1. Make each cell white or black.
2. No two black cells may share an edge.
3. All white cells must be connected to each other through their edges.
4. A white cell with a number and an arrow show how many black cells are in the row or column pointed at by the arrow.
5. Cells with a number and an arrow can be shaded in. If they are, the clue is meaningless, and may or may not be true.
Make sure to read the Japanese-Crossword-Puzzle-Like Puzzle Tips.
Before I start, I'm just going to say, I love this name. It's like the best thing ever.
Anyway, just a few general things: remember to not use a clue unless you know it's white. The way you usually get started is by finding somewhere that there's an impossible clue and filling that in black.
Theorem 1: If a clue is impossible, it is black.
This is so obvious I thought of not mentioning it, but oh well. There are a few different ways this can happen:
Two black cells can't fit there, so the 2 must be black:
If you have something like this as well:
This 0 is already false because there's a black cell already. So:
Theorem 4 of Straight and Arrow applies here if the clue is white (but here will be Theorem 2):
Theorem 2: If an arrow with number N in a white cell 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.
Example:
By this Theorem, these cells are black:
(R3C4 is white because of rule 3)
Theorem 3: If an arrow is not known if it's true or not, but has all cells it needs in its row/column, the first cell in front of the arrow is white.
Why is this? Think of this zero:
The zero has two possibilities, it's true, or it isn't. Here's the possibilities:
They both have that white cell, so we can write this:
It doesn't just work with zeros, though. It also works with bigger numbers, as long as they have all the cells they need:
We also know this because of this theorem:
That's all in the way of theorems, actually. Let's solve the example puzzle:
Where do we start? Well, the two in R3C4 is impossible, and is black by Theorem 1:
We can now use theorem 2 on the two in R2C4:
(Remember, R1C2 is white because of Rule 3)
We can use Theorem 3 on this cell here:
Now that we know that cell is white, the one below it makes this cell black:
That new white zero makes all of these white:
The two in R2C6 uses Theorem 2:
Now the three in R3C3 needs these two black spots:
We can use Theorem 2 on the three in R4C8:
That one in R6C9 is impossible:
Now what? Well, the white cells in the top sections can only go out through R3C2, and the white cells in the bottom right can only go out through R8C7, so both of those cells are white:
But that doesn't help, we can't even use Theorem 2 on them, because they each have six open cells, not five. However, we can use Theorem 3 on this zero:
NOW we can use Theorem 2 on the three in R3C2:
There are three permutations that the two in R4C1 can be. They are:
The last two both choke out a white cell, so the first one is correct:
Now we can use Theorem 2 on the three in R8C7 (it needs two left and has 2*2-1 = 3 white cells left):
The two in R5C4 is impossible, because it can't get two black cells:
Now there's the same situation with the two in R6C5:
The one in R7C5 makes this black cell:
And the last two is impossible:
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