Recognizing $\mathbf {W_2}$ Graphs

Abstract

Let G be a graph. A set $S \subseteq V(G)$ is independent if its elements are pairwise nonadjacent. A vertex $v \in V(G)$ is shedding if for every independent set $S \subseteq V(G) {\setminus } N[v]$ there exists $u \in N(v)$ such that $S \cup \{u\}$ is independent. An independent set S is maximal if it is not contained in another independent set. An independent set S is maximum if the size of every independent set of G is not bigger than |S|. The size of a maximum independent set of G is denoted $\alpha (G)$. A graph G is well-covered if all its maximal independent sets are maximum, i.e. the size of every maximal independent set is $\alpha (G)$. The graph G belongs to class $\mathbf {W_2}$ if every two disjoint independent sets in G are included in two disjoint maximum independent sets. If a graph belongs to the class $\mathbf {W_2}$, then it is well-covered. Finding a maximum independent set in an input graph is a well-known NP-hard problem. Recognizing well-covered graphs is co-NP-complete. Recently, it was proved that deciding whether an input graph belongs to the class $\mathbf {W_2}$ is co-NP-complete. However, when the input is restricted to well-covered graphs, the complexity status of recognizing graphs in $\mathbf {W_2}$ is still not known. In this article, we investigate the connection between shedding vertices and $\mathbf {W_2}$ graphs. On the one hand, we prove that recognizing shedding vertices is co-NP-complete. On the other hand, we find polynomial solutions for restricted cases of the problem. We also supply polynomial characterizations of several families of $\mathbf {W_2}$ graphs.

1 Introduction

1.1 The Classes $\mathbf {W_k}$

An independent set of vertices in a graph G is a set of vertices $S \subseteq V(G)$ whose elements are pairwise nonadjacent. An independent set is maximal if is is not a subset of another independent set. An independent set is maximum if G does not contain an independent set of a higher cardinality. The cardinality of a maximum independent set in G is denoted $\alpha (G)$. A graph G is well-covered if all its maximal independent sets are maximum, i.e., the size of every maximal independent set is $\alpha (G)$.

Problem 1.1

WC

Input: A graph G.

Question: Is G well-covered?

Finding a maximum independent set in an input graph is known to be an NP-complete problem. However, finding a maximal independent set in an input graph can be done polynomially using the greedy algorithm. If the input is restricted to well-covered graphs then the greedy algorithm for finding a maximal independent set always yields a maximum independent set. Hence, finding a maximum independent set in a well-covered graph is a polynomial task. Recognizing well-covered graphs is known to be co-NP-complete [2, 15].

Let k be a positive integer. A graph G belongs to class $\mathbf {W_k}$ if every k pairwise disjoint independent sets in G are included in k pairwise disjoint maximum independent sets [16]. It holds that $\mathbf {W_1} \supseteq \mathbf {W_2} \supseteq \mathbf {W_3} \supseteq \ldots$, where $\mathbf {W_1}$ is the family of all well-covered graphs. The W2 problem and the WCW2 problem are defined as follows.

Problem 1.2

W2

Input: A graph G.

Question: Is $G \in \mathbf {W_2}$ ?

Recently, the complexity status of Problem 1.2 was settled in the following.

Theorem 1.3

[5] For each $k \ge 2$, recognizing $\mathbf {W_k}$ graphs is co-NP-complete.

Problem 1.4

WCW2

Input: A well-covered graph G.

Question: Is $G \in \mathbf {W_2}$ ?

The complexity status of Problem 1.4 is still open.

In [3] graphs with two disjoint maximum independent sets are studied.

Theorem 1.5

[3] Let G be a graph. The following assertions are equivalent:

• 1. G admits two disjoint maximum independent sets.

• 2. There exists a matching M of size $\alpha (G)$ such that G[V(M)] is a bipartite graph.

• 3. There exists a set $A \subseteq V(G)$ such that $G \setminus A$ is a bipartite graph having a perfect matching of size $\alpha (G)$.

A vertex $v \in V(G)$ is shedding if for every independent set $S \subseteq V(G) {\setminus } N[v]$ there exists $u \in N(v)$ such that $S \cup \{u\}$ is independent. Equivalently, v is shedding if there does not exist an independent set in $V(G) \setminus N[v]$ which dominates N(v) [18]. The SHED problem is defined as follows.

Problem 1.6

SHED

Input: A graph G and a vertex $v \in V(G)$.

Question: Is v shedding ?

Let Shed(G) denote the set of all shedding vertices of G.

Theorem 1.7

[9, 10] For every graph G having no isolated vertices, the following assertions are equivalent:

• 1. G is in the class $\mathbf {W_2}$.

• 2. $G \not \approx P_{3}$ and $G \setminus v$ is well-covered, for every $v \in V(G)$.

Theorem 1.8

[9, 10] For every well-covered graph G having no isolated vertices, the following assertions are equivalent:

• 1. G is in the class $\mathbf {W_2}$.

• 2. $G \setminus N[v]$ is in the class $\mathbf {W_2}$, for every $v \in V(G)$.

• 3. $Shed(G) = V(G)$.

Recently, it was shown in [5] that the assumption that the graph is well-covered is necessary for Theorem 1.8. Namely, [5] provides a graph in which all vertices are shedding, but the graph is not well-corered, and therefore it does not belong to the class $\mathbf {W_2}$.

This article finds polynomial solutions for several restricted cases of Problem 1.2 and Problem 1.4. We also prove that Problem 1.6 is co-NP-complete.

1.2 Definitions and Notation

Throughout this article the following notation and definitions are used. Graphs are undirected, simple, and loopless. The number of vertices in a graph G is $n = |V(G)|$, and the number of edges is $m = |E(G)|$. If G contains cycles than the girth of G is the length of a shorests cycle in the graph. Otherwise, the girth of G is infinite.

