# Trichain: A Bottle of Rum

This is a Trichain puzzle. The rules of Trichain are as follows:

1. Numbers must be white, and match the size of their white islands.
2. Each white island can only contain at most one number. Some white islands may have no numbers.
3. Black squares must be part of L-trominoes.
4. Black L-trominoes cannot be adjacent horizontally or vertically, but all of them must be connected diagonally. In other words, one standing on a black square must be able to visit any other black square via Chess King's moves, stepping on black squares only.

Here is an example Trichain puzzle, and its solution:

Here is the real Trichain. Each X represents a different number. Go for it! The completed grid is:

Working in the upper right corner first:

If the upper right corner is shaded, then the L must come down around the 2, leaving R2C8 unshaded. But then R1C8 must not be shaded for the 2 to be satisfied, forcing R2C8 to be shaded, a contradiction. Thus the upper right corner must be open, getting us to: Now look at the 3 in the upper middle:

Its third unshaded square cannot be R3C6, since L-trominos cannot block off three unshaded squares in a line. If we assume the square left of the 3 is shaded, we run out consequences quickly to find that the 3 in the upper left corner is isolated: This is a contradiction, so we must have this square unshaded, and thus the squares to its left and bottom are both shaded. If the corner between these two is shaded, we must have R1C3 and R2C3 unshaded, which leads to a contradiction similar to the previous. This gets us to: Let's briefly look at the bottom left corner:

The unshaded square with the 2 cannot be the one to its right, for if it were, the L capping it on the end would have to cover the 4. Filling this in gives: Look at the middle right:

Look at the 3 at middle right. The square above it cannot be unshaded, for if it were, the square above that would be shaded to avoid the 3 and the X being in the same area, and that would create at least a tetromino above. This then forces R5C9 to be shaded as well, and thus R5C8 to be unshaded.

Focus on the X in R4C9. If R4C8 is unshaded, then R4C7 must be unshaded as well, since an L shading it would cover either R5C6 or R5C8, which are unshaded. But then the rightmost two Xs are forced to be in the same area. Thus R4C8 must be shaded, and consequently R4C7 as well. This gives: Now the lower right:

Look at the square exactly in the middle of the two 3s must be shaded; if it were not, the squares above and below it would have to be shaded since a line of three unshaded squares cannot be isolated. The Ls containing these two shaded squares cannot intersect, so one or the other must isolate the 3 to which it is adjacent. R8C9 is then shaded as well, and R8C8 must be unshaded.

Now bounce back up to the two squares at R4C7 and R4C8, and try to finish the L by assuming R3C8 is shaded. We can follow through a lot of consequences to get to the following diagram: In this diagram, the green squares, the 8, the pink square next to it, and the 3 in the corner must contain 11 clear squares and an L to separate the 8-group from the 3-group. But there are only 13 squares here...a contradiction.

At this point, I'm pretty tired, and need to admit that I case-bashed my way to finish the L between the two 3s in the bottom right. If you try to shade either R9C9 or R9C10, you are forced to cover either R6C7 or R8C7 with an L, and if you try to shade R7C9 the upper 3 cannot be blocked off. So R7C10 it is! Following the basic logic from here gives: Finishing up around the 8:

The square to the left of the 8 cannot be shaded, for if it were, there are only seven squares that can be unshaded in its group while still excluding the 4. We then must shade R6C6, since unshading would connect to the X. This forces R6C5 to be shaded as well. Trying to complete this L with R7C5 forces a too large pentomino around the left side of the 4, so R5C5 must finish this L. To keep the 8 and the leftmost X separate, we must shade R7C4 and R7C5. At this point, there are only a few unforced squares remaining, and the final grid is easy to arrive at.