0 0 0 0 -1 0 0 0 0 0 0 0 0 -1 0 0 0 0 -1 #### 1 2 3 4*-1 0 0 0 5 0 0 0 6 -1 0 0 0 7 * -1 #### 1 2 3 4 -1 0 0 0 5 x 0 0 0 6 -1 0 0 0 7 x x x x -1... #### 1 2 3 4 -1 * 12 0 0 5 x 11 0 0 6 -1> <10 9 8 7 x x x x -1... #### #!/usr/bin/perl -w use strict; # The width of our square # (\$ARGV or 5 if no arguments given): my \$n= (@ARGV,5); # Our square, of size \$n+1 x \$n+1 due to the sentinel values: my @s= ( ( (0)x\$n, -1 )x\$n, (-1)x\$n, -1 ); # The directions we will move (right, down, left, up) # since from \$s[\$p], \$s[\$p+1] is just to the right and # \$s[\$p-\$n-1] is just above: my @d= ( 1, \$n+1, -1, -\$n-1 ); # Our starting direction, an index into @d; 0 for "right", \$d: my \$d= 0; # Our starting position (index into @s); -1 so our first step # to the right will land us at \$s: my \$p= -1; # Our starting value (to be stored into @s); # 0 so we'll enter 1 after our first step: my \$v= 0; # Take a step (\$p +=). If that step lands on an occupied # spot, then we're done. So continue while zero (not true): while( ! \$s[ \$p += \$d[\$d] ] ) { # Store the next value where we just stepped to: \$s[\$p]= ++\$v; # Look where we will step next. # If occupied (not zero, i.e. true)... if( \$s[ \$p + \$d[\$d] ] ) { # ...then switch to the "next" direction in @d # wrapping back to \$d if needed: \$d= (\$d+1) % @d; } } # Print out the 1..\$v values plus \$n newlines represented # by the first \$v+\$n elements of @s: for( @s[ 0 .. \$v+\$n-1 ] ) { # Sentinel values represent newlines: if( \$_ < 0 ) { print \$/; } else { # Otherwise left-justify the value, just # wide enough to hold the largest value, \$v: printf " %".length(\$v)."d", \$_; } } #### my @s= ( ( (0)x\$n, -1 )x\$n, (-1)x\$n, -1 ); #### 0 0 0 0 -1 0 0 0 0 -1 0 0 0 0 -1 0 0 0 0 -1 -1 -1 -1 -1 -1