Asteroidal Sets and Dominating Paths

Abstract

An independent set of three vertices is called an asteroidal triple (AT) if there exists a path between any two of them that avoids the neighborhood of the third. Asteroidal triple-free (AT-free) graphs are very well-studied, but some of their various superclasses are not. We study two of these superclasses: hereditary dominating pair (HDP) graphs and diametral path graphs. We correct a mistake that has appeared in the literature claiming that the class of diametral path graphs are a superclass of HDP. More specifically, we show that a graph with a dominating shortest path does not necessarily contain a dominating diametral path. We say a graph is a strict dominating pair graph if it contains a dominating pair but has no dominating diametral path, and we show structural and algorithmic properties of these graphs. To study properties of HDP graphs, we introduce the notion of spread in asteroidal triples. Given a dominating pair, we show that all paths between this pair meet the common neighborhood of some pair from each asteroidal triple. We use these results to improve the best known time complexity for the recognition of chordal HDP graphs.

5 Appendix

In this appendix, we prove Theorem 1. To simplify the proofs, we introduce the notion of corner vertices, vertices that bear witness to the fact that a diametral path is not dominating but a given pair (a, b) is a dominating pair.

Definition 6

If Q is a non-dominating diametral path in a graph G and (a, b) is a dominating pair, then vertex c is a corner w.r.t. (Q, a) if c is not dominated by Q, an endpoint of Q is adjacent to a, and $c \in N(a)$.

Note that corners are adjacent to elements of the dominating pair, rather than to intermediate vertices on dominating pair paths. Next, we prove that corner vertices are inevitable in strict dominating pair graphs.

Lemma 7

Let G have a dominating pair (a, b) and let $k = d_G(a,b)$, but no dominating diametral path (and thus in particular ${{\,\textrm{diam}\,}}(G) = k + 1$). Let $(d_1,d_2)$ be a diametral pair s.t. $d_1 \in N(a)$ and $d_2 \in N(b)$. There exists a diametral $d_1,d_2$-path Q that contains b and a corner vertex c w.r.t. (Q, a).

Proof

By Corollary 1 we have $d_G(d_1,d_2) = k + 1$. We claim that $d_G(d_1,b) = k$. By Corollary 1, $d_G(d_1,b) \le k + 1$. If $d_G(d_1,b) < k$ then $d_G(d_1,d_2) < k + 1$ because $d_2 \sim b$, a contradiction. Thus, $d_G(d_1,b) = k$. Similarly, $d_G(a,d_2) = k$. Let M be a shortest $d_1,b$-path. By assumption, diametral path $M \mathord {-} d_2$ is not dominating while dominating pair path $a \mathord {-} M$ is dominating. Since $( a \mathord {-} M ) \setminus ( M \mathord {-} d_2 ) = \{a\}$, there exists $c \in N(a)$ s.t. $c \not \sim (M \mathord {-} d_2)$. Thus, we can set $Q = M \mathord {-} d_2$ and we are done (Fig. 7).    $\square$

Fig. 7.

On the left we depict a single corner vertex c. On the right, Lemma 7 is applied to both dominating pair vertices a and b. The thick path is a shortest a, b-path of length k. The double-lined path is Q within the proof. Dotted edges may not exist

The remaining proofs will utilize the existence of corner vertices to resolve various properties of strict dominating pair graphs.

Lemma 8

Under the hypotheses of Lemma 7, let $P_1,P_2$ be diametral $d_1,d_2$-paths containing b and a, respectively. Let $c_1$ and $c_2$ be corner vertices w.r.t. $(P_1,a)$ and $(P_2,b)$, respectively. If $c_1 \sim c_2$ then ${{\,\textrm{diam}\,}}(G) \in \{3,4\}$.

