Bipartization of graphs

TL;DR

This paper introduces a new characterization of bipartite graphs where the domination number equals the size of the smaller part, using a novel graph operation to deepen understanding of domination properties.

Contribution

It presents a new characterization of certain bipartite graphs based on a novel graph operation, advancing the theoretical understanding of domination numbers.

Findings

Characterization of bipartite graphs with domination number equal to the smaller part's size

Introduction of a new graph operation for analyzing domination properties

Theoretical insights into domination number in bipartite graphs

Abstract

A dominating set of a graph $G$ is a set $D\subseteq V_G$ such that every vertex in $V_G-D$ is adjacent to at least one vertex in $D$, and the domination number $\gamma(G)$ of $G$ is the minimum cardinality of a dominating set of $G$. In this paper we provide a new characterization of bipartite graphs whose domination number is equal to the cardinality of its smaller partite set. Our characterization is based upon a new graph operation.

Full text

Bipartization of graphs

Mateusz Miotk, Jerzy Topp, and Paweł Żyliński

University of Gdańsk, 80-952 Gdańsk, Poland

{mmiotk,j.topp,zylinski}@inf.ug.edu.pl

Keywords: Bipartite graph; Bipartization; Domination number.

AMS subject classification: 05C69; 05C76; 05C05.

1 Introduction and notation

For notation and graph theory terminology we in general follow [2]. Specifically, let $G=(V_{G},E_{G})$ be a graph with vertex set $V_{G}$ and edge set $E_{G}$. For a subset $X\subseteq~{}V_{G}$, the subgraph induced by $X$ is denoted by $G[X]$. For simplicity of notation, if $X=\{x_{1},\ldots,x_{k}\}$, we shall write $G[x_{1},\ldots,x_{k}]$ instead of $G[\{x_{1},\ldots,x_{k}\}]$. For a vertex $v$ of $G$, its neighborhood, denoted by $N_{G}(v)$, is the set of all vertices adjacent to $v$, and the cardinality of $N_{G}(v)$, denoted by $\deg_{G}(v)$, is called the degree of $v$. The closed neighborhood of $v$, denoted by $N_{G}[v]$, is the set $N_{G}(v)\cup\{v\}$. In general, the neighborhood of $X\subseteq V_{G}$, denoted by $N_{G}(X)$, is defined to be $\bigcup_{v\in X}N_{G}(v)$, and the closed neighborhood of $X$, denoted by $N_{G}[X]$, is the set $N_{G}(X)\cup X$. A vertex of degree one is called a leaf, and the only neighbor of a leaf is called its support vertex (or simply, its support). A weak support is a vertex adjacent to exactly one leaf. Finally, the set of leaves and the set of supports of $G$ we denoted by $L_{G}$ and $S_{G}$, respectively.

A subset $D$ of $V_{G}$ is said to be a dominating set of a graph $G$ if each vertex belonging to the set $V_{G}-D$ has a neighbor in $D$. The cardinality of a minimum dominating set of $G$ is called the domination number of $G$ and is denoted by $\gamma(G)$. A subset $C\subseteq V_{G}$ is a covering set of $G$ if each edge of $G$ has an end-vertex in $C$. The cardinality of a minimum covering set of $G$ is called the covering number of $G$ and denoted by $\beta(G)$.

It is obvious that if $G=((A,B),E_{G})$ is a bipartite graph, then $\gamma(G)\leq\min\{|A|,|B|\}$. In this paper the set of all bipartite graphs $G=((A,B),E_{G})$ in which $\gamma(G)=\min\{|A|,|B|\}$ is denoted by ${\cal B}$. Some properties of the graphs belonging to the set ${\cal B}$ were observed in the papers [1, 3, 4, 5, 6], where all graphs with the domination number equal to the covering number were characterized. In this paper, inspired by results and constructions of Hartnell and Rall [3], we introduce a new graph operation, called the bipartization of a graph with respect to a function, study basic properties of this operation, and provide a new characterization of the graphs belonging to the set ${\cal B}$ in terms of this new operation.

