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On the Minimum Eccentricity Isometric Cycle Problem
Etienne Birmele, Fabien de Montgolfier, Léo Planche
To cite this version:
Etienne Birmele, Fabien de Montgolfier, Léo Planche. On the Minimum Eccentricity Isometric Cy- cle Problem. Electronic Notes in Theoretical Computer Science, Elsevier, 2019, 346, pp.159-169.
�10.1016/j.entcs.2019.08.015�. �hal-02430756�
Etienne Birmel´ ´ e
1and Fabien de Montgolfier
2and L´ eo Planche
3Laboratoire MAP5, Universit Paris Descartes and CNRS, Sorbonne Paris Cit´e, Paris, France1,3 IRIF, Universit´e Paris Diderot - Paris 72,3
Abstract
In this paper we investigate the Minimum Eccentricity Isometric Cycle (MEIC) problem. Given a graph, this problem consists in finding an isometric cycle with smallest possible eccentricityk. We show that this problem is NP-Hard and we propose a 3-approximation algorithm running inO(n(m+kn)) time.
Keywords: isometric cycle; eccentricity;k-domination;k-covering
1 Introduction
For both graph classification purposes and applications, it is an important issue to summarize a graph into a more simple object such as a tree or a path, or in the case of this article, a cycle. Different constructions and metrics offer such a characterization, for example tree-decompositions and tree-width [8]. Another approach, on which we focus in this article, is to study the problem in terms of domination.
In the path case, the problem consists in finding a path such that every vertex in the graph belongs to or has a neighbor in the path. Several graphs classes were defined in terms of dominating paths. Graphs containing a dominating pair, that is vertices such that every path linking them is dominating are studied in [5]. Graphs such that short dominating paths are present in all induced subgraphs are characterized in [1]. Linear-time algorithms to find dominating paths or dominating vertex pairs were also developed for AT-free graphs [3,4].
Dominating paths do not exist however in every graph. A natural extension of the notion of domination is the notion ofk-coverage (also calledk-domination). For a given integer k, a pathk-covers the graph if every vertex is at distance at most k from the path. The smallest ksuch that a path k-covers the graph is then a metric as desired.
Another formulation for this metric is the notion ofeccentricity. Given a graphG,x∈V(G) andS⊂V(G), we noted(x, S) = mins∈SdG(x, s). The eccentricity ofS, denoted ecc(S), is the smallestk such that for any x∈V(G), dG(x, S)≤k.
Theminimum-eccentricity shortest path (MESP) problem was introduced by Dragan and Leitert [6]. The name is transparent as it consists in finding a shortest path with minimal eccentricity. The study of this problem was originally motivated by its link to the embedding with low distortion of graphs into the line. Dragan and Leitert have shown that this problem is NP-complete, and designed some approximation algorithms, and established its relationship with the Line Embedding problem [6]. The MESP problem is also related with graphs problems arising from biological data set [2].
A problem related to the MESP problem consists in finding an isometric cycle with minimal eccentricity, as defined in [2]. A cycle isisometric if it preserves distances:
1 Email: etienne.birmele@parisdescartes.fr
2 Email: fm@irif.fr
3 Email: leoplanche@gmail.com
Birmel´e, de Montgolfier, Planche
Definition 1.1 [Isometric cycle] Let Gbe a graph and C a cycle of G. dG denotes the distance in G and dC the distance in G[C]. C is an isometric cycle if and only if, for every two vertices u, v of C we have:
dG(u, v) =dC(u, v).
In other words, one of the two paths linking u and v in the cycle is a shortest path in the graph. Note that an isometric cycle is necessarily an induced cycle. The problem we are interested in consists in finding an isometric cycle with minimum eccentricity.
Definition 1.2 [Minimum Eccentricity Isometric Cycle Problem (MEIC)] Given a graphG, find an isometric cycleC such that, for every isometric cycleU,ecc(C)≤ecc(U).
A longest isometric cycle may be computed in O(n4.752log(n)) time [7], which gives a polynomial-time 3-approximation for the MEIC problem [2]. It was also shown that a graph admitting an isometric cycle with low eccentricity admits a cycle embedding of low distortion [2].
