# Convolution (Literally)

Decode the following message.

Hint:

Refer to the title.

Kernel is $$\begin{bmatrix}1&1\\1&1\end{bmatrix}$$.

The answer is Sny confirmed, but can you get a clearer image by reversing the convolution (instead of applying an approximate inverse convolution)?

• Just a wild guess, but is the answer "Sny"? :-) Jan 29 at 15:53
• :-) IHATE15CHAR
– Sny
Jan 29 at 15:56
• It's always Sny. XD Jan 29 at 16:03
• Now the question is how :) Jan 29 at 16:21
• Not my downvote, but perhaps the concept "101 grid/cell-style puzzles whose answer is usually 'Sny'" could use some refinement, especially since most of them seem to be write-ins. There may have been a nice aha effect to your first puzzles, but it has worn thin by now, at least for me. Sorry to be so negative. Your puzzles seem to have their fans and I don't mind that you post them, but I don't really care for them, either. Jan 30 at 5:30

All my time spent watching 3B1B on YouTube has led to this moment (he has a video covering convolutions).

Coding the image into a matrix of color-corresponding values of 0, 1, 2, and 3 (where darker ~ larger number), we can get a matrix like the following:

1 2 2 1 0 0 0 0 0 0 0 0 0
2 3 2 1 0 0 0 0 0 0 0 0 0
2 3 2 1 1 1 2 2 1 1 1 1 1
1 2 3 2 2 2 3 3 2 2 2 2 2
1 2 3 2 2 2 2 2 2 2 3 3 2
1 2 2 1 1 1 1 1 1 1 2 3 2
0 0 0 0 0 0 0 0 0 1 2 3 2
0 0 0 0 0 0 0 0 0 1 2 2 1


Then, assuming a sharpening kernel of:

 0 -1  0
-1  4 -1
0 -1  0


I get an output (assuming the edges are 0 and extending the edges) of:

 0  -1  -2  -2  -1   0   0   0   0   0   0   0   0   0   0
-1   0   2   3   1  -1   0   0   0   0   0   0   0   0   0
-2   2   3   0   0  -2  -1  -2  -2  -1  -1  -1  -1  -1   0
-2   2   3  -1  -2   0  -1   2   2  -1   0   0   0   1  -1
-1  -1  -1   3   0   1   0   3   3   0   1   0   0   3  -2
-1   0   0   3   0   1   1   0   0   1   0   3   2   1  -2
-1   1   3   2  -1   0   0   0   0   0  -2  -1   2   1  -2
0  -1  -2  -2  -1  -1  -1  -1  -1  -2   0   0   3   2  -2
0   0   0   0   0   0   0   0   0  -1   1   3   2   0  -1
0   0   0   0   0   0   0   0   0   0  -1  -2  -2  -1   0


Which, conveniently re-colored in Excel using conditional formatting, looks like:

And finally (somewhat) reads like "S n y"! (As the comments correctly guessed)

• Upvoted, but consider that my image is likely not the input of the convolution but the output of the convolution. The kernel is $\begin{bmatrix}1&1\\1&1\end{bmatrix}$.
– Sny
Jan 30 at 1:10
• I'm a 3b1b fan too :-)
– Sny
Jan 30 at 1:10
• @JaapScherphuis the problem is that we're trying to reverse the convolution, not apply a convolution.
– Sny
Jan 30 at 7:20
• @SnySmartie Yes, and he one Alexander is applying here is the (first order approximate) inverse of yours. Jan 30 at 7:27
• Surely a better inverse can be used, since you can't really tell. Can you link me to a resource about inverses of convolutions? @JaapScherphuis thanks.
– Sny
Jan 30 at 7:35