2 Bipartization of a graph

Let ${\cal K}_{H}$ denote the set of all complete subgraphs of a graph $H$. If $v\in V_{H}$, then the set $\{K\in{\cal K}_{H}\colon v\in V_{K}\}$ is denoted by ${\cal K}_{H}(v)$. If $X\subseteq V_{H}$, then the set $\bigcup_{v\in X}{\cal K}_{H}(v)$ is denoted by ${\cal K}_{H}(X)$, and it is obvious that ${\cal K}_{H}(X)=\{K\in{\cal K}_{H}\colon V_{K}\cap X\not=\emptyset\}$. Let $f\colon{\cal K}_{H}\to\mathbb{N}$ be a function. If $K\in{\cal K}_{H}$, then by ${\cal F}_{K}$ we denote the set $\{(K,1),\ldots,(K,f(K))\}$ if $f(K)\geq 1$, and we let ${\cal F}_{K}=\emptyset$ if $f(K)=0$. By ${\cal K}_{H}^{f}$ we denote the set of all positively $f$-valued complete subgraphs of $H$, that is, ${\cal K}_{H}^{f}=\{K\in{\cal{K}}_{H}\colon f(K)\geq 1\}$.

Definition 1**.**

Let $H$ be a graph and let $f\colon{\cal K}_{H}\to\mathbb{N}$ be a function. The bipartization of $H$ with respect to $f$ is the bipartite graph $B_{f}(H)=((A,B),E_{B_{f}(H)})$ in which $A=V_{H}$, $B=\bigcup_{K\in{\cal K}_{H}}{\cal F}_{K}$, and where a vertex $x\in A$ is adjacent to a vertex $(K,i)\in B$ if and only if $x$ is a vertex of the complete graph $K$ ($$i=1,\ldots,f(K)$$).

Example 1**.**

Fig. 1 presents a graph $H$ (for which ${\cal K}_{H}=\{H[a],H[b],H[c],H[d],H[a,b],$ $H[a,c],H[b,c],H[c,d],H[a,b,c]\}$) and its two bipartizations $B_{f}(H)$ and $B_{g}(H)$ with respect to functions $f,\,g\colon{\cal K}_{H}\to\mathbb{N}$, respectively, where $f(H[a])=1$, $f(H[b])=1$, $f(H[c])=2$, $f(H[d])=0$, $f(H[a,b])=3$, $f(H[a,c])=0$, $f(H[b,c])=2$, $f(H[c,d])=3$, $f(H[a,b,c])=1$, while $g(H[v])=0$ for every vertex $v\in V_{H}$, $g(H[u,v])=1$ for every edge $uv\in E_{H}$, and $g(H[a,b,c])=0$. Observe that $B_{g}(H)$ is the subdivision graph $S(H)$ of $H$ (i.e., the graph obtained from $H$ by inserting a new vertex into each edge of $H$).**

3 Properties of bipartizations of graphs

It is clear from the above definition of the bipartization of a graph with respect to a function that we have the following proposition.

Proposition 1**.**

The bipartization of a graph with respect to a function has the following properties:

• $(1)$

If $B_{f}(H)=((A,B),E_{B_{f}(H)})$ is the bipartization of a graph $H$ with respect to a function $f\colon{\cal K}_{H}\to\mathbb{N}$, then:

• $(\rm a)$

$N_{B_{f}(H)}(v)=\bigcup_{K\in{\cal K}_{H}(v)}{\cal F}_{K}$* if $v\in A$.*

• $(\rm b)$

$N_{B_{f}(H)}(X)=\bigcup_{K\in{\cal K}_{H}(X)}{\cal F}_{K}$* if $X\subseteq A$.*

• $(\rm c)$

$N_{B_{f}(H)}((K,i))=V_{K}$* if $(K,i)\in B$ ($$i=1,\ldots,f(K)$$).*

• $(\rm d)$

$|V_{B_{f}(H)}|=|V_{H}|+\sum_{K\in{\cal K}_{H}}f(K)$* and $|E_{B_{f}(H)}|=\sum_{K\in{\cal K}_{H}}f(K)\,|V_{K}|$.*

• $(2)$

If $H$ is a connected graph and $f\colon{\cal{K}}_{H}\to\mathbb{N}$ is a function such that every edge of $H$ belongs to a positively $f$-valued complete subgraph of $H$, then the bipartization $B_{f}(H)$ is a connected graph.

