Finding a shortest even hole in polynomial time

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A R T I C L E

Finding a shortest even hole in polynomial time

Hou‐Teng Cheong | Hsueh‐I Lu

Department of Computer Science and Information Engineering, National Taiwan University, Taipei, Taiwan

Correspondence

Hsueh‐I Lu, Department of Computer Science and Information Engineering, National Taiwan University, 1 Roosevelt Road, Section 4, Taipei, 106 Taiwan.

Email:hil@csie.ntu.edu.tw

Funding information

Ministry of Science and Technology, Taiwan

Abstract

An even (respectively, odd) hole in a graph is an induced cycle with even (respectively, odd) length that is at least four. Bienstock proved that detecting an even (respectively, odd) hole containing a given vertex is NP‐complete. Conforti, Cornuéjols, Kapoor, and Vušković gave the first known polynomial‐time algorithm to determine whether a graph contains even holes. Chudnovsky, Kawarabayashi, and Seymour esti- mated that Conforti et al.'s algorithm runs inO n( ⁴⁰) time on an n‐vertex graph and reduced the required time toO n( ³¹). Subsequently, da Silva and Vušković, Chang and Lu, and Lai, Lu, and Thorup improved the time to O n( ¹⁹), O n( ¹¹), and O n( )⁹ , respectively. The tractability of determining whether a graph contains odd holes has been open for decades until the algorithm of Chudnovsky, Scott, Seymour, and Spirkl that runs in O n( )⁹ time, which Lai et al. also reduced toO n( )⁸ . By extending Chudnovsky et al.'s techniques for detecting odd holes, Chudnovsky, Scott, and Seymour (respectively) ensured the tractability of finding a long (respectively, shortest) odd hole. They also ensured the NP‐hardness of finding a longest odd hole, whose re- duction also works for finding a longest even hole.

Recently, Cook and Seymour ensured the tractability of finding a long even hole. An intriguing missing piece is the tractability of finding a shortest even hole, left open for 16 years by, for example, Chudnovsky et al. and

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Johnson. We resolve this open problem by augmenting Chudnovsky et al.'s even‐hole detection algorithm into the first known polynomial‐time algorithm, running in O n( ³¹) time, for finding a shortest even hole in ann‐ vertex graph that contains even holes.

K E Y W O R D S

data structure, induced subgraph, polynomial‐time algorithm, shortest even hole

1 | I N T R O D U C T I O N

An even (respectively, odd) hole in a graph is an induced cycle with even (respectively, odd) length that is at least four. Detecting induced subgraphs are fundamentally important problems [8,10,13,14,20,19,28,29,31,34,35,37,38,40]. A most prominent example regarding detecting induced cycles is the seminal strong perfect graph theorem of Chudnovsky, Robertson, Seymour, and Thomas [15], conjectured by Berge in 1960 [3–5], ensuring that the perfection of a graph can be determined by detecting odd holes in the graph and its complement. Chudnovsky, Cornuéjols, Liu, Seymour, and Vušković [9] gave the first known polynomial‐time algorithm, running inO n( )⁹ time, for recognizingn‐vertex perfect graphs. Bienstock [6,7] proved that detecting an odd hole containing a prespecified vertex is NP‐hard in the early 1990s. The tractability of detecting an odd hole remained unknown until the recentO n( )⁹ ‐time algorithm of Chudnovsky, Scott, Seymour, and Spirkl [18]. Lai, Lu, and Thorup [39] implemented Chudnovsky et al.'s algorithm to run inO n( )⁸ time, also leading to anO n( )⁸ ‐time algorithm for recognizing perfect graphs. Chudnovsky, Scott, and Seymour [16] showed that it takes O n( ^20ℓ+40)time to detect an odd hole with length at leastℓ. Chudnovsky, Scott, and Seymour [17] ensured the NP‐hardness of finding a longest odd hole and gave an^{O n}⁽ ¹⁴⁾‐time algorithm to obtain a shortest odd hole, if one exists.

Even holes in graphs have also been extensively studied in the literature [1,22,23,26,27, 30,32,36,42,44]. Vušković [43] gave a comprehensive survey on even‐hole‐free graphs.

According to Bienstock [6,7], detecting an even hole containing a prespecified vertex is also NP‐hard. Conforti, Cornuéjols, Kapoor, and Vušković [21,24] gave the first polynomial‐time algorithm for detecting even holes in ann‐vertex graph, running in^{O n}⁽ ⁴⁰⁾time. Chudnovsky, Kawarabayashi, and Seymour [11] reduced the time toO n( ³¹). Chudnovsky et al. [11] also observed that the time of detecting even holes can be further reduced toO n( ¹⁵) as long as detecting prisms is not too expensive, but this turned out to be NP‐hard [41]. Chudnovsky and Kapadia [10] and Maffray and Trotignon [41, Algorithm 2] devisedO n( ³⁵)‐time and O n( )⁵ ‐ time algorithms for detecting prisms in theta‐free and pyramid‐free graphs, respectively.

