The following approximation algorithm for the Minimal Dominating Set Problem is said by a fellow student to be a 1.5-approximation:

  1. Start with empty set $S$
  2. As long as not all vertices are covered:
    • Add a vertex that is not in $S$ with the most uncovered neighbors including itself.
    • Mark it and all its neighbors as covered
  3. return $S$

I can't find a proof for the approximation ratio however and my own attempts to prove it have failed so far since I lack the right approach to proof something like this.

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    $\begingroup$ Can you provide the source of this claim (link/book/..) $\endgroup$ Commented Mar 11, 2023 at 19:57
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    $\begingroup$ Yes, it would be interesting to know where you got that information. Not only is the claim incorrect, but Dominating Set is APX-hard, which means it's likely that no polynomial-time algorithm is able to approximate this problem to a constant ratio. $\endgroup$
    – Highheath
    Commented Mar 12, 2023 at 6:53
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    $\begingroup$ Cross-posted: cs.stackexchange.com/q/158968/755, math.stackexchange.com/q/4656428/14578. Please do not post the same question on multiple sites. $\endgroup$
    – D.W.
    Commented Mar 12, 2023 at 7:53
  • $\begingroup$ @NarekBojikian it's nothing official, just a fellow students claim, I don't know if he relied on any supposedly reliable source. $\endgroup$ Commented Mar 12, 2023 at 12:53
  • $\begingroup$ @D.W. got, it - I will delete the other question $\endgroup$ Commented Mar 12, 2023 at 12:53

1 Answer 1


This claim is incorrect. The diagram below shows a small counterexample:

A small counterexample

Consider two different runs of the greedy algorithm that break ties in different ways:

  1. Run 1 chooses the 6 red vertices in order from darkest to lightest, then chooses the yellow vertex.
  2. Run 2 chooses the 3 green vertices in order from darkest to lightest, then chooses the yellow vertex.

(Check for yourself that each run is a valid possible execution of the algorithm.) The first run chooses a solution of size 7; the second, an (optimal) solution of size 4. $7/4 = 1.75 > 1.5$, meaning the greedy algorithm is capable of producing a solution that violates the claimed bound.

  • $\begingroup$ Thank you very much. That seems to indicate that the algorithm could be a 2-approximation since your example could be expanded to something similar but with the optimal solution being n and the worst case being $(2n-1)/(n)$, however Highheath states under my question, that a constant ratio approximation is unlikely to exist . I'm afraid, I am still missing an approach to determine and proof the actual approximation ratio. $\endgroup$ Commented Mar 12, 2023 at 13:07
  • $\begingroup$ You're welcome. The Wikipedia page on Dominating Set is helpful. $\endgroup$ Commented Mar 12, 2023 at 13:39

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