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I have this question to answer:

For each node i in an undirected network $G = (N,E)$, let $N(i) = \{j \in N : \{i, j\} \in E\}$ denote the set of neighbors of node $i$ and let $c_e\geq0$ denote the length of edge $e \in E$. For each node $i \in N$, suppose the set $N(i)$ is partitioned into two subsets, $N^+(i)$ and $N^-(i)$ such that $j \in N^+(i)(j \in N^-(i))$ is referred to as a positive (negative) neighbor of $i$. (Note: Regardless of whether $j$ is a positive or negative neighbor of $i$, $i$ can be either a positive or negative neighbor of j.) Consider the problem of finding a maximum-length path $(s =) i(0)–i(1)–···–i(h)(= t)$ in $G$ between two nodes $s ∈ N$ and $t ∈ N$ subject to the following restriction: For each internal node $i(k)(k \in\{1,...,h − 1\})$ on the path, the set $\{i(k − 1),i(k + 1)\}$ must contain exactly one positive neighbor and one negative neighbor of $i(k)$. Prove NP-completeness of the decision problem and state whether it strongly NP-complete or not.

I wonder about the steps of the proof and whether shall I start from the longest path problem or from another problem instance.

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  • $\begingroup$ Can you be more precise about what a "positive" or a "negative" neighbor is? $\endgroup$ – Panzerkroete Jun 12 at 3:59
  • $\begingroup$ It is just a partition (or labeling) in order to satisfy the given constraint where each internal node has two different labeled nodes before and after in the path. $\endgroup$ – Bassem Jun 12 at 13:23
  • $\begingroup$ I am still confused. 1. you say, you have a "network". What do you mean by that? A network data structure (used in the maximum flow problem )or just a undirected weighted graph $G$ which should model a network ? 2. Why do you want to consider $N^+$ and $N^-$? Let $\pi=(v_0,v_1,...v_n)$ be a path. Then by definition each $v_i, 0<i<n$ has exactly one positive/negative neighbor ($v_{i-1}, v_{i+1}$). So why $N^+$ and $N^-$? 3: Can your graph further contain cyles and is your graph connected? $\endgroup$ – Panzerkroete Jun 13 at 13:02
  • $\begingroup$ For the first point, I mean an undirected weighted graph 𝐺 representing a network. For the second point, not necessarily since the partition of neighbors happens before we construct a path .. so for example if a node $i$ has only two neighbors which are both + and $N^{-}(i)$ is empty .. then this node cannot be on the longest path since it violates the constraint. $\endgroup$ – Bassem Jun 13 at 13:32
  • $\begingroup$ I have no information on whether the network contains cycles but let's assume it is connected. $\endgroup$ – Bassem Jun 13 at 13:33
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I think here is a pitfall due to inaccurate notation. Notation for me:

  1. A trail is a sequence of distinct connected edges.
  2. A path is a trail with distinct vertices.

In the problem instance $MNPL$ (maximum neighbor path length) the sets $(G,N^+, N^-)$ and the weight function $c_e$ are given as input(and hereby fixed). Since $N$ is partitioned, for two vertices $u,v$ either $\{u,v\}\notin E$ or $u$ is a positive neighbor of $v$ or vice-versa. Especially $u$ and $v$ cannot be a positive neighbor of each other by definition ($N$ would not be partitioned). Hence $G$ must be a directed graph. Now, for any path $s \rightsquigarrow (u,v,w) \rightsquigarrow t$(not way nor cycle nor walk!) the neighbor-condition is fullfilled.

Maybe this was the real challenge of this exercise?

So, claim: $MNPL$ is $NP$-complete.

Proof: $MNPL \leq_p LPP$ (longest path problem) Consider the reduction $f$. For $(G,N^+,N^-,c_e)$ construct $G'=(V',E',c_e')$ as the following:

  1. $c_e'$ = $c_e$
  2. $V' = V$
  3. $E' = \bigcup_{v \in N^+(u), u \in N} (u,v) \cup \bigcup_{u \in N^-(v), v \in N} (v,u)$

Clear: $f$ is computable and runs in poly-time ($O(|E|)$). By construction and with the observations above $f$ will construct a digraph from $N$ and the neighbor sets.

So $f$ is a reduction and $MNPL \in NP$ is obvious using "guess and check". Since $LPP$ is strongly $NP$-complete, $MNPL$ is as well strongly $NP$-complete. This is no decision problem but a maximization problem.

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  • $\begingroup$ Your idea looks clear to me .. but to prove a problem is NP-complete aren't we supposed to reduce an NP-complete problem to our problem? $\endgroup$ – Bassem Jun 13 at 19:34
  • $\begingroup$ I do not provide a reduction here because I think, that problem is $NP$-complete because you restrictions on the neighbors are to strict. For example, given the complete graph $K_5$ (look here for its appearence)where all edges have weight one. Here, the longest path would be obvious contain all five vertices (no matter which vertix is $s$ or $t$). However for any $s-u-v-w-t$ path $N^+(v)=N^-(v)=\{s,u,w,t\} \neq \{u,w\}$. Hence your neighbor condition is not fulfilled. By that, I cannot reduce to the $HamiltonianPath$ -problem $\endgroup$ – Panzerkroete Jun 13 at 22:38
  • $\begingroup$ This is because a hamiltonian path in for example $G=K_5$ is not a valid longest path in your network graph $G'$. The standard polynomial reduction ($G'$ with $n$ vertices has longest path with length $n-1$ iff. $G$ has a hamiltonian path ) does not work here.The implication Hpath to longest path is not true. So, there is no reduction to any known $NP$-hard problem (at least I could not find one) and so $MNPL$ is not NP-complete. Either your definiton of neighbor is somehow wrong or the problem is really in $P$. Do you have this exercise somewhere and can link the source? $\endgroup$ – Panzerkroete Jun 13 at 22:47
  • $\begingroup$ I have no source for it .. it was provided by a professor as an exercise to prove NP completeness. I was thinking that either Hamiltonian path or TSP problems could be the problems to use for the proof and then I was thinking about constrained shortest path problem but I still couldn't find clear reduction proof. $\endgroup$ – Bassem Jun 14 at 0:58
  • $\begingroup$ @Bassem I added the core argument for reduction to the longest path problem for directed graphs. At a few points a bit more argumentation is needed (like why is the neighbor condition fulfilled by a directed graph, why is $f(w) \in LPP => w \in MNPL$?, why is $MNPL \in NP$?). Give me feedback if you have doubts $\endgroup$ – Panzerkroete Jun 14 at 1:14

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