In this paper, we first show that the MEIC problem is NP-complete. To do so, we propose a reduction of MESP to MEIC. Then we propose a O(n(m+kn))-time 3-approximation algorithm. This is faster than the O(n4.752log(n)) time algorithm proposed in [2] but with the additional constraint that the size of the isometric cycle of minimal excentricity has to be at least 32 times its eccentricity (see Theorem4.6). Notice that finding a minimum-eccentricity isometric cycle is useful only in “donut-shaped” graphs, implying the cycle is long and has short eccentricity.
Definitions and Notations
Through this paper we consider finite connected undirected graphs. For a graphG= (V, E), we usen=|V| and m=|E|to denote the cardinality of the vertex set and the edge set of G. A shortest path between two vertices u and v is a path whose length is minimal among all u, v-paths. Its length (counting edges) is the distance d(u, v). When the precision is needed, we denotedG(u, v) the distance between two verticesuand v in the graphG. Depending on the context, we consider a path either as a sequence, or as a set of vertices. The distance d(v, S) between a vertexv and a set S is the smallest distance between v and a vertex fromS. The eccentricity ecc(S) of a set S is the largest distance betweenS and any vertex ofG. Theball of centerv and radiusr, notedB(v, r), is the set of vertices at distance at mostrfromv. Given two sets of verticesAandB, we noteA\B the set of vertices ofAwhich are not inB.
2 NP-completeness
We propose a polynomial reduction of MESP to MEIC.
Let f be the function which associates to a graph with vertices V(G) = {v1, ...vn} a family of graphs f(G) ={H(i,j)∈{1...n}2}such that for every (i, j)∈ {1...n}2,H(i,j)is the graphGto which a pathPi,j is added of size 2nbetweenvi andvj.
Lemma 2.1 Given a graph G, the shortest path of minimal eccentricity of Gis of eccentricity k if and only if, for any graph off(G), its isometric cycle of minimal eccentricity is of eccentricity at least k.
Proof. Let us show that for every graph Hi,j in f(G), any isometric cycle of minimal eccentricity contains Pi,j and a shortest path of minimal eccentricity inGbetweenvi andvj.
Claim 1 : Any isometric cycle of minimal eccentricity in Hi,j containsPi,j.
Consider any cycle C of G. It does not contain any vertex of Pi,j other than vi and vj. The path Pi,j
being of length 2n, the eccentricity of the cycleC in Hi,j is of size at leastn. Consider now any cycle inHi,j
containingPi,jand a shortest path betweenviandvj. This cycle is isometric and of eccentricity at mostn−2.
Claim 2 : Any isometric cycle of minimal eccentricity in Hi,j containsPi,j and a shortest path of minimal eccentricity in Gbetween vi andvj.
We have already established that such a cycle C contains Pi,j and a path Q in G between vi and vj. Furthermore, Q or Pi,j has to be a shortest path as C is isometric. The distance dG(vi, vj) being at most n−1 andPi,j being of size 2n, Qis thus a shortest path linkingvi andvj. Moreover, as every shortest path between any vertex ofPi,j and any vertex ofGcontains eithervi orvj,ecc(C) =eccG(Q), whereeccG stands for the eccentricity in the graphG. Cbeing of minimal eccentricity among the isometric cycles,Qis of minimal
eccentricity among the shortest paths betweenvi andvj. 2
Lemma 2.1 implies that the solution of MESP inG may be computed by solving MEIC inf(G), and the transformationf is clearly polynomial. MESP being an NP-complete problem [6], it follows that :
Theorem 2.2 The problem MEIC is NP-complete.
2
3 Preliminary results
We define for every cycle C an arbitrary cyclic orientation and for every u, v vertices of the cycle, we note Cuv andCvu the two paths inC linkinguandv. In order to make proofs clearer, we define the operator ’ as follows :
Definition 3.1 Let G be a graph with an isometric cycle C k-dominating G. For every vertex xin G, x0 denotes a vertex ofC at distance at mostkfromx, randomly chosen if several choices are possible.
The following lemma is a crucial preliminary result, which will allow us later to build 3k-dominating cycles if the graph contains an isometric cyclek-dominating it.
Lemma 3.2 Let Gbe a graph with an isometric cycle C k-dominating G, anduandv be any two vertices in G. Every path betweenuandv 2k-dominates eitherCu0,v0 orCv0,u0.
Proof.
Notations used in the proof are illustrated in Figure1.
