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Elijah,
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...

There is an underlying pattern that can be reduced to simple enough instructions to end up with a single array of N2 elements populated iteratively. It didn't take me too long to figure it out once I flattened a couple of starting lists with their corresponding solutions

```01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16
01 02 03 04 12 13 15 05 11 16 15 06 10 09 08 07

01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 2
+4 25
01 02 03 04 05 16 17 18 19 06 15 24 25 20 07 14 23 22 21 08 13 12 11 1
+0 09
You can see the solution below:
```#!/usr/bin/perl
use strict;
use warnings;

my \$n = \$ARGV[0] || 5;
my @spiral = gen_spiral(\$n);

output(@spiral);

sub output {
my @spiral = @_;
my \$n = sqrt @spiral;
my \$cnt;
for ( @spiral ) {
print "\n" if \$cnt++ % \$n == 0;
printf("%-3d ", \$_);
}
}

sub gen_spiral {
my \$n = shift;
my (\$tot, @spiral) = (\$n * \$n, ());
my (\$index, \$cnt, \$l_border, \$r_border, \$fill_size) = (1, 1, 1, \$t
+ot, \$n + 1);

while ( \$cnt <= \$tot ) {

##### Moving right #####

# Step 1, decrease fill size
--\$fill_size;

# Step 2, fill in
for ( 1 .. \$fill_size ) {
\$spiral[ \$index++ ] = \$cnt++;
}
--\$index;

# Step 3, skip by n's
for ( 1 .. \$fill_size - 1 ) {
\$index += \$n;
\$spiral[\$index] = \$cnt++;
}

##### Moving left #####

# Step 1, decrease fill size
--\$fill_size;

# Step 2, fill in
for ( 1 .. \$fill_size ) {
\$spiral[ --\$index ] = \$cnt++;
}

# Step 4, skip by n's
for ( 1 .. \$fill_size - 1 ) {
\$index -= \$n;
\$spiral[\$index] = \$cnt++;
}
++\$index;
}
shift @spiral;  # Arrays are 0 based
return @spiral;
}

Cheers - L~R

In reply to Re: Spiraling integers by Limbic~Region
in thread Spiraling integers by Elijah

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