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Matching: https://en.wikipedia.org/wiki/Matching_(graph_theory)

Vertex Cover: https://en.wikipedia.org/wiki/Vertex_cover

It is easy to see that

$$\texttt{minimum vertex cover} \leq 2 \texttt{ maximum matching}$$ I want to know that for what kind of graphs the equality is hold in the above inequality.

As an instance, $C_3$ is an example.

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Answer. Such a graph $G$ is a disjoint union of odd complete graphs.

Obviously such graphs satisfy the equality $$\texttt{minimum vertex cover} = 2 \texttt{ maximum matching}.\quad (\star)$$

Assume that $G=(V,E)$ satisfies $(\star)$. Denote by $k$ the size of maximal independent set in $G$, then $$k=|V|-\texttt{minimum vertex cover}=|V|-2\cdot\texttt{maximum matching}=\\ \texttt{ minimum number of vertices not covered by a matching}.$$ On the other hand, by Tutte — Berge formula, if $k$ is the minimum number of vertices not covered by a matching, then there exists a subset $U\subset V$ such that $G-U$ has $|U|+k$ odd connected components. If $|U|>0$, then taking a vertex from each component we get an independent set with more than $k$ vertices. Therefore $U=\emptyset$, $G$ has $k$ odd components and if $G$ has also an even component, we again may take an independent set with more than $k$ vertices. Also if one of these connected components $C$ is not a complete graph, we may take to not-connected vertices in $C$ and a vertex from each other component, again having too large independent set. That is.

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  • $\begingroup$ I don't understand why $G$ can't have a independent set larger than $k$. $\endgroup$
    – phantom
    Commented May 2, 2020 at 6:06
  • $\begingroup$ @allfaker By definition of $k$. $\endgroup$ Commented May 2, 2020 at 9:48
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    $\begingroup$ I understand now, I forget the true |𝑉|=𝚖𝚒𝚗𝚒𝚖𝚞𝚖 𝚟𝚎𝚛𝚝𝚎𝚡 𝚌𝚘𝚟𝚎𝚛+𝚖𝚊𝚡𝚒𝚖𝚞𝚖 𝚒𝚗𝚍𝚎𝚙𝚎𝚗𝚍𝚎𝚗𝚝 𝚜𝚎𝚝 $\endgroup$
    – phantom
    Commented May 2, 2020 at 14:52

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