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This is a metapuzzle of Bongard problems. Rules (quoted from here)

In a Bongard problem, you are given 12 pictures, 6 on one page and 6 on the other. The pictures on the left page conform to a rule, and the pictures on the right page conform to a different rule. Furthermore, a picture on one side cannot conform to the rule on the other side. The goal is to determine the rules.

This is the problem:

enter image description here

I use the following reference system:

enter image description here

Feedback is appreciated.

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    $\begingroup$ Is the answer simply.. rot13(Gur yrsg fvqr fubhyq or n inyvq Obatneq chmmyr, juvyr gur evtug fnvq vf vainyvq.)? $\endgroup$
    – athin
    Feb 10 '20 at 3:32
  • $\begingroup$ @athin yes. it is easy if you already know BPs. $\endgroup$
    – JAGO
    Feb 11 '20 at 9:56
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The Bongard problems on the 1 side of the Meta-Bongard are all easy to solve (Numbers are OP's reference system):

1A:

Nothing | Something

1B:

White shape | Black shape

1C:

Square | Non-Square Rectangle

1D:

Big thing | Small thing

1E:

Threesome | Foursome

1F:

Black Circle | White Triangle

The 2 side is significantly harder... (Bold Numbers are Meta-Bongard box, Italic are Bongard)

2A:

Here, the two sides are exactly the same, so there is no possible rule to differentiate them.

2B:

Here, there is an apparent rule: Arranged | Scrambled.
Unfortunately, the order of the pieces cannot be used to distinguish the sides of a Bongard problem, and the pieces themselves are otherwise identical.

2C:

This one is the same as 1C - except that its 2D has been replaced with a Square, breaking the rule and preventing a new rule from being possible (as the shapes are otherwise identical or randomized).

2D:

Here, the two sides are almost, but not quite, identical - shades of 2A! There is an apparent rule here, though: Bigger | Smaller.
Unfortunately, much like in 2B, this sort of comparison isn't a valid way to make a rule, and the "small" things are too close in size to the "big" things to just copy-paste the rule for 1D.

2E:

Here, both sides are just a single dot, randomly placed. There's almost a rule here (Centered | Not Centered), but 2F breaks it.

2F:

Here, both sides are just noise. I'm sure a dedicated computer could tease a rule out of this one, but it won't be human readable and will feel very contrived - best not to bother.

So what's the rule for the Meta-Bongard?

I regret to inform you that I cheated - I decoded @athin's rot13 comment under the question. The rule is Valid | Invalid.
That is, the 1 side are all valid Bongard problems, and the 2 side are all invalid Bongard problems. I had an inkling of this, but 2C threw me for a loop (I couldn't see that it was different from 1C until it was too late).

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  • $\begingroup$ You may have cheated but the most difficult part was rot13(gb cebir gung gur OC ba gur evtug jrer abg inyvq), so thanks for that detailed answer. $\endgroup$
    – xhienne
    Feb 25 at 13:10

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