Let $v \in V(G)$. The set of vertices adjacent to v in G is denoted $N_{G}(v)$. Define $N_{G}[v] = \{v\} \cup N_{G}(v)$. The degree of v in G is $d_{G}(v) = |N_{G}(v)|$. If G is the only graph mentioned in the context, then $N_{G}(v)$, $N_{G}[v]$ and $d_{G}(v)$ are abbreviated to N(v), N[v] and d(v), respectively. For every $S \subseteq V(G)$ denote $N[S] = \cup _{v \in S} N[v]$ and $N(S) = N[S] {\setminus } S$.

Let S and T be sets of vertices of G. Then S dominates T if $T \subseteq N[S]$. If S dominates V(G), then S is a dominating set. A set $S \subseteq V(G)$ is a clique if every two vertices of S are adjacent to each other. A vertex $v \in V(G)$ is simplicial if N[v] is a clique.

2 Recognizing Shedding Vertices

2.1 Co-NP-Completeness

Let ${{\mathcal {X}}}=\{x_{1},\ldots x_{n}\}$ be a set of 0-1 variables. We define the set of literals $L_{{\mathcal {X}}}$ over ${\mathcal {X}}$ by $L_{{\mathcal {X}}} = \{x_{i}, \overline{x_{i}}: i = 1,\ldots ,n\}$, where ${\overline{x}} = 1 - x$ is the negation of x. A truth assignment to ${\mathcal {X}}$ is a mapping $t:{{\mathcal {X}}} \longrightarrow \{0,1\}$ that assigns a value $t(x_{i}) \in \{0,1\}$ to each variable $x_{i} \in {\mathcal {X}}$. We extend t to $L_{{\mathcal {X}}}$ by putting $t(\overline{x_{i}}) = \overline{t(x_{i})}$. A literal $l \in L_{{\mathcal {X}}}$ is true under t if $t(l) = 1$. A clause over ${\mathcal {X}}$ is a conjunction of some literals of $L_{{\mathcal {X}}}$, such that for every variable $x \in {\mathcal {X}}$, the clause contains at most one literal out of x and its negation. Let $\mathcal{C}=\{c_{1},\ldots ,c_{m}\}$ be a set of clauses over ${\mathcal {X}}$. A truth assignment t to ${\mathcal {X}}$ satisfies a clause $c_{j} \in {\mathcal {C}}$ if $c_{j}$ contains at least one true literal under t. The SAT problem, defined as follows, is a well-known NP-complete problem [8].

Problem 2.1

SAT

Input: A set of variables ${{\mathcal {X}}}=\{x_{1}, \ldots ,x_{n}\}$, and a set of clauses ${{\mathcal {C}}}=\{c_{1}, \ldots ,c_{m}\}$ over ${\mathcal {X}}$.

Question: Is there a truth assignment to ${\mathcal {X}}$ which satisfies all clauses of ${\mathcal {C}}$?

Theorem 2.2

[8] The SAT problem is NP-complete.

The WCSHED problem, defined as follows, is closely related to the SHED problem.

Problem 2.3

WCSHED

Input: A well-covered graph G and a vertex $v \in V(G)$.

Question: Is v shedding ?

In the WCSHED problem, the input includes a graph G such that for every $x \in V(G)$ and for every maximal independent set S of $V(G) \setminus N[x]$ it holds that $\alpha (N(x) {\setminus } N(S)) \le 1$. The question is whether for a specific vertex $v \in V(G)$ it holds that $\alpha (N(v) {\setminus } N(S)) = 1$ for every maximal independent set S of $V(G) \setminus N[v]$.

The complexity status of the WCSHED problem is not known. However, the complexity status of the SHED problem is found by Theorem 2.4.

Theorem 2.4

The SHED problem is co-NP-complete, even if its input is restricted to graphs without cycles of length 3.

Proof

Let $I=(G, v)$ be an instance of the SHED problem. An independent set $S \subseteq N_{2}(v)$ which dominates N(v), if it exists, is a witness that v is not a shedding vertex, and I is negative. Therefore, the problem is co-NP. We prove co-NP-completeness by a supplying a polynomial reduction from the SAT problem to the complement of the SHED problem.

Let $I_{1} = ({{\mathcal {X}}}=\{x_{1},\ldots ,x_{n}\}, \mathcal{C}=\{c_{1},\ldots ,c_{m}\})$ be an instance of the SAT problem. Define a graph G as follows.

$$\begin{aligned}V(G) = \{v\} \cup \{v_{i}: 1 \le i \le m\} \cup U,\end{aligned}$$

where $U = \{ u_{i,l}: l \in c_{i}, 1 \le i \le m \}$.

$$\begin{aligned}E(G) = \{vv_{i}: 1 \le i \le m\} \cup \{v_{i}u_{i,l}: l \in c_{i}, 1 \le i \le m\} \cup E(U),\end{aligned}$$

where $E(U) = \{ u_{i,l}u_{j,{\overline{l}}}: l \in c_{i}, {\overline{l}} \in c_{j}, 1 \le i \le m, 1 \le j \le m \}$.

The subgraph induced by U is bipartite, since every edge connects instances of a literal and its negation. Moreover, v is not on a triangle since its neighbors are independent. A vertex $v_{i}$ is not on a triangle since the clause $c_{i}$ does not contain both a literal and its negation. Therefore, G does not contain cycles of length 3.

Let $I_{2} = (G, v)$ be an instance of the complement of the SHED problem. It remains to prove that $I_{1}$ and $I_{2}$ are eqiuvalent.

