4.3 Cleaning an interval graph
4.3.1 Some structural observations
A nonempty set M ⊆ V(G) is a module of G, if for every x ∈ V(G)\M, NG(x) either includes M or is disjoint fromM. A moduleM in Gis complete, ifG[M] is connected and there is no vertex inx∈NG(M) such thatNG(x)⊆NG[M]. Lemma 4.3.2 gives a characteri-zation of the complete modules of an interval graph. For an illustration, see Figure 4.1. Note that{a1}and{a2, a3}are modules ofGthat are not complete. The sets{a1, a2, a3},{b1, b2} and{c1, c2, c3, c4, d1, d2, d3, d4, e1, e2} are examples of complete module characterized by (a) of Lemma 4.3.2, and the set{e1, e2}illustrates the complete modules characterized by (b) of Lemma 4.3.2.
Lemma 4.3.2. Given an interval graph G and a labeled PQ-tree T representing G, some set M ⊆ V(G) is a complete module of G, if and only if one of the following statements holds:
(a)M =R−1(Tz)for somez∈V(T), and ifz is a P-node then R−1(z)6=∅
(b)M =Lq(a, b)for some Q-nodeq∈V(T)having children x1, . . . , xmand some pair (a, b)
with a < b, such that R−1(Txi) = ∅ for each i contained in [a, b], and Lq(a0, b0) = ∅ for each[a0, b0]properly contained in [a, b].
Proof. First, letM be a complete module inG. Let us choose a vertexv∈M such thatR(v) is the closest possible to the root of T. Since G[M] is connected, v is unique, and we also get R−1(TR(v)) ⊇ M. First, suppose that R(v) is a P-node or a leaf. Then v is contained in each clique of F(TR(v)). Thus, if R(x) is in TR(v) for some vertex x, then NG(x) ⊆ NG(v) ⊆ NG[M]. By the completeness of M, we get x ∈ M. Hence, R−1(TR(v)) ⊆ M implyingR−1(TR(v)) =M. Therefore, (a) holds in this case.
Now, suppose thatR(v) is a Q-nodeqwith childrenx1, . . . , xm, and letMq =M∩R−1(q).
Leta= min{Qleftq (w)|w∈Mq} andb= max{Qrightq (w)|w∈Mq}Using the completeness ofM, we can argue again thatR−1(Txh)⊆M for eachhcontained in [a, b] and thatw∈M holds for each w ∈ R−1(q) such that Qq(w) is contained in [a, b]. Thus, if [a, b] = [1, m]
thenM =R−1(Tq), implying that (a) holds. Otherwise, asqis a Q-node, there must exist a vertexu∈R−1(q)\M such thatQq(u) properly intersects [a, b]. Asumust be adjacent to each vertex ofM (asM is a module), we get that R−1(Txh) =∅for everyhin [a, b] that is not contained inQq(u). In particular, we get that eitherR−1(Txa) =∅orR−1(Txb) =∅. We can assume w.l.o.g. thatR−1(Txa) =∅ holds. Thus,Mq−(a)6=∅, and sinceMq−(a)∩M 6=∅, using again that M is a module, we obtain that each w ∈ Mq must start in a and also thatR−1(Txh) =∅for everyhin [a, b]. Note that this impliesMq =M. Now, fromR−1(Txb) =
∅ we get in a similar way that each w ∈ M must end in b, proving Qq(w) = [a, b] for everyv, w∈M. Now, using the completeness ofM and putting together these facts, we get that the conditions of (b) must hold.
For the other direction, it is easy to see that if (a) holds for someM, thenM indeed must be a complete module ofG. Second, ifM =Lq(a, b) for someqand [a, b], thenM is clearly a module, and the remaining conditions of (b) ensure thatM is complete.
We will say that a complete module M is simple, if the conditions in (b) hold for M. Clearly,NG(M) is a clique if and only ifM is not simple, and ifM is simple thenG[M] is a clique. In Figure 4.1,{e1, e2} is a simple complete module.
For a graph H, some set M ⊆V(G) is an occurrence of H in G as a complete module, ifM is a complete module for whichG[M] is isomorphic toH. LetM(H, G) be the set of the occurrences ofH inG as a complete module. Using that each element ofM(H, G) is a subset ofV(G) having size|V(H)|, we obtain the following consequence of Lemma 4.3.2.
