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Santoitchi is yet another genre involving trominoes. It appears to be invented by Inaba Naoki, and then picked up by Bram de Laat, who uses the name Sandwich (blame Google Translate for the possible mistranslation, as the genre apparently has nothing to do with sandwiches). 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:

Rules:

  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, which features only threes and ones (as in the name of the genre). A question mark represents one number between zero (inclusive) and infinity.

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First,

there's only one way the 3 in the bottom left can be adjacent to three shaded cells.
enter image description here
And what about the 3 above it? It can't go up by two to place its other two shaded cells, because that would block off a square on the left wall. So it must go right two: this places two more shaded cells.
enter image description here

Now, since every tromino must have exactly one clue in it,

the unshaded cell at the bottom must be part of the 1 on the right.
enter image description here
And likewise, the cell above it can't be part of the 3, or it wouldn't be able to touch 3 gray cells -- so it must be part of the ? clue instead.
enter image description here
And some more reachability logic gets us here:
enter image description here

Next we can look at a different area of the puzzle:

The 3 in the top left has only one way to satisfy it without putting two shaded cells by the 1 clue.
enter image description here

And finally, one argument finishes it off:

There are 17 clues; this means there will be 13 shaded cells in all. We have 11 of them, so the 3 on the right must touch one of our already-placed cells. This can only be the one in the second row.
And then, we can place one shaded cell with the top 3; the bottom 3 only has one place its other shaded cell can go, and the puzzle is solved!
enter image description here

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  • 1
    $\begingroup$ Correct! The logic for the 3 at R6C2 is slightly easier if you solve the top left corner first. Also, nice job for spotting the counting logic so quickly! $\endgroup$ – Bubbler Oct 28 at 2:25
  • $\begingroup$ I figured it was you when I scrolled through the length. $\endgroup$ – ention everyone Oct 28 at 2:46

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