graphs whose powers are chordal and graphs whose powers are interval graphs

Graphs Whose Powers areChordal and Graphs WhosePowers are Interval Graphs

Carsten FlotowMATHEMATISCHES SEMINAR DER UNIVERSITAT HAMBURG BUNDESSTR. 55,

20146 HAMBURG, GERMANY

ABSTRACT

The main theorem of this paper gives a forbidden induced subgraph condition on G thatis sufficient for chordality of Gm. This theorem is a generalization of a theorem of Bala-krishnan and Paulraja who had provided this only for m = 2.

We also give a forbidden subgraph condition on G that is sufficient for chordality ofG2m.

Similar conditions on G that are sufficient for Gm being an interval graph are alsoobtained.

In addition it is easy to see, that no family of forbidden (induced) subgraphs of G isnecessary for Gm being chordal or interval graph. c© 1997 John Wiley & Sons, Inc.

INTRODUCTION

We consider finite, simple, undirected graphsGwith vertex set V (G) and edge setE(G).A graphG = (V,E) is chordal, if each cycle Cn (that is the graph with vertex set {v1, . . . , vn} and edgeset {(v1, v2), (v2, v3), . . . , (vn−1, vn), (vn, v1)}) of length n ≥ 4 has a chord (i.e., there exists anedge (vi, vj) with j 6= i ± 1, addition modulo n). It is well known [6], that interval graphs canbe characterized as chordal graphs, which do not contain an asteroidal triple (i.e., three verticesu, v, w such that each pair of them is connected by a path that doesn't pass through the third andany of its neighbors).

The distance of two vertices u, v ∈ V (G) in G, denoted by dG(u, v), is the length of a shortestpath from u to v in G. Let m be a positive integer, the mth power of a graph G = (V,E), denotedby Gm, is the graph with vertex set V, where two vertices are adjacent in Gm iff dG(u, v) ≤ m.

P5 + a is a path on five vertices v1, . . . , v5, where the edge (v2, v4) is added (E(P5) :={(v1, v2), (v2, v3), . . . , (v4, v5)}).

Journal of Graph Theory Vol. 24, No. 4, 323 330 (1997)c© 1997 John Wiley & Sons, Inc. CCC 0364-9024/97/040323-08

324 JOURNAL OF GRAPH THEORY

Balakrishnan and Paulraja obtained the following [1]

Theorem . If G has no induced subgraph isomorphic to K1,3, P5 + a or Cn, n ≥ 6 then G2 ischordal.

We will show, that G2 can be replaced by Gm if we replace Cn by Ck, k ≥ 2(m + 1), seeTheorem 1. For the case m = 2 (resp. m = 3) it is possible to allow P5 + a as an inducedsubgraph, but then Ck, k ≥ 4 (resp. k ≥ 6) must be forbidden instead (Theorems 2 and 3).

If m is even, we may allow induced K1,3, P5 + a in G, but we have to forbid all cycles(including non-induced ones!) greater than five (Theorem 4).

Since Gm has no asteroidal triple if G has none [7] we derive some corollaries about in-terval graphs: In each theorem we also forbid an asteroidal triple in G, concluding Gm is aninterval graph. A forbidden induced subgraph condition on G is also possible. This is given inTheorem 5.

It is easy to see, that no family of forbidden subgraphs of G is necessary for chordality of Gm,that means the existence of a theorem of the following type is impossible:

If Gm is chordal, then G has no subgraphs of the family (Gi)i∈I .To see this, define G as follows: V (G) :=

⋃i∈I V (Gi) ∪ {v}, E(G) :=

⋃i∈I E(Gi)

∪ {(v, x) : x ∈ ⋃i∈I V (Gi)}. G contains each Gi as an induced subgraph, but Gm,m ≥ 2

is complete and therefore chordal.We close the introduction by giving an application of our results: An often studied topic in

graph theory is to determine graph parameters, (e.g. ω(G) := maximum size of a clique, χ(G) :chromatic number) of graphs. It is well known, that these problems are NP-hard in the generalcase but polynomial solvable for chordal graphs. Let ωk(G) := maximum size of a set of verticesof G, the distance between each pair of which is at most k and χk(G) := minimum number ofcolors, that are necessary to color the vertices in such a way, that two vertices with distance ≤ khave distinct colors. There are no efficient algorithms known, to determine ωk(G) and χk(G) inthe general case. But if we know that Gk is chordal, then ωk(G) = ω(Gk), χk(G) = χ(Gk) andsimilar parameters are computable in polynomial time.

GRAPHS WHOSE POWERS ARE CHORDAL

A generalization of the above mentioned theorem of Balakrishnan and Paulraja is

Theorem 1. If G has no induced K1,3, P5 + a and Ck, k ≥ 2(m + 1), then Gm is chordal.

