# Norinori: QR-Code-ish

This is an entry for Fortnightly Topic Challenge #44: Introduce a new grid deduction genre to the community.

Here is a standard Norinori puzzle.

Rules of Norinori:

1. Shade exactly two cells in each bordered region.
2. Each shaded cell must be orthogonally adjacent to exactly one other shaded cell, making a domino-like block.
3. Each domino-like block may be in the same or may be in a different region.

The completed grid:

Step-by-step solution:

To start, note that for any L-tromino in a norinori puzzle, the cell outside of thi tromino which would make this tromino into a 2x2-square cannot be shaded (otherwise we would run into a contradiction with the fact that two of the cells in the tromino have to be shaded). This gives the following initial deductions:

Now the top left 2x2-square can be completed. Furthermore, the Z-tetromino to the left of this square can only be completed in one way:

We can make some easy deductions:

The single shaded cell on the left can only be completed in one way without blocking off all cells in the region below it:

Next, note that the 2x2-region in the lower left corner can only be filled in one way without blocking off to many cells in the adjacent I-trominos:

Again we can make some easy deductions:

If we now look at the two large regions in the middle of the bottom of the grid, note that every shades cell in the left-most of these regions needs to be adjacent to a shaded cell in the right-most region. This accounts for every shaded cell in the right-most region, so in particular we know that a lot of the cells of this region cannot be shaded:

Some more straight-forward deductions:

Now we can look at the remaining L-tromino: if the central cell of this tromino would be empty, then the resulting two shaded dominos would block off too many cells of the 2x2-region below the tromino. So this cell needs to be shaded. Using this we can easily fill out some more cells:

Next consider the top-right cell of the bottom-right 2x2-region: this cell cannot be unshaded, since otherwise the resulting domino at the bottom of this region would block off the L-tetromino below the 2x2-region:

To continue, look at the border marked red in the following figure:

Both above and below this border, we need to shade an even number of cells. This means that no domino can cross this border: then the number of cells on each side would become odd (note that we cannot have multiple dominos crossing this border at the same time). Furthermore, not both cells directly below the border can be shaded, because this would block off too many cells in the region above the border. Combining all this information, there is only one remaining way to shade 2 cells in the I-tromino below the border:

The remainder of the grid can be completed in a fairly straight-forward manner:

• Great job! The logics are all on-point! :) Commented Nov 30, 2020 at 22:45