Duality and Hereditary K\"onig-Egerv\'ary Set-systems

TL;DR

This paper characterizes hereditary K"onig-Egerváry set-systems, providing a new criterion based on set differences, which enhances understanding of the structure of these special set-systems and their relation to K"onig-Egerváry graphs.

Contribution

It introduces a novel characterization of hereditary K"onig-Egerváry set-systems using set difference equalities, extending the existing graph-theoretic framework.

Findings

Characterization of HKE set-systems via set difference equality

Equivalence between HKE set-systems and a specific set difference condition

Application of the characterization in related graph theory research

Abstract

A K\"onig-Egerv\'ary graph is a graph $G$ satisfying $\alpha(G)+\mu(G)=|V(G)|$, where $\alpha(G)$ is the cardinality of a maximum independent set and $\mu(G)$ is the matching number of $G$. Such graphs are those that admit a matching between $V(G)-\bigcup \Gamma$ and $\bigcap \Gamma$ where $\Gamma$ is a set-system comprised of maximum independent sets satisfying $|\bigcup \Gamma'|+|\bigcap \Gamma'|=2\alpha(G)$ for every set-system $\Gamma' \subseteq \Gamma$; in order to improve this characterization of a K\"onig-Egerv\'ary graph, we characterize \emph{hereditary K\"onig-Egerv\'ary set-systems} (HKE set-systems, here after). An \emph{HKE} set-system is a set-system, $F$, such that for some positive integer, $\alpha$, the equality $|\bigcup \Gamma|+|\bigcap \Gamma|=2\alpha$ holds for every non-empty subset, $\Gamma$, of $F$. We prove the following theorem: Let $F$ be a set-system. $F$ is an HKE set-system if and only if the equality $|\bigcap \Gamma_1-\bigcup \Gamma_2|=|\bigcap \Gamma_2-\bigcup \Gamma_1|$ holds for every two non-empty disjoint subsets, $\Gamma_1,\Gamma_2$ of $F$. This theorem is applied in \cite{hke},\cite{broken}.

Equations

$$|\bigcup F|+|\bigcap F|=2\alpha (F).$$

$$|\bigcup \Gamma |+|\bigcap \Gamma |=2\alpha$$

$$|\bigcup \Gamma |+|\bigcap \Gamma |=|A\cup B|+|A\cap B|=|A|+|B|=2\alpha (F).$$

$$|\bigcap {\Gamma }_1-\bigcup {\Gamma }_2|=|\bigcap {\Gamma }_2-\bigcup {\Gamma }_1|.$$

$$|A^{\ast }-\bigcup {\Gamma }_2|=|\bigcap {\Gamma }_2-A^{\ast }|,$$

$$|\bigcup {\Gamma }_2\cup A^{\ast }|-|\bigcup {\Gamma }_2|=|\bigcap {\Gamma }_2|-|\bigcap {\Gamma }_2\cap A^{\ast }|,$$

$$|\bigcup {\Gamma }_2\cup A^{\ast }|+|\bigcap {\Gamma }_2\cap A^{\ast }|=|\bigcap {\Gamma }_2|+|\bigcup {\Gamma }_2|.$$

$$\bigcap ({\Gamma }_1-\{A^{\ast }\})=\{x:x\in A\text{ for every }A\in {\Gamma }_1\text{ with }A\ne A^{\ast }\},$$

$$\bigcup ({\Gamma }_1-\{A^{\ast }\})=\{x:x\in A\text{ for some }A\in {\Gamma }_1\text{ with }A\ne A^{\ast }\}$$

$$\bigcap ({\Gamma }_1\cup \{A^{\ast }\})=A^{\ast }\cap \bigcap {\Gamma }_1.$$

$$|\bigcap {\Gamma }_1-\bigcup {\Gamma }_2|=|\bigcap ({\Gamma }_1-\{A^{\ast }\})-\bigcup {\Gamma }_2|-|\bigcap ({\Gamma }_1-\{A^{\ast }\})-\bigcup ({\Gamma }_2\cup \{A^{\ast }\})|.$$

$$|\bigcap ({\Gamma }_1-\{A^{\ast }\})-\bigcup {\Gamma }_2|=|\bigcap {\Gamma }_2-\bigcup ({\Gamma }_1-\{A^{\ast }\})|$$

