enter image description here As with my previous puzzle I made the 12 "rooms" larger but kept the same rules. I'm also adding the link to the rules for those that don't know the puzzles Click me!

And have fun!


the black circles are only cosmetic, please ignore them.


There was a minor mistake that made the puzzle a bit inelegant (the solution was not changed). Now it is fixed, I also removed the incorrect solution that was found; this is the last edit, I will eventually add the answer if no one can find it.

Hint 2:

Some numbers don't play well with others.

  • $\begingroup$ This one is really frustrating me. Twice I think I've got it and then found one card that doesn't meet the rule... $\endgroup$
    – IanF1
    Commented Jun 14, 2018 at 10:04
  • 1
    $\begingroup$ @IanF1 Take it easy, this one's a jerk $\endgroup$ Commented Jun 14, 2018 at 16:40
  • $\begingroup$ The key has got to be the difference between the bottom left of each set $\endgroup$
    – IanF1
    Commented Jun 14, 2018 at 16:42
  • $\begingroup$ Could we get a hint for this? It's been over a year with no one making progress. $\endgroup$
    – JS1
    Commented Oct 31, 2019 at 5:26

3 Answers 3



dsq The count of intersections between red and black lines is even at left and odd at right.
We have 0;2;2;0;2;12 at left and 1;3;1;1;3;1 at right.
I'm really not sure about it because of the 5th left picture...

  • 1
    $\begingroup$ Nice thinking, but not the answer. Also, due to the ambiguity of the "balls" it's hard to connect them in a definite way. +1 anyway $\endgroup$ Commented Apr 26, 2018 at 12:36
  • 1
    $\begingroup$ @SilverCookies What do you mean by "ambiguity?" $\endgroup$ Commented Nov 4, 2019 at 4:36

This seems to work

At most $2$ shoelaces may intersect and non-intersecting shoelaces may not have the highest and lowest endpoints or the rightmost and leftmost endpoints.


I think the answer could be that

The line segment on the left only intersect with 1 line segment (either itself or another). Some of the line segment on the right intersect with more than 1 segment.

  • $\begingroup$ Not the answer. Also, not allowed under the rules: the pattern on the right cannot appear in the left boxes, not even once $\endgroup$ Commented Jun 11, 2018 at 12:44

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