Proof

It is easy to check that $k \ge 2$. Notice that there exists a path $R = \langle a,c_1,c_2,b \rangle$ of length 3. If R is induced, then $k = 3$ and thus ${{\,\textrm{diam}\,}}(G) = k + 1 = 4$. If R is not induced, then $k < 3$.    $\square$

Next, we consider the effect that corner vertices have on the asteroidal number of the graph. To prepare for the theorem, we resolve general consequences of having a corner vertex that does not belong to a diametral pair.

Lemma 9

Under the hypotheses of Lemma 7, let c be a corner vertex w.r.t. (Q, a) where Q is a diametral $d_1,d_2$-path that contains b. The following hold:

• 1. If R is a shortest c, b-path and $d_G(c,b) < k$, then $N[d_1] \cap R = \emptyset$.

• 2. If $G \setminus N[d_1]$ disconnects c from b, then $(c,d_2)$ is a diametral pair. Moreover, if $R'$ is any shortest $c,d_2$-path in G, then $d_1$ is adjacent to the first vertex on $R'$ following c.

Proof

Assume for the sake of contradiction that $v \in N[d_1] \cap R$. Since $v \ne c$ the walk $d_1 \mathord {-} R[v,b] \mathord {-} d_2$ has length strictly less than $k + 1$, a contradiction.

For the second half, let $R'$ be a shortest $c,d_2$-path and suppose that $d_G(c,d_2) \le k$. Since $R' \mathord {-} b$ is a c, b-walk, there exists w on $R' \mathord {-} b$ such that $w \in N[d_1]$. Note that $w \ne c$ by definition and $w \ne b$ since $k \ge 2$. The path $R'' = d_1 \mathord {-} R'[w,d_2]$ has length strictly less than $k + 1$, a contradiction. Thus $(c,d_2)$ is diametral and $R'$ has length $k + 1$. Note that the same argument shows that w cannot occur later than the first vertex on $R'$ following c.    $\square$

We can see that Lemma 9 is symmetric with respect to a and b for a given dominating pair (a, b).

To simplify the following proof, we define new relationship notation. Let (a, b) be a dominating pair and let $(d_1,d_2)$ be a diametral pair such that $d_1 \in N(a)$ and $d_2 \in N(b)$. Let c be a corner vertex with respect to (P, a) where P is a diametral $d_1,d_2$-path that contains b. In that case, we say that $d_1 \prec ^P c$ (Fig. 8).

Lemma 10

Let G satisfy the hypotheses of Lemma 7 and let ${{\,\textrm{diam}\,}}(G) > 4$. There exist $c,c' \in N(a)$ with a shortest c, b-path P and a shortest $c',b$-path $P'$ s.t. $N[c] \cap P' = N[c'] \cap P = \emptyset$.

Proof

By Corollary 1, we let $(d_1,d_2)$ be a diametral pair s.t. $d_1 \in N(a)$ and $d_2 \in N(b)$. We let $c_0 = d_1$ and $P_0$ be a diametral $d_1,d_2$-path that contains b. We will construct a sequence of distinct corner vertices $c_1,c_2, \dots$ in N(a), and diametral paths $P_1,P_2, \dots$ such that $P_i$ is a $c_i,d_2$-path and $c_{i+1}$ is a corner vertex w.r.t. $(P_i,a)$. Moreover, our sequence will satisfy the condition that for each $c_i$, set $N[c_i]$ meets every shortest $c_{i+1},b$-path.

To be precise, given $c_0 \prec ^{P_0} c_1 \prec ^{P_1} \dots \prec ^{P_{i-1}} c_i \prec ^{P_i}$ we let $c_{i+1}$ be a corner vertex w.r.t. $(P_i,a)$, as promised by Lemma 7, distinct from $c_0, \dots , c_i$. If there is no corner vertex distinct from the earlier ones, the process terminates at the path $P_i$.