Later, da Silva and Vušković [27] improved the time of detecting even holes inG toO n( ¹⁹). Chang and Lu [8] further reduced the running time toO n( ¹¹). The best currently known algorithm for detecting even holes, due to Lai, Lu, and Thorup [39], runs inO n( )⁹ time. Very recently, Cook and Seymour [25] announced anO n( ^9ℓ+3)‐time algorithm for detecting even holes with length at least ℓ. Following the approach of Chudnovsky et al. [17] for the NP‐hardness of longest odd hole, one can verify that the longest even hole problem is

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NP‐hard by reducing from the problem of determining whether theⁿ‐vertex graph^Gadmits a Hamiltonianuw‐path for two given vertices^uandw: For the graphH obtained fromG by subdividing each edge once and then adding a pathuvw, a longest even hole of H has 2n vertices if and only ifGadmits a Hamiltonianuw‐path. As displayed in Table1, the complexity of finding a shortest even hole, open for 16 years (see, e.g., [11, page 86] and [33, page 166]), became the only missing piece. As summarized in the following theorem, we resolve this open problem by augmenting Chudnovsky et al.'s even‐hole detection algorithm into the first known polynomial‐time algorithm, running in^{O n}⁽ ³¹⁾ time, for finding a shortest even hole in ann‐ vertex graph that contains even holes.

Theorem 1. For anyn‐vertex graph^G, it takesO n( ³¹)time to either obtain a shortest even hole ofG or ensure thatG contains no even hole.

Our shortest‐even‐hole algorithm is based upon Chudnovsky et al.'s techniques [11] that lead to theirO n( ³¹)‐time algorithm for detecting even holes. To our surprise, their techniques for detecting even holes suffice to resolve the problem that they left open in the first paragraph of their paper, writing“the complexity of finding the shortest even hole in a graph is still open as far as we know.” We thought that perhaps their more general setting of allowing for weighted graphs caused them to overlook the possibility of further pushing for a shortest even hole, for example, in their Lemma 5.3. It is very kind of Professor Seymour to tell us the background after seeing our submission:

“… As you know, with Maria Chudnovsky and Ken Kawarabayashi, we three published a paper in 2005 to test in poly time whether a graph had an even hole. Most of the paper would work to also find the shortest even hole, but there was one step that did not: we tested for prisms, and stopped if we found a prism; that guaranteed that there was an even hole, but would not give us the shortest. The paper was written up like that, and in the introduction we said that the problem of finding a shortest even hole remained open.

Before submission, we noticed that we didn't need to test for prisms, we could do it another way, and we replaced the prism‐testing step with the alternative. We noticed shortly afterwards that now all the steps in the paper worked to find a shortest even hole. (I don't remember whether we noticed this before or after submission or before or after publication.) At the time, we didn't think this was of great interest. But in any case, the paper was published with the old introduction, where we said the shortest even hole problem was still open. (I don't remember

T A B L E 1 The state‐of‐the‐art of hole detection

Odd hole Even hole

Containing a vertex NP‐hard [6,7] NP‐hard [6,7]

Longest NP‐hard [17] NP‐hard [17]

At least ℓ edges O n( ^20ℓ+40)[16] O n( ^9ℓ+3)[25]

Detection O n( )⁸ [39] O n( )⁹ [39]

Shortest O n( ¹⁴)[17] O n( ³¹)[this paper]

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whether we just forgot to change the intro, or didn't notice the solution until it was too late to change the paper.)

And then the fact that we knew how to solve this faded from our minds; I wrote a paper recently (with Linda Cook), in which I said the shortest even hole problem was open.…”

We are happy to have rediscovered the interestingly forgotten result.

2 | P R O V I N G T H E O R E M 1

We first reduce Theorem1to Lemmas4and5via Lemmas2and3and then prove Lemmas4 and5in Sections 2.1and2.2.

Lemma 2 (Lai, Lu, and Thorup [39, Theorem 1.6]). For any n‐vertex graph^G, it takes O n( )⁹ time to determine whetherG contains even holes.

