What Jan Burse did not note in the original question and why this is of interest is that while the specific case is to find the answer for 6 items in a set, the more general problem of finding the minimum of a superpermutation is an unanswered question in maths. What makes this of value to a Prolog community is that while the answers for the values 1,2,3,4, and 5 are known.
there is also a proof for a lower bound (4chan proof)
n! + (n − 1)! + (n − 2)! + n − 3.
and it is suspected (YouTube) (Site) (paper) that this is an upper bound
L₂(n) = n! + (n − 1)! + (n − 2)! + (n − 3)! + n − 3
and possibly even (YouTube) (link)
L₃(n) = n! + (n − 1)! + (n − 2)! + (n − 3)! + n − 4
and that amateur people in math and computer science can work on this by finding examples that are equal to or lower than the value predicted in the proof.
||4chan lower bound
||Current Best Minimal Superpermutation Length
||Known to be minimum
||Current Best Minimal Superpermutation
In the StackOverflow discussion Jan B. asked
could be a nice example for tabling maybe?
While I don’t actively work on the problem, like many other problems I read up enough on it to have a working vocabulary so that when I read more about it I understand the problem, then I put the problem on the so called back burner of the brain.
When I think about this as a Hamilton path and the kernel/extensions structures it seems there could be a correspondence between the kernel/extensions and tabling.
Also when I recently watched this video about about neural networks, it seemed there was a similarity to simplifying a problem by reducing the given set to a more canonical form, but what and how they are related I don’t know.
Posting here not to start a discussion, but to give feedback for others who read this and say that yes I agree that looking at tabling to help solve the problem is a path I would put on my short list.