Which set of arrows replace the question mark in the image below?



Read in zigzag

  • $\begingroup$ Is this a question you created or is it from some other site? $\endgroup$ – Kevin L Aug 25 '18 at 8:04
  • 3
    $\begingroup$ @KevinL It's a question I created. $\endgroup$ – u_ndefined Aug 25 '18 at 8:13
  • $\begingroup$ Oh nevermind then :) $\endgroup$ – Kevin L Aug 25 '18 at 8:27

My guess is



In the first column the inner two arrows change only every other line. The second column is similar except offset by one. This means that the fourth and fifth row should be the same and E is the only option where that is true.


I'd say the answer is



The number of vertical arrows is always odd on the left side whereas it is always even on the right side.
The number of green arrows is always even on the left side whereas it is always odd on the right side.
The last arrow is always on the same color on the left and on the right.

F) is the only pattern with an even number of vertical arrows, an odd number of green ones and that ends with a green one.

  • $\begingroup$ I would say the same answer, since (ROT13) vg fgnegf jvgu gur fnzr glcr bs neebj nf R). Abgr gung guvf vf n cnggrea, nf O) fgnegf jvgu gur fnzr neebj nf N) naq Q) fgnegf jvgu gur fnzr neebj nf P) (rvgure jvgu fnzr be qvssrerag pbybhef) :D $\endgroup$ – Mr Pie Aug 25 '18 at 12:30
  • $\begingroup$ I don't get your comparison between A and B, C and D, E and F. But if you mean that rot13(gur svefg neebj orgjrra gur yrsg naq evtug pbyhza vf nyjnlf va gur fnzr qverpgvba), you are right, the actual explanation may be much simpler than mine. $\endgroup$ – xhienne Aug 25 '18 at 13:22
  • $\begingroup$ yeah that's what I meant.... ${}$*reads his own comment*${}$ .... I'm tired :\ $\endgroup$ – Mr Pie Aug 25 '18 at 14:14
  • $\begingroup$ :-) Maybe you should have posted it as an alternate answer as JonMark did $\endgroup$ – xhienne Aug 25 '18 at 14:16
  • $\begingroup$ Nice reasoning but not the answer $\endgroup$ – u_ndefined Aug 25 '18 at 14:31



the first symbol in the left boxes is the same as the first symbol in the right boxes, by row.

  • $\begingroup$ Not the answer, sorry $\endgroup$ – u_ndefined Aug 25 '18 at 14:57

I think the answer is



There are boxes containing 1,2,3,4 Green Marks and 1,2,3 Red Marks. The only missing pattern is 4 Red Marks.

P.S. Not a solid reasoning, but the only one I could see at the moment.


Okay, I have an hypothesis

F or E


I noticed something thanks to the hint, is that the rule change if the line is odd or even. Let's say that the table is lined from 1 to 5, and columned from a to b. The tuples inside get indices i, j, k, l.
1ai doesn't change in 1bi.
1aj change of orientation in 1bj
1ak change of color in 1bk.
1al doesn't change.

2ai change of color in 2bi.
2aj change of direction in 2bj, but also changes color.
2ak change of direction in 2bk, but also changes color.
2al change of direction in 2bl.

3ai doesn't change in 3bi.
3aj changes orientation in 3bj.
3al change color in 3bl.
3ak doesn't change.
My theory is thus that if the line is odd, one has to change orientation so the next changes color. If the even lines, one has to change color so the next change orientation.
The last line is the fifth, so odd. So it needs to change orientation so the other changes color. The only possibilities are E) or F). Sadly, I wasn't able to determine which one of these two (if it was clockwise/anti clockwise for the changes, I'd say F, but it's not regular).


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