$$|\bigcap ({\Gamma }_1-\{A^{\ast }\})-\bigcup ({\Gamma }_2\cup \{A^{\ast }\})|=|\bigcap ({\Gamma }_2\cup \{A^{\ast }\})-\bigcup ({\Gamma }_1-\{A^{\ast }\})|.$$

$$|\bigcap {\Gamma }_1-\bigcup {\Gamma }_2|=|\bigcap {\Gamma }_2-\bigcup ({\Gamma }_1-\{A^{\ast }\})|-|\bigcap ({\Gamma }_2\cup \{A^{\ast }\})-\bigcup ({\Gamma }_1-\{A^{\ast }\})|.$$

$$|\bigcap {\Gamma }_1-\bigcup {\Gamma }_2|=|\bigcap {\Gamma }_2-\bigcup {\Gamma }_1|.$$

$$|\bigcup \Gamma |+|\bigcap \Gamma |=2|D|,$$

$$|\bigcup \Gamma -D|=|D-\bigcap \Gamma |.$$

$$\sum _{⟨{\Gamma }_1,{\Gamma }_2⟩\in H}|\bigcap {\Gamma }_1-\bigcup {\Gamma }_2|=\sum _{⟨{\Gamma }_1,{\Gamma }_2⟩\in H}|\bigcap {\Gamma }_2-\bigcup {\Gamma }_1|.$$

$$|\bigcup \Gamma -D|=\sum _{⟨{\Gamma }_1,{\Gamma }_2⟩\in H}|\bigcap {\Gamma }_1-\bigcup {\Gamma }_2|$$

$$|D-\bigcap \Gamma |=\sum _{⟨{\Gamma }_1,{\Gamma }_2⟩\in H}|\bigcap {\Gamma }_2-\bigcup {\Gamma }_1|.$$

$$\bigcup \Gamma -D=\bigcup _{⟨{\Gamma }_1,{\Gamma }_2⟩\in H}(\bigcap {\Gamma }_1-\bigcup {\Gamma }_2),$$

$$|\bigcup \Gamma -D|=\sum _{⟨{\Gamma }_1,{\Gamma }_2⟩\in H}|\bigcap {\Gamma }_1-\bigcup {\Gamma }_2|,$$

$$\sum _{⟨{\Gamma }_1^{+},{\Gamma }_2^{+}⟩\in H}|\bigcap {\Gamma }_1^{+}-\bigcup {\Gamma }_2^{+}|=\sum _{⟨{\Gamma }_1^{+},{\Gamma }_2^{+}⟩\in H}|\bigcap {\Gamma }_2^{+}-\bigcup {\Gamma }_1^{+}|.$$

$$|\bigcap {\Gamma }_1-\bigcup {\Gamma }_2|=\sum _{⟨{\Gamma }_1^{+},{\Gamma }_2^{+}⟩\in H}|\bigcap {\Gamma }_1^{+}-\bigcup {\Gamma }_2^{+}|$$

$$|\bigcap {\Gamma }_2-\bigcup {\Gamma }_1|=\sum _{⟨{\Gamma }_1^{+},{\Gamma }_2^{+}⟩\in H}|\bigcap {\Gamma }_2^{+}-\bigcup {\Gamma }_1^{+}|,$$

$$\bigcap {\Gamma }_1-\bigcup {\Gamma }_2=\bigcup _{⟨{\Gamma }_1^{+},{\Gamma }_2^{+}⟩\in H}(\bigcap {\Gamma }_1^{+}-\bigcup {\Gamma }_2^{+})$$

$$|\bigcap {\Gamma }_1-\bigcup {\Gamma }_2|=\sum _{⟨{\Gamma }_1^{+},{\Gamma }_2^{+}⟩\in H}|\bigcap {\Gamma }_1^{+}-\bigcup {\Gamma }_2^{+}|,$$

$$\alpha =\frac{|\bigcup F|+|\bigcap F|}{2}.$$

$$P_1=\{\{{\Gamma }_1,{\Gamma }_2\}\in P:D\in {\Gamma }_1\}$$

$$P_2=\{\{{\Gamma }_1,{\Gamma }_2\}\in P:D\notin {\Gamma }_1\}.$$

Full text

An HKE set-system is a set-system, $F$, such that for some positive integer, $\alpha$, the equality $|\bigcup\Gamma|+|\bigcap\Gamma|=2\alpha$ holds for every non-empty subset, $\Gamma$, of $F$.