• $(3)$

If $H$ is a graph and $f,\,g\colon{\cal K}_{H}\to\mathbb{N}$ are functions such that $f(K)\geq g(K)$ for every $K\in{\cal K}_{H}$, then the graph $B_{g}(H)$ is an induced subgraph of $B_{f}(H)$.

Our study of properties of bipartizations we begin by showing that every bipartite graph is the bipartization of some graph with respect to some function.

Theorem 1**.**

For every bipartite graph $G=((A,B),E_{G})$ there exist a graph $H$ and a function $f\colon{\cal K}_{H}\to\mathbb{N}$ such that $G=B_{f}(H)$.

Proof.

We say that vertices $x$ and $y$ of $G$ are similar if $N_{G}(x)=N_{G}(y)$. It is obvious that this similarity is an equivalence relation on $B$ (as well as on $A$ and $A\cup B$). Let $B_{1},\ldots,B_{l}$ be the equivalence classes of this relation on $B$, say $B_{i}=\{b_{1}^{i},b_{2}^{i},\ldots,b_{k_{i}}^{i}\}$ for $i=1,\ldots,l$. It follows from properties of the equivalence classes that $|B_{1}|+\ldots+|B_{l}|=|B|$, $N_{G}(b_{1}^{i})=N_{G}(x)$ for every $x\in B_{i}$, and $N_{G}(b_{1}^{i})\not=N_{G}(b_{1}^{j})$ if $i,j\in\{1,\ldots,l\}$ and $i\not=j$.

Now, let $H=(V_{H},E_{H})$ be a graph in which $V_{H}=A$ and two vertices $x$ and $y$ are adjacent in $H$ if and only if they are at distance two apart from each other in $G$. Let ${\cal K}_{H}$ be the set of all complete subgraphs of $H$, and let $f\colon{\cal K}_{H}\to\mathbb{N}$ be a function such that $f(K)=|\{b\in B\colon N_{G}(b)=V_{K}\}|$ for $K\in{\cal K}_{H}$. Next, let $K_{i}$ be the induced subgraph $H[N_{G}(b_{1}^{i})]$ of $H$. It follows from the definition of $H$ that $K_{i}$ is a complete subgraph of $H$. In addition, from the definition of $f$ and from properties of the classes $B_{1},\ldots,B_{l}$, it follows that $f(K_{i})=|B_{i}|>0$ ($i=1,\ldots,l$), and $f(K)=0$ if $K\in{\cal K}_{H}-\{K_{1},\ldots,K_{l}\}$. Consequently, ${\cal K}_{H}^{f}=\{K_{1},\ldots,K_{l}\}$.

Finally, consider the bipartite graph $B_{f}(H)=((X,Y),E_{B_{f}(H)})$ in which $X=V_{H}=A$, $Y=\bigcup_{K\in{\cal K}_{H}}{\cal F}_{K}=\bigcup_{K\in{\cal K}_{H}^{f}}{\cal F}_{K}=\bigcup_{i=1}^{l}\{(K_{i},1),\ldots,(K_{i},k_{i})\}$, and where $N_{B_{f}(H)}((K_{i},j))$ $=V_{K_{i}}=N_{G}(b_{1}^{i})$ for every $(K_{i},j)\in Y$. Now, one can observe that the function $\varphi\colon A\cup B\to X\cup Y$, where $\varphi(x)=x$ if $x\in A$, and $\varphi(b_{j}^{i})=(K_{i},j)$ if $b_{j}^{i}\in B$, is an isomorphism between graphs $G$ and $B_{f}(H)$. ∎

We have proved that a bipartite graph $G=((A,B),E_{G})$ is the bipartization $B_{f}(H)$ of a graph $H=(V_{H},E_{H})$ (in which $V_{H}=A$ and $E_{H}=\{xy\colon\,x,y\in A\,\,\mbox{and}\,\,d_{G}(x,y)=2\}$) with respect to a function $f\colon{\cal K}_{H}\to\mathbb{N}$, where $f(K)=|\{b\in B\colon N_{G}(b)=V_{K}\}|$ for $K\in{\cal K}_{H}$. The same graph $G$ is also the bipartization $B_{g}(F)$ of a graph $F=(V_{F},E_{F})$ (in which $V_{F}=B$ and $E_{F}=\{xy\colon\,x,y\in B\,\,\mbox{and}\,\,d_{G}(x,y)=2\}$) with respect to a function $g\colon{\cal K}_{F}\to\mathbb{N}$, where $g(K)=|\{a\in A\colon N_{G}(a)=V_{K}\}|$ for $K\in{\cal K}_{F}$. Consequently, every bipartite graph may be the bipartization of two non-isomorphic graphs.