Suppose $I_{1}$ is positive. There exists a truth assignment, t, to ${\mathcal {X}}$ which satisfies all clauses of ${\mathcal {C}}$. Let $S = \{ u_{i,l}: l \in c_{i}, 1 \le i \le m, t(l) = 1 \}$. The fact that S does not contain both a literal and its negation implies that S is independent. The fact that t satisfies all clauses of $\mathcal C$ implies that S dominates $N(v) = \{v_{i}: 1 \le i \le m\}$. Therefore, v is not shedding, and $I_{2}$ is positive.

On the other hand, if $I_{2}$ is positive there exists an independent set $S \subseteq N_{2}(v)$ which dominates N(v). Let t be a truth assignment such that $t(l)=1$ for every literal l which has an instance $u_{i,l} \in S$ for some $1 \le i \le m$. This assignment is valid since S does not contain instances of both a literal and its negation. The fact that S dominates $N(v) = \{v_{i}: 1 \le i \le m\}$ implies that all clauses are satisfied by t. Therefore, $I_{1}$ is positive. $\square$

Example 2.5

$I_{1} = ({{\mathcal {X}}},{{\mathcal {C}}} )$ is an instance of the SAT problem, where ${{\mathcal {X}}} = \{x_{1}, x_{2}, x_{3}, x_{4}\}$,

$$\begin{aligned}{{\mathcal {C}}} = \{c_{1} = (x_{1}, \overline{x_{2}}, x_{3}), c_{2}=(x_{2},x_{4}), c_{3}=(\overline{x_{1}}, \overline{x_{3}}, \overline{x_{4}}), c_{4}=(x_{1}, \overline{x_{2}}, \overline{x_{3}}, \overline{x_{4}})\}\end{aligned}$$

$I_{2} = (G, v)$ is an equivalent instance of the complement of the SHED problem, where G is the graph shown in Fig. 1.

The satisfying truth assignment $t(x_{1}) = t(x_{2}) = 1$, $t(x_{3}) = t(x_{4})=0$ corresponds to the independent set $\{u_{1,x_{1}}, u_{2,x_{2}}, u_{3,\overline{x_{3}}},u_{3,\overline{x_{4}}}, u_{4,x_{1}}, u_{4, \overline{x_{3}}}, u_{4,\overline{x_{4}}}\} \subseteq N_{2}(v)$ which dominates N(v).

Fig. 1

The graph G

2.2 Non-Shedding Vertices and Relating Edges

Let G be a graph and $xy \in E(G)$. Then xy is relating if there exists an independent set $S \subseteq V(G) {\setminus } N[\{x,y\}]$ such that each of $S \cup \{x\}$ and $S \cup \{y\}$ is a maximal independent set of G. The RE problem, defined as follows, is known to be NP-complete.

Problem 2.6

RE

Input: A graph G and an edge $e \in E(G)$.

Question: Is e relating ?

Theorem 2.7

[1] The RE problem is NP-complete.

Non-shedding verices and relating edges are closely related notions. A witness for their existence is an independent set of vertices, which dominates all vertices of the graph except the non-shedding vertex or the endpoints of the relating edge. Moreover, one can prove that SHED problem is co-NP-complete by a polynomial reduction from the RE problem to the complement of the SHED problem. We do not present the reduction and the proof of its correctness because this result is a restricted case of Theorem 2.4, in which cycles of length 3 are forbidden. Also Theorem 2.8 shows the connection between non-shedding vertices and relating edges.

Theorem 2.8

Let G be a graph without cycles of lengths 4, 5 and 6, and $xy \in E(G)$. Suppose $N(x) \cap N(y) = \varnothing$, $d(x) \ge 2$ and $d(y) \ge 2$. The following assertions are equivalent:

• 1. None of x and y is a shedding vertex.

• 2. xy is a relating edge.

Proof

$1 \implies 2$ Since x is not shedding, there exists an independent set $S_{x} \subseteq N_{2}(x)$ which dominates N(x). Similarly, there exists an independent set $S_{y} \subseteq N_{2}(y)$ which dominates N(y). The fact that G does not contain cycles of lengths 4 implies that $S_{x} \cap N(y) = \varnothing$ and $S_{y} \cap N(x) = \varnothing$. Define $S = S_{x} \cup S_{y}$. Then $S \cap N(\{x,y\}) = \varnothing$. Assume, on the contrary, that S is not independent. Then there exists $x'' \in S_{x} \setminus N(y)$ and $y'' \in S_{y} \setminus N(x)$ such that $x''y'' \in E(G)$. Let $x' \in N(x'') \cap N(x)$ and $y' \in N(y'') \cap N(y)$. Then $(x'',x',x,y,y',y'')$ is a cycle of length 6, which is a contradiction. Therefore, S is independent. Let $S^{*}$ be a maximal independent set of $G \setminus N[\{x,y\}]$ which contains S. Then each of $S^{*} \cup \{x\}$ and $S^{*} \cup \{y\}$ is a maximal independent set of G. Therefore, xy is a relating edge.

$2 \implies 1$ There exists an independent set S such that each of $S \cup \{x\}$ and $S \cup \{y\}$ is a maximal independent set of G. The fact that $S \cup \{x\}$ is independent implies that $S \cap N(x) = \varnothing$. The fact that $S \cup \{y\}$ is dominating and $N(x) \cap N(y) = \varnothing$ implies that $N(x) {\setminus } \{y\}$ is dominated by $S \cap N_{2}(x)$. Similarly, $S \cap N(y) = \varnothing$ and $N(y) {\setminus } \{x\}$ is dominated by $S \cap N_{2}(y)$.