LetP be a path betweenuandv. Suppose thatP does not 2k-dominate some vertexb on the pathCu0,v0 and consider any vertexainCv0,u0. We haveu0 (resp. v0) in the pathCa,b (resp. Cb,a).
Thenuis at distance at mostkofCa,bandv is at distance at mostkofCb,a. Moreover, as every vertex of Gis at distance at most kof one of those two paths, there exist cand dthat are adjacent vertices inP such thatc0 is in Ca,b andd0 in Cb,a.
As d(c0, d0)≤d(c0, c) +d(c, d) +d(d, d0)≤2k+ 1 and C is an isometric cycle, either Cc0,d0 or Cd0,c0 is of length at most 2k+ 1 and is thus 2k-dominated by {c, d}. Furthermore b andaare not in the same subpath of C betweenc0 and d0, hence eithera orb is 2k-dominated by{c, d}. As b cannot be 2k-dominated byP it follows that ais 2k-dominated by{c, d} hence byP.
The previous claim being true for every ainCv0,u0, the lemma follows. 2
u
v u0
v0
b c0 a d0
c d
≤k
≤k
C P
a
b
m a0
b0
m0
|P|
4
C P0
P
Fig. 1. Notations used in Lemma3.2(left) and in Lemma4.1(right)
4 Approximation in O(n(m + kn)) time
In this section we develop a new approximation algorithm for the MEIC problem, with a better complexity compared to [2] (Theorem 4) but requiring a graph with a longer isometric cycle.
Consider a graph with an isometric cycle of eccentricityk. Algorithm1(F irstCycle) first computes a cycle U0 of eccentricity at most 3k but non necessarily isometric. Its size is then iteratively reduced while keeping an eccentricity of at most 3k. The algorithm ends when the last computed cycle is isometric.
Birmel´e, de Montgolfier, Planche
1 FirstCycle Input: A graphG Output: A cycleU0
2 Letabe any vertex ofG
3 Computeb a vertex furthest froma
4 ComputeP a shortest path fromatob
5 Letmbe a vertex in the middle of P
6 If aandb are connected inG0=G\B(m,b|P|4 c)then
7 LetP0 be a shortest path betweenaandbin G0
8 Letc be the vertex ofPa,m∪P0 furthest froma
9 Letdbe the vertex ofPm,b∪P0 furthest fromb
10 ReturnPc,d∪Pc,d0
11 Return an error ”MEIC too short with respect tok”
Algorithm 1:Pseudocode of the algorithmF irstCycle Lemma 4.1 Let Gbe a graph with a k-dominating isometric cycleC.
If C is of size at least26k+ 6thenF irstCycle(G) returns a cycleU0 2k-dominatingC. Furthermore U0 is of size at most|C|+ 4kand at least|C| −2k−2.
Proof.
Notations used in the proof are illustrated in Figure1.
IfF irstCycle(G) returns a cycleU0, it is the union of two pathsP andP0. Letmbe a vertex in the middle ofP (randomly chosen among the two possible ones if —P— is odd) and assume w.l.o.g. thatm0 is in Ca0b0.
Claim 1: P is of length at least12k.
Leta0 be a vertex ofC furthest froma0.
ecc(a)≥d(a, a0)≥d(a0, a0)−k≥ bC
2c −k≥12k+ 3 As bis a vertex ofGfurthest froma, we have the result.
Claim 2: B(m,b|P|4 c)does not disconnectafrom b.
Consider any vertexy in Cb0a0, asm0 is inCa0b0,
d(m, y)≥d(m0, y)−d(m, m0)≥min(d(m0, a0), d(m0, b0))−k But
d(m0, a0)≥d(a, m)−d(a, a0)−d(m, m0)≥ b|P|
2 c −2k d(m0, b0)≥d(b, m)−d(b, b0)−d(m, m0)≥ b|P|
2 c −2k so that
d(m, y)≥ b|P|
2 c −3k≥ b|P|
4 c+b|P|
4 c −3k >b|P|
4 c Cb0a0 is therefore at distance at leastb|P|4 cofm. Furthermore,
d(a, m) =b|P|
2 c ≥ b|P|
4 c+ 3k d(b, m) =b|P|
2 c ≥ b|P| 4 c+ 3k
It follows that the paths of lengthk fromato a0 andb tob0 are at distance greater thanb|P4|c+ 2kofm.
