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Can you figure out how the patterns in the picture were generated, and determine what the hidden pattern at the bottom right should look like?

This is a visual puzzle, and the solution can be determined by simply looking at the image. But if you'd die for a hint, you might be on the right track.

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I believe Ivo Beckers has the right answer but not for the intended reason. Rather, the idea is to

take the faces of a dice, in order 1-6, and for each one join the dots that are present to the positions where dots could be but aren't.

So, in particular,

since the "1" and "6" faces are complements, they feature line segments from the central dot to the six positions present in the "6".

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  • $\begingroup$ This is what I thought when I first saw the puzzle, but I couldn't make it work for some of the examples; I think I've figured out that my problem was a difference in expectations, explained here. Thought I'd mention it, in case anyone else is similarly stymied. Note that the link is a BIG hint if you don't want to be spoiled. $\endgroup$ – 1006a Oct 2 '16 at 5:17
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It will look like this:

enter image description here

Because in a single row you overlap the first two images and the third is the result of that, with double lines being removed.

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  • $\begingroup$ Good observation! But we have to explain how all the other patterns were generated. Look at the hints in the puzzle. $\endgroup$ – GOTO 0 Sep 30 '16 at 8:29
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The missing figure is...

...the same as the first (top left) figure.

In each row, the middle figure shows...

...the differences between the figures on its left and right.

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If presence of line is 1 and absence of line is 0. And each row of figures is an equation than

Figure 1 (Exclusive or or Exclusive disjunction) Figure 2 (Equals) Figure 3

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  • $\begingroup$ Welcome to Puzzling! Could you expand on this answer a bit? I'm not quite sure how you get from "presence of line is 1 and absence of line is 0" to what's in the spoilertag. $\endgroup$ – Rand al'Thor Sep 30 '16 at 16:41
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    $\begingroup$ This is the exactly same answer that Ivo Beckers and humn came up with, only expressed in mathematical jargon. $\endgroup$ – Paul Sinclair Sep 30 '16 at 20:38
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This is simply an xor operation. Each straight line is a 1 and therefore they cancel. If there is a line in the left figure and no corresponding line in the right figure, the line remains and vice versa.

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  • $\begingroup$ Welcome to Puzzling! Can you explain why you think this is the answer? Answers without explanation are deleted. $\endgroup$ – Deusovi Sep 30 '16 at 22:19

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