Since $d(y) \ge 2$, there exists $y' \in N(y) \setminus \{x\}$. The fact that G does not contain cycles of length 5 implies that $y'$ is not adjacent to $S \cap N_{2}(x)$. Therefore, $S_{x} = ( S \cap N_{2}(x) ) \cup \{y'\}$ is a witness that x is not shedding. Similarly, there exists $x' \in N(x) \setminus \{y\}$, and $S_{y} = ( S \cap N_{2}(y) ) \cup \{x'\}$ is a witness that y is not shedding. $\square$

We show that each of the conditions of Theorem 2.8 is necessary. If it does not hold then also the conclusion of Theorem 2.8 does not hold.

• If $N(x) \cap N(y) \ne \varnothing$, then the conclusion of Theorem 2.8 does not necessarily hold. For example, G is a union of a path $(x_{2},x_{1},x,y,y_{1},y_{2})$ and a triangle (x, z, y). In this case xy is a relating edge, while x and y are shedding.

• If y is a leaf, then the conclusion of Theorem 2.8 does not hold. For example, G is a path (a, b, x, y). In this case xy is a relating edge, while y is not shedding.

• If G contains a cycle of length 4, then the conclusion of Theorem 2.8 does not necessarily hold. For example, if G is a union of a path $(x_{2},x_{1},x,y,y_{1},y_{2})$ and a cycle $(x',x,y,y')$, then xy not relating while x and y are not shedding vertices.

• If G is a copy of $C_{5}$ and $xy \in E(G)$, then the conclusion of Theorem 2.8 does not hold. The edge xy is relating, while x and y are shedding vertices.

• If G contains a cycle of length 6, then the conclusion of Theorem 2.8 does not necessarily hold. For example, if G is a union of cycles $(x''_{1},x'_{1},x,y,y'_{1},y''_{1})$ and $(x''_{2},x'_{2},x,y,y'_{2},y''_{2})$, then xy is not relating while x and y are not shedding vertices.

3 Hereditary Families of Graphs

Let P be a property of graphs. Then P is a hereditary property if for every graph G which satisfies P all induced subgraphs of G satisfy P. Similarly, a hereditary family of graphs is a family F such that if $G \in F$ then all induced subgraphs of G belong to F. For example, claw-free graphs and bipartite graphs are hereditary families, while the set of connected graphs is not a hereditary family.

Theorem 3.1

Let F be a hereditary family of graphs, such that there exists an O(f(n))-time algorithm for deciding whether an input graph $G \in F$ is well-covered. Then deciding for an input graph $G \in F$ whether it belongs to $\mathbf {W_2}$ can be done in O(nf(n)) time.

Proof

If $G \approx P_{3}$ or G contains isolated vertices then obviously G is not in $\mathbf {W_2}$.

Suppose $G \not \approx P_{3}$ and G does not contain isolated vertices. By Theorem 1.7, G belongs to $\mathbf {W_2}$ if and only if for every vertex $v \in V(G)$ the graph $G \setminus v$ is well-covered. Hence, an algorithm which decides whether a graph belongs to $\mathbf {W_2}$ invokes n times the algorithm for recognizing well-covered graphs. $\square$

Theorem 3.2

Let F be a hereditary family of graphs. Assume that there exists an O(f(n))-time algorithm which receives as its input a graph $G \in F$ and a vertex $v \in V(G)$, and decides whether $v \in Shed(G)$. Then deciding for an input well-covered graph $G \in F$ whether it belongs to $\mathbf {W_2}$ can be done in O(nf(n)) time.

Proof

If G contains isolated vertices then obviously G is not in $\mathbf {W_2}$. Suppose G does not contain isolated vertices. By Theorem 1.8, G belongs to $\mathbf {W_2}$ if and only if $Shed(G) = V(G)$. Hence, an algorithm which decides whether G belongs to $\mathbf {W_2}$ invokes n times the algorithm for recognizing shedding vertices. $\square$

4 Claw-Free Graphs

This section considers the family of claw-free graphs. It supplies polynomial algorithms for recognizing shedding vertices and $\mathbf {W_2}$ graphs.

Theorem 4.1

There exists an $O(n^3)$ time algorithm which receives as its input a claw-free graph, G, and a vertex $v \in V(G)$, and decides whether v is shedding.

Proof

Define a weight function $w:N_{2}(v) \longrightarrow \mathbb {Z}$ by $w(x) = |N(x) \cap N(v)|$, for every $x \in N_{2}(v)$. Let S be an independent set of $N_{2}(v)$. We prove that distinct vertices of S dominate distinct vertices of N(v). Suppose on the contrary that $s_{1} \in S$ and $s_{2} \in S$ have a common neighbor $v' \in N(v)$. Then $G[\{v',v,s_{1},s_{2}\}]$ is a claw, which is a contradiction. Hence, $w(S) = \sum _{s \in S} w(s) = \sum _{s \in S} |N(s) \cap N(v)| = |N(S) \cap N(v)|$. If follows that the weight of every independent set of $N_{2}(v)$ can not exceed |N(v)|.

Invoke an algorithm for finding a maximum weight independent set in a claw-free graph. First such an algorithm is due to Minty [14], while the best known one with the complexity $O(n^3)$ may be found in [4]. The algorithm outputs a set $S^{*} \subseteq N_{2}(v)$. If $w(S^{*}) = |N(v)|$ then $S^{*}$ is an independent set of $N_{2}(v)$ which dominates N(v), and, therefore, v is not shedding. However, if $w(S^{*}) < |N(v)|$ then there does not exist an independent set of $N_{2}(v)$ which dominates N(v), and v is shedding. Deciding whether v is shedding can be done in $O(n^3)$ time. $\square$

Corollary 4.2

There exists an $O(n^4)$ time algorithm which receives a well-covered claw-free graph, and decides whether it belongs to $\mathbf {W_2}$.

