create an oriented cycle along the outer edges and due to Lemma 2.2.1 the complement of this graph does not have a bilateral AT-free ordering.
On the other hand, one can use Lemma 2.2.11 to construct the altered knotting graph K∗[G] which is given in Figure 2.8b. This graph does not contain an odd cycle, implying that Gis strong asteroid free.
(a) Complement of a graph which is strong as-teroid free, but not bilateral AT-free.
(b) The according altered knotting graph where the grey vertices are only depicted in one of the slices of the cycle.
This resolves Question 2.1.6, by showing that there are both strong asteroid free graphs that are not bilateral AT-free (Figure 2.8a) and bilateral AT-free graphs are not strong asteroid free (Figure 2.1b).
Corollary 2.2.14. The classes of bilateral AT-free graphs and strong asteroid free graphs are not comparable.
to use three sweeps of BFS to compute a bilateral AT-free order, also proving that claw-free AT-free graphs are bilateral AT-free.
As an overview of the proof, we will show that the first application of BFS ends in an admissible vertex. Using this fact, we can then proceed to show that a BFS starting in this admissible vertex yields an AT-free order. Finally, using this order as a tie-breaker in a third run of BFS outputs a bilateral AT-free order.
Lemma 2.3.1. Let G be claw-free and AT-free. Then the last vertex of a BFS is ad-missible.
Proof. Let s be the first and z the last vertex of the BFS and let k := distG(s, z).
Suppose there are a, b ∈ V such that z ∈ I[a, b]. As G is AT-free, at least one of a or b must be in the last layer LkG(s) of the BFS, without loss of generality this is a. If distG(s, b) < distG(s, z), then Nsk−1(a) ⊆ Nsk−1(z), as otherwise there is a z-avoiding a-b-path. If distG(s, b) = distG(s, a) = distG(s, z), then either Nsk−1(a) ⊆ Nsk−1(z) or Nsk−1(b)⊆Nsk−1(z), asG is AT-free, and without loss of generality we can assume this to be true fora. Therefore, aand z have a common neighbour cin Lk−1G (s). If cis not the start vertex of the BFS, thenchas a neighbour dinLk−2G and a, z, c, d form a claw.
Ifcis the start vertex, then bmust also be adjacent toc and a, b, c, dform a claw.
For the second step in this proof, we need a technical lemma which characterises the distance layers given by a BFS starting in an admissible vertex.
Lemma 2.3.2. Let G be a claw-free, AT-free graph and let s ∈ V be admissible in G and t eccentric with respect to s. Then all but the first distance layers of s, i.e., L0G(s), L2G(s), . . . , LkG(s), with k = eccG(s), are cliques and s and t form a dominating pair.
Proof. For L0G(s) this is obvious. Let i ≥ 2 and suppose there are a, b ∈ LiG(s) with ab /∈E. As s is admissible, without loss of generality Nsi−1(a) ⊆Nsi−1(b). Therefore, a andb have a common neighbourc∈Li−1G . Thisc in turn has a neighbourd∈Li−2G and a, b, c, dform a claw, which is a contradiction to the assumption.
As any path P between sand thas one vertex from each distance layer LiG(s) and s is adjacent to all vertices in L1G(s), they must form a dominating pair.
The following statement is a simple application of Lemma 2.3.2.
Lemma 2.3.3. Let G be an AT-free, claw-free graph. Then a BFS starting in an admissible vertex yields an AT-free order that is a monotone dominating pair order.
Proof. Let σ be such a BFS on G starting in an admissible vertex s. Suppose z ∈ Idom(a, b) and a, b≺σ z. We can assume that a,b and z do not have the same distance tos(otherwise we can construct a claw as above). AsGis AT-free, on the other hand, at least one ofaorbmust be in the same layer asz. Without loss of generality we can assume that b and z are in the same layer LiG(s) and a is in layer LjG(s) with j < i.
As b and z are independent of each other, they must be in the first layer of the BFS.
As a cannot be the start vertex (it is not adjacent to the other two); a contradiction.
Lemma 2.3.2 states thatσ must be a monotone dominating pair order.
Given a dominating pair, we will need to compare the distances of any two no-adjacent vertices to this dominating pair.
Lemma 2.3.4. Let G= (V, E)be a connected graph with a dominating pairsandt. Let u and v be two vertices with uv /∈E and distG(s, u) < distG(s, v). Then distG(t, u) ≥ distG(t, v).