The verticesaandbare thus connected inG0. This claim implies thatF irstCycle(G) returns a cycle.
Claim 3: U0 2k-dominates C.
Without loss of generality, assume that m0 is inCc0d0. Let us first show that Pcm 2k-dominates Cc0m0. To do so, assume that it is not the case. Then, by Lemma3.2,Pcm 2k-dominates Cm0c0. The vertex m0 being in Cc0d0, it follows thatd0 is inCm0c0 and that there exists a vertex xinPcm at distance at most 2k fromd0. It
4
follows that,
|Pcm|=d(c, x) +d(x, m)
|Pcm| ≥d(c, d)−3k+d(d, m)−3k d(c, m)≥d(c, m) +d(m, d) +d(d, m)−6k
6k≥2d(d, m)
As dis a vertex ofP0, it is at distance more thanb|P|4 cfromm, it follows, 6k≥ b|P|
2 c
This contradicts the fact that P is of size larger than 12k, hencePcm 2k-dominates Cc0m0. Similarly, we show thatPmd 2k-dominates Cm0d0, it follows thatP 2k-dominatesCc0d0.
Lemma 3.2 moreover implies thatP0 2k-dominates either Cc0d0 or Cd0c0. But, asb|P|4 cis greater than 3k, m0 cannot be 2k-dominated by a vertex in G0. Asm0 is on Cc0d0, P0 2k-dominates Cd0c0. It follows that U0 2k-dominatesC.
Claim 4: |C| −6k−2≤ |U0| ≤ |C|+ 4k
|P| ≤ |Ca0b0|+ 2k as P is a shortest path between a and b. Similarly, as P0 is a shortest path in G0,
|P0| ≤ |Cb0a0|+ 2k. Consequently,|P|+|P0| ≤ |C|+ 4k.
The other inequality is a direct result of Lemma4.2.
Claim 5: U0 does not contain the same vertex twice
Assume that a vertex xappears more than once in U0. As both P and P0 are shortest path, they each containxat most (and exactly) once. If xis inPc,m then it is more distant fromathancand it should have been selected at line8instead ofc. This leads to a contradiction. The same contradiction appears if we assume xinPm,d.
2 Lemma 4.2 Let Gbe a graph with an isometric cycleC k-dominatingG. Letaandbbe two vertices, possibly a=b, linked by a path P 2k-dominatingCa0,b0. Then,|P| ≥ |Ca0b0| −6k−2.
In particular, ifU is a cycle2k-dominating C,U is of size at least|C| −6k−2.
Proof. Consider any two verticesaandband a path P linking them and 2k-dominatingCa0,b0.
Letmbe the middle ofCa0,b0 andz a vertex ofP which is 2k-dominatingm. Ca0,mbeing of length at most d|C|2 e,|Ca0,m| ≤d(a0, m) + 1. Similarly,|Cm,b0| ≤d(m, b0) + 1.
It follows that
|Ca0b0| ≤d(a0, m) +d(m, b0) + 2≤(k+|Pa,z|+ 2k) + (k+|Pz,b|+ 2k) + 2≤ |Pa,b|+ 6k+ 2
The affirmation concerning the cycle follows by choosing a=b. 2
The function F irstCycle creates a cycleU0, not necessarily isometric, 3k-dominating the graph. Starting withU0, the size of the cycle is iteratively decreased by theN extCycleprocedure described by Algorithm 2, while keeping the 3k-domination property.
1 NextCycle
Input: A graphGand a cycleUi Output: A cycleUi+1
2 fora∈Ui do
3 If S={b| dUi(a, b) =b|U2i|cand dG(a, b)< dUi(a, b)} 6=∅ then
4 ComputeQa shortest path betweenaandb
5 Compute the eccentricity ofQ∪Ua,bi andQ∪Ub,ai
6 Return the set with the lowest eccentricity
7 ReturnUi
Algorithm 2:Pseudocode of the algorithmN extCycle
Lemma 4.3 Let G be a graph containing an isometric cycle C, k-dominating G. For every path P, there exists a subpathI(P)ofC such that:
Birmel´e, de Montgolfier, Planche
• Every vertex of C at distance at most kof P is inI(P)and the extremities ofI(P)are such vertices;
• P 2k-dominates I(P)
k
k
C
P I(P)
Fig. 2. Illustration of Lemma4.3
Proof.
