Subject:
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Re: A recursion question/test...
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Newsgroups:
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lugnet.robotics
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Date:
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Wed, 12 May 1999 21:05:25 GMT
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Viewed:
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1001 times
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JR Conlin <jrconlin@email.com> wrote:
> Iterative:
> my $result = $lowBoundry;
> while ($result++ <=$highBoundry);
> return $result;
If you're not going to make it recursive, why not:
$result = $n * ($n + 1) / 2;
That is effectively the sum of all numbers from 0 to n.
-Kekoa
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Message is in Reply To:
 | | Re: A recursion question/test...
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| (...) Well, you didn't specify if the numbers existed in a list or were iterative, and you didn't specify language so: Smalltalk: ^list sum. or ^(1..n) sum. sorry.. if you'll excuse Perl... List: foreach $num (@list) {$result += $num;} return (...) (26 years ago, 12-May-99, to lugnet.robotics)
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