The step above alternates with the one we describe now, that of finding a path to add to $c_0 \prec ^{P_0} c_1 \prec ^{P_1} \dots \prec ^{P_{i-1}} c_i \prec ^{P_i} c_{i+1}$. At this point there are two ways at which we might be done. If $d_G(c_{i+1},b) < k$ then by the definition of a corner vertex and Lemma 9 we have $N[c_{i+1}] \cap P_{i} = \emptyset$ and $N[c_{i}] \cap P_{i+1} = \emptyset$, so we are done, letting $c = c_{i+1}$, $c' = c_{i}$, $P' = P_{i} \setminus \{d_2\}$, and $P = P_{i+1} \setminus \{d_2\}$. Also, if $N[c_{i}]$ does not meet every shortest $c_{i+1},b$-path, then there exists a $c_{i+1},b$-path R of length k in $G \setminus N[c_i]$. Now $c = c_{i+1}$, $c' = c_{i}$, $P' = P_{i} \setminus \{d_2\}$, and $P = R$ satisfy the conclusion of the lemma.

On the other hand if $d_G(c_{i+1},b) = k$ and $N[c_i]$ does meet every shortest $c_i,b$-path, then by Lemma 7 there exists a diametral $c_{i+1},d_2$-path $P_{i+1}$ that contains b which we add to the end of our sequence.

If in our construction we never succeeded in producing the required $c,c',P,P'$ then the construction must have terminated because we could not find a corner vertex $c_{f+1}$ distinct from all the earlier $c_i$. Thus we have constructed

$$ c_0 \prec ^{P_0} c_1 \prec ^{P_1} c_2 \prec ^{P_2} \cdots \prec ^{P_{f-1}} c_f \prec ^{P_{f}} . $$

By Lemma 7 there is a corner vertex $c_{f+1}$ w.r.t. $(P_f,a)$ and so, by assumption, there exists h in the set $\{0,1, \dots , f-1\}$ such that $c_{f+1} = c_h$. Suppose that $h = f - 1$. Then we have $(c_{f-1} = c_h) \sim P_f$, a contradiction. Otherwise, we have $h < f - 1$. We will prove that this leads to a contradiction.

Let $P_h' = c_h \mathord {-} (P_{h+1} \setminus \{c_{h+1}\})$ and let $s,s'$ be the first vertices in $P_h, P_h'$ following $c_h$. We will prove by reverse-induction that for all $h + 1 < j \le f$ that $c_j \sim s$ and $c_j \sim s'$. This holds for $j = f$ since $c_{f+1} = c_h$ and by construction $N[c_{f}]$ meets every shortest $c_{f+1},b$-path.

Now suppose that $h+1 < j < f$. By the inductive hypothesis $c_{j+1} \mathord {-} (P_h\setminus \{c_h\})$ is a diametral $c_{j+1}, d_2$-path, and thus, since $N[c_j]$ meets every diametral $c_{j+1}, d_2$-path at its second vertex, $c_j \sim s$. Similarly, since $c_{j+1} \mathord {-} (P_h' \setminus \{c_h\})$ is diametral, $c_j \sim s'$ (Fig. 8)

Fig. 8.

Depicting Lemma 10 where $c_3 \prec ^{P_3} c_0$, so $h = 0$. The thick path is $P_h \setminus \{d_2\}$ and the double-lined path is $P_{h'} \setminus \{d_2\}$

Finally, when $j = h + 2$ we reach a contradiction. We must have $c_{h+2} \not \sim s'$ since $c_{h+1}$ is a corner w.r.t. $(P_{h+1},a)$ and $s' \in P_{h+1}$. This contradiction establishes that the construction must have terminated through one of the conditions that give us appropriate $c,c',P,P'$.    $\square$

Lemma 10 is also symmetric with respect to either vertex in a given dominating pair. Applying the lemma twice, we find there are four vertices such that the removal of the closed neighborhood of any one of them does not disconnect the remaining three. These four vertices form an asteroidal set of size 4. Thus, we complete the proof of Theorem 1.