Swapping and handshakes with neighbours in a circle $N$ people are at a party and decide to play a cooperative game. They begin by standing in a circle. The game proceeds in turns. In each turn, one person is chosen to perform one of the following actions:


Shake hands with someone adjacent to them
Swap positions with someone adjacent to them


The game ends when every pair of players has shaken hands, and the aim of the game is to minimize the number of swaps required. What is the optimal strategy?

I've tried working through some small examples (e.g. $N = 3,4,5,6$ require $0,1,3,5$ swaps minimum but I can't really find a pattern in the strategy).
Suppose $\{a_1, a_2, ...,a_n\}$ people play this game. The pairs of neighbours are

$$ \{a_1,a_2\} $$
$$ \{a_2, a_3\} $$
$$ \vdots $$
$$ \{a_n, a_1\} $$

Which gives $n$ pairs of neighbours. Swapping with a neighbour perturbs the set by

$$ \{a_1, a_2, ... a_{n-k+1}, a_{n-k}, ..., a_n\} $$

So that, now, $a_{n-k+1}$ can shake with $a_{n-k}$ (which presumably they already have) and (more importantly) shake with $a_{n-k-1}$

In order to shake with everyone, the person originally  at location $n-k$ must migrate (by swapping) around the ring to position $n-k+3$ (and I assume this person shakes with everyone along the way).

We need a total of $\frac{n(n-1)}{2}$ handshakes, and each swap allows us to shake hands with 2 new people (one with the person who swapped, and one with the person who got swapped). 

I will propose that $\left\lceil \frac{n(n-1)}{4} \right\rceil $ swaps will guarantee everyone has shaken hands, but I have not come up with a minimum expression.

Perhaps someone might take over.
Suppose $\{a_1, a_2, ...,a_n\}$ people play this game. The pairs of neighbours are

$$ \{a_1,a_2\} $$
$$ \{a_2, a_3\} $$
$$ \vdots $$
$$ \{a_n, a_1\} $$

Which gives $n$ pairs of neighbours. Swapping with a neighbour perturbs the set by

$$ \{a_1, a_2, ... a_{n-k+1}, a_{n-k}, ..., a_n\} $$

So that, now, $a_{n-k+1}$ can shake with $a_{n-k}$ (which presumably they already have) and (more importantly) shake with $a_{n-k-1}$

In order to shake with everyone, the person originally  at location $n-k$ must migrate (by swapping) around the ring to position $n-k+3$ (and I assume this person shakes with everyone along the way).

We need a total of $\frac{n(n-1)}{2}$ handshakes, and each swap allows us to shake hands with 2 new people (one with the person who swapped, and one with the person who got swapped). 

I will propose that $\left\lceil \frac{n(n-1)}{4} \right\rceil $ swaps will guarantee everyone has shaken hands, but I have not come up with a minimum expression.

Perhaps someone might take over.