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Let $G$ be a graph with $\binom{n}{2}$ vertices, each labeled with a different unordered couple $\{a,b\}$ such that $a,b \in [n] = \{1,2,\ldots,n\}, a \not= b$.

For any vertex with label $\{a,b\}$ and $m = \lfloor (n-1)/2 \rfloor$ different elements $a_1, \ldots, a_m \in [n]$, $a_1, \ldots, a_m \not= a,b$, and $m$ different elements $b_1, \ldots, b_m \in [n]$, $b_1, \ldots, b_m \not= a,b$, not required to be different from $a_1, \ldots, a_m$, we know that there exists at least one edge $\{\{a,a_i\},\{b,b_i\}\}, 1 \le i \le m$ in $G$.

Can we compute or estimate the minimum number of edges that $G$ must have?

Modeling the problem with an Integer Linear Program I have computed the following minimum values for $n=3,4,5,6,7$ respectively: $3,15,32,83,138$.

My actual final goal is to find the minimum number of colors for a vertex coloring of the graph, but maybe this question is easier and we could use the minimum number of edges to bound the number of colors (see here).

If $a,a_1, \ldots, a_m$ and $b,b_1, \ldots, b_m$ are sets of a union-closed family and an edge $\{\{a,a_i\},\{b,b_i\}\}$ means $a \cup a_i \not= b \cup b_i$, then the required property is necessary for a counterexample of the union-closed sets conjecture, otherwise we would have $a \setminus b \subseteq b_1 \cap \cdots \cap b_m \cap a = \emptyset$ and $b \setminus a \subseteq a_1 \cap \cdots \cap a_m \cap b = \emptyset$ and then $a = b$, a contradiction. If we could prove that a coloring for $G$ can be done only with at least $n+1$ colors, then we would prove the union-closed sets conjecture: the number of sets of the family must be greater or equal than the number of colors. The argument as it is may not be sufficient, because e.g. for $n=7$ the inequality $138 \le \binom{n}{2}^2(k-1)/2k$ gives just $k \ge 2$ but there are some possible improvements, like e.g. we can allow $b_i = b$ and $a_i = a$.

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    $\begingroup$ Are you asking for the minimum number of edges such that all graphs $G$ with that number of edges will satisfy the property? Alternatively, are you asking for the minimum number of edges such that there exists a graph $G$ with that number of edges that satisfies the property? $\endgroup$ Commented Aug 25 at 0:00
  • $\begingroup$ @TheOtherTerry the minimum number of edges such that there exists a graph $G$ with that number of edges that satisfy the property, or in other words among all $G$ satisfying the property what is the minimum number of edges. Maybe it's better if I replace "can" with "must" now. $\endgroup$ Commented Aug 25 at 5:53
  • $\begingroup$ Ooops I misread the problem (missed that the index $i$ must be the same for $a_i$ and $b_i$) $\endgroup$ Commented Aug 25 at 10:20
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    $\begingroup$ Then the answer is of course of order $n^4$, since for fixed $a, b$ we must have $cn^2$ edges between $(a, \cdot) $ and $(b, \cdot)$. $\endgroup$ Commented Aug 25 at 10:29
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    $\begingroup$ Yes, and I get $138$ for $n=7$. $\endgroup$ Commented Aug 25 at 21:34

1 Answer 1

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For the sake of simplicity we identify the vertices of $G$ with their labels.

For the lower bound for distinct elements $a,b$ of $[n]$ consider the bipartite graph $G_{\{a,b\}}$, with the parts $V_a=\{\{a,c\}:c\in [n]\setminus\{a,b\}\}$ and $V_b=\{\{b,c\}:c\in [n]\setminus\{a,b\}\}$, where any vertices $v\in V_a$ and $u\in V_b$ are adjacent iff they are not adjacent in $G$. It is required that $G_{\{a,b\}}$ has no perfect matching of size $m$. By Kőnig's theorem, $G_{\{a,b\}}$ has a vertex cover of size at most $m-1$. So $G_{\{a,b\}}$ has at most $$(m-1)\deg G_{\{a,b\}}\le (m-1)(n-2)$$ edges. Thus there are at least $$|V_a||V_b|-(m-1)(n-2)=(n-m-1)(n-2)$$ edges between $V_a$ and $V_b$ in $G$. Summing on all ${n\choose 2}$ pairs $\{a,b\}$ of distinct elements of $[n]$ and observing that each edge of $G$ is accounted in at most $4$ such pairs, we obtain that $G$ has at least ${n\choose 2}(n-m-1)(n-2)/4$ edges. We can improve the latter bound a bit to $${n\choose 2}\left((n-m-1)+\frac{(n-m-1)(n-3)}4\right)=$$ $$\frac{(n-m-1)(n-1)n(n+1)}8,$$ because between $V_a$ and $V_b$ in $G$ there are at least $n-2-(m-1)=n-m-1$ edges of the form $\{\{a,c\},\{b,c\}\}$ for some $c\in [n]\setminus\{a,b\}$, which are accounted only once in the sum.

For the upper bound consider an auxiliary graph $H$ on the vertex set $[n]$ such that each vertex of $H$ has degree at least $n-m$. The straightforward saturating algorithm constructs such $H$ with $e\le \left\lceil \frac{n(n-m)}2\right\rceil$ edges. To provide the edges of the graph $G$, we begin from a graph with no edges and for each edge $\{a,b\}$ of $H$ we connect the vertex $\{a,b\}$ of $G$ with each other vertex of $G$. It is easy to see that $G$ satisfies the required condition and has ${e\choose 2}+e\left({n\choose 2}-e\right)=\frac{n^2e-ne-e^2-e}2$ edges.

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