We prove the following theorem: Let $F$ be a set-system. $F$ is an HKE set-system if and only if the equality $|\bigcap\Gamma_{1}-\bigcup\Gamma_{2}|=|\bigcap\Gamma_{2}-\bigcup\Gamma_{1}|$ holds for every two non-empty disjoint subsets, $\Gamma_{1},\Gamma_{2}$ of $F$.

This theorem is applied in [2],[1].

1. Introduction

In this section we give the basic definitions and motivate the study of HKE set-systems.

For a uniform set-system, $F$, we denote by $\alpha(F)$ the cardinality of a set in $F$. We write $\alpha$, when $F$ is clear from the context.

The following definition contradicts the definition of a König-Egerváry set-system in [3].

Definition 1.1**.**

Let $F$ be a uniform set-system. $F$ is said to be a König-Egerváry set-system (KE set-system in short), if the following equality holds:

[TABLE]

Definition 1.2**.**

An HKE set-system is a set-system, $F$, such that for some positive integer, $\alpha$, the equality

[TABLE]

holds for every non-empty subset, $\Gamma$, of $F$.

Proposition 1.3**.**

Every HKE set-system is a uniform set-system. So a set-system $F$ is HKE if and only if each subset $\Gamma$ of $F$ is KE.

Proof.

Let $F$ be an HKE set-system and let $A\in F$. By Definition 1.2, where we substitute $\Gamma=\{A\}$, we have $|A|=\alpha$. So $F$ is a uniform set-system and $\alpha=\alpha(F)$. ∎

Proposition 1.4**.**

Let $F$ be a uniform set-system. If $|F|\leq 2$ then it is an HKE set-system.

Proof.

It is clear when $|F|=1$. So assume $|F|=2$, $F=\{A,B\}$. Take a non-empty sub-set-system $\Gamma$ of $F$. Without loss of generality, $\Gamma=F$. So

[TABLE]

∎

Theorem 1.5 and Propositions 1.6,1.7 exemplifies the usefullness of HKE set-systems in the study of König-Egerváry graphs.

The following theorem is a restatement of [3, Theorem 2.6] in our notation.

Theorem 1.5**.**

$G$* is a König-Egerváry graph if and only if there is a matching between $V(G)-\bigcup\Gamma$ and $\bigcap\Gamma$, where $\Gamma$ is an HKE set-system comprised of maximum independent sets.*

Proposition 1.6**.**

Let $G$ be a KE graph. Then $\Omega(G)$ is an HKE set-system.

Proof.

By [4, Theorem 3.6] and [4, Corollary 2.8]. ∎

Proposition 1.7**.**

Every KE set-system that is comprised of maximum independent sets of some graph is an HKE set-system.

Proof.

By [4, Corollaries 2.7 and 2.9]. ∎

2. HKE set-systems and duality

In this section, we characterize the HKE set-systems; consequently, we get a new characterization of a König-Egerváry graph. Proposition 2.2 is a weak version of Theorem 2.5, where we add the assumption, that the set-system is uniform.

In order to state Proposition 2.2, Theorem 2.5 and Corollary 2.6, we present the following equality:

Equality 2.1**.**

[TABLE]

Proposition 2.2**.**

Let $F$ be a uniform set-system.

The following are equivalent:

1. (1)

$F$* is an HKE set-system.* 2. (2)

Equality 2.1 holds for every two non-empty disjoint sub-set-systems, $\Gamma_{1},\Gamma_{2}$ of $F$, 3. (3)

Equality 2.1 holds for every two non-empty disjoint sub-set-systems, $\Gamma_{1},\Gamma_{2}$ of $F$ with $\Gamma_{1}\cup\Gamma_{2}=F$.

The argument of Proposition 2.2 is based on the following exercise:

Exercise 2.3**.**

Assume that $\{A,B,C,D\}$ is an HKE set-system (so in particular $\{A,B,C\}$ is an HKE set-system). Prove:

1. (1)

$|A-B-C|=|B\cap C-A|$. A clue: $A-B-C=(A\cup B\cup C)-(B\cup C)$ and $B\cap C-A=(B\cap C)-(A\cap B\cap C)$. 2. (2)

$|A\cap B-C-D|=|C\cap D-A-B|$. A clue: $A\cap B-C-D=(A-C-D)-(A-B-C-D)$. Apply Clause (1).

We now prove Proposition 2.2.

Proof.

$(1)\Rightarrow(2):$ We prove it by induction on $r=:|\Gamma_{1}|$.

