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Consider a graph $G=(V,E)$ with $V=\{\{a,b\}: a,b \in [n], a \not= b\}$, or equivalently $\ V:=\binom {[n]}2\ $ for short.

Assume also that $G$ is a cluster graph (a disjoint union of complete graphs) with at most $n$ clusters.

For every distinct elements $a,b$ of $[n]$ consider a bipartite subgraph $G_{\{a,b\}}$ of $G$, with parts $V_a=\{\{a,c\}:c\in [n]\setminus\{a,b\}\}$ and $V_b=\{\{b,c\}:c\in [n]\setminus\{a,b\}\}$. It is required that $G_{\{a,b\}}$ has no perfect matching of size $m$.

QUESTION: I would like to get an upper bound for the number of edges in $G$. In particular, can we prove or disprove that the upper bound is $< \binom{n}{3}$ when $m = \lfloor (n-1)/2 \rfloor$?

This problem is the dual of this one, because the graph $G$ there is the complement of $G$ here. Also, here we have added the "cluster graph" requirement above.

The problem can be rephrased as follows: among all commutative magmas $S$ of order $n$, where there does not exist $a, b \in S = \{s_1, \ldots, s_n\}$ with $a \ne b$, and two $m$-element subsets $X = \{x_1, \ldots, x_m\}$ and $Y = \{y_1, ... ,y_m\}$ of $S \setminus \{a,b\}$ such that $ax_i = by_i$ for each $i=1, \ldots m$, and defined $q_1, \ldots, q_n$ the number of occurrences of $s_1, \ldots, s_n$ respectively in the "half" Cayley table for different operands only, we want to find an upper bound for $\sum_{i=1}^n \binom{q_i}{2}$. The problem is related to this conjecture, and in particular to the case of the union-closed sets conjecture. Proving that the upper bound is $< \binom{n}{3}$ when $m = \lfloor (n-1)/2 \rfloor$ would prove it, due to this other proof. Maybe instead of finding an upper bound for $\sum_{i=1}^n \binom{q_i}{2}$ we could try to seek a sequence $q_1^{\star}\ge\ldots\ge q_n^{\star}$ that majorizes all feasible $q_1\ge\ldots\ge q_n$.

Without the "cluster graph" requirement we know from the linked question answer that:

$$|E| \le \binom{\binom{n}{2}}{2} - \frac{(n-m-1)(n-1)n(n+1)}8$$

and using an Integer Linear Program for the case $m = \lfloor (n-1)/2 \rfloor$ we know that the maximum possible values for $|E|$ are $0,0,13,22,72$ respectively for $n=3,4,5,6,7$.

Adding the new "cluster graph" requirement, the maximum possible values for $|E|$, obtained with an updated ILP, are $0,8,12$ respectively for $n=3,5,6$ and the problem is infeasible for $n=4$. We have $8 \lt \binom{5}{3}$ and $12 \lt \binom{6}{3}$.

Now I describe my attempt, following the answer to the linked question. Let $C_1, \ldots, C_n$ be the (possibly empty) clusters of $G$. Let $G_{\{a,b\}}^{(i)}$, subgraphs of $C_i$, $1 \le i \le n$, be defined as in the second paragraph. Their parts are $V_a^{(i)}$ and $V_b^{(i)}$ of size $x_a^{(i)}$ and $x_b^{(i)}$ respectively. Let $y_i = \min(x_a^{(i)},x_b^{(i)})$, $1 \le i \le n$. We want to find:

$$\max \sum_{i=1}^n \binom{x_a^{(i)}+x_b^{(i)}}{2}$$

subject to:

$$\sum_{i=1}^n y_i \le m-1 \\ \sum_{i=1}^n x_a^{(i)} = n-2 \\ \sum_{i=1}^n x_b^{(i)} = n-2 $$

@RobPratt answer regarding this IQP shows how to solve it, however the maximum values are worse than the maximum number of edges of $G_{a,b}$ in the answer to the linked question, i.e. $(m−1)(n−2)$, therefore my attempt does not work.

Is it possible to get an analytical bound to the question above?

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  • $\begingroup$ What are optimal solutions to your IQP for small $n$? $\endgroup$ Commented Aug 31 at 17:20
  • $\begingroup$ How does $m$ enter to the question? $\endgroup$ Commented Sep 3 at 9:31
  • $\begingroup$ @OlegOrlov I have fixed it now. $\endgroup$ Commented Sep 3 at 9:39
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    $\begingroup$ @WlodAA no, vertices are only of the form $\{a,b\}$ with $a,b \in [n]$ and $a \not= b$. This is somewhat implicit in the set notation, however, I am editing it to make it clearer. $\endgroup$ Commented Sep 9 at 12:39
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    $\begingroup$ Thank you. I needed to have such a clear notation in order to read the rest of your Q. $\endgroup$ Commented Sep 9 at 12:42

2 Answers 2

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For your IQP, I get the following optimal solutions for small $n$.

