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Normal Chocolate Banana rules apply.


  1. Paint some of the cells black.

  2. Black cells linked orthogonally must always form a rectangle (or a square).

  3. Non-blacked cells linked orthogonally must not form a rectangle (or a square).

  4. A number indicates the area (the number of cells) of the orthogonally region containing this cell.

Click on the image to use the online solver and view the rules.

  • $\begingroup$ They were right, I just didn't quite like the way it was written $\endgroup$
    – Sny
    Commented Apr 22 at 23:53

1 Answer 1


The 51 seems pretty unusual:

It has to be unshaded, since a shaded rectangle of size 51 would have a side of length at least 17, which can't fit in this grid. Now since this is unshaded, and its area must contain either R1C9 or R2C10 (or probably both), the 1 in R2C9 must be shaded. This means that the cells orthogonally adjacent to R2C9 must be unshaded, which forces R3C8 to be shaded. Thus its orthogonal neighbors must be unshaded as well.

Looking in the upper left corner:

The 1 in R2C2 must be shaded. If it were not, then R1C2 and R2C1 would have to be shaded, which would force either a non-rectangular shaded region if R1C1 were shaded, or a rectangular unshaded region if not. This now forces R2C2s orthogonal neighbors to be unshaded, and specifically forces R2C3 and R3C2 to be shaded. Now by counting, R3C3 must be shaded as well. This shading must extend either left or down, which forces R4C4 to be unshaded. The grid thus far:

Initial progress

Let's do a little thinking about the clues on the diagonal:

If a clue on the diagonal is shaded, then it cannot be contained in a 1-wide rectangle with any cell north or west of the clue. If clue X were contained in such a shaded rectangle, then it would have X unshaded squares on its north or west side, which would contain an X-1 clue. In particular, this forces the 7 clue to be unshaded. Were it shaded, since 7 is prime it would have to be in a 1x7 rectangle, which by its placement in the grid must contain either the cell north or west of it.

Another look at the 51 clue:

There are only 45 cells strictly above the diagonal, so the 51 must reach around through either R1C1 or R10C10, possibly both. If R10C10 is in this region, then the 8 clue must be shaded, and by the above must be in a 2x4 rectangle, which would force R10C9 to be shaded. Thus the 51 clue cannot get all of its extra cells going around the bottom right corner, so it must go around the top left. This shows R1C1, R1C3 and R3C1 are all unshaded.

Let's look at the 6 clue:

Suppose the 6 were unshaded. Then we would have both R7C8 and R8C7 shaded to separate the unshaded 6 from the unshaded 7. The unshaded 7 region must contain a cell orthogonally adjacent to the 8 clue, forcing it to be shaded, but any 8-cell rectangle containing the 8 must also contain one of R7C8 or R8C7, making the region non-rectangular. So the 6 is shaded. The shaded region on the 6 must contain a cell orthogonally adjacent to the 5-clue, forcing it to be unshaded. This now forces the 4-clue to be shaded.

Keep picking at the 4-clue:

Actually, the 4 clue can't be in a 1x4 rectangle at all. We've already seen that such a rectangle cannot contain the cell either north or west of the 4, but even if it flows strictly south or east, it creates a line of 4 unshaded cells that abuts the unshaded 3-clue. So the rectangle containing the 4 clue must be a 2x2 square. Let's tackle the case it goes northeast, to cells R4C5, R4C6, R5C5 and R5C6. If this occurs, then R6C7 must be shaded, for if not the unshaded region on the 5 clue would have at least 6 cells. As this shaded cell is adjacent to the 6-clue, it forces the shaded region on the 6-clue to be a 2x3 rectangle from R6C7 down to R7C9. This puts an unshaded cell adjacent to the 8-clue, forcing it to be shaded, and it must lie on a 2x4 rectangle which includes R9C7 down to R10C9. This leaves no space for the unshaded 7-region: it can't go left without running into the unshaded 5-region, so it must include R8C10, but it cannot get a 7th cell without adding at least one more. A long-ish contradiction, but it shows the 4-region goes southwest. This forces the region on the 2-clue and the 3-region as well, which gives us some knock-on deductions. The grid thus far:


Looking up:

We already noted that the 51-region has to escape around the upper left, so R1C4 and R1C5 are unshaded. Now we note R7C6 must be unshaded, completing the 5-region. If not, the 6-clue must be a 2x3 rectangle from R7C6 to R8C9, forcing R6C7 and R7C8 to be shaded. R6C8 must be shaded, else the 5 and 7 regions would merge, forcing R5C7 to be unshaded. Now the only way to get one more cell in the 5 region forces at least 2, giving 6 total cells, which cannot be. Thus R7C6 is unshaded, completing the 5-region, and forcing shaded cells around it, and in particular forcing the location of the 6-region. The 8-region is forced to be a 2x4 rectangle in the lower right corner, and the knock-on unshaded squares give all the cells of the 7-region. The grid thus far:

More progress

Hopefully coming home with the 51-clue:

R3C10 and R1C8 need to be unshaded: the former to prevent an unshaded rectangle, the latter to let the 51-clue out. And we actually didn't even need to think that hard, because there are 49 cells either shaded, or unshaded in other smaller regions, so all of the remaining cells need to be unshaded to satisfy the 51-clue. The final grid:
Final grid


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