Case a: $r=1$, so $\Gamma_{1}=\{A^{*}\}$ for some set $A^{*}$. In this case, we apply the idea of Exercise 2.3(1).

We should prove that

[TABLE]

namely,

[TABLE]

or equivalently,

[TABLE]

But by Clause $(1)$, each side of this equality equals $2\alpha$.

Case a: $r>1$. In this case, we apply the idea of Exercise 2.3(2). We fix $A^{*}\in\Gamma_{1}$. First we write three trivial equalities, for convenience:

[TABLE]

[TABLE]

and

[TABLE]

We now begin the computation.

[TABLE]

The right side of this equality is a subtraction of two summands. Since $|\Gamma_{1}-\{A^{*}\}|<|\Gamma_{1}|$, we may apply the induction hypothesis on each summand:

[TABLE]

and

[TABLE]

By the three last equalities we get:

[TABLE]

So

[TABLE]

Equality 2.1 is proved, so Clause $(2)$ is proved.

$(2)\Rightarrow(1):$ Let $\Gamma$ be a non-empty subset of $F$. Fix $D\in\Gamma$. Since $F$ is a uniform set-system, $|D|=\alpha$ (this is the unique place where we use the assumption that $F$ is a uniform set-system, but we eliminate this assumption later). Therefore it is enough to prove that

[TABLE]

or equivalently,

[TABLE]

Let $H$ be the set of ordered pairs $\langle\Gamma_{1},\Gamma_{2}\rangle$ of non-empty disjoint subsets of $\Gamma$ such that $\Gamma_{1}\cup\Gamma_{2}=\Gamma$ and $D\in\Gamma_{2}\}$.

By Clause (2),

[TABLE]

So it is enough to prove the following two equalities:

[TABLE]

and

[TABLE]

Since their proofs are dual, we prove the first equality only.

[TABLE]

(on the one hand, if $x\in\bigcup\Gamma-D$ then for $\Gamma_{1}=\{A\in\Gamma:x\in A\}$ and $\Gamma_{2}=\{A\in\Gamma:x\notin A\}$ we have $x\in\bigcap\Gamma_{1}-\bigcup\Gamma_{2}$ and $\langle\Gamma_{1},\Gamma_{2}\rangle\in H$. On the other hand, assume that $x\in\bigcap\Gamma_{1}-\bigcup\Gamma_{2}$ for some $\langle\Gamma_{1},\Gamma_{2}\rangle\in H$. Then $x\in\bigcup\Gamma$ (because $x\in\bigcap\Gamma_{1}$ and $\emptyset\neq\Gamma_{1}\subseteq\Gamma$) and $x\notin D$ (because $x\notin\bigcup\Gamma_{2}$ and $D\subseteq\bigcup\Gamma_{2}$). So $x\in\bigcup\Gamma-D$). Therefore

[TABLE]

because this is a sum of cardinalities of disjoint sets (if $\langle\Gamma_{1},\Gamma_{2}\rangle$ and $\langle\Gamma_{3},\Gamma_{4}\rangle$ are two different pairs in $H$ then there is no element $x\in(\bigcap\Gamma_{1}-\bigcup\Gamma_{2})\cap(\bigcap\Gamma_{3}-\bigcup\Gamma_{4})$. Otherwise, take $A\in\Gamma_{1}-\Gamma_{3}$ (or vice versa). So $A\in\Gamma_{4}$. Hence, $x\in\bigcap\Gamma_{1}\subseteq A$ and $x\notin\bigcup\Gamma_{4}\supseteq A$, a contradiction).

The implication $(2)\Rightarrow(1)$ is proved.

Since Clause $(3)$ is a private case of Clause $(2)$, it remains to prove $(3)\Rightarrow(2)$. Let $\Gamma_{1},\Gamma_{2}$ be two non-empty disjoint subsets of $F$. We should prove Equality 2.1 for these $\Gamma_{1}$ and $\Gamma_{2}$, without assuming $\Gamma_{1}\cup\Gamma_{2}=F$. Let $H$ be the set of disjoint pairs $\langle\Gamma_{1}^{+},\Gamma_{2}^{+}\rangle$ of $F$ such that $\Gamma_{1}\subseteq\Gamma_{1}^{+}$, $\Gamma_{2}\subseteq\Gamma_{2}^{+}$ and $\Gamma_{1}^{+}\cup\Gamma_{2}^{+}=F$.