Proof

Follows immediately from Theorem 4.1 and Theorem 3.2. $\square$

Deciding whether an input claw-free graph is well-covered can be done in $O(m^{\frac{3}{2}}n^{3})$ time [12]. Therefore, Theorem 3.1 implies that recognizing $\mathbf {W_2}$ claw-free graphs can be completed in $O(m^{\frac{3}{2}}n^{4})$ time. However, Theorem 4.3 supplies a more efficient algorithm which solves this problem.

Theorem 4.3

There exists an $O(m^{\frac{3}{2}}n^{3})$ time algorithm for recognizing $\mathbf {W_2}$ claw-free graphs.

Proof

Invoke the algorithm of [12] for deciding whether the input graph G is well-covered. If G is not well-covered then $G \not \in \mathbf {W_2}$. If G is well-covered, invoke the algorithm of Corollary 4.2 to decide whether G belongs to $\mathbf {W_2}$.

The complexity of recognizing $\mathbf {W_2}$ claw-free graphs is $O(m^{\frac{3}{2}}n^{3} + n^4) = O(m^{\frac{3}{2}}n^{3})$. $\square$

5 Graphs Without Cycles of Length 5

5.1 Shedding Vertices

Theorem 5.1

[13] Let G be a graph without cycles of length 5, and $v \in V(G)$. Then the following holds.

• 1. Every maximal independent set of $N_{2}(v)$ dominates $N(v) \cap N(N_{2}(v))$.

• 2. If $N_{2}(v)$ dominates N(v) then every maximal independent set of $N_{2}(v)$ dominates N(v).

Corollary 5.2

Let G be a graph without cycles of length 5. Let $v \in V(G)$. Then v is shedding if and only if $N_{2}(v)$ does not dominate N(v).

Corollary 5.3

The following problem can be solved in $O(n+m)$ time. Input: A graph G without cycles of length 5, and a vertex $v \in V(G)$. Question: Is v shedding ?

Proof

By Corollary 5.2, v is shedding if and only if $N_{2}(v)$ does not dominate N(v). Since one can decide in $O(n+m)$ time whether $N_{2}(v)$ dominates N(v), it is possible to decide in the same complexity whether v is a shedding vertex. $\square$

Corollary 5.4

The following problem can be solved in $O(n(n+m))$ time. Input: A well-covered graph G without cycles of length 5. Question: Is G in the class $\mathbf {W_2}$ ?

Proof

Follows immediately from Theorem 3.2 and Corollary 5.3. $\square$

5.2 Graphs Without Cycles of Lengths 3 and 5

Theorem 5.5

Let $G \in \mathbf {W_2}$ such that $G \not \approx K_{2}$. Let $v \in V(G)$. Then at least one of the following holds.

• 1. v is on a triangle $(v,v_{1},v_{2})$ such that $\{v_{1},v_{2}\}$ is not dominated by $N_{2}(v)$.

• 2. v is on a $C_{5}$.

Proof

The fact that $G \in \mathbf {W_2}$ and $G \not \approx K_{2}$ implies that G does not contain leaves. Two cases are considered.

case 1: $N_{2}(v)$ does not dominate N(v) There exists $v_{1} \in N(v)$ which is not dominated by $N_{2}(v)$. Since $v_{1}$ is not a leaf, there exists $v_{2} \in N(v) \cap N(v_{1})$. Condition 1 holds.

case 2: $N_{2}(v)$ dominates N(v). Let $N(v) = \{v_{1},\ldots ,v_{k}\}$. For every $1 \le i \le k$ let $v'_{i} \in N_{2}(v) \cap N(v_{i})$. Since v is a shedding vertex, $\{v'_{1},\ldots ,v'_{k}\}$ is not independent. Hence, $(v'_{i},v'_{j}) \in E(G)$ for some $1 \le i < j \le k$. Therefore, $(v,v_{i},v'_{i},v'_{j},v_{j})$ is a $C_{5}$ which contains v. Condition 2 holds. $\square$

Corollary 5.6

Let G be a graph without cycles of lengths 3 and 5. Then $G \in \mathbf {W_2}$ if and only if $G \approx K_{2}$.

5.3 Graphs Without Cycles of Lengths 4 and 5

In [7], the following definitions are used. A graph G belongs to the family F if there exists $\{x_{1},x_{2},\ldots ,x_{k}\} \subseteq V(G)$, where for each $1 \le i \le k$ it holds that $x_{i}$ is simplicial, $|N[x_{i}]| \le 3$, and $\{ N[x_{i}]: i = 1, 2, \ldots , k\}$ is a partition of V(G).

Theorem 5.7

[7] Let G be a connected well-covered graph containing neither $C_{4}$ nor $C_{5}$ as a subgraph. Then

• 1. G contains a shedding vertex or $G \approx K_{1}$ if and only if $G \in F$.

• 2. Otherwise, G is isomorphic to $C_{7}$ or $T_{10}$. (See Fig. 2.)

Fig. 2

The graph $T_{10}$

Theorem 5.8 is the main result of this section.

Theorem 5.8

Let G be a connected graph containing neither $C_{4}$ nor $C_{5}$ as a subgraph. The following conditions are equivalent.

• 1. $G \in \mathbf {W_2}$

• 2. $G \approx K_{2}$ or there exists a partition $\{T_{1}, T_{2}, \ldots , T_{k}\}$ of V(G), such that $G[T_{i}]$ is a triangle with at least 2 simplicial vertices, for each $1 \le i \le k$.

