual: 7 Delta: +4.086%
Insert 948108745: Predicted 8 Actual: 8 Delta: 8.996%
Insert 954455113: Predicted 9 Actual: 9 Delta: 4.244%
Insert 967783959: Predicted 10 Actual: 10 Delta: 7.404%
Insert 988381482: Predicted 13 Actual: 11 Delta: 17.487%
Insert 992691311: Predicted 13 Actual: 12 Delta: 13.814%
Insert 995935158: Predicted 14 Actual: 13 Delta: 9.624%
Insert 1011301141: Predicted 16 Actual: 14 Delta: 16.457%
Insert 1013742872: Predicted 17 Actual: 15 Delta: 12.616%
Insert 1022258193: Predicted 18 Actual: 16 Delta: 14.242%
Insert 1031874023: Predicted 20 Actual: 17 Delta: 16.989%
Insert 1034026887: Predicted 20 Actual: 18 Delta: 13.909%
Insert 1036254774: Predicted 21 Actual: 19 Delta: 11.051%
Insert 1037064360: Predicted 21 Actual: 20 Delta: 7.093%
Insert 1037193945: Predicted 21 Actual: 21 Delta: 2.569%
Insert 1037309710: Predicted 21 Actual: 22 Delta: +1.957%
...( ~ 52000 ellided }
Insert 2375752842: Predicted 52209 Actual: 52277 Delta: +0.130%
Insert 2375755671: Predicted 52209 Actual: 52278 Delta: +0.131%
Insert 2375756509: Predicted 52209 Actual: 52279 Delta: +0.133%
Insert 2375760656: Predicted 52210 Actual: 52280 Delta: +0.133%
Insert 2375763928: Predicted 52211 Actual: 52281 Delta: +0.134%
Insert 2375785238: Predicted 52215 Actual: 52282 Delta: +0.128%
Insert 2375788721: Predicted 52215 Actual: 52283 Delta: +0.128%
Insert 2375789878: Predicted 52216 Actual: 52284 Delta: +0.130%
Insert 2375790896: Predicted 52216 Actual: 52285 Delta: +0.131%
Insert 2375798283: Predicted 52217 Actual: 52286 Delta: +0.131%
And that still includes the possibility  though I believe it to be remote  that there are 1 or more actual dups in there.
My only regret now is that I wish I'd allowed the run to go to the 3/4 point instead of stopping it half way. The number of false positive would still have been easily manageable for the second pass: C:\test>rbtcsbloomprobb 32 10 30e8
N=4294967296, V=10, X=30e8 integral(30e8)=12580199467.653, integral(0)
+=12579822861.8159
Expected collisions: 376605.837186813
That is an order of magnitude less that my own crude attempt was predicting, hence why I stopped the run.
The only way I can repay you is to assure you that I will do my very best to try and understand the formula  which I do not currently.
And of course, give you credit, when they world comes knocking at my door for my eminently patentable  at least if you take Apple as your guide  algorithm :)
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".
 [reply] [d/l] [select] 

I will do my very best to try and understand the formula
Perhaps this will help some.
Below is some fairly simple code that does the precise calculations of the odds for a collision in a single hash (and compares those calculations with the formula roboticus proposed). I came up with a simpler implementation than I expected to. I didn't even try to implement this at first (only a little because I expected it to be more cumbersome, but) mostly because I knew it would be impractical for computing odds for such a large number of insertions. It consumes O($inserts) for memory and O($inserts**2) for CPU.
$= 1;
my( $b )= ( @ARGV, 2**32 ); # Total number of bits in the hash.
my $i= 1; # Number of insertions done so far.
my @odds = 1; # $odds[$s] == Odds of their being $s+1 bi
+ts set
while( 1 ) { # Just hit CtrlC when you've seen enough
my $exp= $b*( 1  exp($i/$b) );
my $avg = 0;
$avg += $_ for map $odds[$_]*($_+1), 0..$#odds;
my $err = sprintf "%.6f%%", 100*($exp$avg)/$avg;
print "$i inserts, $err: avg=$avg exp=$exp bits set\n"
if $i =~ /^\d0*$/;
# Update @odds to in preparation for the next value of $i:
for my $s ( reverse 0..$#odds ) {
$odds[$s+1] += $odds[$s]*($b$s1)/$b;
$odds[$s] *= ($s+1)/$b;
}
$i++;
}
$i tracks the number of insertions done so far. $odds[$s] represents the odds of there being $s+1 bits set in the hash (after $i insertions). $avg is an average of these values of $s (1..@odds) weighted by the odds. But, more importantly, it is also the odds of getting a singlehash collision when inserting (after $i insertions) except multiplied by $bits ($b). I multiple it and 1exp($i/$b) by $b to normalize to the expected number of set bits instead of the odds of a collision because humans have a much easier time identifying a number that is "close to 14" than a number that is "close to 14/2**32".
$odds[1] turns out to exactly match (successively) the values from birthday problem. For low numbers of $inserts, this swamps the calculation of $avg (the other terms just don't add up to a significant addition), which is part of why I was computing it for some values in my first reply. (Since you asked about that privately.)
I have yet to refresh my memory of the exact power series expansion of exp($x), so what follows is actually half guesses, but I'm pretty confident of them based on vague memory and observed behavior.
For large $bits, 1exp($inserts/$bits) ends up being close to 1/$bits because 1exp($inserts/$bits) expands (well, "can be expanded") to a power series where 1/$bits is the first term and the next term depends on 1/$bits**2 which is so much smaller that it doesn't matter much (and nor do any of the subsequent terms, even when added together).
On the other hand, for large values of $inserts, 1exp($inserts/$bits) is close to $avg because the formula for $avg matches the first $inserts terms of the power series expansion.
I hope the simple code makes it easy for you to see how these calculations match the odds I described above. But don't hesitate to ask questions if the correspondence doesn't seem clear to you. Running the code shows how my calculations match roboticus' formula. Looking up (one of) the power series expansions for computing exp($x) should match the values being computed for @odds, though there might be some manipulation required to make the match apparent (based on previous times I've done such work decades ago).
 [reply] [d/l] [select] 




