Sierpinskij (Re: Binomial Expansion)

[larsen]

A pattern? Yes, of course!
Here what you can obtain if you paint a Pascal Triangle.

Update I think this link could be amusing: Pascal's Triangle

#!/usr/bin/perl

use strict;

my $n = 10;

for my $i (0 .. $n) {
    
    for my $j (0 .. $i) {
        my $coefficient = nCr($i, $j);
        if ($coefficient % 2) {
            print '#';
        } else {
            print ' ';
        }
    }
    
    print "\n";
}

# returns n!/r!(n-r)!
sub nCr
{
    my $n=shift;
    my $r=shift;

    return int nFactorial($n) / int nFactorial($r) * int nFactorial($n
+-$r);
}

# like the name says, n!
sub nFactorial
{
    my $n=shift;
    my $product = 1;

    while($n>0)
    {
        $product *= $n--;
    }

    return  $product;
}

[tye]

I thought part of the point of Pascal's triangle was to show that you don't have to compute factorials:

my @line= (1);
while( 1 ) {
    print "@line\n";
    push @line, 0;
    for(  reverse 1..$#line  ) {
        $line[$_] += $line[$_-1];
    }
}

Be sure to pipe the output to more.
        - tye (but my friends call me "Tye")

[rpc]

Not only is it Pacal's triangle, it's also a Sirpinski Gasket if you associate one character/color to even numbers, and another to odd.

Math rox.

Update: Ugh i am not awake yet. Forgive my redundancy.

[oha]

i found that code some times ago, unfortunately i don't know the author. i post here, maybe someone don't know.

#!/usr/bin/perl -l
s--@{[(gE^Ge)=~/[^g^e]/g]}[g^e]x((!!+~~g^e^g^e)<<pop).!gE-ge,
s-[^ge^ge]-s,,,,s,@{[(g^';').(e^'?')]},(G^'/').(E^'|')^Ge,ge,
print,s,(?<=/[^g^e])[^g^e][^g^e],$&^(G^'/').(E^'|')^gE,ge-ge