A counting problem. There are $n ≥ 3$ married couples attending a daily couples
therapy group. Each attendee is assigned to one of two round tables, so that
no one sits at the same table with his/her spouse. The order of seating at
each table remains fixed once and for all.
Initially, $s$ of the attendees have contracted a contagious disease. For
each person P, consider P’s two neighbors at the table, as well as P’s spouse
(who, of course, sits at the other table). Each day, if at least two of these
three people are sick, then P gets sick too, and remains sick forever.
Eventually everyone gets sick. Across all possible seating arrangements,
what is the smallest possible value of $s$?
Try: By PHP and simple reasoning I got $s=n+1$. Is it true? Im surprised at the simplicity of the answer.
 A: In the graph with an edge between spouses and neighbours, colour an edge blue if exactly one of its vertices is sick and green otherwise. Thus a healthy vertex gets sick if it has two blue edges, and by getting sick all three edges change colour. Thus every newly infected person decreases the number of blue edges by at least one. Incidentally, the last person getting infected (we may assume wlog. that concurrent infections take place milliseconds apart) will turn three blue edges green.
If we start with $s$ infected people, there are at most $3s$ blue edges, hence at most $(3s-3)+1$ additional people can be infected. We conclude that $4s-2\ge 2n $, or
$$ s\ge\left\lceil\frac{n+1}{2}\right\rceil=\begin{cases}m+1&n=2m\\m+1&n=2m+1\end{cases}.$$
Now let's see if this bound can always be achieved:


*

*For $n=2m$ even, infect every second man and one of the women. 
Then after one day, all men are sick. From then on, the sickness will spread step y step among the women.

*For $n=2m+1$ odd, again infect all even-numbered men $M_2,M_4,\ldots, M_{2m}$ and also one woman $W_1$. After one day, the men alone will infect $M_2,M_3,\ldots, M_{2m-1},M_{2m}=M_{n-1}$. Additionally, $W_1$ and $M_2$ infect $M_1$. After that, $M_n$ is infected from $M_1$ and $M_{n-1}$, and the infection among the women occurs as above.


So indeed, is suffices to infect $$\left\lceil\frac{n+1}{2}\right\rceil $$
"strategically" positioned people originally. 
Remark: The number does not depend on the seat arrangement!
