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Trichain is a new 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 title is a pun in two ways. One is that I designed this puzzle to not have any 1s or 2s (and therefore only using higher numbers), after Deusovi pointed in chat that the puzzle type seems to heavily rely on them. You can find the other while solving it :)

  • $\begingroup$ In retrospect, I was a little hasty when writing this one. Though the puzzle as a whole has a unique solution, I did overlook a specific branch at the reasoning around 7, resulting in a much longer, unintended backtracking. $\endgroup$
    – Bubbler
    Commented Oct 8, 2020 at 10:11

1 Answer 1


The solution to the puzzle is as follows:


Steps to solving:

Firstly, notice that the 3 cannot be a straight line otherwise we would have to have a wall of 3 consecutive shaded cells.
Step 1

Filling in some cells adjacent to trimonoes and the 3 gives us this:
Step 2

To avoid a shaded L-tetromino, R10C5 must be shaded. If that is part of a top-left corner L-trimono, the 7 region is too large. So R10C6 is also shaded. Step 3

Now, consider what happens if R9C5 is shaded. A few more cells must be shaded to block off the 7 as shown:
Step 4.1
We get some more cells from the central 3:
Step 4.2
If R7C7 was shaded, then the right 3 would become an X-pentomino. So R9C7 is shaded instead. However, this results in an I-trimono being shaded to prevent the 4 from becoming an F-pentomino:
Step 4.3
Thus, R9C5 is not shaded, and R9C6 therefore has to be. This gives another shaded cell to block off the 7. R9C8 also has to be shaded, otherwise the 4 becomes an L-tetromino which would result in an adjacent shaded I-trimono.
Step 4.4

If R8C8-9 were shaded, the 4 would either have only 1 cell or at least 11 cells, depending on whether there is a trimono in the corner. Step 5.1
If R8C7-8 were shaded, the 4 would either have 3 or 9 cells, depending on whether there is a trimono in the corner.
Step 5.2
Thus, R10C8-9 are shaded. (Also, R6C6 is shaded to prevent the 3 from becoming a V-pentomino.) Then if R9C10 were shaded the resultant trimono would make a one-cell 4 region, so it must be unshaded.
Step 5.3

If the 4 were an S-tetromino, it would join with the unshaded R8C8 to become a W-pentomino. So it is instead a T-tetromino, as shown:
Step 6

Several trimonoes are now forced, to block off the 3 regions:
Step 7

Now, we need to block off the 4 so that it doesn't become a 7-cell snake:
Step 8

If R3C5 were shaded, to avoid the 5 from becoming a 7-cell tree we need to shade R2C6 which would then form an at least 4-cell shaded block.
Step 9.1
So R3C5 is unshaded.
Step 9.2

If R2C5 were unshaded, note that since either R7C1 or R7C2 are shaded the 5 must escape down or left, and it would thus connect to form a 7-cell tree.
Step 10.1
So R2C5 is shaded. Also, R3C3 is shaded to complete a trimono.
Step 10.2

If R1C4-5 are shaded, then to stop the 5 from becoming a P-hexomino, R3C6 is also shaded. But then the resultant trimono blocks off the 5 to a region of at most 4 cells.
Step 11.1
If R1C5 and R2C6 are shaded, then the trimono to the right of 5 needs R1C7 shaded too, blocking the 5 into a monomino.
Step 11.2
Thus, R2-3C6 are shaded, forcing R1C7 shaded in the trimono to the 5's right as well.
Step 11.3

Finally, the 5 must form an N-pentomino, and the last trimono is forced:
Step 12

  • 1
    $\begingroup$ Although you got it right in the end, you had a few leaps in the logic. E.g. At step 2, you can't be sure that the tromino on the left of middle 3 makes a left turn. Also at step 3, there are more possibilities for the mid-bottom tromino, though all of them are eliminated by the 4 (which is IMO the hardest part of this puzzle). $\endgroup$
    – Bubbler
    Commented Oct 8, 2020 at 7:55
  • 1
    $\begingroup$ @Bubbler Ah yes, I see what you're referring to. I don't have much time right now, but I'll see if I can resolve the gaps in the logic later. $\endgroup$
    – boboquack
    Commented Oct 8, 2020 at 8:12
  • 2
    $\begingroup$ @Bubbler Should all be fixed now :) $\endgroup$
    – boboquack
    Commented Oct 8, 2020 at 9:23

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