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Re^4: Understanding Split and Join

by ferreira (Chaplain)
on Jan 02, 2007 at 11:01 UTC ( #592528=note: print w/replies, xml ) Need Help??

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  1. or download this
    $ perl -e '@a = split //, 'abc'; print "@a"'
    a b c
    $ perl -e '@a = split / (?#) /x, 'abc'; print "@a"'
    a b c

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[oiskuu]: You want to draw the path? $iter = combinations( $TOP_HEIGHT .. $NODE_HEIGHT, $pos); then get the path from the iter, level change at depth x, for(..) $pos += $level_change; something like that
[oiskuu]: yergh, <code> tags... [$top_height .. $node_height]
[Eily]: Discipulus by demonstrate I suppose you mean simulate with perl?
[Eily]: the demonstration itself is fairly easy. The number of paths on a node is the sum of the numbers of paths to the two nodes above (or one node above on the edges of the triangle)
[Eily]: since the value of a node is also the sum from the two nodes above, you just have to demonstrate that the equality is true at the top of the triangle
[Discipulus]: yes Eily, thanks oiskuu but i dont get it.. ;=( maybe I'll ask a SOPW
[LanX]: all combinations with same amount of left and right?
[Eily]: if you want to store in a structure with the coordinates as key, arrays might do, since the keys are going to be 0..n
[LanX]: (Pascale path)
[Eily]: paths like that

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