Proof

$1 \implies 2$ Suppose that $G \in \mathbf {W_2}$. By Theorem 1.8, $V(G) = Shed(G)$. Therefore, G is isomorphic neither to $K_{1}$ nor to $C_{7}$ nor to $T_{10}$. By Theorem 5.7, $G \in F$. Hence, there exists $\{x_{1},x_{2},\ldots ,x_{k}\} \subseteq V(G)$, where for each $1 \le i \le k$ it holds that $x_{i}$ is simplicial, $|N[x_{i}]| \le 3$, and $\{ N[x_{i}]: i = 1, 2, \ldots , k\}$ is a partition of V(G).

If $|N[x_{i}]| = 1$ then $x_{i}$ is an isolated vertex, which is a contradiction. If $|N[x_{i}]| = 2$ then $x_{i}$ is a leaf, and therefore $G \approx K_{2}$. In the remaining case $|N[x_{i}]| = 3$ for every $1 \le i \le k$. Let $y_{i}$ and $z_{i}$ be the neighbors of $x_{i}$, and assume on the contrary that both are not simplicial. There exists $y'_{i} \in N(y_{i}) {\setminus } \{x_{i}, z_{i}\}$ and $z'_{i} \in N(z_{i}) {\setminus } \{x_{i}, y_{i}\}$. if $y'_{i} = z'_{i}$ then $(y'_{i}, y_{i}, x_{i}, z_{i})$ is a $C_{4}$, which is a contradiction. If $y'_{i} \in N(z'_{i}),$ then $(z'_{i}, y'_{i}, y_{i}, x_{i}, z_{i})$ is a $C_{5}$, which is a contradiction, again. Therefore, $y'_{i}$ and $z'_{i}$ are distinct and non-adjacent.

Define $A = \{x_{i}\}$ and $B = \{y'_{i},z'_{i}\}$. Clearly, A and B are independent and disjoint. Therefore, they are contained in two disjoint maximal independent sets, $A^{*}$ and $B^{*}$, respectively. However, $x_{i} \not \in B^{*}$ since $x_{i} \in A^{*}$. Moreover, $y_{i}$ and $z_{i}$, are not in $B^{*}$, because they are dominated by $\{y'_{i}, z'_{i}\} \subseteq B^{*}$. Therefore, $x_{i} \not \in N[B^{*}]$, which is a contradiction. At least one of $y_{i}$ and $z_{i}$ is simplicial.

$2 \implies 1$ Suppose that Condition 2 holds and $G \not \approx K_{2}$. Then $|S \cap T_{i}| \le 1$ for every independent set $S \subseteq V(G)$ and for every $1 \le i \le k$. Moreover, every dominating independent set contains exactly k vertices. If A and B are disjoint independent sets of G, then there exist $A \subseteq A^{*}$ and $B \subseteq B^{*}$ such that $A^{*}$ and $B^{*}$ are disjoint maximal independent sets of G. Therefore, $G \in \mathbf {W_2}$. $\square$

6 Graphs with Girth at Least 5

In [6] well-covered graphs with girth at least 5 are characterized. The following notation and definitions are used. A leaf is a vertex of degree 1. A steam is a vertex adjacent to a leaf. An edge connecting a leaf and a steam is called pendant. A 5-cycle C is called basic if C does not contain two adjacent vertices of degree 3 or more. A graph G belongs to the family ${\textbf{P}}{\textbf{C}}$ if V(G) can be partitioned into two subsets, P and C, such that the following holds.

• P contains the vertices adjacent to the pendant edges.

• The pendant edges are a perfect matching of P.

• C contains the vertices of the basic 5-cycles.

• The basic 5-cycles are a partition of C.

Theorem 6.1

[6] Let G be a graph with girth 5 or more. Then the following conditions are equivalent.

• 1. G is well-covered and contains a shedding vertex.

• 2. G belongs to the family ${\textbf{P}}{\textbf{C}}$.

Theorem 6.2 is the main result of this section.

Theorem 6.2

The only graphs with girth 5 or more in the class $\mathbf {W_2}$ are $C_{5}$ and $K_{2}$.

Proof

Let G be a graph with girth 5 or more in the class $\mathbf {W_2}$. Obviously, G is well-covered. By Theorem 1.8, $Shed(G) = V(G)$. By Theorem 6.1G belongs to the family ${\textbf{P}}{\textbf{C}}$. The only graph in $\mathbf {W_2}$ which contains leaves is $K_{2}$. Suppose that G does not contain leaves and belongs to $\mathbf {W_2}$. Then $P=\varnothing$.

Let $C=(v_{1},\ldots ,v_{5})$ be a basic cycle in G, and assume on the contrary that $d(v_{1}) \ge 3$. Then $d(v_{2})=d(v_{5})=2$. Let $v_{1}' \in N(v_{1}) \setminus \{v_{2},v_{5}\}$. Define $A=\{v_{1}'\}$ and $B=\{v_{2},v_{5}\}$. Since A and B are disjoint and independent, they are contained in two disjoint maximal independent sets, $A^{*}$ and $B^{*}$, of G. Since $v_{2}$ and $v_{5}$ belong to $B^{*}$, they do not belong to $A^{*}$. Hence, $v_{2}$ and $v_{5}$ are neighbors of vertices belonging to $A^{*}$. However, $v_{1} \not \in A^{*}$. Hence, $\{v_{3}, v_{4}\} \subseteq A^{*}$. Consequently, $A^{*}$ is not independent, which is a contradiction.

Therefore, the degree of every vertex of C is 2, and $G \approx C_{5}$. $\square$

Theorem 6.3

There exists an O(n) algorithm which solves the following problem.

Input: A connected well-covered graph G with girth 5 or more and a vertex $v \in V(G)$.

Question: Is v is shedding.

Proof

Let $I=(G,v)$ be an instance of the problem. By Theorem 6.1, if G does not belong to the family ${\textbf{P}}{\textbf{C}}$ then v is not shedding. Suppose G belongs to the family ${\textbf{P}}{\textbf{C}}$. Four cases are possible.