Proof. Assandtform a dominating pair, we see that for every vertexx∈V it holds that distG(s, x) + distG(x, t)≤distG(s, t) + 2. Suppose distG(u, t)<distG(v, t). This implies distG(s, t) + 2≤distG(s, u) + distG(u, t) + 2≤distG(s, v) + distG(v, t)≤distG(s, t) + 2.
Therefore, distG(s, u) + distG(u, t) = distG(s, t) and there must be a shortests-t-path P that containsu. As P is dominating anduv /∈E; this is a contradiction to distG(s, u)<
distG(s, v) and distG(t, u)<distG(t, v).
Using all of the above lemmas, we can finally show the main theorem of this section.
Theorem 2.3.5. Let G be a claw-free AT-free graph. Then G has a bilateral AT-free ordering that is a monotone dominating pair order and this order can be found in linear time.
Proof. Let y be an admissible vertex of G and σ1 := BF S(G, y) as well as σ2 = BF S+(G, σ1). We have already shown that s:= σ1(n) =σ2(1) is admissible and that both σ1 and σ2 are AT-free orders that are monotone dominating pair orders. Thus, suppose there is a z ∈ Idom(a, b) such that z ≺σ2 a, b. Again we can assume that 1 = distG(s, a) = distG(s, z)<distG(s, b).
Case 1 (distG(y, b)>distG(y, z)): We also know that distG(s, b)>distG(s, z); this is a contradiction to the statement of Lemma 2.3.4.
Case 2 (distG(y, b)<distG(y, z)): Then b≺σ1 z which in turn implies z ≺σ1 a; this is a contradiction, as it would imply that a≺σ2 z, becausea, z∈L1G(s).
Case 3 (distG(y, b) = distG(y, z)): As before, we can assume that 1 = distG(y, b) = distG(y, z)<distG(y, a) and thus z≺σ1 a; again a contradiction.
Corollary 2.3.6. Any claw-free AT-free graph is bilateral AT-free.
2.3.2 AT-free Graphs without Bad Claws
In the proof of Lemma 2.3.3 we can see that the main obstacles are triples of vertices a, b, z ∈ V(G) with z ∈ Idom(a, b) that form the prongs of a claw. Therefore, it is not necessary to forbid all induced claws in the graph to avoid these difficulties. Rather, it suffices to forbid just those claws which have this property with regard to the intervals.
This justifies the following definition:
Definition 2.3.7. Let G be a graph and leta, b, z, c∈V induce a claw with base c. We will call such a claw a bad claw, if z∈Idom(a, b).
1
2 3
4
5 6
7 a0 a
b0 b z0 z
BFS: σ1: (1,2,3, a0,4,5, a,6,7, z0, b0, b, z) BFS(σ1): σ2: (z,7,6,3, b, a,1,2,5,4, a0, b0, z0) BFS(σ2): σ3: (z0,4,5,2, b0, a0,1,3, a,6,7, b, z)
Figure 2.9: A bad-claw-free graph for which BFS does not yield an AT-free order
It seems reasonable to expect that by forbidding such bad claws we will be able to get similar results to the ones above. On the other hand, there are examples of AT-free bad-claw-free graphs for which the above procedure does not yield either an AT-free order nor a bilateral AT-free ordering (see Figure 2.9). In particular, Lemma 2.3.1 does not hold in general for these graphs. Therefore, we will use LBFS which guarantees us an admissible vertex as its end-vertex.
Lemma 2.3.8 (Corneil et al. [40]). Let G= (V, E) be an AT-free graph and let σ be an ordering of V produced by an LBFS. Then the vertex t:=σ(n) is admissible in G.
In fact, the properties of LBFS even make up for the absence of the strong struc-tural property of Lemma 2.3.2 and we can prove analogues to both Lemma 2.3.3 and Theorem 2.3.5.
Lemma 2.3.9. Let G be AT-free and bad-claw-free. Then an LBFS starting in an admissible vertex yields an AT-free order that is a monotone dominating pair order.
Proof. Letσbe an LBFS order starting in an admissible vertexs. Supposez∈Idom(a, b) and a, b ≺σ z. Without loss of generality, we see that i:= distG(s, b) = distG(s, z), as Gis AT-free. For that same reason eitherNsi−1(b)⊆Nsi−1(z) orNsi−1(a) ⊆Nsi−1(z) or both.
Now suppose distG(s, a) =i. Assis admissible, anda,b andzare independent, they must have a common neighbour cwith distG(s, c) =i−1 and therefore a,b and z and cform a bad claw; a contradiction.