Example 2**.**

Fig. 2 depicts the bipartite graph $G$ which is the bipartization of the non-isomorphic graphs $H$ and $F$ with respect to functions $\overline{f}\colon{\cal K}_{H}\to\mathbb{N}$ and $\overline{g}\colon{\cal K}_{F}\to\mathbb{N}$, respectively, which non-zero values are displayed in the figure. **

It is obvious from Theorem 1 that every tree is a bipartization. We are now interested in providing a simple characterization of graphs $H$ and functions $f\colon{\cal K}_{H}\to\mathbb{N}$ for which the bipartization $B_{f}(H)$ is a tree. We begin with the following notation: An alternating sequence of vertices and complete graphs $(v_{0},F_{1},v_{1},\ldots,v_{k-1},F_{k},v_{k})$ is said to be a positively $f$-valued complete $v_{0}-v_{k}$ path if $v_{i-1}v_{i}$ is an edge in the complete graph $F_{i}$ for $i=1,\ldots,k$. We now have the following two useful lemmas.

Lemma 1**.**

Let $H$ be a connected graph, and let $f\colon{\cal{K}}_{H}\to\mathbb{N}$ be a function. If there are two vertices $u$ and $v$ and two distinct internally vertex-disjoint positively $f$-valued complete $u-v$ paths in $H$, then the bipartization $B_{f}(H)$ contains a cycle.

Proof.

If $(v_{0}=u,F_{1},v_{1},\ldots,v_{m-1},F_{m},v_{m}=v)$ and $(v_{0}^{\prime}=u,F_{1}^{\prime},v_{1}^{\prime},\ldots,v_{n-1}^{\prime},F_{n}^{\prime},v_{n}^{\prime}=v)$ are distinct internally vertex-disjoint positively $f$-valued complete $u-v$ paths in $H$, then $(v_{0},(F_{1},1),v_{1},\ldots,v_{m-1},(F_{m},1),v_{m})$ and $(v_{0}^{\prime},(F_{1}^{\prime},1),v_{1}^{\prime},\ldots,v_{n-1}^{\prime},(F_{n}^{\prime},1),v_{n}^{\prime})$ are distinct $u-v$ paths in $B_{f}(H)$, and so they generate at least one cycle in $B_{f}(H)$. ∎

Let us recall first that a maximal connected subgraph without a cutvertex is called a block. A graph $H$ is said to be a block graph if each block of $H$ is a complete graph. The next lemma is probably known, therefore we omit its easy inductive proof.

Lemma 2**.**

If ${\cal S}$ is the set of all blocks of a graph $H$, then $\sum\limits_{B\in{\cal S}}\left(|V_{B}|-1\right)=|V_{H}|-1$.

Now we are ready for a characterization of graphs which bipartizations (with respect to some functions) are trees.

Theorem 2**.**

Let $H$ be a connected graph, and let $f\colon{\cal{K}}_{H}\to\mathbb{N}$ be a function such that every edge of $H$ belongs to some positively $f$-valued complete subgraph of $H$. Then the bipartization $B_{f}(H)$ is a tree if and only if the following conditions hold:

• $(1)$

$f(K)\leq 1$* for every non-trivial complete subgraph $K$ of $H$.*

• $(2)$

$H$* is a block graph.*

• $(3)$

For a non-trivial complete subgraph $K$ of $H$ is $f(K)=1$ if and only if $K$ is a block of $H$.

Proof.

Assume that $B_{f}(H)$ is a tree. The statement (1) is obvious, for if there were a non-trivial complete subgraph $K$ of $H$ for which $f(K)\geq 2$, then for any two vertices $u$ and $v$ belonging to $K$, the sequence $(u,(K,1),v,(K,2),u)$ would be a cycle in $B_{f}(H)$.

