Previous: Trichain: Going Higher

Trichain is a puzzle type I invented. It is somewhat similar to Nurikabe, but it uses a chain of L-trominoes instead of a continuous wall.

Example puzzle with solution:


  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.

Now, solve the following puzzle. The question marks indicate unknown numbers, which may or may not be distinct.

This one introduces classic Nurikabe-like logic, and the beginning moves at the center. Also, it is my first serious try at a symmetric design (except for the example puzzle), and I found that unknown numbers work well in this genre.


1 Answer 1


The completed grid:



Putting the blocks between clues separated by a single square, focus on the 6 in the middle. The square above it must be unshaded, since the L cannot go to either side. This forces the squares to either side to be blocked, and we finish the Ls in the only possible way. Filling in the simple deductions from here, we force the square right of the 3 to be unshaded, which completes its group. Again doing the simple deductions, we complete the group of 6 in the middle, which we can also close off. Finishing the simple deductions, we end up with:


Now look at the 14:

Note that the open group of 14 cannot use the square to the right of the 11, so the 14 must escape to the right of this square. This forces the square right of the 11 to be shaded, as well as the one above it. We can also complete the L right of the 14, and unshade adjacent to it. Now counting squares in the group of 14, we see we have to have both squares R2C10 and R4C10 in this group. Since all of the row from R2C7 to R2C10 is unshaded, we cannot fit an L in row 1, so R1C8 through R1C10 must be unshaded. This gives us 14 squares in the group, so we close it off with existing partial Ls, and finish the simple deductions. The grid thus far:


Now look at the group of 11:

We must block R2C4 to separate this group from the one below, and this L must also include R2C3 and R3C3, otherwise the group of 11 will be pinched off too soon. Ensuring the squares around this L are unshaded, we now see that R1C2 must be unshaded to continue filling the 11 block, which forces R1C1 and R2C1 to be unshaded as well. This fills out the group of 11, and we have some more simple deductions from there. The grid thus far:


The lower left corner:

The trickiest bit is to look at the square above the 6. If this square is unshaded, then simple deductions leave us in this position:


The only shading we can do in the lower left corner is to either bring the top L into the top left corner, or to put an L in the corner below the 6. The latter clearly cannot happen, since it would cut the 6s area to at most three unshaded squares. If we bring in the top L, then this corner has 5 squares, and we must shade all of them, but this gives us 7 squares. Thus we cannot do any additional shading in this corner, which forces it to have at least 8 squares. So the square above the 6 must be shaded.

If we do not finish this L to the left, we have to put an L in the corner below the 6 to prevent the 6s group from meeting with the large group coming down the left side. Combined with an L needed to separate the 6 group from the question mark a knight's move away, the 6 can be in a group of at most three squares. This forces the square under the question mark to be unshaded. Following the simple deductions that occur, we restrict the group of 6 to a small corner area which can only fill up to size if the bounding Ls are finished as far away from the corner as possible. Finishing the additional simple deductions, our grid looks like:


Finishing up:

In the lower right corner, note the square right of the 7 cannot be shaded, since it forces the 7 to escape down where it is quickly pinched off by the Ls needed to separate its group from the question mark. This forces the square above the question mark to be shaded, and also R7C10 to be unshaded, which brings the number of unshaded squares in 7s group up to 5. If the square beneath the 7 is unshaded, then this group grows to 7, forcing both R9C7 and R9C8 to be shaded, creating a shaded tetromino. Once we know this R8C8 is shaded, the rest of the grid falls with simple deductions.

  • $\begingroup$ The solution is correct :) $\endgroup$
    – Bubbler
    Commented Oct 13, 2020 at 0:40
  • $\begingroup$ Nice expressive write-up. Did you enjoy the 6 and 7 at the corners? $\endgroup$
    – Bubbler
    Commented Oct 13, 2020 at 2:31
  • $\begingroup$ @Bubbler Absolutely...the counting logic was a nice change of pace. Were those the intended deductions? $\endgroup$ Commented Oct 13, 2020 at 12:13
  • $\begingroup$ Yes, your steps are exactly the same as mine, modulo the order of completed corners. $\endgroup$
    – Bubbler
    Commented Oct 13, 2020 at 12:21
  • $\begingroup$ Great puzzle, @Bubbler...please keep 'em coming! $\endgroup$ Commented Oct 13, 2020 at 12:22

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