Proof. Assume Gm is not chordal. Then there exists in G a closed walk v1 · · ·P1 · · · v2 · · ·P2

· · · v3 · · · · · · vk · · ·Pk · · · v1, k ≥ 4,wherePi is a shortest path fromvi tovi+1 inG anddG(vi, vi+1)≤ m, dG(vi, vj) ≥ m + 1 ∀j 6= i, i± 1 (addition modulo k).

Claim 1. If j 6= i, i± 1 then Pi and Pj have no vertex in common:

FIGURE 1.

GRAPHS WITH CHORDAL AND INTERVAL POWERS 325

FIGURE 2.

If j 6= i± 2 a common vertex would imply dG(va, vb) ≤ m for a ∈ {i, i+ 1} and b ∈ (j, j + 1),a contradiction. Now let j = i + 2 and x be the common vertex. Let da denote the lengthof the subpath of Pi or Pj from va to x. Obviously di + di+1 ≤ m, di+2 + di+3 ≤ m anddi + di+2 ≥ m + 1, di+1 + di+3 ≥ m + 1, a contradiction.

If Pi and Pi+1 have common vertices, w.l.o.g. we assume Pi = Q1, Q, and Pi+1 = Q,Q2,where Q1 := vi, w

i1, . . . , w

ik−1, Q := wi

k, . . . , wili, vi+1, Q := vi+1, w

il , . . . , w

ik (Q backwards),

Q2 := wi+1li−k+2, . . . , w

i+1li+1

, vi+2 and Q1 ∩Q2 = ∅, see Figure 1.This assumption is possible, because Pi, Pi+1 are shortest: If Pi and Pi+1 have vertices x, y

in common, the subpaths of Pi and Pi+1 from x to y must have the same lengths. Therefore thex-y-subpath of Pi+1 can be replaced by the x-y-subpath of Pi. This replacing operation resultsin Figure 1. Next, it will be shown that claim 2 holds.

Claim 2. Pi and Pi+1 cannot have a common edge.Assume that Pi and Pi+1 have a common edge (see Figure 2). Since wi

k and its 3 neighbors donot induce K1,3 and Pi, Pi+1 are shortest, the dotted edge of Figure 2 must exist.

Now let wix ∈ Q1, w

i+1y ∈ Q2 and (wi

x, wi+1y ) ∈ E(G) (e.g., the dotted edge). Consider the

path vi, . . . , wix, w

i+1y , . . . , vi+2 with length l > m. Then we have the following

Lemma. There exist wiα ∈ Q1, w

i+1β ∈ Q2, (w

iα, w

i+1β ) ∈ E(G), such that the path vi, . . . ,

wiα, w

i+1β , . . . , vi+2 has length < l.

Proof of the Lemma. Suppose wix, w

i+1y and wi

k form a triangle (see Figure 3), then this triangleand w,w′ would induce P5 + a, so one of the dotted edges has to occur. (Note, that an edge(w,wi

k) is impossible, since Pi, Pi+1 are shortest.) This dotted edge will create a path from vi tovi+2 of length < l as claimed in the Lemma.

So w.l.o.g. we may assume that (wi+1y , wi

k) 6∈ E (see in Figure 4).Since wi+1

y , wi+1y−1, w

i+1y+1, w

ix do not induce a K1,3, we suppose w.l.o.g. that (wi

x, wi+1y−1) ∈ E

(dotted). Now the thick edges together with the dotted one form P5 + a. Since this P5 + amay not be induced, the existence of the edge (wi

x−1, wi+1y−1) (that is the sinuous line) follows.

Therefore the existence of a path vi, . . . , wix, w

i+1y , . . . , vi+2 implies the existence of the path

FIGURE 3.


FIGURE 4.

vi, . . . , wix−1w

i+1y−1, . . . , vi+2 (over the sinuous line) with the same length. This path also implies

the existence of further paths vi, . . . , wix−p, w

i+1y−p, . . . , vi+2 with the same length. This finishes,

when wix−p = vi or wi+1

y−p = wik. In the first case, the length of the path vi, w

i+1y−p, . . . , vi+2 is

less than or equal to the length of Pi+1, and therefore ≤ m. In the second case, the existence ofan edge (wi

x−p, wik) follows, but than Pi is not a shortest path, contradiction. So the Lemma is

proved.Repeated use of this Lemma provides a path from vi to vi+2 of length≤ m, thus dG(vi, vi+2) ≤

m, which is impossible, so Claim 2 is settled.Due to Claim 2 we may assume that no Pi and Pi+1 have a common edge. Then

⋃ki=1 Pi is

a cycle of length ≥ 2(m + 1) (since dG(v1, v3) ≥ m + 1). By hypothesis this cycle must havea chord. Obviously no chord of the form (vi, vj), (vi, w

jx), (w

ix, w

iy) is possible, so each chord is

of the form (wix, w

jy), i 6= j.

