Hot diggity! I didn't read the puzzle definition closely enough!

Well, I could easily fix that by updating valid(). Give me a few minutes and I'll update this node with a solution that solves the puzzle accurately.

Update: hum... I'm getting protection faults :( I think I can't use regexps in valid() when it's called from within a regexp.

Update: Fine, I won't use regexp in valid(). What follows is my updated solution. The only difference the var $regsz and valid() has an additional check.

Original
========
___1__74_
_5__9__32
__67__9__
4__8_____
_2_____1_
_____9__5
__4__73__
73__2__6_
_65__4___

Solution
========
392185746
857496132
146732958
479851623
528673419
613249875
284567391
731928564
965314287
[download]

I had a play with your code to try and fix the problem, but headed off in a much less elegant direction than you did by adding a $grid_c vector and another nested loop.

It got rather messy rather quickly. :-(

---

Perl is Huffman encoded by design.

Yeah, there are diminishing returns to adding more shortcuts.

Technically, $grid_v is not needed at all.

return if index(substr($grid_v, $x*$size, $size), $n) >= 0;
[download]

can be written as

foreach my $y_ (0 .. $size-1) {
   return if substr($grid_h, $y_*$size+$x, 1) eq $n;
}
[download]

I just thought the former would be faster, even though we must constantly make copies of $grid_v in addition to $grid_h.