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Given a graph G and an integer k , recognize whether G contains dominating set X with no more than k vertices. And that is by finding a propositional formula ϕG,k that is only satisfiable if and only if there exists a dominating set with no more than k vertices, so basically reducing it to SAT-Problem.

The solution to this problem is supposed to be similar to reducing clique to SAT. Here is how that looks like: https://blog.computationalcomplexity.org/2006/12/reductions-to-sat.html

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    $\begingroup$ What is your question? $\endgroup$
    – Nathaniel
    Jun 11 at 13:32
  • $\begingroup$ We are a question-and-answer site, so we require you to articulate a specific question about your situation. It might also be helpful to share with us the context where you encountered this task. Finally, we discourage questions that simply state an exercise-style task out of context and expect us to solve it. You might find this page helpful in improving your question. $\endgroup$
    – D.W.
    Jun 11 at 20:20
  • $\begingroup$ Right, sorry. The question is how to do it? How to reduce the dominating set problem to SAT $\endgroup$
    – Matjaž
    Jun 11 at 21:07
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Take a look at the Cook Levin theorem, that shows a reduction from any $NP$ problem to $SAT$. Since $Dominating-Set\in NP$, this is a reduction $Dominating-Set\le_p SAT$.

In the case I didn't understand your question and you wanted a reduction $SAT\le_p Dominating-Set$, consider creating the following graph: add a node for each variable $x_i$ and add another node for its negation $\overline{x_i}$, and add an edge between the two nodes. For each such variable, add another node $h_i$ that has an edge to $x_i$ and $\overline{x_i}$.

Additionally, for each clause add a node $c_j$ and add an edge between it to any literal in it.

Now we want to find a dominating set of size $k$, where $k$ is the number of variables in the SAT instance.

For more information on this, take a look here.


I hope this helped!

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  • $\begingroup$ Right, I actually somehow missed that. Thanks! $\endgroup$
    – Juho
    Jun 12 at 21:06

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