Re^5: One Zero variants_without

With 435 zeroes and 343 ones, you'll never live long enough to see it finish, regardless of the method used to generate the sequence. A simple way to figure out how many permutations a list will generate is with

my $num_zeros = 3; my $num_ones = 2;
use Math::BigInt;
sub fac {
    my $result = Math::BigInt->new("1");
    my $n = Math::BigInt->new(shift());
    while($n){ $result *= $n--; }
    return $result;
}
my $num_permutations = fac($num_ones + $num_zeros ) / (fac($num_ones) 
+* fac($num_zeros));
print("Number of permutations of a ($num_zeros, $num_ones) list is $nu
+m_permutations\n");'
[download]

A list with 453 zeros and 343 ones will have roughly 1.962e230 different permutations -- are you sure you need all of them?

Comment on Re^5: One Zero variants_without_repetition Download Code

Replies are listed 'Best First'.
Re^6: One Zero variants_without_repetition (logue) by tye (Sage) on Aug 08, 2007 at 05:58 UTC
I like this way of computing it better, for several mostly unimportant reasons. :) `#!/usr/bin/perl -w use strict; print countPairPermutations( @ARGV ), $/; sub countPairPermutations { my( $x, $y )= @_; ( $x, $y )= ( $y, $x ) if $y < $x; my $log= 0; for my $i ( 1 .. $x ) { $log += log( $y+$i ) - log( $i ); } my $exp= int( $log / log(10) ); my $mant= exp( $log - log(10)$exp ); $mant= sprintf "%.f", 6 < $exp ? 3 : 8, $mant; $mant .= $exp ? "e" . $exp : ""; return $mant if 6 < $exp; return 0 + $mant; }` [download] My code appears to agree with your code (though I didn't count the number of digits output by your code) but they both disagree with your node, reporting that there are 5.807e234 different permutations (not 1.962e230). Update: Or, using my prototype module: `#!/usr/bin/perl -w use strict; require Math::BigPositiveOkayPrecision; print countPairPermutations( @ARGV ), $/; sub countPairPermutations { my( $x, $y )= @_; ( $x, $y )= ( $y, $x ) if $y < $x; my $p= Math::BigPositiveOkayPrecision->new( 1 ); for my $i ( 1 .. $x ) { $p= $p * ( $y+$i ) / $i; } return $p; }` [download] - tye	[reply] [d/l] [select]

Replies are listed 'Best First'.

Re^6: One Zero variants_without_repetition (logue)
by tye (Sage) on Aug 08, 2007 at 05:58 UTC

I like this way of computing it better, for several mostly unimportant reasons. :)

#!/usr/bin/perl -w
use strict;

print countPairPermutations( @ARGV ), $/;

sub countPairPermutations
{
    my( $x, $y )= @_;
    ( $x, $y )= ( $y, $x )
        if  $y < $x;
    my $log= 0;
    for my $i (  1 .. $x  ) {
        $log += log( $y+$i ) - log( $i );
    }
    my $exp= int( $log / log(10) );
    my $mant= exp( $log - log(10)*$exp );
    $mant= sprintf "%.*f", 6 < $exp ? 3 : 8, $mant;
    $mant .=  $exp  ?  "e" . $exp  :  "";
    return $mant
        if  6 < $exp;
    return 0 + $mant;
}
[download]

My code appears to agree with your code (though I didn't count the number of digits output by your code) but they both disagree with your node, reporting that there are 5.807e234 different permutations (not 1.962e230).

Update: Or, using my prototype module:

#!/usr/bin/perl -w
use strict;
require Math::BigPositiveOkayPrecision;

print countPairPermutations( @ARGV ), $/;

sub countPairPermutations
{
    my( $x, $y )= @_;
    ( $x, $y )= ( $y, $x )
        if  $y < $x;
    my $p= Math::BigPositiveOkayPrecision->new( 1 );
    for my $i (  1 .. $x  ) {
        $p= $p * ( $y+$i ) / $i;
    }
    return $p;
}
[download]

- tye

[reply]
[d/l]
[select]