comparing two binary numbers

[Johnson Lau]

Dear all,

I need to compare two binary numbers and need perl to return the
number of matching bits.

For example:

$aaa = "10111100";
$bbb = "00101100";

In this case, the number of matching bits is 6.

I know I could split the strings and compare the bits one by one.
However, is there any faster functions for this? I need to compare
millions of such strings with 1000 bits for each.

Thanks a lot!!

Johnson Lau

[Jialin Li]

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I think what you need is counting how many ones in the xnor calculation of
two binary numbers
I cannot find xnor in Perl, only using xor, then counting how many ones,
that is the number of different bits

As the binary numbers are of 1000 bits, I think you should use Bit::Vector.
here is the code:

use strict;
use warnings;

use Bit::Vector;

use constant NSIZE => 8;

my $aaa = "10111100";
my $bbb = "00101100";

my $a = Bit::Vector->new(NSIZE);
my $b = Bit::Vector->new(NSIZE);
my $c = Bit::Vector->new(NSIZE);

$a->from_Bin($aaa);
$b->from_Bin($bbb);
$c->Xor($a, $b);
my $count = ( $c->to_Bin() =~ tr/1//);
print $count;

__END__

I did not benchmark to test the speed benefits of this method.
~

[Dr.Ruud]

"Johnson Lau" schreef:

> I need to compare two binary numbers and need perl to return the
> number of matching bits.
>
> For example:
>
> $aaa = "10111100";
> $bbb = "00101100";
>
> In this case, the number of matching bits is 6.

perl -Mstrict -Mwarnings -le'
my $b1 = "10111100";
my $b2 = "00101100";
$b2 =~ tr/01/10/;

my $n = ($b1 & $b2) =~ tr/0/0/;
print $n;
'
6

To derive the length, a prepared hash (with 256 keys) will be faster.

perl -Mstrict -Mwarnings -le'
my $b1 = oct "0b" . "10111100";
my $b2 = oct "0b" . "00101100";

my $n = sprintf("%08b", $b1 ^ $b2) =~ tr/0/0/;
print $n;
'

To derive the length, a prepared array (with 256 elements) is much
faster.

--
Affijn, Ruud

"Gewoon is een tijger."

[Sisyphus]

On May 11, 3:09 pm, johnson...@cuhk.edu.hk (Johnson Lau) wrote:
> Dear all,
>
> I need to compare two binary numbers and need perl to return the
> number of matching bits.
>
> For example:
>
> $aaa = "10111100";
> $bbb = "00101100";
>
> In this case, the number of matching bits is 6.
>

use strict;
use warnings;

use Math::GMPz qw(:mpz);

my $str1 = "10111100";
my $str2 = "00101100";

my $wanted = length($str1) - Rmpz_popcount(Math::GMPz->new($str1, 2) ^
Math::GMPz->new($str2, 2));
print $wanted, "\n";

Cheers,
Rob

[Chris Charley]

After reading the others' replies, I believe this xor operation will do what
you want;

#!/usr/bin/perl
use strict;
use warnings;

my $aaa = "10111100";
my $bbb = "00101100";

my $same = ($aaa ^ $bbb) =~ tr/\0//;

print $same;

Chris

[John W. Krahn]

Johnson Lau wrote:
> Dear all,

Hello,

> I need to compare two binary numbers and need perl to return the
> number of matching bits.
>
> For example:
>
> $aaa = "10111100";
> $bbb = "00101100";

Those aren't binary numbers, they are strings.

perldoc perlnumber

> In this case, the number of matching bits is 6.
>
> I know I could split the strings and compare the bits one by one.
> However, is there any faster functions for this? I need to compare
> millions of such strings with 1000 bits for each.

If you actually had "binary" numbers then you could use unpack( '%32b*',
$var ) to count bits.

perldoc perlpacktut

John
--
Perl isn't a toolbox, but a small machine shop where you
can special-order certain sorts of tools at low cost and
in short order. -- Larry Wall

[Dr.Ruud]

"John W. Krahn" schreef:
> Johnson Lau:

>> I need to compare two binary numbers and need perl to return the
>> number of matching bits. For example:
>> $aaa = "10111100";
>> $bbb = "00101100";
>
> Those aren't binary numbers, they are strings.

I don't mind reading "binary number" as "some base-two representation of
a number".

The word "number" is often used for an encoded stream of digits.