There is a hidden message inside the solution of the puzzle. An answer will only be accepted if it correctly identifies said message.

enter image description here

P.S. I promise this is my last Chocolate Banana puzzle of the week. I'm getting carried away :) I'll find something else.

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.

  5. The solution is unique except its reflection. You do not need this information to solve the puzzle.

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

  • 3
    $\begingroup$ All these same puzzles within such a short space of time are making us go bananas! $\endgroup$
    – PDT
    Commented Apr 24 at 13:41
  • 1
    $\begingroup$ (and chocolate. yummy.) $\endgroup$
    – Sny
    Commented Apr 24 at 13:51

1 Answer 1


Start with the easy stuff:

All the 1 and 2 clues must be shaded since they must be rectangles. We get some easy knock-on unshaded cells, both from limiting the shaded regions, and from not having single unshaded cells. In addition, the 11-clues have to be unshaded, since no 1x11 rectangle will fit in the grid. Similarly, the 9-clue must also be unshaded, and it must be separated from the 11-clues by shaded blocks. This satisfies the 2-clue, and forces the 3 and 4 clues at left to be shaded. The grid thus far:



The 3 and 4 shaded clues must be separated by an unshaded cell, which forces the 4 to be a 1x4 rectangle, and the knock-ons resolved the 2-clue above it. The unshaded 11 next to the 4 must escape up, forcing the shaded 3 to go north. The expanded 11-region now forces the adjacent 3 and 6 clues to be shaded, and the 3 is actually resolved. This now forces the other 3 clue at bottom to be shaded, and the diagonal shaded 2 forces it go strictly up. The 4 at bottom is now forced to be shaded, and its region must be a 2x2. This also forces the 2-clue at right. The grid thus far:


The 9-clue:

must continue out and completes, so the rest of the lower left corner must be a shaded 1x3.

At right:

The unshaded 14-clue must escape, and this forces the 5 to be shaded. With the location of the shaded 2, it must go down, forcing the 2 as well. The lower right corner must come escape to not be a rectangle, which forces the bottom row, third from right to be shaded to satisfy the 14. The grid thus far:


Upper right:

The corner must be unshaded to avoid an unshaded rectangle. Now the 14-clue in the top row must be unshaded, since no 1x14 or 2x7 rectangle can fit. The shaded 6 now plays its part. No 1x6 rectangle can fit, so it must be a 2x3. If it is 3 rows and 2 columns, then the 14-region hooks up with the 11-region, so the 6 must be 2 rows and 3 columns. Now the 2-clue to the left cannot go up, for it would both force the 4 to be unshaded and hook it up to the 14-region, so the 2 must go left. This forces the 4 to be shaded, since it already abuts four unshaded squares. The grid thus far:


Finishing up:

Now the 4-clue could go left or right to complete. But if it goes left, it will overload the 14-region, so it must go right. To finish, the 14 needs the cells in the upper right, so the third from left in the top row must be unshaded, and this completes the 14-region, so the cell to its left must be shaded. The final grid:


The final message is:

SNY, the OP’s previous username. enter image description here

  • $\begingroup$ Curses, forgot the message part. Will delete if someone pips me. $\endgroup$ Commented Apr 24 at 14:04
  • $\begingroup$ You pipped me! But since you weren’t here maybe a few months ago and are unfamiliar with the OP’s old puzzles or previous user name the message is SNY! That was his old name. $\endgroup$
    – PDT
    Commented Apr 24 at 14:22
  • 1
    $\begingroup$ I could edit it in if you want I feel this is a little unfair as a message to a lot of people. $\endgroup$
    – PDT
    Commented Apr 24 at 14:24
  • $\begingroup$ Damn I hesitated to post SNY because it resembled OP's name, but eventually settled on finding pi in the solution (fyi you can see 3,1,5,1,4) x) $\endgroup$
    – Fluorine
    Commented Apr 24 at 14:27
  • $\begingroup$ I will be changing my name back to sny as soon as I can :) @Fluorine where did u find pi? $\endgroup$
    – Sny
    Commented Apr 24 at 14:35

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