The 5th in the series, yet another advanced one. What two figures (one in range a-i and the other in range 1-9) are the odd ones that should be switched to restore both patterns, and why?

created by myself


switching f with 2 is not the correct solution.

  • $\begingroup$ With that amount of detail, I think a higher resolution version would be appreciated. $\endgroup$ Commented Mar 1, 2018 at 18:29
  • $\begingroup$ I see your point. $\endgroup$ Commented Mar 1, 2018 at 18:38
  • $\begingroup$ If there is some reason (beyond "that wasn't what I was thinking of") why the existing answers do not solve this puzzle, it's not apparent at all what part of the puzzle actually invalidates those responses. Because it seems they should be at least as valid as any other answer you might have in mind, this may be "too broad" — you may need to update the puzzle to make sure incorrect responses are demonstrably invalid. $\endgroup$
    – Rubio
    Commented Mar 3, 2018 at 0:48
  • $\begingroup$ The main reason is that patterns exists in everything, simple to complex, and the existing answers does not create the best patterns available among the 81 possible alternatives. "Best" as in most (as in amount of) properties following a set of rules. I havn't checked my answer against all 80 alternatives, but mine "beats" the 2 existing answers. I can list, in detail, both pattern setups as soon as the correct answer shows up. $\endgroup$ Commented Mar 3, 2018 at 6:28
  • $\begingroup$ Every part is an odd one in some perspective, the crucial thing here is two-sided: Finding what parts is the most odd ones compared to what they are replaced with, its about minimizing the oddness. $\endgroup$ Commented Mar 3, 2018 at 6:52

3 Answers 3


I think the answer is:

switching b with 6


In each figure there are always 3 pictograms which are the same at the same place (here I counted in filled and non-filled objects as the non-filled is part of the filled one) only the b and 6 are in the wrong place. After switching the new b, 2 and f, new 6 pairs restore the pattern.

  • $\begingroup$ A very good and close to correct answer but not the one I had in mind when designing this $\endgroup$ Commented Mar 2, 2018 at 21:04

My answer:

D and 8


maintains symmetry of the centre icons (left to right for a-i, top to bottom for 1-9) and of the filled and unfilled circles (vice versa), makes the orientation of the semicircles on the left, and arrows on the right match, keeps the pointed arrow pattern, and doesn't break the x,+ pattern.

  • $\begingroup$ Very good reasoning, and about as close to my intended correct answer you can be without nailing any of the two I had in mind when creating this. I guess one can argue if your suggestion beats mine or not $\endgroup$ Commented Mar 2, 2018 at 20:44

I have two answers in my question - I will go with the first one. I am sorry for my english that I couldn't explain it that well.

I think the answer is

Switching i and 1

because if they would get switched

- the top object would fit (black square to black square and white top with black bottom to white top with black bottom). Nevertheless this got me thinking because (when it's not switched) the top object in the last box is "in" the top object in the first box - which would fit nicely and wouldn't make it odd. Nevertheless - the task is to find the MOST odd box. - the "inner object" will fit nicely in to the whole picture. For instance on the fig a-i) the boxes 1,3,5,7 and 9 (odd numbers - intended?) will have the same inner object - the same happens in the fig. 1-9) - the last object in the boxes will match (pluses in fig. a-i) and not-filled-squares in fig. 1-9)) enter image description here

Another answer could be to

exchange f and 2


that's a bit obvious. But b,d,f and h and 2,4,6 and 8 fit extremely well (the pattern that and the circles and the "middle thing"). enter image description here

  • $\begingroup$ Read again the hint under the question. $\endgroup$ Commented Mar 5, 2018 at 9:12

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