User blog:Wythagoras/Friendship function

Friendship bracket function
We denote a person with n friends with fn.

With (fk 1 ,fk 2 ,fk 3 ,fk 4, ... fk n ), we denote that these n people each know each other. To clarify, when drawn as a graph, the graph is complete.

Friend(k)
For Friend(k), we have a sequence of friend bracket functions. The nth element can have at most n+k persons, and the nth element can't be a friendship minor of an earlier element. It can have multiple brackets.

Friendship minor
Let the earlier element be expressed as

(fk 1 ,fk 2 ,fk 3 ,fk 4, ... fk t 1 )(fk t 1+1  ,fk t 1+2  ,fk t 1+3  ,fk t 1+4  , ... fk t 2  )...(fk t n+1  ,fk t n+2  ,fk t n+3  ,fk t n+4  , ... fk t n+1  )

Let the other element be expressed as

(fk 1' ,fk 2' ,fk 3' ,fk 4', ... fk t 1' ' )(fk t 1'+1 ' ,fk t 1'+2 ' ,fk t 1'+3 ' ,fk t 1'+4 ' , ... fk t 2' ' )...(fk t n'+1 ' ,fk t n'+2 ' ,fk t n'+3 ' ,fk t n'+4 ' , ... fk t n+1' ' )

For those who can't read the accents very well, every k has an accent and every t has an accent.

An element is a friendship minor of an earlier element if: In informal terms:
 * tq = tq' for all q
 * kq ≤ kq' for all q
 * Each bracket has the same number of persons
 * Each person has at least as many friends as the person on the equal place.

Number of persons
The number of persons is lower than you might expect. In (f1,f3)(f1,f3)(f1,f3) there are 6 people on the first look. But you can have a person knowing 3 others, each knowing no-one else. So this are 4 persons.

Formally, we look for the smallest graph Q such that for every bracket (fk 1 ,fk 2 ,fk 3 ,fk 4, ... fk n ) in the element, we can find a minor of Q that has valences k1, k2, ... kn-1, kn and a minor is of the complete graph Kn. The number of vertices q in the graph Q is the number of persons in the element of the sequence.