$n=3$, maximum $0$: \begin{matrix} i & x_a & x_b & y \\ \hline 1 &0 &0 &0 \\ 2 &0 &1 &0 \\ 3 &1 &0 &0 \\ \end{matrix}

$n=4$, maximum $2$: \begin{matrix} i & x_a & x_b & y \\ \hline 1 &0 &0 &0 \\ 2 &0 &0 &0 \\ 3 &0 &2 &0 \\ 4 &2 &0 &0 \\ \end{matrix}

$n=5$, maximum $7$: \begin{matrix} i & x_a & x_b & y \\ \hline 1 &0 &0 &0 \\ 2 &1 &3 &1 \\ 3 &2 &0 &0 \\ 4 &0 &0 &0 \\ 5 &0 &0 &0 \\ \end{matrix}

$n=6$, maximum $13$: \begin{matrix} i & x_a & x_b & y \\ \hline 1 &1 &4 &1 \\ 2 &0 &0 &0 \\ 3 &3 &0 &0 \\ 4 &0 &0 &0 \\ 5 &0 &0 &0 \\ 6 &0 &0 &0 \\ \end{matrix}

You can solve these problems very quickly via integer linear programming by linearizing the quadratic objective as follows. For $i\in\{1,\dots,n\}$ and $k\in\{0,\dots,2n-4\}$, let binary decision variable $z_{ik}$ indicate whether $x_a^{(i)}+x_b^{(i)}=k$. The problem is then to maximize $$\sum_{i=1}^n \sum_{k=0}^{2n-4} \binom{k}{2} z_{ik}$$ subject to the original constraints and the following additional linear constraints: \begin{align} \sum_{k=0}^{2n-4} z_{ik} &= 1 &&\text{for $i\in\{1,\dots,n\}$} \\ \sum_{k=0}^{2n-4} k z_{ik} &= x_a^{(i)}+x_b^{(i)} &&\text{for $i\in\{1,\dots,n\}$} \end{align}

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  • $\begingroup$ I have fixed an error in my IQP formulation. It is still running forever in my tests. Maybe your approach above can succeed with the new formulation as well. $\endgroup$ Commented Sep 1 at 7:40
  • $\begingroup$ @FabiusWiesner I updated the results for your new IQP. $\endgroup$ Commented Sep 1 at 15:11
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This is a partial answer.

For the final purpose of my question, the upper bound is not going to be useful. I have found a way to model a more efficient Integer Quadratic Program and found that the maximum possible values for $|E|$ when $n=7$ and $m=\lfloor (n-1)/2 \rfloor = 3$ is $50 \gt \binom{n}{3}$ obtained from the cluster graph formed by the complete graphs with the following sets of vertices:

$$ \begin{align} & \{\{1,2\},\{1,4\},\{1,5\},\{2,4\},\{2,5\},\{3,4\},\{4,5\},\{4,6\},\{4,7\},\{6,7\}\}, \\ & \{\{3,5\},\{5,6\}\}, \\ & \{\{2,3\},\{2,7\}\}, \\ & \{\{2,6\},\{3,6\}\}, \\ & \{\{3,7\},\{5,7\}\}, \\ & \{\{1,6\},\{1,7\}\}, \\ & \{\{1,3\}\} \end{align} $$

Even modeling an IQP requiring that the sizes $k_1 \ge \ldots \ge k_n$ of the complete graphs majorize $n-1 \ge n-2 \ge \ldots \ge 1 \ge 0$ we can find a maximum $|E|$ for $n=7$ and $m=\lfloor (n-1)/2 \rfloor = 3$ of value $43 \gt \binom{n}{3}$, and the following sets of vertices for its complete graphs:

$$ \begin{align} & \{\{1,2\},\{1,3\},\{1,4\},\{1,5\},\{1,7\},\{2,7\},\{4,7\}\}, \\ & \{\{1,6\},\{2,4\},\{3,4\},\{3,6\},\{4,5\},\{4,6\}\}, \\ & \{\{3,5\},\{5,6\},\{5,7\},\{6,7\}\}, \\ & \{\{2,3\},\{3,7\}\}, \\ & \{\{2,5\}\}, \\ & \{\{2,6\}\}, \\ & \{\} \end{align} $$

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