http://www.perlmonks.org?node_id=487200

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.