The pattern game

[KyriakosCH]

Maybe we can have a pattern-based quiz?

In the following, you have to find why the pictures on the left are a different group from those in the right.



The above is from a relatively famous book. The common trait in all the patterns in that book is that they are actually very easy to see, but problems can arise due to not instantly knowing which information is relevant.

[Danvzare]

Oh that's easy. The pictures on the left are different from the pictures on the right, because they're not the same pictures. 
The pattern is that they all use simple shapes.
Now give me a hard one! 

(Is it the answer you're looking for? Of course not. Is it a correct answer? Most definitely.  )

[Mandle]

Took me a few minutes but I think I have it solved:

Spoiler

In the left-hand group two black squares of the same size exist in the same set. In the right-hand group they do not.

[close]

[KyriakosCH]

^_^

Thank you both for playing! Mandle is (more) correct.
It is problem 58, from Bongard's patterns book: https://www.foundalis.com/res/bps/bongard/p058.htm

Mandle, you can post a new pattern quiz if you feel like it. Or if you don't, someone else can.

[Mandle]

Here's a better wording of the initial question that makes easy answers like "They are all different." or "The line down the middle divides them" or whatever:

"What is the logical rule that distinguishes the sets of shapes on the left from the sets of shapes on the right? The rule means that no set of shapes from the left side can be on the right side, or vice versa."

[KyriakosCH]

https://www.youtube.com/watch?v=hjR4cWR3I58

This is very simple, but I agree with the uploader that it was nicely phrased.

[KyriakosCH]

Ok, same question. By which logic do the six arrangements on the left differ from those in the right?

[KyriakosCH]

Any guesses? This is (once again) a conjunction (in the first quiz it was black^same size ≠ black^different size), although with three parts instead of two.
It would be nice if this doesn't die immediately, but I tried