Case 1: v is a leaf. Obviously, v is not shedding.

Case 2: v is a neighbor of a leaf. Obvously, v is shedding.

Case 3: v is on a basic cycle, C, and $d(v)=2$. Let $N(v) = \{v_{1}, v_{2}\}$. The vertex v is shedding if and only if $N(\{v_{1},v_{2}\}) \setminus V(C) \ne \varnothing$.

Case 4: v is on a basic cycle, C, and $d(v) \ge 3$. There exist two vertices, $v_{1}$ and $v_{2}$, in C which are adjacent to v. It holds that $d(v_{1}) = d(v_{2}) = 2$. There does not exist an independent set of $N_{2}(v)$ which dominates $\{v_{1},v_{2}\}$. Therefore, v is shedding.

In [6] a finite list of all connected well-covered graphs with girth at least 5 and without shedding vertices is presented. Hence, deciding whether G belongs to the family ${\textbf{P}}{\textbf{C}}$ can be completed in O(1) time. Deciding which case holds for v can be done in O(n) time. If Case 3 holds then O(n) time is needed to decide whether the condition $N(\{v_{1},v_{2}\}) {\setminus } V(C) \ne \varnothing$ holds. Otherwise, deciding whether v is shedding can be done in O(1) time. The total complexity of the algorithm is O(n). $\square$

7 Well-Covered Graphs Without Cycles of Lengths 4 and 6

Theorem 7.1

The following problem can be solved in $O(n(n+m))$ time. Input: A graph G without cycles of lengths 4 and 6, and a vertex $v \in V(G)$. Question: Is v shedding ?

Proof

Every vertex of $N_{2}(v)$ is adjacent to exactly one vertex of N(v), or otherwise G contains a $C_{4}$. Every component of $N_{2}(v)$ contains at most 2 vertices, or otherwise G contains either a $C_{4}$ or a $C_{6}$. Let $A_{1},\ldots ,A_{k}$ be the components of $N_{2}(v)$.

The following algorithm decides whether v is shedding. It constructs a flow network $F_{v} = \{ G_{F} = (V_{F}, E_{F}), s \in V_{F}, t \in V_{F}, c:E_{F}\longrightarrow {\mathbb {R}} \}$. A similar construction was used in [11].

Define $V_{F} = N(v) \cup N_{2}(v) \cup \{a_{1},\ldots ,a_{k}, s, t\}$, where $a_{1},\ldots ,a_{k}, s, t$ are new vertices, s and t are the source and sink of the network, respectively. Denote $A = \{a_{1}, \ldots , a_{k}\}$.

The directed edges $E_{F}$ are:

• the directed edges from s to each vertex of N(v).

• all directed edges $v_{1}v_{2}$ s.t. $v_{1}\in N(v)$, $v_{2} \in N_{2}(v)$ and $v_{1}v_{2}\in E$.

• the directed edges $za_{i}$, for each $1 \le i \le k$ and for each $z \in A_{i}$.

• the directed edges $a_{i}t$, for each $1\le i\le k$.

Let $c \equiv 1$. Invoke Ford and Fulkerson’s algorithm for finding a maximum flow $f:E_{F}\longrightarrow {\mathbb {R}}$ in the network. The flow in a vertex $x \in V_{F}$ is defined by: $\Sigma _{(u,x) \in E_{F}} f(u,x)$. Let $S_{v}$ be the set of vertices in $N_{2}(v)$ in which there is a positive flow. One can prove that $S_{v}$ is a maximum independent set of $N_{2}(v)$, Ford and Fulkerson’s algorithm terminates after $|S_{v}|$ iterations, and $|f| = |S_{v}| = |N(v) \cap N(S_{v})|$. For more details see [11].

If $|S_{v}| = |N(v)|$ then $S_{v}$ dominates N(v), and, therefore, v is not shedding. However, if $|S_{v}| < |N(v)|$ then $S_{v}$ does not dominate N(v), and there does not exist an independent set of $N_{2}(v)$ which dominates N(v). In this case the algorithm terminates announcing that v is shedding.

Every iteration of Ford and Fulkerson’s algorithm takes $O(n+m)$ time. In the restricted case of this context, the algorithm of Ford and Fulkerson terminates after O(n) iterations. Hence, total complexity of this algorithm is $O(n(n+m))$. $\square$

Corollary 7.2

The following problem can be solved in $O(n^{2}(n+m))$ time. Input: A well-covered graph G without cycles of lengths 4 and 6. Question: Is G in the class $\mathbf {W_2}$ ?

Proof

Follows immediately from Theorems 3.2 and 7.1. $\square$

Note that the complexity status of recognizing well-covered graphs without cycles of lengths 4 and 6 is not known. If recognizing well-covered graphs without cycles of lengths 4 and 6 can be done polynomially then recognizing graphs in $\mathbf {W_2}$ without cycles of lengths 4 and 6 is a polynomial task, too.

8 $\alpha (G) = k$

This section considers graphs with $\alpha (G) = k$. It supplies polynomial algorithms for recognizing well-covered graphs, shedding vertices and $\mathbf {W_2}$ graphs.

Theorem 8.1

Let $k \ge 2$. The following problem can be solved in $O(n^{k-1})$ time. Input: A graph G with $\alpha (G) = k$. Question: Is G well-covered?

Proof

The following algorithm solves the problem. For every set $S \subseteq V(G)$ such that $|S| < k$, decide whether S is independent and dominates V(G). Once a dominating independent set is found, the algorithm terminates announcing that G is not well-covered. However, if all sets of size smaller than k were checked, and none of them is dominating and independent then G is well-covered.

