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.
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