1. Draw a single loop that connects the centers of the grid cells.
2. The loop may only travel horizontally or vertically, not diagonally (so all turns are right angles).
3. The loop may only turn at the centers of the grid cells.
4. The loop may not cross itself or branch off.
5. The loop may not go through any cells with a number in it.
6. Any cell that does not have an arrow or is part of the loop must be shaded black.
7. No other cells may be black.
8. No two black cells may share an edge.
9. A cell with a number and an arrow show how many black cells are in the row or column pointed at by the arrow.
Japanese Crossword meets the Loop. Well, there's not really much Japanese Crossword type stuff here, it's mostly just loop stuff with arrows. You should read the Loop Tips before going on here.
(I, uh, don't usually use x's in this puzzle. You can if you want, but I'm not doing it here.)
Alright, now that you've done that, here's the first thing:
Mark if a cell isn't black, even if you don't know how the loop will pass through it.
I usually mark it by putting a dot in the middle, because then the loop will pass through the dot and hide it. Remember, if there's a zero pointing somewhere, all cells in front of it in the line will be dots. Also, if a number already has as many black cells as it needs, all other cells in that line will be dots. As an example of all of this:
R1C2 and R2C2 will be dots because of that 0, R2C3 will be a dot because it's next to a black cell, and R3C4 and R4C4 will be dots because the one already has a black cell (and R3C4 is next to a black cell anyway):
(I don't usually make the dots that big, but you couldn't see them otherwise)
Remember rule 6. If there is a part that can not be a part of the loop and is not a number, you make it black. If there is an arrow one away from an empty corner, that corner is black.
Remember that numbers stop the loop. Numbers and black cells act really similarly other than the fact that you can have a black cell next to a number.
Theorem 1: If a dot has only two exits, the loop must go through both of those.
Consider this:
This makes these dots:
By this theorem, R1C1 and R1C3 both get a part of the loop:
The theorem also extends R3C1:
We can actually extend that line even further:
And, notice R2C2 couldn't possibly be a part of the line (this would be more obvious if we were using x's), and thus must be black:
I'm mainly doing all of this to show how you do these puzzles. Only the beginning was showing the theorem.
Theorem 2: If two empty cells are orthogonally adjacent and they only have two ways to escape, the two empty cells are a part of the loop.
Say there was also this zero in the last example:
Think of R3C4 and R4C4. The two empty cells are orthogonally adjacent, and only have two escapes, both to the left. This follows the theorem, and they must be a part of the loop:
(If you want to see why this theorem is true, try solving this puzzle without using those two spots in the loop.)
Theorem 3: If two blocked cells are two away orthogonally (including diagonally adjacent as two away orthogonally), and the cell(s) in between them are empty, all empty cells that are two away orthogonally from both of the blocked cells must be part of the loop.
That's a doozy! I don't think anybody would get that without pictures, so here you go. There are two different ways this works. One is (using zeros as my blocked cells, though they really could be anything):
All of these cells must be part of the loop:
But, be careful with how you use it. If you started with:
You would only get:
(You could, however, use Theorem 2 to make those same cells dots.)
It also works here:
In fact, even in an empty grid you can still figure out this:
Because the edges act like blocked cells and at the corners there are two blocked cells diagonal to each other, this works.
Theorem 4: If an arrow with number N points in a direction where there are 2*N-1 empty spots in a row, every other cell from the arrow to the wall must be black.
This is like Theorem 1 of Room and Reason. An example:
N=2 two in this case, and 2*2-1 = 3. Well, there are 3 empty cells between that two and the wall in a row, so we do this:
I would write some dots, but this isn't solvable so I won't bother.
For an example puzzle, here's the example from the top of the post:
I'm going to do some unnecessary things just to show them off. First, I will use Theorem 2 on R9C1 and R9C2 (this puzzle is 9x9, so that's just the bottom left corner):
I could have done it at several places down there, but I chose there. Writing that line and extending it:
I'm just going to write everywhere you can use Theorem 3. That Theorem doesn't have to be used in this puzzle to solve it, but I think it's a good idea to show it off.
I THINK that's everything.
Now, Theorem 4 on the three in R5C4:
We can now use Theorem 1 on R4C9, R5C6, R5C8, and R6C9:
With this new information, we can use it again on R4C7 and R6C7:
...And again on R4C5 and R6C5:
These lines extend:
I just noticed that zero on the bottom, so there's these dots:
Now by Theorem 1 again:
The one at R3C4 has only one empty spot left to go to. So, by Theorem 4, there's a black cell here:
Extending a couple lines:
Theorem 1:
That line now has to go out for a while:
These two cells can't be in the loop so must be black:
The two in R1C2 has one black cell needed and one empty cell still, so it needs this:
The two in R5C2 only has two black cells to go to:
All of these lines need to get made because of Theorem 1 (and extending other lines):
While we're here, there's this thing that's black for multiple reasons:
The two in R2C7 has one spot left, as does the three in R9C9:
Now we extend some lines, some normally, some by Theorem 1, and one by not wanting closed loops:
The three in R8C2 has one spot left:
Extending a couple more lines:
Now we need to use Theorem 2 on this last bit:
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