Thursday, April 28, 2016

Powers of Graphs Are Almost Always Hard

Early in 2011, I spent a few months as an exchange student in the University of Melbourne. Although I was writing my thesis at the time, I was looking for topics for a small-scale research, as I had to turn in some written assignment as part of the exchange program. I can't remember why, but I thought a bit about powers of graphs, and pondered about their relation to hardness. Since it turned out to be an easy question to solve, it wasn't enough for the written assignment (let alone a research paper), but I do have a warm spot for it, so here it is:

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Background: Powers of Graphs 

Given a simple graph (we restrict our discussion to this type for brevity) $G(V, E)$, we define $G^2$ by adding an edge between every two vertices $v,j\in V$ if there exists some $w\in V$ such that $\{v, w\} \in E$, and $\{u, w\} \in E$.
In other words - if there was a path of length 2 between $v$ and $u$ in $G$, then there's an edge between them in $G^2$. This works similarly for $G^3$ (where paths of length 3 or less in $G$ become edges) and so on. 
Clearly, if the longest distance in the graph is of length $d$ (in which case we say that $d$ is the diameter of $G$), then $G^d$ is simply a clique, since all paths in $G$ are at most of length $d$.

Background - Independent Set

Given a graph $G(V, E)$, an independent set of vertices is some $U\subseteq V$ such that $u,v\in U \Rightarrow \{u,v\}\notin E$. Finding a maximum independent set (i.e. largest in size, not to be confused with maximal which means it is not strictly a subset of any other independent set) in a graph is known to be NP-hard, and also hard to approximate.

So here's the question:

 In a clique, every vertex is a maximum (and also maximal) independent set, so for every simple graph $G$ with diameter $d$, solving the independent set problem for $G^d$ is easy: just choose any vertex you like, and you're done. This raises the question - if a problem is in general hard on $G$, but easy on $G^d$, how hard is it for $G^k$ where $k < d$? I mean, at which point does this very hard problem becomes trivial? 

I'll save you the suspense - independent set is generally hard on $G^{d-1}$, so the problem is not all that interesting, and the hardness of independent set on powers of graphs is basically: hard, hard, hard, ... , hard, trivial. Not an exciting result, I admit, but let's prove it anyway.

Proof by Reduction

Proof sketch: given a graph $G$, we'll construct a graph $H$, that is only slightly larger than $G$, has the same diameter, and is such that solving the independent set problem for $H^{d-1}$ will give us the solution for $G$.

First, the case of an even $d$, which is simpler:
Denote $G$'s vertices by $v_1, v_2, ... , v_n$, we construct $H$ by taking $n$ disconnected vertices denoted $u_1, u_2, ..., u_n$ - we'll call those the 'old' vertices - and adding $n\cdot (\frac d 2 - 1)$ 'new' vertices to them in the following manner:
1. Add one special vertex $c$.
2. For every $i : 1\leq i \leq n$, connect $u_i$ to $c$ by a path of length $\frac d 2$ using $\frac d 2 - 1$ of the 'new' vertices (all of these paths should be disjoint, and that's okay, since we have exactly $n\cdot (\frac d 2 - 1)$ of these new vertices). Denote the vertices of the path from $c$ to $u_i$: $c\rightarrow u_i ^1 \rightarrow u_i ^ 2 \rightarrow ... \rightarrow u_i ^ {\frac d 2 - 2} \rightarrow u_i ^{\frac d 2 - 1} \rightarrow u_i$.
3. For every $\{v_k, v_j\} \in E$ that form an edge in $G$, add to $H$ the edge $\{u_k ^1, u_j ^1\}$.
 $H$ is thus sort of a big star graph, having $c$ in the middle, with some additional edges due to step (3) above.

A few observations:
1. For every two vertices of the original graph $\{v_l,v_t\}\notin E$, the shortest path from $u_l$ to $u_j$ in $H$ goes through $c$ and is of length $d$. It follows that the diameter of $H$ is at least $d$.
2. In $H^{d-1}$, all of the new vertices form a clique, and each of them is connected by an edge to each of the old vertices. The diameter of $H$ is therefore at most $d$, and so it is exactly $d$. 
3. It follows that a maximum independent set in $H^{d-1}$ cannot contain any of the new vertices.
4. $\{u_j, u_k\}$ is an edge in $H^{d-1}$ if and only if $\{v_j, v_k\}$ is an edge in $G$, so every independent set in $G$ correlates to an independent set of 'old' vertices in $H^{d-1}$ and vice versa.  
Putting all these observations together boils down to the fact tha every maximum independent in $H^{d-1}$ correlates to a maximum independent set in $G$ and vice versa.

What about an odd $d$? well, instead of a single vertex serving as the center, we use a clique of n vertices that serves as center of $H$, and everything else works very much the same.

Note that a slight tweak can be used to show that $H^{d-2}$ is no easier, and so on.

So, alas, independent set is hard on all but the $d$-th power of a graph, in which case it is trivial.

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