I've been trying to solve this problem for a long time.
Problem
Let $R$ be a cycle with $2n$ nodes and assume there are $2k$ particles performing a simple random walk in this ring (i.e., they have an equal probability of moving clockwise or counter-clocwise). Also, all the particles move at the same time. If, in the end of any round, two particles meet at the same node, both desappear. We can consider also that the particles' positions have all the same parity (i.e., in any given round, the distances between any pair of particles are all even.)
The problem is to find the expected number of rounds it takes for all of them desappear. We are given the initial configuration of the graph, and of course we have the distance between all consecutive pairs of particles.
Simple Case
In the case we have $k=1$ there is an easy solution. We can take the initial distance $d_0$ between the two particles and consider the random walk performed by $\frac{d}{2}$, we have that it is going to reach the absorbing states $0$ and $n$ in an expected number of rounds given by the recurrence relation:
$E_i = 1 + \frac{1}{4}E_{i-1} + \frac{1}{4}E_{i+1} +\frac{1}{2}E_{i}$
where $E_0 = 0 \text{ and } E_n = 0$
We have te particular solution $E_i = -2i^2$ and the general solution is of the form:
$E_i = A1^i + Bi1^i - 2i^2\Rightarrow A = 0 \text{ and } B = 2n$
Therefore we have:
$E_i = 2(i(n-i))$
Note that this make perfect sense since the simple random walk gives us $i(n-i)$ as the expected number of steps and in this case we have that the distance remains the same in 50% of the cases, so the steps should be te double.
Conclusion
In conclusion, even being able to solve the simple case, I wasn't able to generalise this method nor find a new one for the general case.
Thanks for all the help in advance.