in reply to Re: [OT] The statistics of hashing. in thread [OT] The statistics of hashing.
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:
$ cat ana_2.pl
#!/usr/bin/perl
#
# ana_2.pl N V X
#
# N=vector size (bits), V=number of vectors, X=sample number
#
use strict;
use warnings;
use feature ':5.10';
my $n=shift; $n = 1<<$n;
my $v=shift;
my $x=shift;
my ($exp1, $exp2, $exp3);
given ($v) {
when ( 1) { $exp1=ex_1($n, $x), $exp2=ex_1($n, 0) }
when ( 2) { $exp1=ex_2($n, $x), $exp2=ex_2($n, 0) }
when ( 3) { $exp1=ex_3($n, $x), $exp2=ex_3($n, 0) }
when ( 4) { $exp1=ex_4($n, $x), $exp2=ex_4($n, 0) }
when (10) { $exp1=ex_10($n, $x), $exp2=ex_10($n, 0) }
default { die "Need symbolic integral form for $v vectors!\n"; }
}
$exp3 = $exp1$exp2;
print "N=$n, V=$v, X=$x integral($x)=$exp1, integral(0)=$exp2\n";
print "Expected collisions: $exp3\n";
sub ex_1 {
my ($N, $X) = @_;
return $N *exp( $X/$N)
+ $X;
}
sub ex_2 {
my ($N, $X) = @_;
return $N/2*exp(2*$X/$N)
+ 2*$N *exp( $X/$N)
+ $X;
}
sub ex_3 {
my ($N, $X) = @_;
return $N/3*exp(3*$X/$N)
 3*$N/2*exp(2*$X/$N)
+ 3*$N *exp( $X/$N)
+ $X;
}
sub ex_4 {
my ($N, $X) = @_;
return $N/4*exp(4*$X/$N)
+ 4*$N/3*exp(3*$X/$N)
 3*$N *exp(2*$X/$N)
+ 4*$N *exp( $X/$N)
+ $X;
}
sub ex_10 {
my ($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;
}
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.
Re^3: [OT] The statistics of hashing.(SOLVED)
by BrowserUk (Pope) on Apr 02, 2012 at 23:59 UTC

Thank you roboticus.
I ran your algorithm against the results of my 60hour run, and the correlation is simply stunning!
#! 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] 

Perhaps this will help some.
I seriously hope this will not offend you, but suspect it will.
Simply put, your post does not help me at all.
I am a programmer, not a mathematician, but given a formula, in a form I can understand(*), I am perfectly capable of implementing that formula in code. And perfectly capable of coding a few loops and print statements in order to investigate its properties.
What I have a problem with  as evidently you do too  is deriving those formula.
Like you (according to your own words above; there is nothing accusatory here), my knowledge of calculus is confined to the coursework I did at college some {mumble mumble} decades ago. Whilst I retain an understanding of the principles of integeration; and recall some of its uses, the details are shrouded in a cloud of disuse.
Use it or lose it, is a very current, and very applicable aphorism.
The direction my career has taken me means that I've had no more than a couple of occasions when calculus would have been useful. And on both those occasions, I succeeded in finding "a man that can", who could provide me with an understandable formula, and thus, I achieved my goal without having to relive the history of mathematics.
(*) A big part of the problem is that mathematicians not only have a nomenclature  which is necessary  the also have 'historical conventions'  which are not; and the latter are the absolute bane of the layperson's life in trying to understand the mathematician's output.
There you are, happily following along when reach a text that goes something like this:
We may think intuatively of the Riemann sum: Ʃ^{b}_{a} f(x) dx
as the infinite sum: f(x_{0})dx + f(x_{1})dx + ... + f(x_{H  1})dx + f(x_{H})(b  x_{H})
Where did H come from? Where did a disappear to? Is H (by convention) == to b  a?
For the answer to this and other questions, tune in next week ..... to the last 400 (or sometimes 4000) years of the history of math
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] 

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Re^3: [OT] The statistics of hashing.
by BrowserUk (Pope) on Apr 03, 2012 at 12:39 UTC

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] 