Choose i, j with a chord from Pi to Pj . We first assume Pi ∩ Pj = ∅, the other case is tosettle analogously. Let wi

x be the vertex in Pi with the lowest index x having a chord to Pj . Inthe family of all chords from wi

x to Pj let (wix, w

jy) be the one with the greatest index y (see

Figure 5).Sincewj

y, wjy−1, w

jy+1, w

ix do not induceK1,3, the dotted edge has to exist. But nowwi

x−1, wix,

wjy−1, w

jy, w

jy+1, induce P5 + a, a contradiction. So Gm is chordal.

Remark 1. For the case m = 2 Theorem 1 is exactly the Theorem of Balakrishnan andPaulraja, [1].

Remark 2. The condition, which is sufficient for chordality ofGm is also sufficient for chordal-ity of Gm+l for each positive integer l.

FIGURE 5.


FIGURE 6.

In the case ofm = 2, 3 the proof of Theorem 1 can be modified to obtain the next two theorems.The proof of Theorem 2 is omitted, since this theorem is a consequence of a theorem in [5], whichsays, that the square of a chordal graph G is chordal, iff each trampoline of G is suspended (i.e.,if {x1, . . . , xn, y1, . . . , yn} induces a trampoline, where {x1, . . . , xn} is complete, {y1, . . . , yn}independent and yi is exactly adjacent to xi and xi+1 (addition modulo n), then another vertex zhas to exist, which is adjacent to yk and yl, where k 6= l ± 1).

Theorem 2. If G has no induced K1,3 and Ck, k ≥ 4, then G2 is chordal.

Theorem 3. If G has no induced K1,3 and Ck, k ≥ 6, then G3 is chordal.

Proof. SupposeG3 is not chordal. We use the same terminology as in the proof of Theorem 1.Again Pi and Pi+1 cannot share a common edge since in that case Pi and Pi+1 must have length3 and Pi, Pi+1 have exactly one common edge (otherwise dG(vi, vi+2) ≤ 3, a contradiction).But then K1,3 occurs as an induced subgraph (otherwise again dG(vi, vi+2) ≤ 3).

Since Pi and Pi+1 cannot share a common edge⋃k

i=1 Pi is a cycle of length ≥ 6. This cyclemust have a chord of the form (wi

x, wjy), i 6= j. W.l.o.g. j = i + 1 or j = i + 2 (otherwise

dG(va, vb) ≤ 3 for a ∈ {i, i + 1}, b ∈ {j, j + 1} would be a contradiction). Obviously it is notpossible, that Pi and Pj both have length 2, so we assume Pi has length 3. Now, there are leftonly a few possibilities for positions of chords, see Figure 6 (note dG(vi, vj+1) ≥ 4).

In (2), (4) an induced K1,3 arises. Thus the chords of (1) and (3) are the only possible ones.Replacing the wi

li− vj − wi+1

1 -paths by the chords (wili, wi+1

1 ) (whenever exist), we have aninduced cycle of length ≥ 6, a contradiction. So G3 must be chordal.

Remark 3. The condition that is sufficient for chordality of G2 (Theorem 2) is sufficient forchordality of Gm for each positive integer m: For m = 1, 2 that's trivial; since chordality of Gm

implies chordality of Gm+2 [3] the remark follows by induction.

Remark 4. The condition of Theorem 3 is sufficient for chordality of all odd powers of G, butnot sufficient for chordality of G4, an example is the graph of Figure 7.

FIGURE 7.


FIGURE 8.

Theorem 4. If G has no subgraph isomorphic to Ck, k ≥ 6, then all even powers of G arechordal.

Proof. SupposeG2m is not chordal, then we havek ≥ 4 vertices v1, . . . , vk with the propertiesmentioned in the proof of Theorem 1 (terminology as there). In CasePi andPi+1 have no commonedge,

⋃ki=1 Pi is a cycle of length ≥ 2(2m + 1) ≥ 6, a contradiction.

In Case Pi and Pi+1 have a common edge, we just know Pi ∩Pj = ∅ ∀j 6= i, i± 1. Thereforewe have the situation of Figure 8 (for k = 4, but for k ≥ 5 analogously).

If bi, bi+1 have length 1, then by ai + bi + ai+1 ≤ 2m, ai+1 + bi+1 + ai+2 ≤ 2m and ai +bi + bi+1 + ai+2 ≥ 2m + 1 follows ai+1 ≤ m− 1.

Assume bi = 1 ∀i, then ai ≤ m − 1 ∀i and so dG(v1, v3) ≤ 2m, a contradiction. Now letb1 ≥ 2. Assume b2 = b3 = b4 = 1, then a3, a4 ≤ m− 1. Since a2 + b2 + b3 + a4 ≥ 2m + 1, itfollows a2 ≥ m, analogously a1 > m. But then a1 + b1 + a2 ≥ 2m + 1, a contradiction again.