The number of sets which are checked is $O(n^{k-1})$. For each such a set, deciding whether it is independent and dominating takes $O(k^{2})$ time, which is constant for every fixed k. Therefore, deciding whether G is well-covered can be done in $O(n^{k-1})$ time. $\square$

Theorem 8.2

Let $k \ge 2$. The following problem can be solved in $O(n^{k})$ time. Input: A graph G with $\alpha (G) = k$ and a vertex $v \in V(G)$. Question: Is v a shedding vertex?

Proof

Since $\alpha (G) = k$, an independent set of $N_{2}(v)$ contains at most $k-1$ vertices.

The following algorithm solves the problem. For every set $S \subseteq N_{2}(v)$ such that $|S| \le k-1$, decide whether S is independent and dominates N(v). Once an independent set of $N_{2}(v)$ which dominates N(v) was found, the algorithm terminates announcing that v is not shedding. If all possible sets were checked, and none of them is independent and dominates N(v) then v is shedding.

The number of sets which are checked is $O(n^{k-1})$. For each such a set, deciding whether it is independent and dominates N(v) takes O(n) time. Therefore, deciding whether v is shedding can be done in $O(n^{k})$ time. $\square$

It follows from Theorem 3.1 and Theorem 8.1 that recognizing $\mathbf {W_2}$ graphs with $\alpha (G) = k$ can be done in $O(n^{k})$ time. Theorem 8.3 finds another algorithm which solves this problem with the same complexity.

Theorem 8.3

Let $k \ge 2$. The following problem can be solved in $O(n^{k})$ time. Input: A graph G with $\alpha (G) = k$. Question: Is $G \in \mathbf {W_2}$ ?

Proof

If G contains isolated vertices then obviously $G \not \in \mathbf {W_2}$. Otherwise, check all sets of vertices with size smaller than k. For each such a set, S, if S is independent and dominates V(G), then G is not well-covered, and therefore it is not in $\mathbf {W_2}$. If S is independent and $N[S] = V(G) {\setminus } \{v\}$ for some $v \in V(G)$, then v is not shedding, and therefore $G \not \in \mathbf {W_2}$. However, if $|V(G) {\setminus } N[S]| \ge 2$ for every independent set S with size smaller than k, then G is well-covered and all vertices are shedding. By Theorem 1.8, G belongs to $\mathbf {W_2}$.

This algorithm considers $O(n^{k-1})$ sets. Each set is checked in O(n) time. Hence, the algorithm terminates in $O(n^{k})$ time. $\square$

9 Conclusions

Let $\mathcal {G}(\widehat{C_{i_{1}}},\ldots ,\widehat{C_{i_{k}}})$ be the family of all graphs which do not contain $C_{i_{1}}$,...,$C_{i_{k}}$. The following table presents complexity results concerning the five major problems presented in this paper. The empty table cells correspond to unsolved cases.

Input	WC	W2	WCW2	SHED	WCSHED
general	co-NPC [2, 15]	co-NPC [5]		co-NPC Th. 2.4
claw-free	P [17]	P Th. 4.3	P Cor. 4.2	P Th. 4.1	P Th. 4.1
$\mathcal {G}(\widehat{C_{3}})$				co-NPC Th. 2.4
$\mathcal {G}(\widehat{C_{3}},\widehat{C_{4}})$	P [6]	P Th. 6.2	P Th. 6.2		P Th. 6.3
$\mathcal {G}(\widehat{C_{5}})$			P Cor. 5.4	P Cor. 5.3	P Cor. 5.3
$\mathcal {G}(\widehat{C_{3}},\widehat{C_{5}})$		P Cor. 5.6	P Cor. 5.6	P Cor. 5.3	P Cor. 5.3
$\mathcal {G}(\widehat{C_{4}},\widehat{C_{5}})$	P [7]	P Th. 5.8	P Th. 5.8	P Cor. 5.3	P Cor. 5.3
$\mathcal {G}(\widehat{C_{4}},\widehat{C_{6}})$			P Cor. 7.2	P Th. 7.1	P Th. 7.1

10 Addendum

A vertex $v \in V(G)$ is shedding if and only if for every independent set $S \subseteq N_{2}(v)$ it holds that $N(v) {\setminus } N(S) \ne \varnothing$. A restricted case of a shedding vertex is a vertex v such that for every maximal independent set S of $V(G) {\setminus } N[v]$ it holds that $\alpha (N(v) {\setminus } N(S)) = 1$. Assume that G is in the class $\mathbf {W_2}$, $v \in V(G)$, and S is a maximal independent set of $V(G) \setminus N[v]$. Then $H = G[V(G) {\setminus } N[S]] = G[N[v] {\setminus } N(S)]$ is well-covered. However, $\{v\}$ is a maximal independent set of H. Hence, H is a clique. Therefore, Theorem 1.8 is an instance of Theorem 10.1.

Theorem 10.1

For every well-covered graph G the following assertions are equivalent:

• 1. G is in the class $\mathbf {W_2}$.

• 2. $\alpha (N(v) \setminus N(S)) = 1$ for every vertex $v \in V(G)$ and for every maximal independent set S of $V(G) \setminus N[v]$.

In the arXive version of this paper we have conjectured that Theorem 10.1 holds for general graphs.

Conjecture 10.2

For every graph G the following assertions are equivalent:

• 1. G is in the class $\mathbf {W_2}$.

• 2. $\alpha (N(v) \setminus N(S)) = 1$ for every vertex $v \in V(G)$ and for every maximal independent set S of $V(G) \setminus N[v]$.

Recently, Conjecture 10.2 was refuted in [5], providing a graph where all its vertices are shedding, while the graph is not well-corered.