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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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