Thanks. As may be (becoming) obvious, much of that is over my head for now :)
However, Now I know how to derive the constants, I have this which I can substitute for your ex_4() & ex_10() by supplying the power as the first argument. Its output matches those two exactly for all the argument combinations I've tried:
sub PTn {
my @row;
for( 1 .. shift ) {
push @row, 1;
$row [$_] += $row [$_  1] for reverse 1 .. @row  2;
}
return @row;
}
sub expN {
my( $P, $N, $X ) = @_;
my $xDIVn = $X / $N;
my @coefs = PTn( $P+1 );
my $rv = 0;
my $toggle = 1;
for my $p ( reverse 1 .. $P ) {
$rv += $toggle * $coefs[ $p ] * $N / $p * exp( $p * $xDIVn );
$toggle *= 1;
}
return $rv + $X;
}
That allowed me to investigate the affect of using fewer or more hashes without having to hand code a new function for each.
And the results are quite interesting. This shows that each new (pair?) of hashes added does increase the discrimination substantially, though the gains obviously fall off fairly rapidly. But the really interesting part are the numbers for odd numbers of hashes: C:\test>rbtcsformverify H=10 B=32
Using N x 2**32 vectors:
12 : 24 0 24 0 24
+ 0 24 0 24 0
16 : 32 0 31 0 31
+ 0 31 0 31 0
24 : 48 0 48 0 48
+ 0 48 0 48 0
32 : 64 0 64 0 64
+ 0 64 0 64 0
48 : 95 0 95 0 95
+ 0 95 0 96 0
64 : 127 0 127 0 127
+ 0 128 0 128 0
96 : 191 0 191 0 192
+ 0 192 0 192 0
128 : 255 0 256 0 256
+ 0 256 0 256 0
192 : 383 0 383 0 383
+ 0 384 0 384 0
256 : 511 0 512 0 512
+ 0 512 0 512 0
384 : 767 0 768 0 768
+ 0 768 0 768 0
512 : 1023 0 1024 0 1024
+ 0 1024 0 1024 0
768 : 1535 0 1536 0 1536
+ 0 1536 0 1536 0
1024 : 2047 0 2048 0 2048
+ 0 2048 0 2048 0
1536 : 3071 0 3072 0 3072
+ 0 3072 0 3072 0
2048 : 4095 0 4096 0 4096
+ 0 4096 0 4096 0
3072 : 6143 0 6144 0 6144
+ 0 6144 0 6144 0
4096 : 8191 0 8192 0 8192
+ 0 8192 0 8192 0
6144 : 12287 0 12288 0 12288
+ 0 12288 0 12288 0
8192 : 16383 0 16384 0 16383
+ 0 16383 0 16384 0
12288 : 24575 0 24576 0 24576
+ 0 24576 0 24576 0
16384 : 32767 0 32767 0 32767
+ 0 32767 0 32768 0
24576 : 49151 0 49151 0 49152
+ 0 49152 0 49152 0
32768 : 65535 0 65536 0 65536
+ 0 65536 0 65536 0
49152 : 98303 0 98304 0 98304
+ 0 98303 0 98303 0
65536 : 131071 0 131072 0 131071
+ 0 131072 0 131072 0
98304 : 196606 0 196608 0 196607
+ 0 196608 0 196608 0
131072 : 262142 0 262144 0 262144
+ 0 262143 0 262144 0
196608 : 393211 0 393216 0 393216
+ 0 393215 0 393216 0
262144 : 524280 0 524288 0 524287
+ 0 524287 0 524288 0
393216 : 786414 0 786431 0 786431
+ 0 786432 0 786432 0
524288 : 1048544 0 1048575 0 1048576
+ 0 1048576 0 1048576 0
786432 : 1572792 0 1572863 0 1572864
+ 0 1572864 0 1572864 0
1048576 : 2097024 0 2097151 0 2097151
+ 0 2097152 0 2097152 0
1572864 : 3145440 0 3145727 0 3145728
+ 0 3145727 0 3145728 0
2097152 : 4193792 0 4194303 0 4194304
+ 0 4194304 0 4194304 0
3145728 : 6290304 0 6291455 0 6291455
+ 0 6291455 0 6291455 0
4194304 : 8386560 1 8388607 0 8388608
+ 0 8388608 0 8388608 0
6291456 : 12578306 4 12582911 0 12582912
+ 0 12582912 0 12582912 0
8388608 : 16769029 10 16777215 0 16777216
+ 0 16777215 0 16777216 0
12582912 : 25147409 35 25165823 0 25165823
+ 0 25165824 0 25165824 0
16777216 : 33521706 85 33554431 0 33554431
+ 0 33554431 0 33554432 0
25165824 : 50258063 286 50331646 0 50331647
+ 0 50331648 0 50331648 0
33554432 : 66978132 678 67108860 0 67108863
