I like Method 2

OR

Method 4 :
Enter the "number of turns" along marked arrows. Answer will be 6,6 for the example.

Method 5, still keeping the same reasoning as my post in Loop without All Cells. I/O for Inside or Outside; row 1 would be OIOOIO.

To get it without need of shading insides, simply start from left/top/wherever you need to start counting. Begin from "very far" outside the loop; it's outside. Everytime you cross a segment, it changes the inside/outside.

For example, you begin from somewhere outside the grid but in row 1, to the left of the grid, and moving right. When you get to the grid, you haven't encountered any segment, so first square is outside. The second square requires passing a segment so it's inside. The third square requires passing a segment again so it's outside again, etc.

This is identical to marking on which columns the vertical segments of row 1 are on, with leftmost column being column 0/1 (your choice). This method can't be applied if the loop can intersect itself (MellowMelon's Crosslink?), but it's rare enough to justify having some sort of exception for those kind of puzzles.

In principle, I like this method. But for a bigger grid, it would mean lot of typing. So I would like to suggest to compress the answer key to 11211 (ah, not compressed much in this case, but in general it should be). For Column 4, the answer key will be 312 instead of OOOIOO.

It also gets rid of any "English" letters. It can easily be applied to other types e.g. Tapa, Fillomino as well.

ETA : If you are worried about LMI submission engine not being able detect variable length answer keys, we should be able to upgrade the engine to check "sum of digits" to be same as "size of the grid" in this case.

Edited by debmohanty 2012-12-14 6:34 AM