**Puzzle 111:** As John mentioned, without any extra equations, \\(F\\) may map arrows to any function. Since \\(F(\textrm{Person})\\) is a set of four people, and \\(F(\textrm{BestFriend})\\) is an endomorphism on this set, we have \\(4^4\\) choices of function, and hence \\(4^4\\) databases.

**Puzzle 112:** With this equation, \\(F(\textrm{BestFriend})\\) must be an _involution_. Involutions are special kinds of permutations, i.e. those of degree at most two. [OEIS A000085](https://oeis.org/A000085) tells us there are ten involutions on a set of four elements. (We can enumerate these using cycle notation: \\((1)(2)(3)(4), (12)(3)(4), (13)(2)(4), (14)(2)(3), (23)(1)(4), (24)(1)(3), (34)(1)(2), (12)(34), (13)(24), (14)(23)\\). Can you see how I produced these?)

**Puzzle 113:** With this equation, \\(F(\textrm{BestFriend})\\) must be a permutation of degree at most three. This means that its cycle decomposition must _only_ consist of cycles of order either 1 or 3; adding a 2 in the mix would force us to take the least common multiple of the degrees, forcing us to either degree 2 or 6, which is not what we want. Again, [OEIS A001470](https://oeis.org/A001470) tells us there are nine such permutations on a set of four elements: \\((1)(2)(3)(4), (123)(4), (132)(4), (124)(3), (142)(3), (134)(2), (143)(2), (234)(1), (243)(1)\\).