Therefore, we can assume that j := distG(s, a) < i. With the above we see that Nsi−1(b) ⊆Nsi−1(z) and there is a b-avoiding a-z-path P. Let x be the σ-last vertex of P. As b≺σ zσ x, due to Theorem 1.5.8 the vertexbmust see everys-x-path and thus also everyx-a-path; a contradiction. Thus, every LBFS starting in an admissible vertex yields an AT-free order.
Finally, Theorem 1.5.8 states that every LBFS order of an AT-free graph starting in an admissible vertex is a monotone dominating pair order.
Analogously to claw-free AT-free graphs, a final third sweep of LBFS yields the desired bilateral AT-ordering.
Theorem 2.3.10. LetGbe an AT-free graph that does not have a bad claw as an induced subgraph. Then G has a bilateral AT-free ordering that is a monotone dominating pair order and such an order can be found in linear time.
1 2
3 4
a
b c z
LBFS: σ1: (1,2,4, z,3, b, a, c) LBFS(σ1): σ2: (c, a, b, z,4,3,2,1) LBFS(σ2): σ3: (1,2,3,4, z, b, a, c)
Figure 2.10: Example of a graph with a bad claw. On the right, one can see that the secondσ2 is not an AT-free order andσ3is not a bilateral AT-free order. In fact, this is an example of an AT-free graph that does not possess a bilateral AT-free ordering.
Proof. Let σ1 be an order given by LBFS starting in an arbitrary vertex of G. Let σ2:=LBF S+(G, σ1) withy:=σ1(n) =σ2(1) and let σ3 :=LBF S+(G, σ2).
We have already shown that y and s:=σ2(n) = σ3(1) are admissible and that both σ2 and σ3 are AT-free orders.
Suppose σ3 is not a bilateral AT-free ordering. Then there is a z ∈ Idom(a, b) with z ≺σ3 a, b, as σ3 is AT-free. Let a, z and b be such that z is σ2-leftmost with this property. Without loss of generality we can assume thataandzhave the same distance tos, due to sbeing admissible. Also,a, zand b cannot all have the same distance tos, as this would imply a bad claw. Thus, distG(s, z) = distG(s, a)<distG(s, b).
Case 1 (distG(z, y)<distG(b, y)): We also know that distG(s, b)>distG(s, z) and that y andsform a dominating pair; this is a contradiction to the statement of Lemma 2.3.4.
Case 2 (dist(b, y)<distG(z, y)): As σ2 is AT-free, we can deduce that b ≺σ2 z ≺σ2 a.
Let z0 be the σ3-first vertex in the connected component C of Γσz,a3. Due to the Prior Path Theorem there is a z0-z path P in C such that P avoids a. As z ≺σ3 a we can assume that a ≺σ2 z0. If z0b /∈ E we see that z0 ≺σ3 a ≺σ3 b and z0 ∈ Idom(a, b); a contradiction to the choice of z. Therefore, let z0b ∈E. With distG(b, y) <distG(a, y) this implies that l := distG(z0, y) ≤ distG(a, y). Because z0b ∈ E and y is admissible Nyl−1(a)⊂Nyl−1(z0); a contradiction to a≺σ2 z0.
Case 3 (distG(z, y) = distG(b, y)): AsGis bad-claw-free, distG(a, y)6= distG(b, y). Sup-pose distG(a, y)<distG(b, y). As distG(s, a)<distG(s, b) and yand sare a dominating pair; a contradiction to Lemma 2.3.4. Suppose distG(a, y) >distG(b, y). Then z≺σ2 a and with the argument from Case 2, which uses the Prior Path Theorem, we receive a contradiction.
Corollary 2.3.11. Let G = (V, E) be an AT-free graph that does not contain a bad claw. Then G is bilateral AT-free.
These results indicate that a linear time algorithm to construct AT-free orders could
also exist for the general case of AT-free graphs. However, none of the techniques used for the (bad-)claw-free graphs can be transferred. Corneil and Stacho [38] already showed that there are AT-free graphs which do not possess AT-free orders that are also LBFS orders. In addition, Figure 2.10 shows a graph which does not possess a bilateral AT-free ordering. Therefore, it will be necessary to use a different search algorithm, possibly a BFS-derivative based on BFSconv. We summarise these suppositions in the following conjecture.
Conjecture 2.3.12. Let G= (V, E) be an AT-free graph. There is a linear time algo-rithm that computes an AT-free (BFS) order.