+ 0 67108864 0 67108864 0
50331648 : 100369532 2283 100663276 0 100663295
+ 0 100663296 0 100663296 0
67108864 : 133696160 5397 134217665 0 134217727
+ 0 134217728 0 134217728 0
100663296 : 200156106 18111 201326276 5 201326591
+ 0 201326591 0 201326592 0
134217728 : 266359979 42681 268434469 24 268435455
+ 0 268435455 0 268435455 0
201326592 : 398007464 142383 402648282 179 402653177
+ 0 402653183 0 402653184 0
268435456 : 528654369 333608 536855705 738 536870874
+ 1 536870911 0 536870911 0
402653184 : 787008255 1100214 805232175 5328 805305969
+ 30 805306365 0 805306367 0
536870912 : 1041542872 2548692 1073515796 21337 1073739728
+ 211 1073741802 2 1073741823 0
805306368 : 1539620713 8218836 1609548824 146448 1610591774
+ 3082 1610612273 70 1610612725 1
1073741824 : 2023785226 18623993 2144354575 558473 2147380065
+ 19731 2147479815 755 2147483497 30
1610612736 : 2953695056 57532294 3207472286 3485537 3220308576
+ 247526 3221157361 19018 3221220099 1532
2147483648 : 3837421596 125076314 4257101987 12129165 4290939237
+ 1371778 4294491373 167491 4294907677 21417
3221225472 : 5487393872 357203949 6295545197 63685472 6413893311
+13112112 6436324179 2901774 6441061714 670967
4294967296 : 7009904423 721946381 8229596479 188903097 8487725572
+56541428 8558136669 18112153 8579512265 6047518
I suspect it is a coding error on my behalf, but I guess it could be a quirk of the numbers?
Here's a set using 1 .. 16 2**16 bit hashes: C:\test>rbtcsformverify H=16 B=16
Using N x 2**16 vectors:
12 : 23 0 23 0 24 0 24 0
+ 24 0 23 0 23 0 23 0
16 : 31 0 31 0 31 0 32 0
+ 32 0 32 0 32 0 32 0
24 : 47 0 47 0 48 0 47 0
+ 48 0 47 0 47 0 47 0
32 : 63 0 63 0 64 0 64 0
+ 64 0 64 0 64 0 64 0
48 : 95 0 95 0 95 0 95 0
+ 95 0 96 0 96 0 96 0
64 : 127 0 127 0 128 0 128 0
+ 128 0 127 0 128 0 127 0
96 : 191 0 191 0 192 0 192 0
+ 192 0 192 0 191 0 191 0
128 : 255 0 255 0 256 0 255 0
+ 256 0 256 0 256 0 256 0
192 : 383 0 383 0 383 0 384 0
+ 384 0 384 0 384 0 384 0
256 : 511 0 511 0 511 0 511 0
+ 512 0 511 0 512 0 512 0
384 : 766 0 767 0 767 0 768 0
+ 768 0 767 0 768 0 767 0
512 : 1022 0 1023 0 1023 0 1024 0
+ 1024 0 1024 0 1024 0 1024 0
768 : 1531 0 1535 0 1535 0 1536 0
+ 1536 0 1536 0 1536 0 1536 0
1024 : 2040 0 2047 0 2047 0 2048 0
+ 2048 0 2048 0 2048 0 2048 0
1536 : 3054 0 3071 0 3071 0 3071 0
+ 3072 0 3072 0 3072 0 3072 0
2048 : 4064 0 4095 0 4095 0 4095 0
+ 4095 0 4096 0 4096 0 4096 0
3072 : 6073 2 6143 0 6143 0 6143 0
+ 6144 0 6144 0 6144 0 6144 0
4096 : 8066 5 8191 0 8191 0 8191 0
+ 8191 0 8191 0 8191 0 8192 0
6144 : 12008 16 12286 0 12287 0 12287 0
+12287 0 12287 0 12288 0 12287 0
8192 : 15892 38 16380 0 16383 0 16383 0
+16383 0 16383 0 16383 0 16383 0
12288 : 23492 125 24559 2 24575 0 24575 0
+24575 0 24575 0 24575 0 24576 0
16384 : 30880 284 32720 8 32766 0 32767 0
+32767 0 32767 0 32767 0 32767 0
24576 : 45069 877 48942 53 49138 3 49150 0
+49151 0 49151 0 49151 0 49151 0
32768 : 58554 1908 64958 185 65474 20 65528 2
+65535 0 65535 0 65535 0 65535 0
49152 : 83730 5450 96062 971 97868 200 98210 44
+98282 10 98299 2 98302 0 98303 0
65536 : 106962 11016 125573 2882 129512 862 130586 276 1
+30912 92 131018 31 131053 11 131065 3
Sorry for the wrapping.
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".
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