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I came across a challenge at a site I visit regularly and decided to give it a go.

The challenge was to write a program (in any language) to get a spiraling printout of integers based upon the user input. For example if the user input "5" then the output would be:

1   2   3   4   5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9

. . . . n
.
.
.
. . . . 2n-1 and so on...

I decided to of course write it in perl and used a two dimensional array method. Here is the code I came up with:

#!/usr/bin/perl -w use strict; print "Enter n: "; chomp(my $n = <>); my @shiza; my $firstD = my $secondD = 0; my $cnt = my $inc = 1; while ($cnt <= ($n*$n)) { while (!$shiza[$firstD][$secondD] && $secondD < $n && $inc == 1) { $shiza[$firstD][$secondD] = $cnt; $secondD++ unless($shiza[$firstD][$secondD+1] || ($secondD+1) >= + $n); $cnt++; } $firstD++; while (!$shiza[$firstD][$secondD] && $firstD < $n && $inc == 1) { $shiza[$firstD][$secondD] = $cnt; $firstD++ unless($shiza[$firstD+1][$secondD] || ($firstD+1) >= $ +n); $cnt++; } $secondD--; $inc = 0; while (!$shiza[$firstD][$secondD] && $secondD >= 0 && $inc == 0) { $shiza[$firstD][$secondD] = $cnt; $secondD-- unless($shiza[$firstD][$secondD-1] || ($secondD-1) < +0); $cnt++; } $firstD--; while (!$shiza[$firstD][$secondD] && $firstD >= 0 && $inc == 0) { $shiza[$firstD][$secondD] = $cnt; $firstD-- unless($shiza[$firstD-1][$secondD] || ($firstD-1) < 0) +; $cnt++; } $secondD++; $inc = 1; } for (@shiza) { for (@{$_}) { print "$_\t" if($_); } print "\n"; }
I am sure there is a mathematical equation that would simplify this but I was spending too much time with my TI-89 Titanium trying to figure it out so I just decided to write it assigning the values to the two dimensional way in a spiraling manner also.

In reply to Spiraling integers by Elijah

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