Suppose now that $H$ is not a block graph. Then there exists a block in $H$, say $B$, which is not a complete graph. Thus in $B$ there exists a cycle such that not all its chords belong to $B$. Let $C=(v_{0},v_{1},\ldots,v_{l},v_{0})$ be a shortest such cycle in $B$. Then $l\geq 3$ and we distinguish two cases. If $C$ is chordless, then, by Lemma 1, $B_{f}(H)$ contains a cycle. Thus assume that $C$ has a chord. We may assume that $v_{0}$ is an end-vertex of a chord of $C$, and then let $k$ be the smallest integer such that $v_{0}v_{k}$ is a chord of $C$. Now the choice of $C$ implies that the vertices $v_{0},v_{1},\ldots,v_{k}$ are mutually adjacent, and therefore, $k=2$. Similarly, $v_{0},v_{k},\ldots,v_{l}$ are mutually adjacent, and so we must have $l=3$. Consequently, $C=(v_{0},v_{1},v_{2},v_{3},v_{0})$ and $v_{0}v_{2}$ is the only chord of $C$. Now it is obvious that there are at least two $v_{0}-v_{2}$ positively $f$-valued complete paths in $H$. From this and from Lemma 1 it follows that the bipartition $B_{f}(H)$ contains a cycle. This contradiction completes the proof of the statement (2).

Let $B$ be a block of $H$. We have already proved that $B$ is a complete graph. Let $B^{\prime}$ be a proper non-trivial complete subgraph of $B$. To prove (3), it suffices to observe that $f(B^{\prime})=0$. On the contrary, suppose that $f(B^{\prime})\not=0$. We now choose two distinct vertices $v$ and $u$ belonging to $B^{\prime}$, and a vertex $w$ belonging to $B$ but not to $B^{\prime}$. This clearly forces that there are at least two $v-u$ positively $f$-valued complete paths in $H$. Consequently, by Lemma 1, $B_{f}(H)$ contains a cycle, and this contradiction completes the proof of the statement (3).

Assume now that the conditions (1)–(3) are satisfied for $H$ and $f$. Since end-vertices of $B_{f}(H)$, corresponding to positively $f$-valued one-vertex complete subgraphs of $H$, are not important to our study of tree-like structure of $B_{f}(H)$, we can assume without loss of generality that $f(H[v])=0$ for every vertex $v\in V_{H}$. Consequently, $H$ is a block graph and $f(K)=1$ for every block $K$ of $H$, while $f(K^{\prime})=0$ for every other complete subgraph $K^{\prime}$ of $H$. It remains to prove that $B_{f}(H)$ is a tree. Since $B_{f}(H)$ is a connected graph, it suffices to show that $|E_{B_{f}(H)}|=|V_{B_{f}(H)}|-1$. Let ${\cal S}$ be the set of all blocks of $H$. Then ${\cal K}_{H}^{f}={\cal S}$, $|V_{B_{f}(H)}|=|V_{H}|+\sum_{K\in{\cal K}_{H}^{f}}\!f(K)=|V_{H}|+|{\cal S}|$, and $|E_{B_{f}(H)}|=\sum_{K\in{\cal K}_{H}^{f}}\!f(K)|V_{K}|=\sum_{K\in{\cal S}}|V_{K}|=\sum_{K\in{\cal S}}(|V_{K}|-1)+|{\cal S}|$. Now, since $\sum_{K\in{\cal S}}(|V_{K}|-1)=|V_{H}|-1$ (by Lemma 2), we finally have $|E_{B_{f}(H)}|=(|V_{H}|-1)+|{\cal S}|=(|V_{H}|+|{\cal S}|)-1=|V_{B_{f}(H)}|-1$. ∎

Corollary 1**.**

For every connected graph $H$, there exists a function $f\colon{\cal K}_{H}\to\mathbb{N}$ such that the bipartization $B_{f}(H)$ is a tree.

Proof.

Let $F$ be a spanning block graph of $H$ and let $f\colon{\cal K}_{F}\rightarrow\{0,1\}$ be a function such that $f(K)=1$ if and only if $K$ is a block of $F$. Clearly, $f$ satisfies the conditions (1)–(3) of Theorem 2, and so the bipartization $B_{f}(H)$ is a tree. ∎

Example 3**.**

Fig. 2 shows the tree $G$ which is the bipartization of two block graphs $H$ and $F$ with respect to functions $\overline{f}$ and $\overline{g}$, respectively, which non-zero values are listed in the same figure.

