Re: compute paths in Pascal's triangle (aka Tartaglia's one)

[tybalt89]

#!/usr/bin/perl

# http://perlmonks.org/?node_id=1211497

use strict;
use warnings;

sub up
  {
  my ($x, $y) = split /-/, $_[0];
  $x && $y and up( $x - 1 . '-' . ($y - 1), @_ );
  $x and up( $x - 1 . "-$y", @_ );
  $x || $y or print "@_\n";
  }

up( '3-1' );

[LanX]

So what if the start is not "0-0" ? ;-)

[tybalt89]

Here's the functional version Re: compute paths in Pascal's triangle (aka Tartaglia's one) modified to allow a top other than 0-0.

#!/usr/bin/perl

# http://perlmonks.org/?node_id=1211497

use strict;
use warnings;

sub up
  {
  my ($row, $col) = split /-/, $_[0];
  my ($startrow, $startcol) = split /-/, $_[-1];
  return $_[0] eq $_[-1] ? "@_[0..@_-2]\n" :
    ($row - $startrow > 0 && $col - $startcol > 0 &&
      up( ~-$row . '-' . ~-$col, @_ ) ) .
    ($row - $startrow > $col - $startcol &&
      up( ~-$row . '-' . $col, @_ ) );
  }

print up( '3-1', '1-0' ); # bottom, top

[LanX]

I think you could just have sticked with your first program, and solve it by a coordinate transformation.

3-1 to 1-0 is essentially the same like 2-1 to 0-0

               0-0
            1-0   1-1
         2-0   2-1   2-2
      3-0   3-1   3-2   3-3
   4-0   4-1   4-2   4-3   4-4
5-0   5-1   5-2   5-3   5-4   5-5

Though you would have ended with more readable code ... nah forget it! ;-)

[tybalt89]

That would be an entirely different problem altogether </Airplane joke>

[LanX]

Come on, that's not much harder than golfing with onions on an airplane!

PS: Zero Hour! Vs Airplane!