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A recursive solution:

use strict; use warnings; my \$n = shift; my \$count = 1; my @matrix; push @matrix, [(0)] for (1..\$n); buildRing (0, \$n - 1); my \$width = length sprintf ("%s", '' . \$count); my \$format = "%\${width}d"; print "" . (join " ", map {sprintf \$format, \$_} @{\$matrix[\$_]}) . "\n" + for (0..(\$n-1)); sub buildRing { my (\$first, \$last) = @_; \$matrix[\$first][\$_] = \$count++ for (\$first..\$last); \$matrix[\$_][\$last] = \$count++ for ((\$first+1)..\$last); \$matrix[\$last][\$last-\$_+\$first-1] = \$count++ for (\$first..\$last-1); for (\$_ = --\$last; \$_ > \$first; --\$_) {\$matrix[\$_][\$first] = \$count++} \$matrix[\$first][\$last] = \$count if (++\$first == \$last); buildRing (\$first, \$last) if (\$first<\$last); }
use strict; use warnings; my \$n = shift || 5; my (\$count, \$first, \$last) = (1, 0, \$n - 1); my @matrix; push @matrix, [(0)] for (1..\$n); { \$matrix[\$first][\$_] = \$count++ for (\$first..\$last); \$matrix[\$_][\$last] = \$count++ for ((\$first+1)..\$last); \$matrix[\$last][\$last-\$_+\$first-1] = \$count++ for (\$first..\$last-1); for (\$_ = --\$last; \$_ > \$first; --\$_) {\$matrix[\$_][\$first] = \$count++} \$matrix[\$first][\$last] = \$count if (++\$first == \$last); redo unless \$first >= \$last; } my \$width = (int log \$count) - 1; print "" . (join " ", map {sprintf "%\${width}d", \$_} @{\$matrix[\$_]}) . + "\n" for (0..(\$n-1));
\$n=shift;(\$c,\$f,\$l,@m)=(1,0,--\$n); {\$m[\$f][\$_]=\$c++for(\$f..\$l);\$m[\$_][\$l]=\$c++for(\$f+1..\$l); \$m[\$l][\$l-\$_+\$f-1]=\$c++for(\$f..\$l-1);for(\$_=--\$l;\$_>\$f;--\$_){\$m[\$_][\$f +]=\$c++} \$m[\$f][\$l]=\$c if(++\$f==\$l);redo unless\$f>\$l}my\$w=(int log\$c); print"".(join"",map{sprintf"%\${w}d",\$_}@{\$m[\$_]})."\n"for(0..\$n)
Update: added (smaller) non-recursive version and golf version

Perl is Huffman encoded by design.

In reply to Re: Spiraling integers by GrandFather
in thread Spiraling integers by Elijah

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