Maximum number of edges such that $\nu(G) < \frac{n}{2}$

Question

Given an even integer $n$. I want to find the largest number of edges in a $n$-vertex graph such that the matching number is strictly less than $\frac{n}{2}$. I believe that the maximum is obtained by taking $G = K_{n-1} \cup \{v\}$. Since $n-1$ is odd we have $\nu(G) = \nu(K_{n-1}) = \frac{n}{2}-1$. Number of edges in $G$ would be number of edges in $K_{n-1}$, that is $\binom{n-1}{2}$. However, I have not been able to prove that this is in fact the largest number of edges. I have tried to do this by induction on the number of vertices but without much luck. Any help would be appreciated.

Answer 1

I think you basically already solved the problem.

You have shown an example with $\binom{n-1}{2}$ edges and $\nu(G) = \frac{n}{2} - 1$.

We only need to show that for any $n$-vertex graph with $\binom{n-1}{2} + 1$ edges, we can find a $\frac{n}{2}$ matching.

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Divide the cases w.r.t. the size of the maximum clique.

Max. clique size $= n - 1$

Basically you have considered this, when adding any edges to $K_{n-1} \cup \{v\}$, we can find a $\frac{n}{2}$ matching.

Max. clique size $= n - t \leq n - 2$ with $t \geq 2$

We have a clique $K_{n-t}$ and $t$ remaining nodes. The number of remaining edges outside $K_{n-t}$ is $$ \binom{n-1}{2} + 1 - \binom{n-t}{2} = (t-1)(n-t) + \frac{t^2 - 3t + 4}{2}. $$ Since by the assumption, $K_{n-t}$ is the maximum clique, each of the remaining $t$ nodes can only have $\leq n - t - 1$ neighbors in the $K_{n-t}$.

Let $V_{K}$ be the set of the nodes in the $K_{n-t}$, let $V_t$ be the set of the remaining $t$ nodes, and let $E_t$ be the set of the remaining edges between $K_{n-t}$ and $V_t$. It is easy to see that we only need to show that we can find a $V_t$-perfect matching in the bipartite graph $G = (V_t, V_{K}, E_t)$, where $|E_t| > (t-1)(n-t) + \frac{t^2 - 3t + 4}{2} - \binom{t}{2} = (t-1)(n-t-1) - 1$ (it is "$>$" instead of "$\geq$", because it is easy to see that $V_t$ cannot form a clique too, otherwise we can easily find a $\frac{n}{2}$ matching).

By Hall's marriage theorem, we CANNOT find such a $V_t$-perfect matching iff there exists a subset $V' \subseteq V_t$ such that $\text{Nbr}(V') \cap V_K < |V'|$. Asumme such a $V'$ exists and let $s$ be the size of a smallest $|V'|$. It implies that $(t-1)(n-t-1) \leq |E_t| \leq s(s-1) + (t-s)(n-t-1)$, which is equivalent to $$ s \geq n - t - 1. $$ This is only possible when $t \geq n - t - 1$, i.e., $t \geq \frac{n}{2}$ (since $n$ is even). I think it should be easy to start from here and show that the clique number must be larger than $n - t \leq \frac{n}{2}$.

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Update

After a second thought, I think it would be much easier to prove that we must have a Hamiltonian cycle (more precisely, "hamiltonian cycle" minus "an edge").

• See, e.g., Maximum number of edges in a non-Hamiltonian graph