# Mutually exclusive replacement paths, an existing problem?

A replacement path is a simple path allocated to an edge $e \in G$ that connects the endpoints of $e$ in $G \setminus \{e\}$.

The problem

An undirected graph $G$ is given and the task is to allocate a replacement path $P_e$ to each edge $e \in G$ such that for any pair of edges $e_1$ and $e_2$ it holds: if $e_2 \in P_{e_1}$ then $e_1 \not\in P_{e_2}$. In words, no two edges use each other in their replacement path. I call the paths $P_{e_1}$ and $P_{e_2}$ mutually exclusive.

The path length is not bounded and they can share edges. So any path allocation that satisfies the mutual exclusion would be good.

Now the question: is there already any problem closely related to this?

One could reduce this problem to Maximum Independent Sets problem: a vertex for every $(e,P_e)$ and edges connecting conflicting pairs (i.e. pairs violating the ME). However, in order to get a polynomial size instance we need to restrict the model so that each edge gets a constant number of candidate paths. However, this is not desired because the MIS instances seem to be NP-Hard (or not?).

• @NealYoung Thanks for your comment! It should be clear now. Jul 2, 2018 at 13:56
• Is such a path allocation feasible in any graph? (Any valid example?) Jul 4, 2018 at 6:18
• A necessary condition is that $G$ is 3-edge-connected. A sufficient condition is that $G$ contains 3 edge-disjoint spanning trees. I think the latter condition is not necessary, though as $K_4$ does not have 3 edge-disjoint spanning trees, but does admit a set of replacement paths: $P_{ac}=(a,d,c)$, $P_{ad}=(a,b,d)$, $P_{cd}=(c,b,a,d)$, $P_{ab}=(a,c,b)$, $P_{bc}=(c, a, d, b)$, $P_{bd}=(b, a, c, d)$. Jul 4, 2018 at 15:20
• Good point @NealYoung, I considered k_4 but didn't try it on paper (I thought it is not feasible there). Jul 5, 2018 at 9:28
• Can anybody find an example of a 3-edge connected graph that doesn't admit a set of replacement paths? Jul 5, 2018 at 14:27

I found a replacement path allocation for $$𝐾_𝑛$$ in a more general setting. Consider an edge $$𝑒_1$$ that uses edge $$𝑒_2$$ on its path. The allocation induces a dependency from $$𝑒_1$$ to $$𝑒_2$$. Mutual exclusion defined above corresponds to a cycle of 2 such dependencies. Now consider the more general setting: a cycle of $$𝑛$$ dependencies induced by a path allocation on $$𝐾_𝑛$$. Assume vertices are $$0,…,𝑛−1$$. On edges $$𝑖,𝑗$$, the path allocation $$𝑖,𝑗+1,0,𝑖+1,𝑗$$ induces cycles of dependencies no shorter than $$𝑛$$. In my application it means that $$𝐾_𝑛$$ can tolerate up to $$𝑛−1$$ edge failures (i.e. edge removal).