4 Graphs belonging to the family ${\cal B}$

In this section, we provide an alternative characterization of all bipartite graphs whose domination number is equal to the cardinality of its smaller partite set, that is, we prove that a graph $G$ belongs to the class ${\cal B}$ if and only if $G$ is some bipartization of a graph. For that purpose, we need the following lemma.

Lemma 3**.**

[4]* Let $G=((A,B),E_{G})$ be a connected bipartite graph with $1\leq|A|\leq|B|$. Then the following statements are equivalent:*

• $(1)$

$\gamma(G)=|A|$.

• $(2)$

$\gamma(G)=\beta(G)=|A|$.

• $(3)$

$G$* has the following two properties:*

• (a)

Each support vertex of $G$ belonging to $B$ is a weak support and each of its non-leaf neighbors is a support.

• (b)

If $x$ and $y$ are vertices belonging to $A-(L_{G}\cup S_{G})$ and $d_{G}(x,y)=2$, then there are at least two vertices $\overline{x}$ and $\overline{y}$ in $B$ such that $N_{G}(\overline{x})=N_{G}(\overline{y})=\{x,y\}$.

We are ready to establish our main theorem that provides an alternative characterization of the graphs belonging to ${\cal B}$ in terms of the bipartization of a graph.

Theorem 3**.**

Let $G=((A,B),E_{G})$ be a connected bipartite graph with $1\leq|A|\leq|B|$. Then $\gamma(G)=|A|$ if and only if $G$ is the bipartization $B_{f}(H)$ of a connected graph $H$ with respect to a non-zero function $f\colon{\cal K}_{H}\to\mathbb{N}$ and $f$ has the following two properties:

• $(1)$

If $uv\in E_{H}$ and $f(H[u,v])=0$, then $f(H^{\prime})>0$ for some complete subgraph $H^{\prime}$ of $H$ containing the edge $uv$.

• $(2)$

If $uv\in E_{H}$ and $f(H[u])=f(H[v])=0$, then $f(H[u,v])\geq 2$.

Proof.

Assume first that $\gamma(G)=|A|$. Then $G$ has the properties (3a) and (3b) of Lemma 3. Let $H=(V_{H},E_{H})$ be a graph in which $V_{H}=A$ and $E_{H}=\{xy\colon x,y\in A\textrm{ and }d_{G}(x,y)=2\}$, and let $f\colon{\cal K}_{H}\to\mathbb{N}$ be a function such that $f(K)=|\{x\in B\colon N_{G}(x)=V_{K}\}|$ for each $K\in{\cal K}_{H}$. Then $G$ is the bipartization $B_{f}(H)$ of $H$ with respect to $f$, as we have shown in the proof of Theorem 1. It is obvious that if $H=K_{1}$, then ${\cal K}_{H}=\{H\}$ and it must be $f(H)\geq 1$ (as otherwise $G=B_{f}(H)$ would be a graph of order one). Thus assume that $H$ is non-trivial. Now it remains to prove that $f$ has the properties (1) and (2).

Let $uv$ be an edge of $H$ such that $f(H[u,v])=0$. Suppose on the contrary that $f(H^{\prime})=0$ for every complete subgraph $H^{\prime}$ containing the edge $uv$. Then the vertices $u$ and $v$ do not share a neighbor in $B_{f}(H)=G$, so $d_{G}(u,v)>2$ and $uv$ is not an edge in $H$, a contradiction. This proves the property (1).

Now let $uv$ be an edge of $H$ such that $f(H[u])=f(H[v])=0$. From these assumptions it follows that $d_{G}(u,v)=2$ and neither $u$ nor $v$ is a support vertex in $G=B_{f}(H)$. Now we shall prove that none of the vertices $u$ and $v$ is a leaf in $G$. First, because $u,v\in A$ and they have a common neighbor, it follows from the first part of the property (3a) of Lemma 3 that at least one of the vertices $u$ and $v$ is not a leaf in $G$. Suppose now that exactly one of the vertices $u$ and $v$ is a leaf in $G$, say $u$ is a leaf. Then it follows from the second part of the property (3a) of Lemma 3 that $v$ is a support vertex in $G=B_{f}(H)$ and, therefore, $f(H[v])>0$, a contradiction. Consequently, both $u$ and $v$ are elements of $A-N_{G}[L_{G}]$. Thus, since $d_{G}(u,v)=2$, the property (3b) of Lemma 3 implies that there are at least two vertices $\bar{u},\bar{v}\in B$ such that $N_{G}(\bar{u})=N_{G}(\bar{u})=\{u,v\}$. Therefore $f(H[u,v])=|\{x\in B\colon N_{G}(x)=\{u,v\}\}|\geq|\{\bar{u},\bar{v}\}|=2$ and this proves the property (2).

