Let $G$ be a simple undirected graph. Recall that $S \subseteq V(G)$ is a dominating set of $G$ if every vertex of $v \in V(G) \setminus S$ has a neighbour in $S.$

It is well known that it is NP complete to determine the minimal dominating set of $G$ and that is usually done via the vertex-cover problem.

As the title suggests, I would like to reduce dominating set to the max clique problem. One can of course use the standard reductions to do so, but I would like to find a as "small as possible" reduction. That is I would like to be able to reduce a graph $G$ to a graph $\widehat{G}$ that is as small as possible and so that $\gamma(G)$ can be efficiently determined from $\omega(\widehat{G}).$ Here $\gamma$ and $\omega$ are the size of the dominating set and clique number respectively.

Can someone suggest such a reduction?


1 Answer 1


There are good reasons to expect that there is no polynomial time reduction that takes as input a graph $G$ and outputs a graph $\hat{G}$ such that $\omega(\hat{G})$ depends only on $\gamma(G)$. In particular the Clique problem is complete for $W[1]$ while Dominating Set is complete for $W[2]$, see the Wikipedia page for Parameterized Complexity. A reduction from Dominating Set to Clique where the size of the clique only depends on the size of the dominating set would imply that $W[2] = W[1]$, which is considered unlikely.

On the other hand it is not too hard to make a reduction where $\omega(\hat{G})$ depends only on $\gamma(G)$ and $|V(G)|$. For every edge $uv \in G$ make two vertices $[u,v]$ and $[v,u]$ in $\hat{G}$. For two vertices $[u_1,v_1]$ and $[u_2,v_2]$ of $\hat{G}$ we put an edge between them if $u_1 \neq u_2$, $u_1 \neq v_2$ and $u_2 \neq v_1$. We claim that $\omega(\hat{G}) = |V(G)| - \gamma(G)$.

The intuition is that the vertex $[u,v]$ is interpreted as ``$u$ is dominated by $v$''. Then each vertex of $G$ not in the dominating set should choose exactly one vertex in the dominating set to dominate it. Someone who has chosen to be dominated may then not be a dominator.

Consider a dominating set $S$ in $G$. We make a set $\hat{S}$ as follows: for each vertex $u \notin S$, pick an arbitrary neighbor $v \in S$ of $u$ and add $[u,v]$ to $\hat{S}$. Clearly $\hat{S}$ is a clique of size $|V(G)| - |S|$.

On the other hand consider a clique $\hat{S}$ in $\hat{G}$. Let $S$ be the set of vertices $u$ in $V(G)$ such that there does not exist a $v \in V(G)$ such that $[u,v] \in \hat{S}$. Observe now that $|S| \geq |V(G)| - |\hat{S}|$, since each element of $\hat{S}$ rules out at most one vertex of $G$. Consider a vertex $u \notin S$. There must be some vertex $v \in V(G)$ such that $[u,v] \in \hat{S}$. We know that $uv \in E(G)$, it remains to show that $v \in S$. Suppose not, then there must be some $v' \in V(G)$ such that $[v,v'] \in \hat{S}$. But $[u,v]$ and $[v,v']$ are non-adjacent in $\hat{G}$, contradicting that $\hat{S}$ is a clique. So $S$ is a dominating set.


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