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This is a three-dimensional nonogram. The first ten squares depict the layers of a $10\times10\times10$ cube. The final square gives the enumerations for the Z-axis, top to bottom. Some rules:

  • If a row or column has no numbers, that means that the composition of that row/column is unknown. Note that this is different than the usual practice in nonograms where a missing number means that the row/column has no shaded cells at all. In our case, such rows can be explicitly marked with a zero.
  • The Z-axis enumerations are all single digits, so e.g. 11 means two stretches of one, not one stretch of eleven.
  • The thin black lines in the middle are just visual aids. They're not meaningful to the puzzle itself.

enter image description here

What day is it?

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  • $\begingroup$ A picture equals a thousand... grids? +1ed $\endgroup$
    – Omega Krypton
    Commented Jul 4, 2019 at 11:04
  • $\begingroup$ Given what the correct answer is, this question deserves way more upvotes! ;) $\endgroup$
    – Mr Pie
    Commented Jul 9, 2019 at 17:11

1 Answer 1

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There might still be some mistakes, but I think this is the solution:

![enter image description here

The image is not what I previously thought, which was

A firework show for the 4th of July
enter image description here

It is actually

A tree with presents, for Christmas day enter image description here

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  • $\begingroup$ Nice job! A couple of "pixels" are off (e.g. row 5, column 7 currently has 10 shaded cells instead of 6&2), but overall that looks like the correct image. What I was trying to draw was a rot13(Puevfgznf gerr jvgu cerfragf haqrearngu), but I won't blame anyone for not seeing that from the final picture :D The animated image is certainly convincing, even though it wasn't what was intended. $\endgroup$
    – Jafe
    Commented Jul 3, 2019 at 19:29
  • $\begingroup$ That makes a lot more sense. It seems obvious now but I think I would have had a hard time guessing it. I have tried to fix up the errors, it is hard work to check 100 vertical columns :) $\endgroup$
    – Jay
    Commented Jul 4, 2019 at 5:06

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