By Clause $(3)$,

[TABLE]

So it remains to prove the following two equalities:

[TABLE]

and

[TABLE]

Since their proofs are dual, we prove the first equality only.

[TABLE]

(On the one hand, if $x\in\bigcap\Gamma_{1}-\bigcup\Gamma_{2}$ then for $\Gamma_{1}=\{A\in\Gamma:x\in A\}$ and $\Gamma_{2}=\{A\in\Gamma:x\notin A\}$, we have $x\in\bigcap\Gamma_{1}^{+}-\bigcup\Gamma_{2}^{+}$ and the pair $\langle\Gamma_{1}^{+},\Gamma_{2}^{+}\rangle$ belongs to $H$. On the other hand, if $x\in\bigcap\Gamma_{1}^{+}-\bigcup\Gamma_{2}^{+}$ for some $\langle\Gamma_{1}^{+},\Gamma_{2}^{+}\rangle\in H$ then $x\in\bigcap\Gamma_{1}^{+}\subseteq\bigcap\Gamma_{1}$ and $x\notin\bigcup\Gamma_{2}^{+}\supseteq\bigcup\Gamma_{2}$. Hence, $x\in\bigcap\Gamma_{1}-\bigcup\Gamma_{2}$). Therefore

[TABLE]

because it is a sum of disjoint sets. ∎

The following proposition eliminates the assumption that $F$ is a uniform set-system.

Proposition 2.4**.**

Clause (3) of Proposition 2.2 implies that $F$ is a uniform set-system.

Proof.

Define

[TABLE]

Let $D\in F$. We prove that $|D|=\alpha$. Let $P$ denote the family of partitions $\{\Gamma_{1},\Gamma_{2}\}$ of $F$ into two non-empty subsets.

Every element in $\bigcup F$ is in $\bigcap\Gamma_{1}-\bigcup\Gamma_{2}$ for some partition $\{\Gamma_{1},\Gamma_{2}\}\in P$ or in $\bigcap F$.

Let

[TABLE]

and

[TABLE]

Define

[TABLE]

and

[TABLE]

By Clause $(3)$ of Proposition 2.2, we have $x=y$.

It is easy to check the following three equalities:

1. (1)

$|\bigcup F|=x+y+|\bigcap F|=2x+|\bigcap F|$, 2. (2)

$|D|=x+|\bigcap F|$ and 3. (3)

$|\bigcup F|+|\bigcap F|=2\alpha$ (by the definition of $\alpha$).

By Equalities (1)-(3), $|D|=\alpha$. Since $D$ is an arbitrary set in $F$, $F$ is a uniform set-system. ∎

Theorem 2.5**.**

Let $F$ be a set-system.

The following are equivalent:

1. (1)

$F$* is an HKE set-system.* 2. (2)

Equality 2.1 holds for every two non-empty disjoint sub-set-systems, $\Gamma_{1},\Gamma_{2}$ of $F$, 3. (3)

Equality 2.1 holds for every two non-empty disjoint sub-set-systems, $\Gamma_{1},\Gamma_{2}$ of $F$ with $\Gamma_{1}\cup\Gamma_{2}=F$.

Proof.

By Proposition 2.2, it is enough to prove that each clause implies that $F$ is a uniform set-system. By Proposition 1.3, Clause $(1)$ implies that $F$ is a uniform set-system. By Proposition 2.4 Clause $(3)$ implies that $F$ is a uniform set-system. But Clause $(2)$ implies Clause $(3)$. ∎

Corollary 2.6**.**

Let $G$ be a graph. The following are equivalent:

1. (1)

$G$* is a KE graph.* 2. (2)

For some non-empty HKE set-system $F\subseteq\Omega(G)$, there is a matching $M:V[G]-\bigcup F\to\bigcap F$ and Equality 2.1 holds for every two non-empty disjoint sub-set-systems, $\Gamma_{1},\Gamma_{2}$ of $F$. 3. (3)

For some non-empty HKE set-system $F\subseteq\Omega(G)$, there is a matching $M:V[G]-\bigcup F\to\bigcap F$ and Equality 2.1 holds for every two non-empty disjoint sub-set-systems, $\Gamma_{1},\Gamma_{2}$ of $F$ with $\Gamma_{1}\cup\Gamma_{2}=F$.

Proof.

By Theorem 2.5 and Theorem 1.5. ∎