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Re: Pascal triange...

by ton (Friar)
on Jun 19, 2002 at 08:46 UTC ( #175591=note: print w/ replies, xml ) Need Help??


in reply to Pascal's triangle...

Hey kiat,
I noticed that your algorithm skips the "1 1" row. I made some changes to fix that bug:

sub pascal { my ($rows) = @_; print "1\n"; for (my $outer = 1; $outer < $rows; $outer++) { my $inner = $outer; print "1"; for ($i = 1; $i < $inner; $i++) { my $denominator = factorial($i)*(factorial($inner-$i)); my $pascalnum = factorial($inner)/$denominator if ($denominator +!= 0); print " $pascalnum" if ($outer > 1); } print " 1\n"; } }
I also think that using the Binomial Theorem when printing out an entire Pascal's triangle is inefficient. I would rather add numbers from the previous row, as this is much, much faster for large numbers of rows. So I wrote up some code to do this:
sub ton_pascal { my $rows = shift; my $last_row = [ ]; my $this_row = [ 1 ]; last unless ($rows > 0); print "1\n"; for (my $i = 1; $i < $rows; ++$i) { $last_row = $this_row; $this_row = [ 1 ]; for (my $j = 1; $j < $i; $j++) { push(@$this_row, $last_row->[$j - 1] + $last_row->[$j]); } push(@$this_row, 1); print join(' ', @$this_row) . "\n"; } }
I'm using array references instead of arrays for extra speed (when we copy the results of $this_row into $last_row), but the references could be removed without affecting the algorithm.

Hope this was helpful... I had fun coding up ton_pascal, so thanks for the problem!

-Ton
-----
Be bloody, bold, and resolute; laugh to scorn
The power of man...


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Re: Re: Pascal triangle...
by kiat (Vicar) on Jun 19, 2002 at 10:52 UTC
    Hello! Ton,

    Thanks for the correction on the double '1' !

    I just ran your code and it works perfectly - it's not only more elegant but also more efficient. It helps me see how the same problem can be solved by looking at it another way. Thanks!!

    kiat
Re: Pascal triange...
by Abigail-II (Bishop) on Jun 19, 2002 at 11:48 UTC
    No need to copy rows. Here's a much smaller function:
    sub pascal { my @row; foreach (1 .. shift) { push @row => 1; $row [$_] += $row [$_ - 1] for reverse 1 .. @row - 2; print "@row\n"; } }

    Abigail

      Hello! Abigail-II,

      I like your smaller function (it's really elegant) but I don't understand how it works, even though it's only a few lines. Could you explain the parts to me?

      Thanks in anticipation :)

      kiat
        It's not hard to see how it works. First thing to realize that one way of calculating the next line in the triangle is to take the previous line twice, shift one of the line one position to the right, and then add the elements piecewise. For instance:
            1  3  3  1
               1  3  3  1
            ------------- +
            1  4  6  4  1
        
        But if you look carefully, you see that to each element, we add the element to the left of it, except for the two elements on the far ends - which will both be one (a "new" 1 on the right, and the one of the left remains "as is").

        And that's how the program works. We first add a new element (with the push), then for each element, we add the preceding one. We have to work backwards of course, which is achieved by the reverse.

        Abigail

      Here's a variation on the theme. The difference is that this version doesn't change @row using pushes. It's going to be sized right the first time.
      sub pascal { my @row = (0) x $_ [0]; $row [0] = 1; foreach (1 .. shift) { print "@row[0 .. $_ - 1]\n"; $row [$_] += $row [$_ - 1] for reverse 1 .. @row; } }

      Abigail

        Maybe we can add space padding to your function :)
        sub pascal { my $max = shift or return; my @row = (0) x $max; $row[0] = 1; foreach (1 .. $max) { print " " x ($max - $_),"@row[0 .. $_ - 1]\n"; $row[$_] += $row[$_ - 1] for reverse 1 .. @row; } }

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