Proposition 4.3.3. For a graph H, the elements M(H, G) are pairwise disjoint.
Moreover, if the graph H is not a clique, then none of the occurrences of H in G as a complete module can be simple, so each set in M(H, G) must be of the form R−1(Tz) for some non-leaf nodezofT. This yields that the sets inM(H, G) are independent (where two vertex sets in a graph are independent if there is no edge between them). Lemma 4.3.4 below states some observations about what happens to a set of disjoint and independent complete modules in a graph after adding or deleting a vertex.
Lemma 4.3.4. Suppose that s∈V(G).
(1) IfM1, . . . , M`are disjoint independent complete modules inG−s, thenMi is a complete module inGfor at least `−4 indices i∈[`].
(2) If M1, . . . , M` are disjoint independent complete modules in G, then Mi is a complete module inG−sfor at least `−4indices i∈[`].
Proof. As Mi andMj are independent ifi6=j, we can assume thatM1left≤M1right <· · ·<
M`left≤M`right. Recall that eachMiis connected by the definition of a complete module. We
4.3. Cleaning an interval graph 53
s1 s2
s3
s4 s5
s6
s7
Mi
Mi−1 Mi+1
Figure 4.4: Mi−1, Mi, and Mi+1 illustrate complete modules of G−S. The set Mi is untouched bys1,s2, and s3, but this is not true for any vertex sj,j≥4.
say thatMi isuntouched (bys), if eithersleft≤Miright−1 andsright≥Mi+1left, orsright< Miright−1 , orsleft> Mi+1left. (See also Figure 4.4.) IfMaandMbare the first and the last one, respectively, among the setsM1, . . . , M` that have a vertex adjacent tos, then eachMi except forMa−1, Ma,Mb, andMb+1 must be untouched bys.
To see (1), we show that if a complete module ofG−sis untouched, then it is a complete module ofG. So assume that Mi is untouched. Clearly, s /∈ Mi. Since eitherNG(s) ⊇Mi
orNG(s)∩Mi=∅,Miremains to be a module inG. Also, ifs∈NG(Mi), thensmust have a neighbor inMi−1 andMi+1. Thus, NG(s)6⊆NG[Mi], so the completeness of Mi in G−s implies its completeness inGas well.
To prove (2), suppose that Mi is an untouched complete module inG. Clearly,Mi is a module inG−sas well, and sinces /∈Mi,Miremains connected inG−s. Letxbe a vertex in NG(Mi). By the completeness ofMi,x is adjacent to some vertexy /∈NG[Mi]. Suppose thatxdoes not have a neighbor outsideNG−s[Mi] inG−s. This can only happen ify=s.
Now, sincey /∈NG[Mi] andMiis untouched bys,xmust also be adjacent to a vertex ofMi−1
or Mi+1. Thus,xhas a neighbor inV(G−s)\NG−s[Mi], proving the completeness ofMi. AsMiis untouched for at least`−4 indices i∈[`], the statement follows.
In the case whenH is a clique andK is an occurrence of H in Gas a complete module, we get that either K = R−1(`) for some leaf` ∈ V(T), or K is simple, i.e. K = Lq(a, b) for some Q-node q ∈ V(T) and some block [a, b]. In the latter case, Lemma 4.3.5 states a useful observation about the block [a, b]. This lemma uses the following definition: we say that a complete module K of G is h-short, if either K =R−1(`) for some leaf ` ∈ V(T), or K = Lq(a, b) for some Q-node q ∈ V(T) and some block [a, b] with b−a ≤ h. The sets{e1, e2} and{b1, b2} are 2-short complete modules ofGin Figure 4.1.
Lemma 4.3.5. If K is a complete module in Gsuch that G[K] is a clique butK is not h-short, then |NG(K)| ≥2(h+ 1).