Notations used in the proof are ilustrated with figure2.
Letu0 andv0 be two vertices ofCsuch thatuandv both belong toP and such thatP 2k-dominatesCu0v0 (such a pair exists as we may always picku0 =v0). Assume that there exists a vertexw0 not inCu0v0 and such that w is in P. By Lemma 3.2, P 2k-dominates eitherCu0w0 or Cw0u0. In the first case,Cu0v0 is a subset of Cu0,w0. In the second case,P 2k-dominates Cw0u0 andCu0v0 henceCw0v0. Asv0 is inCu0w0 and u0 is inCw0v0, the sizes ofCu0w0 andCw0v0 are strictly larger than the one ofCu0v0. The length ofC being finite, there exists a pair of vertices verifying the property and such that the considered arc is maximal. 2 Lemma 4.4 Assume that|C| ≥32k+ 7. LetU be a cycle of size at most|C|+ 4kwhich3k-dominates Gand letaandb be two vertices of U.
Then, Ca0b0 andCb0a0 are both included in either I(Ua,b), or in I(Ub,a).
Proof. Assume that this is not the case. By symmetry, assume thatCa0b0 is neither included in I(Ua,b), nor inI(Ub,a).
Let w01 be the extremity of I(Ua,b) such that Ca0w0
1 ⊂ Ca0b0 and w02 the extremity of I(Ub,a) such that Ca0w02 ⊂Ca0b0. By symmetry, assume thatCa0w02 ⊂Ca0w01 and let us notew0=w01. There existsw∈Ua,b such thatd(w, w0) =k.
Case 1: d(w0, b0)≤4k+ 1
In this case, |Cb0,w0| ≥ |C| −4k−2. By definition of w0,Cb0,w0 ⊂I(Uw,b) hence is 2k-dominated by Uw,b. The Lemma4.2then implies that|Uw,b| ≥ |Cb0,w0| −6k−2≥ |C| −10k−4.
By the definition of w0, we also have that Ca0,w0 ⊂I(Ua,w). Furthermore Cb0,a0 ⊂I(Ub,a) by Lemma 3.2 as, by hypothesis,I(Ub,a) does not containCa0,b0. Therefore,Cb0,w0 ⊂I(Ub,w) and by the Lemma4.2,|Ub,w| ≥
|Cb0,w0| −6k−2≥ |C| −10k−4.
Finally, |U| =|Ub,w|+|Uw,b| ≥ 2|C| −20k−8. As |U| ≤ |C|+ 4k, it implies that |C| ≤24k+ 8, which contradicts our hypothesis.
Case 2: d(w0, b0)≥4k+ 1
Let z be the vertex ofCa0b0 at distance 4k+ 1 of w0. U is 3k-dominating Gby hypothesis, thereforez is at distance at most 3k of a vertex y of U. The fact that d(w0, z)>4k then implies that y0 is not inCa0,w0. We now have to study three different sub-cases, corresponding to y ∈Ua,w, y ∈Uw,b and y ∈ Ub,a, but the arguments are very similar.
6
Let y ∈Ua,w, as shown on the right of Figure 3. Let us decomposeU as U =Uy,w∪(Uw,b∪Ub,y). By definition ofw0,I(Uy,w) containsCy0w0. Therefore|Uy,w| ≥ |Cy0,w0| −6k−2≥ |C| −d(y0, z)−d(z, w0)−6k−2≥
|C| −14k−3.
Moreover, I(Uw,b) containsCb0,w0 by definition ofw0 andI(Ub,y) containsCy0,b0 by definition ofw01. Con- sequently,I(Uw,b∪Ub,y) containsCy0,w0 and|Uw,b∪Ub,y| ≥ |C| −14k−3.
Finally, 2(|C| −14k−3)≤ |C|+ 4k, meaning that|C| ≤32k+ 6, which is a contradiction.
Lety∈Uw,b. Let us decomposeU asU =Uw,y∪(Uy,b∪Ub,a∪Ua,w).
By definition of w0,I(Uw,y) containsCy0w0, therefore|Uw,y| ≥ |Cy0,w0| −6k−2≥ |C| −d(y0, z)−d(z, w0)− 6k−2≥ |C| −14k−3.
