7
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enter image description here

I made this image code a while back. The answer is a meaningful word in the English language.

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4
  • $\begingroup$ Is the title part of the puzzle? $\endgroup$
    – WhatsUp
    Commented Aug 5, 2022 at 5:50
  • 6
    $\begingroup$ For solvers: 8, 16, 24 $\endgroup$ Commented Aug 5, 2022 at 5:52
  • $\begingroup$ @WhatsUp No, it isn't $\endgroup$ Commented Aug 5, 2022 at 6:29
  • 4
    $\begingroup$ Then it is probably better to give your question a more meaningful title. $\endgroup$
    – WhatsUp
    Commented Aug 5, 2022 at 14:02

1 Answer 1

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The first step is to make the image into a much more readable format that gets rid of any unnecessary information. After dividing up the grid as such, an interesting pattern emerges.
Divided Grid

Each square quadrant has exactly one shaded pixel in one of its four corners. Therefore, each square can be assigned a number from 0 to 3 based on the pixel's position.
0 3 2 1 0 3
1 1 0 1 0 1
0 3 2 0 0 1
2 3 0 3 0 2
This pattern actually contains a 0 every four numbers, and the total number of numbers is divisible by four, so it makes sense to reallot the numbers in groups of four.
0 3 2 1
0 3 1 1
0 1 0 1
0 3 2 0
0 1 2 3
0 3 0 2

[Thanks to user39583 for figuring out this next step!]
It would make sense that, since each line has 8 bits of information, it would be one 8 bit binary character per line. All ASCII letters begin with "01" in binary, so it would make sense that "0" would map to "01". The rest of the mapping can be deduced as follows:
0 -> 01
1 -> 11
2 -> 00
3 -> 10
This gives the following binary strings: 01100011
01101111
01110111
01100001
01110010
01100100
This decodes to the final answer (in lowercase), COWARD.

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2
  • $\begingroup$ Have you tried counting going clockwise or counterclockwise? $\endgroup$
    – Auride
    Commented Aug 5, 2022 at 15:26
  • $\begingroup$ @user39583 Nice, updating answer now! $\endgroup$
    – SeptaCube
    Commented Aug 5, 2022 at 16:00

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