With the 'cancelling out' you've performed on the constants in your ex_*() subs, I couldn't see the pattern by which they were derived.
Undoing that, I now see that they come (directly or not) from Pascal's Triangle.
I guess at some point I should review some teaching materials to understand why those constants are used here, but for now it is enough to know that I can now write a generic ex_*() function (generator). Thanks.
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".
 [reply] [d/l] 

BrowserUk:
In the iterative solution, we're accumulating f(x)=(1e^(x/N))^h over the range of x=0 .. NumSamples. That's a rough form of computing the definite integral of the expression. Integrating over three variables (x, N, h) would be a pain, so I treated N and h as constants.
So first, we multiply out our f(x) expression to remove the exponent. So using h as 2, we get:
f(x) = 1  2e^(x/N) + e^(2x/N)
integ(f(x)) = integ( 1  2e^(x/N) + e^(2x/N) )
= integ( 1 )  2*(integ(e^(x/N))) + integ(e^(2x/N))
since:
integ(1) = x + C
integ(e^(bx)) = (1/b)e^(bx) + C
integ(f(x)) = [ x+C ] + [ (2N)e^(x/N) + C ] + [ (N/2)e^(2x/N) + C
+]
Computing a definite integral over a range is simply integ(f(x)) evaluated at the upper limit less the value evaluated at the lower limit. This causes the C terms to cancel.
Pascal's triangle comes out because we've got (a+b)^n, and when we multiply it out, we get the binomial expansion which is where the coefficients come into play.
One point I should mention: You don't have to use 0 as the lower bound. If you wanted the number of collisions you'd experience from sample A to sample B, just evaluate integ(f(B))integ(f(A)). By using A=0 we compute the number of collisions for the entire run.
...roboticus
When your only tool is a hammer, all problems look like your thumb.  [reply] [d/l] 



Re^2: [OT] The statistics of hashing.
by roboticus (Chancellor) on Apr 02, 2012 at 14:07 UTC

Oh, something I forgot to mention: I tried using a constant number of bits but varying the vector size/quantity to see how things scaled. In other words, I compared:
vec size  # vectors 
10,000  1 
5,000  2 
3,333  3 
2,500  4 
2,000  5 
1,000  10 
I found more smaller vectors works better until the number of samples matches the number if bits in the smaller vector. Plotting the functions:
(1exp(x/1000))^10
(1exp(x/2000))^5
(1exp(x/2500))^4
(1exp(x/3333))^3
1exp(x/10000)
using a graphing calculator shows that's where the curves cross:
...roboticus
When your only tool is a hammer, all problems look like your thumb.  [reply] [d/l] 
Re^2: [OT] The statistics of hashing. (accuracy)
by tye (Sage) on Apr 02, 2012 at 21:55 UTC

I should review my transcendental function definitions as power series so I can recognize these correlations in future. I knew how to precisely measure the odds but had no clue how to search to see if there were any way to collapse it to some transcendental function (the exact measurement is quite impractical for very large numbers of inserts).
But having 1exp($inserts/$slots) shown to me, I was able to play with that approximation to see how accurate it is. For this application, it is actually an impressive fit.
If 1exp(1/$slots) starts out very close to 1/$slots (the correct starting value for the odds of a singlebit collision), then the accuracy never decreases as $inserts grows. For small $slots, 1exp(1/$slots) is not very accurate but 1exp($inserts/$slots) slowly becomes more and more accurate as $inserts increases.
For $slots == 2**32, 1exp($inserts/$slots) starts out (at 1==$inserts) so accurate that a 'double' can't even measure the error. So you can consider the calculations based on it "perfect" rather than an approximation. :)
 [reply] 