Furthermore, I(Uy,b) containsCy0,b0 by definition ofw20, I(Ub,a) containsCb0,a0 by hypothesis andI(Ua,w) containsCa0,w0 by definition ofw0, thereforeI(Uy,b∪Ub,a∪Ua,w) containsCy0,w0. Consequently, its length is also greater that|C| −14k−2.
It implies again that |C| ≤32k+ 6, which is a contradiction.
Finally, assume now thaty∈Ub,a. Let us decomposeU asU = (Uw,b∪Ub,y)∪(Uy,a∪Ua,w).
I(Uw,b) contains Cb0,w0 by definition of w0 and I(Ub,y) contains Cy0,b0 by definition of w01. Consequently, I(Uw,b∪Ub,y) containsCy0,w0 and|Uw,b∪Ub,y| ≥ |C| −14k−3.
Furthermore,I(Uy,a) containsCy0,a0 as it does not containCa0,y0 by hypothesis andI(Ua,w) containsCa0,w0
by definition ofw0. It follows thatI(Uy,a∪Ua,w) contains Cy0,w0 and|Uy,a∪Ua,w| ≥ |C| −14k−3.
Once again, it implies that |C| ≤32k+ 6, leading to a contradiction. 2
a
b w0
w02 w
a0
b0
C
I(Ua,b) U
a b
w0 w20 w
a0 b0
y0 y
z C
I(Ua,b) U
Fig. 3. Notations used in the proof of Lemma4.4, incase 1on the left andcase 2on the right
Lemma 4.5 Let Gbe a graph with an isometric cycle C k-dominating Gand of size at least32k+ 7.
Let Ui be a cycle such that:
• Ui 2k-dominates C.
• |C| −6k−2≤ |Ui| ≤ |C|+ 4k
ThenUi+1 =N extCycle(Ui)has the same properties and |Ui+1|<|Ui|.
Proof. The inequalities concerning the length are straightforward given Lemma4.2and the fact thatUi+1is obtained by replacing an arc ofUi with a path of smaller length.
To prove the 2k-domination, consider the path Q linking the vertices a and b of Ui as selected by the algorithm. By Lemma3.2, one can assume w.l.o.g. that I(Q) contains Ca0,b0 and therefore 2k-dominates it.
Furthermore, Lemma4.4implies thatCb0,a0 is included inI(Ua,bi ) orI(Ub,ai ). By symmetry , assume that it is included inI(Ua,bi ).
Then,I(Q∪Ua,bi ) =C, and it follows thatQ∪Ua,bi 2k-dominates Cand is of eccentricity at most 3k. The property is verified ifUi+1=Q∪Ua,bi .
If Ui+1=Q∪Ub,ai ,Q∪Ub,ai also 3k-dominatesG as it is of a lower eccentricity. Lemma4.4then implies that eitherUb,ai orQ2k-dominatesCb0a0, and thereforeQ∪Ub,ai 2k-dominatesC. 2
Birmel´e, de Montgolfier, Planche
1 ApproximationCycle Input: A graphG Output: A cycle
2 U =F irstCycle(G)
3 V =N extCycle(G, U)
4 WhileV 6=U do
5 U =V
6 V =N extCycle(G, V)
7 ReturnV
Algorithm 3:Pseudocode of the algorithmApproximationCycle
The algorithm F irstCycle requires to compute three BFS trees, its complexity is O(m). The complexity bottleneck is testing the If condition on n2 vertex pairs in Line 3 of N extCycle. The distances dG inside Gmay be precomputed in O(nm) time using n BFS, while evaluating the distancedUi inside the cycle take constant time if the vertices are numbered. Testing this condition therefore takes O(n2) time. When a pair (a, b) is found the block until Line6 takesO(m) time, using three BFS again, and is done once. Algorithm6 (N extCycle) thus takesO(n2) time.
We know by Lemma 4.5that the length of the cycles of the sequence Ui strictly decreases and is at least
|C|−6k−2. As|U0| ≤ |C|+4k, it follows that the sequence converges after at most one execution ofF irstCycle and at most 10k+ 2 executions of N extCycle. Not forgetting theO(nm)-time distance precomputation, we have :
Theorem 4.6 Let a graph with an isometric cycle of eccentricitykand of size`≥32k+ 7, an isometric cycle of eccentricity at most3kmay be computed in O(n(m+kn))time.
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