Santoitchi is yet another genre involving trominoes. The name seems to mean Three-and-One in Japanese with a deliberate misspelling (イッチ instead of the usual イチ for "one").

Here is an example puzzle with solution:


  1. Shade some cells. The shaded cells are not allowed to share an edge.
  2. Divide the unshaded cells into trominoes (contiguous group of three cells).
  3. Each tromino must contain exactly one number.
  4. The number indicates how many shaded cells share an edge with the region. (Not to be confused with "how many edges of the region are shared with shaded cells")

Now solve the following puzzle. A question mark represents one number between zero (inclusive) and infinity.

Sorry for puzzle change. This one is designed around a "key deduction" I had in mind. However, shortly after posting the original puzzle, I realized that there's a rather trivial, unintended solving path. The revised one eliminates the trivial path (hopefully; at least I checked but couldn't find any) and will force you to find the "key deduction".

  • $\begingroup$ Realistically, zero and six (not infinity), of course. Anything higher than that would break other rules. $\endgroup$ – Darrel Hoffman Nov 13 '20 at 19:05

The solution:


The “key deduction” involves

finding how many cells are unshaded, i.e. part of trominoes, and how many are shaded. There are 23 numbers on the grid, meaning that there are 23 trominoes on the grid and hence 69 unshaded cells. The grid has a total of 77 cells, so the remaining 77 - 69 = 8 must be shaded cells.

This allows us to make our first breakthrough:

Note that 8 is also the total number of 2s on the grid. Every tromino with a 2 in it must border exactly 2 shaded cells, and the only way we can make that work on this grid is if every shaded cell bordered exactly two of these trominos. This forces us to draw the 2 trominos in the upper and lower center as lines with the 2s in the middle, since any other tromino placement would make it impossible for every shaded cell to be shared. Continuing along the edge of the grid using the shared cell requirement gives us this initial step:
(Note also that we separate numbered cells with edges as every tromino can contain only one number.)

The remainder of the puzzle is fairly straightforward:

There is only one way to draw the tromino for the 0 in R5C7, and after that, there is only one way that cell R6C8 can be part of a tromino, and so on. Also, cell R2C2 has to belong to the ? tromino in R3C2, and this tromino must also contain cell R2C3 otherwise it will be unreachable. And cell R2C6 has to belong to the ? tromino in R3C4:

And we can finally finish it off:

The 0 tromino in R3C4 can only be drawn one way. This forces the rest of the trominos to be drawn as such, giving our final solution:

(Let me know if any steps need to be elaborated on further - after the key deduction the rest of the deductions seemed simple, but there may be something non-obvious I missed.)

  • $\begingroup$ The solution is correct. I wanted the solvers to notice that, after the "key deduction" part, rot13(gur ?f zhfg or nyy barf fvapr lbh unir 8 rkcbfrq funqrq fdhnerf naq 8 ?f, naq ab ? pna pbire gjb funqrq fdhnerf ng bapr), but yeah, the center part being easy was pretty much unavoidable. $\endgroup$ – Bubbler Nov 13 '20 at 5:04

An alternative explanation of the "key deduction" step, for strict logicians:

As HTM noted, the grid must have exactly 8 shaded cells, since we have 7 × 11 = 77 cells in total and 23 × 3 = 69 cells are covered by trominoes.

Observe the positions of the 2's, especially the four 2's at the corners. Notice that a shaded cell anywhere cannot share borders with two different 2's at the corners. This means all 8 shaded cells must border with one of those 2's. The same can be said for the four 2's at the sides.

This condition can be satisfied in two ways: create four pairs of 2's and let each pair share two shaded cells each, or create a giant ring around the board. But the former cannot be satisfied because the 2's on the long side are too far away from both corners, so it must be the latter. The result is the following, as presented in HTM's solution:


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