Thus at least two of the bi-paths have length ≥ 2. Then⋃k

i=1 bi is a cycle of length ≥ 6, acontradiction, so this case is also settled.

Remark 5. Obviously the sufficient condition of Theorem 4 is not sufficient for chordality ofodd powers G2m+1 of G, as can be seen in Figure 9.

GRAPHS WHOSE POWERS ARE INTERVAL GRAPHS

Lemma (Raychaudhuri [7]). If Gk−1 contains no asteroidal triple, then Gk contains no aster-oidal triple.

Using this lemma and the theorems of the previous section, we obtain some corollaries:

Corollary 1. If G has no induced K1,3, P5 + a and Ck, k ≥ 2(m+ 1) and no asteroidal triple,then Gm is an interval graph.

FIGURE 9.


FIGURE 10.

Corollary 2. If G has no induced K1,3 and Ck, k ≥ 6 and no asteroidal triple, then G3 is aninterval graph.

Since the class C of interval graphs is strongly closed under the operation of power (i.e.,Gm−1 ∈ C ⇒ Gm ∈ C, see [4], [7]), we have

Corollary 3. If G has no subgraph isomorphic to Ck, k ≥ 6 and no asteroidal triple, then allpowers Gm,m ≥ 2 of G are interval graphs.

The reader may think, it would be more beautiful to allow asteroidal triples and forbid anotherinduced subgraph. If that's the case, you may enjoy

Theorem 5. If G has no induced K1,3, P5 + a,Ck, k ≥ 2(m + 1) and no induced Hm+1

(Fig. 10), then Gm is an interval graph.

Proof. It suffices to show, that Gm has no asteroidal triple. Assume {v1, v2, v3} is one, thendG(vi, vj) ≥ m+1∀i 6= j.W.l.o.g.,G is connected, thus a closed walk v1 · · ·P1 · · · v2 · · ·P2 · · · v3

· · ·P3 · · · v1 exists, where Pi := vi, wi1, . . . , w

ili, vi+1, with li ≥ m is a shortest vi − vi+1-path.

If Pi and Pi+1 have more than one vertex in common, we make the same assumption as in theproof of Theorem 1.

Case 1. Pi and Pi+1 have a common edge.Surprisingly, this case is impossible as in the proof of Theorem 1.Case 2. P1 ∪ P2 ∪ P3 is a cycle.This cycle has length ≥ 3(m + 1), so it is not an induced cycle. Obviously no chords of the

form (vi, vj), (wix, w

iy) are possible.

If no chords of the form (vi, wi+1x ) exist, it's possible to continue as in the proof of Theorem

1 (below the proof of Claim 2), note vj := vi+1 then.Now consider the case, there are chords of the form (vi, w

i+1x ). Let S := (wi+1

x )x∈I be thefamily of vertices on Pi+1 that are neighbors of vi. All elements of S have distance ≥ m fromvi+1 and vi+2 (since dG(vi, vi+1), dG(vi, vi+2) ≥ m + 1), see Figure 11.

If there exists one element in S without neighbors in S, an induced K1,3 occurs, thus eachelement of S has at least one neighbor in S. If |S| = 2, an induced P5 + a occurs, if |S| = 3 aninduced Hm+1 occurs. The case |S| ≥ 4 is settled in Figure 12.

Thus case 2 is also settled.

FIGURE 11.


FIGURE 12.

ACKNOWLEDGMENT

The results of this paper will form part of my dissertation written under the supervision of Dr.Erich Prisner.

References

[1] R. Balakrishnan and P. Paulraja, Graphs whose squares are chordal, J. Indian Pure Appl. Math. 12(2)(1981), 193–194.

[2] R. Balakrishnan and P. Paulraja, Powers of chordal graphs, J. Austrl. Math. Soc. (Ser A) 35 (1983),211–217.

[3] P. Duchet, Classical perfect graphs, Ann. Discrete Math. 21 (1984), 67–96.

[4] C. Flotow, On powers of m-trapezoid graphs, Discrete Appl. Math. 63 (1995) 187–192.

[5] R. Laskar and P. Shier, On powers and centres of chordal graphs, Discrete Appl. Math. 6 (1983),139–147.

[6] C. B. Lekkerkerker and J. C. Boland, Representation of a finite graph by a set of intervals on the realline, Fund. Math. 51 (1962), 45–64.

[7] A. Raychaudhuri, On powers of interval graphs and unit interval graphs, Congressus Numerantium 59(1987), 235–242.

Received March 1, 1994

graphs whose powers are chordal and graphs whose powers are interval graphs

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