# Finding the most flexible of all 35 hexominoes

Of all 35 hexominoes, which is (or are) the most flexible of all, that is, the one (or ones) that can be converted into the most other hexominoes just by cutting out one of its component squares (thus obtaining a pentamino) and glueing it elsewhere?

In the case of pentominoes, the answer to the corresponding question is the Y, F, and P pentominoes, each of which can be converted into 10 of the 11 other pentominoes.

• (I've downvoted because I don't think this is a good puzzle (or even a puzzle at all, IMO); there is no intentional solve path designed into it, and the only way to figure out the answer seems to be tediously enumerating all the options.) – Deusovi Mar 23 '20 at 17:06
• Must the middle step be a pentamino? I.e. must the remaining squares be connected before I attach the cut out square? – Jens Mar 23 '20 at 21:15
• @Jens: Yes, they must remain connected and be a pentomino. – Bernardo Recamán Santos Mar 24 '20 at 13:44

This one:

 +--+--+
|  |  |
+--+--+
|  |  |
+--+--+
|  |
+--+
|  |
+--+

As it can be converted into

28 of the other 34 hexominoes via the L, P, Y and/or Z pentominoes.
7 via Z
5 via L
5 via P
4 via Y
4 via L or Y
1 via P or Y
1 via P or Z
1 via L or Z

• Hmm. I only got 25. Can you show the 28? – Jens Mar 23 '20 at 22:12
• @Jens I cannot do images from my phone, but I added a list of counts by pentomino – Daniel Mathias Mar 23 '20 at 22:27
• Redid my counting and found the 28. Nice job! – Jens Mar 23 '20 at 22:38

This graph (beautifully drawn by Freddy Barrera: https://puzzling.stackexchange.com/users/36719/freddy-barrera) shows, for each hexomino, all the hexominoes it can be converted to. Notice that the graph is hamiltonian, and that the P hexomino has the largest degree (28), attained by no other hexomino.