Assume now that $H$ is a connected graph, and $f\colon{\cal K}_{H}\to\mathbb{N}$ is a non-zero function having the properties (1) and (2). We shall prove that in the bipartization $B_{f}(H)=((A,B),E_{B_{f}(H)})$, where $A=V_{H}$ and $B=\bigcup_{K\in{\cal K}_{H}}{\cal F}_{K}$, is $|A|\leq|B|$ and $\gamma(B_{f}(H))=|A|$. This is obvious if $H$ is a graph of order 1. Thus assume that $H$ is a graph of order at least 2. From the property (1) it follows that $B_{f}(H)$ is a connected graph. We first prove the inequality $|A|\leq|B|$. To prove this, it suffices to show that $B_{f}(H)$ has an $A$-saturating matching. We begin by dividing $A=V_{H}$ into two subsets $V_{H}^{1}=\{v\in V_{H}\colon f(H[v])\geq 1\}$ and $V_{H}^{0}=\{v\in V_{H}\colon f(H[v])=0\}$. It is obvious that the edge-set $M^{1}=\{v(H[v],1)\colon v\in V_{H}^{1}\}$ is a $V_{H}^{1}$-saturating matching in $B_{f}(H)$. Next, we order the set $V_{H}^{0}$ in an arbitrary way, say $V_{H}^{0}=\{v_{1},\ldots,v_{n}\}$. Now, depending on this order, we consecutively choose edges $e_{1},\ldots,e_{n}$ in such a way that $M^{1}\cup\{e_{1},\ldots,e_{i}\}$ is a $(V_{H}^{1}\cup\{v_{1},\ldots,v_{i}\})$-saturating matching in $B_{f}(H)$.

Assume that we have already chosen a $(V_{H}^{1}\cup\{v_{1},\ldots,v_{i-1}\})$-saturating matching $M^{1}\cup\{e_{1},\ldots,e_{i-1}\}$ in $B_{f}(H)$, and consider the next vertex $v_{i}\in V_{H}^{0}$. If $N_{H}(v_{i})\cap V_{H}^{0}\not=\emptyset$, say $v_{j}\in N_{H}(v_{i})\cap V_{H}^{0}$, then $f(H[v_{j}])=0$ and therefore $f(H[v_{i},v_{j}])\geq 2$ (by the property (2)) and the edge $e_{i}=v_{i}(H[v_{i},v_{j}],1)$ if $j>i$ ($e_{i}=v_{i}(H[v_{i},v_{j}],2)$ if $j<i$) together with $M^{1}\cup\{e_{1},\ldots,e_{i-1}\}$ form a $(V_{H}^{1}\cup\{v_{1},\ldots,v_{i}\})$-saturating matching in $B_{f}(H)$. Thus assume that $N_{H}(v_{i})\subseteq V_{H}^{1}$. Let $v$ be a neighbor of $v_{i}$ in $H$. If $f(H[v_{i},v])\geq 1$, then the edge $e_{i}=v_{i}(H[v_{i},v],1)$ has the desired property. Finally, if $f(H[v_{i},v])=0$, then $f(H^{\prime})>0$ for some complete subgraph $H^{\prime}$ of $H$ containing the edge $v_{i}v$ (by the property (1)) and in this case the edge $e_{i}=v_{i}(H^{\prime},1)$ has the desired property (as $N_{H}(v_{i})\subseteq V_{H}^{1}$). Repeating this procedure as many times as needed, an $A$-saturating matching in $B_{f}(H)$ can be obtained.

