Proving that the shortest simple path problem between two vertices $s$ and $t$ in a graph is NP-complete

Question

How to show that the shortest simple path problem between two vertices $s$ and $t$ (finding a minimum weight path between $s$ and $t$) in a graph is NP-complete? I saw the following proof in a combinatorial optimization lecture, which I didn't understood (I stressed the moment that I didn't understand).

Let $P_1$ be the Hamiltonian path problem:

The Hamiltonian path problem and the Hamiltonian cycle problem are problems of determining whether a Hamiltonian path (a path in an undirected or directed graph that visits each vertex exactly once) or a Hamiltonian cycle exists in a given graph (whether directed or undirected). Both problems are NP-complete. From Wikipedia.

Does it exists an Hamiltonian path in $G$?

Let $P_2$ be the shortest path problem in a directed graph.

If $G$ is the graph within which we search such a Hamiltonian path, we transform $G$ into $\hat G$, replacing each edge $(i,j)$ with two edges $(i,j)$ and $(j,i)$.

For each edge $\{i,j\}$ in $G$, we erase $(i,j)$ and $(j,i)$ from $\hat{G}$, give a weight of $-1$ to all remaining edges, and calculate the shortest (simple) path from $i$ to $j$. If the path length is $-(n-1)$, then this is a Hamiltonian path in $G$. If we found no such path going over all edges, then $G$ has no Hamiltonian path.

• Why is is the case that if the path has length $-(n-1)$ then it constitutes a Hamiltonian path in $G$?
• Why if not such path has length $-(n-1)$ then $G$ has no Hamiltonian path?

Maybe if you were kind to help me understand with a visual example I would better understand?

Last but not least, how did we proves that Hamiltonian path is NP-complete?

Answer 1

In fact, it does not prove NP-completeness (see a list of our related reference questions).

For the shortest path problem (SPP) to be NP-complete, it is crucial you allow negative edge weights. Then, there is a simple polynomial-time reduction from the Hamiltonian path problem to SPP. In other words, you take an arbitrary instance $I$ of the Hamiltonian path problem, and construct an instance $I'$ of SPP such that $I$ has a solution if and only if $I'$ has a solution. To make the reduction work, you only need to set up the edge weights in $I'$ suitably.

Answer 2

Suppose there is a simple path of weight $-(n-1)$ from $i$ to $j$. Since each edge has weight $-1$, this path must contain $n-1$ edges, and so it corresponds to a Hamiltonian path in the original graph.

Conversely, suppose that the original graph has a Hamiltonian path, say starting at $i$ and ending at $j$. This path leads to a directed path of weight $-(n-1)$ in $\hat{G}$ which doesn't use the edges $(i,j),(j,i)$, and so it will be found when considering the edge $\{i,j\}$.

Answer 3

Given an HP (Hamiltonian path) instance ($G=(V,E),s,t$), we define a Shortest-Simple-Path instance ($G,s,t,w,L$) as follows.

We simply let the weight $w(e)$ of every edge $e∈E$ be $−1$ and let $L=−(n−1)$. ($n=|V|$ is the number of vertices in $G$.) So, in this instance, we need to know whether there is a simple path in $G$ from $s$ to $t$ of weight at most $−(n−1)$ or not.

As all weights are $−1$, this is the same as whether there is a simple path in $G$ from $s$ to $t$ with at least $n−1$ edges or not. Such a path must be a Hamiltonian path in $G$ from $s$ to $t$ as $G$ contains only $n$ vertices. Thus, $G$ contains a Hamiltonian path from $s$ to $t$ if and only if the shortest simple path from $s$ to $t$ in $G$ has cost at most $−(n−1)$. This proves HP $≤_P$ Shortest-Simple-Path.