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I've been figuring for this quite some time. Hope someone can help me out.

enter image description here

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    $\begingroup$ You might want to re-upload the picture in a bigger size, as it appears to me upon zooming in, that some lines are bold while others are less so. The rotation of the central star is probably of significant importance as well. $\endgroup$ – Tarius Apr 16 '17 at 9:48
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    $\begingroup$ I can't help but be reminded of the countdown timer in Predator. $\endgroup$ – Ian MacDonald May 2 '17 at 21:50
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D

because:

Reading either top to bottom or left to right, if two diagonals are present, keep them, otherwise nothing. If exactly one straight is present, keep it, otherwise nothing. That is, the diagonal's are AND'd and the straights are XOR'd.

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I can construct a...

valid union of vectors using very basic movements (black and grey moves 90° back and forth, yellow and purple moves 135°, blue moves 90° and red moves 45°).

See this

enter image description here

This leads to

alternative A.

The solution is probably not unique, but this is one which maps to the set possible solutions.

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  • $\begingroup$ this seems overly complex. Moreover, it doesnt have any consistency for columns, only for rows. $\endgroup$ – supinf May 3 '17 at 15:02
  • $\begingroup$ @supinf Sure, I totally agree. I don't think this is an intended solution. $\endgroup$ – Carl Löndahl May 3 '17 at 17:36
  • $\begingroup$ this doesn't take into account the rotation of the center star $\endgroup$ – lPlant Jun 8 '17 at 20:51
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Answer:

D. Reason: (from left to right in any particular row) For the third image to have an arm whose position is a multiple of $\frac{\pi}{2} \text{ rad}$ the only one of the previous two images must have an arm in that position. If both of them have the arm in that position, then there will be no arm in that position in the third image. For any other arm whose position must be at $k = \frac{n\pi}{3}$ such that it is not a multiple of $\frac{\pi}{2}$ in the third image, both the previous images must have the arm at that position.

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