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### Re: Pascal triange...

by Abigail-II (Bishop)
 on Jun 19, 2002 at 11:48 UTC ( #175631=note: print w/replies, xml ) Need Help??

in reply to Re: Pascal triange...

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

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Re: Pascal triange...
by Abigail-II (Bishop) on Jun 19, 2002 at 17:04 UTC
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

```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;
}
}

Re: Re: Pascal triange...
by kiat (Vicar) on Jun 19, 2002 at 13:40 UTC
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

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