Proof. By the conditions of the lemma, we know thatK=Lq(a, b) for some Q-nodeq∈V(T) with childrenx1, . . . , xmand some block [a, b] such thatb−a≥h+ 1. By the completeness ofK, we get thatR−1(Txh) =∅for anyhcontained in [a, b], soM+(h) andM−(h) cannot be empty. Taking these sets for allhin [a, b], with the exception of the setsM+(a) andM−(b), we get 2(b−a)≥2(h+ 1) nonempty sets that are pairwise disjoint, each containing some vertex ofNG(K). This implies the boundNG(K)≥2(h+ 1).
Observe that if two differenth-short complete modulesK1 andK2inGare not indepen-dent, then K1 = Lq(a, b) and K2 =Lq(c, d) must hold for some Q-nodeq in T and some blocks [a, b] and [c, d] that properly intersect each other. Now, if b−a≤h, then there can be at most 2h such blocks [c, d] for which these conditions hold. This implies that given a h-short complete module K, there can be at most 2hdifferent h-short complete modules ofGneighboringK (but not equal toK). It is also easy to see that the maximum number
of pairwise neighboringh-short complete modules in a graph is at most h+ 1. Making use of these facts, Lemma 4.3.6 states some results abouth-short complete modules of a graph in a similar fashion as Lemma 4.3.4. As opposed to Lemma 4.3.4, here we do not require the complete modules to be independent.
Lemma 4.3.6. Suppose that s∈V(G).
(1) If M1, . . . , M` are disjoint h-short complete modules in G−s, then Mi is a complete module inGfor at least `−(3h+ 5)indices i∈[`].
(2) IfM1, . . . , M` are disjoint h-short complete modules inG, thenMi is a complete module inG−sfor at least `−(4h+ 3)indices i∈[`].
Proof. The proof relies on the observation that there can only be a few indices i such that eithersleft orsright lies within [Mileft, Miright].
To see (1), suppose that Mi is not a h-short complete module in Gfor some i. Clearly, G[Mi] is connected. First, assume that Mi is not a module because there are some x, y ∈ Mi such that s is adjacent to x but not to y. In this case, either xleft < sright < yleft or yright < sleft < xright. It is not hard to see that this implies that there can be at most two such modulesMi. Now, assume thatMi is a module, but is not complete. This implies thatMi ⊆NG(s)⊆NG[Mi] is true. Note that ifj 6=i thenMj ⊆NG(s)⊆NG[Mj] is only possible ifMi andMj are neighboring. Thus, there can be at mosth+ 1 such indicesi.
Finally, if Mi is complete module in G but it is not h-short, then the number of max-imal cliques containing the vertices of Mi must be more in G than in G−s, implying that either Mileft < sleft ≤ Miright or Mileft ≤ sright ≤ Miright. As Mileft < sleft ≤ Miright andMjleft< sleft≤Mjright can only hold simultaneously ifMi andMj are neighboring, and such a statement is also true for the latter condition, we get that there can be at most 2(h+1) indicesifor whichMi ish-short inG−sbut not inG. Summing up these facts, we obtain that there can be at most 2 + (h+ 1) + 2(h+ 1) = 3h+ 5 indices i for which Mi is not ah-short complete module inG.
To prove (2), notice that eachMi remains a module in G−sas well. Observe also that ifs /∈Mi, thenMi remains connected in G−s. By the disjointness of the setsM1, . . . , M`, each of them is connected inG−sexcept for at most one. Suppose that Mi1,Mi2, andMi3
are independent, and for eachj ∈ {1,2,3}, Mij is a connected module in G−s but it is not complete. This means that there are verticesx1, x2, and x3 such that xj ∈NG(Mij), butNG(xj)⊆NG[Mij] for eachj. By the completeness of these modules inG, this implies that each ofx1,x2, andx3 are adjacent to s, ands /∈NG[Mij] for anyj. But this can only hold if somexj is adjacent to each vertex ofMij0 for somej 6=j0, and sinceMij andMij0 are independent, this contradicts the assumption thatNG(xj) ⊆NG[Mij]. Thus, there cannot exist such indices i1, i2 andi3, implying that we can fix two indices j andj0 such that for anyMi that is a connected module inG−sbut not complete,Miis neighboring eitherMj
orMj0, implying that there can be at most 2(2h) + 2 such indicesi. To finish, observe that ifMi is a complete module in G−s, then it must be h-short, as the deletion of s cannot increase the number of maximal cliques that containMi.