To complete the proof, it remains to show that $\gamma(B_{f}(H))=|A|$. In a standard way, suppose to the contrary that $\gamma(B_{f}(H))<|A|$. Let $D$ be a minimum dominating set of $B_{f}(H)$ with $|D\cap A|$ as large as possible. Since $\gamma(B_{f}(H))=|D|$, the inequality $\gamma(B_{f}(H))<~{}|A|$ implies that $|A-D|>|D\cap B|\geq 1$. In addition, since $|D\cap A|$ is as large as possible, the set $V_{H}^{1}$ $(=\{v\in V_{H}\colon f(H[v])\geq 1\})$ is a subset of $D\cap A$, while $A-D$ is a subset of $V_{H}^{0}$ $(=\{v\in V_{H}\colon f(H[v])=0\})$. Now, because $|A-D|>|D\cap B|$ and each vertex of $A-D$ has a neighbor in $D\cap B$, the pigeonhole principle implies that there are two vertices $x$ and $y$ in $A-D$ which are adjacent to the same vertex in $D\cap B$. Hence, $x$ and $y$ are adjacent in $H$ (by the definition of $B_{f}(H)$). Now, since $f(H[x])=f(H[y])=0$, the property (2) implies that $f(H[x,y])\geq 2$. Next, since $N_{B_{f}(H)}((H[x,y],1))=N_{B_{f}(H)}((H[x,y],2))=\{x,y\}$ and $\{x,y\}\cap D=\emptyset$, the vertices $(H[x,y],1)$ and $(H[x,y],2)$ belong to $D\cap B$. Consequently, it is easy to observe that the set $D^{\prime}=(D-\{(H[x,y],1),(H[x,y],2)\})\cup\{x,y\}$ is a dominating set of $B_{f}(H)$, which is impossible as $|D^{\prime}|=|D|$ and $|D^{\prime}\cap A|>|D\cap A|$. This completes the proof. ∎

Example 4**.**

The graph $H$ and the function $f\colon{\cal K}_{H}\to\mathbb{N}$ given in Example 1 have the properties (1) and (2) of Theorem 3 and therefore the bipartization $B_{f}(H)$ belongs to the family ${\cal B}$, that is, $\gamma(B_{f}(H))=|A|$, where $A$ is the smaller of two partite sets of $B_{f}(H)$ shown in Fig. 1.

The graph $F$ and the function $\overline{g}$ given in Fig. 2 do not satisfy the condition (2) of Theorem 3. However, the bipartization $G=B_{\overline{g}}(F)$ is a graph belonging to the family ${\cal B}$ since $G$ is also the bipartization $B_{\overline{f}}(H)$, with $H$ and $\overline{f}$ given in Fig. 2 and possessing properties (1) and (2) of Theorem 3.

It is obvious that the complete bipartite graph $K_{m,n}$ is the bipartization of the complete graph $K_{m}$ (resp. $K_{n}$) with respect to the function $f\colon{\cal K}_{K_{m}}\to\{0,n\}$, where $f(K)=0$ if and only if $K\in{\cal K}_{K_{m}}-\{K_{m}\}$ (resp. $g\colon{\cal K}_{K_{n}}\to\{0,m\}$, where $g(K)=0$ if and only if $K\in{\cal K}_{K_{n}}-\{K_{n}\}$). It is also evident that if $\min\{m,n\}\geq 3$, then $K_{m,n}$ does not belong to the family ${\cal B}$ (as $\gamma(K_{m,n})=2<\min\{m,n\}$), and neither $K_{m}$ and $f$ nor $K_{n}$ and $g$ possess the property (2) of Theorem 3.

Finally, as an immediate consequence of Theorems 2 and 3 we have the following simple characterization of trees in which the domination number is equal to the size of a smaller of its partite sets. All such trees are bipartizations of block graphs.

Corollary 2**.**

Let $T=((A,B),E_{T})$ be a tree in which $1\leq|A|\leq|B|$. Then $\gamma(T)=|A|$ if and only if $T$ is the bipartization $B_{f}(H)$ of a block graph $H$ with respect to a non-zero function $f\colon{\cal K}_{H}\to\mathbb{N}$ and $f$ has the following two properties:

• $(1)$

$f(K)=1$* if $K$ is a block of $H$, and $f(K^{\prime})=0$ if $K^{\prime}$ is a non-trivial complete subgraph of $H$ which is not a block of $H$.*

• $(2)$

$\max\{f(H[u]),f(H[v])\}\geq 1$* for every edge $uv$ of $H$ *$($or, equivalently, the set $\{v\in V_{H}\colon f(H[v])\geq 1\}$ is a covering set of H$$).