Let i k[ , ] for integers i and k consist of the integers j withi≤ ≤j k. Let S denote the cardinality of setS. Let G denote the number of edges of graphG. LetV G( ) consist of the vertices of graphG. For any vertex setS, let G S[ ] denote the subgraph ofG induced byS. For any subgraph H ofG, letG H[ ] =G V H[ ( )]. Letd u vG( , )for verticesuandvof graphGdenote the distance ofu andv inG. Auv‐path is a path with end‐vertices^u andv.

LetG be a graph containing even holes. LetC be a shortest even hole ofG. LetP be a uv‐path of^G for distinct and nonadjacent verticesuandvofC.PisC‐good if the union of Pand auv‐path of C remains a shortest even hole of^G.PisC‐bad if^P is notC‐good. ^Pis C‐shallow if

P d u vC( , ) − 1

  ≥ (1)

andG P[ ∪C2]for theuv‐paths^C¹andC2 ofC with   C1 ≤ C2 is a hole.P is aC‐shortcut if

P d u v P C

2 ( , ) and <

4 .

  C    

≤ ≤ (2)

Observe that if P is aC‐good C‐shortcut, then  ^P ⁼^{d u v}C^{( , )}. C is good in G if each C‐shortcut in^GisC‐good. C is bad in^GifC is not good inG.Pis a worstC‐shortcut in^GifPis aC‐bad C‐shortcut such that either (i)   ^P ⁼ ^P^′ andd u vC( , )≥d u vC( ′, ′) or (ii)   P < P′ holds for eachC‐bad C‐shortcut^{u v}^{′ ′}‐path ^P′inG. Observe thatC is bad inG if and only if there is a worstC‐shortcut in^G. A graphG is shallow if there is a shortest even holeC of G such that each worst C‐shortcut is C‐shallow. A graph ^G is antishallow if each worst C‐shortcut for each bad shortest even hole C of ^G is not C‐shallow. Observe that if ^G is shallow and antishallow, thenG contains a good shortest even hole.

Lemma 3 (Chudnovsky et al. [11, Lemma 4.5]). For anyn‐vertex graph^G that contains even holes, it takesO n( ²⁵)time to obtain induced subgraphsG1, …, Gr ofGwithr=O n( ²³) such that aGiwithi∈[1, ]r is shallow and contains a shortest even hole ofG.

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Lemma 4. For anyn‐vertex graph^G, it takesO n( )⁶ time to obtain a subgraphC ofGsuch that(i)C is a shortest even hole ofG or(ii)G is antishallow.

A graphGis long ifGdoes not contain any even hole with at most 22 vertices. A graphGis badifG does not contain any good shortest even hole.

Lemma 5. For anyn‐vertex long graph^G, it takesO n( )⁸ time to obtain a subgraphC ofG such that (i)C is a shortest even hole ofG or(ii)G is bad.

Proof of Theorem1. By Lemma 2, we may assume thatG contains even holes. Spend O n( ²²)time to either obtain a shortest even hole ofGor ensure thatGis long. IfGis long, then apply Lemma 3 inO n( ²⁵) time to obtain induced subgraphsG1, …, Gr of G with r=O n( ²³) such thatGi for an (unknown) index i∈[1, ]r is shallow and contains a(n unknown) shortest even holeC ofG. SinceG is long, so is eachGi. For each j∈[1, ]r, apply Lemmas4and5onGjin overall( ( ) +O n⁶ O n( ))⁸ ⋅O n( ²³) =O n( ³¹)time to obtain subgraphs Cj and D_j of Gj such thatCj or D_j is a shortest even hole of Gj unlessGj is antishallow and bad. Finally, we report aCjor D_jthat is an even hole ofGjwhose number of edges is minimized over all j∈[1, ]r. SinceGiis shallow,Gicannot be antishallow and bad. Thus, at least one ofCiandDiis a shortest even hole ofGi, which has to be a shortest

even hole ofG byC_i ⊆G_i. □

It remains to prove Lemmas4 and5.

2.1 | Proving Lemma 4

For any vertex subset U of a graphG, letG −U=G V G[ ( )⧹U].

Lemma 6 (Chudnovsky et al. [11, Lemma 5.1]). Let C be a shortest even hole ofG. Letu andv be distinct vertices ofC. Letuv‐path ^PofG be aC‐shallow worst C‐shortcut. Let^C¹ andC2 be theuv‐paths of C with   ^C1 ≤ ^C2 . If x (respectively, y) is the neighbor ofu (respectively, v) inC1andC3is the xy‐path of^C¹, then the following statements hold:

1. If R_uv is auv‐path of^G, then  P ≤ Ruv.

2. IfR_uv is auv‐path of^G with  P = Ruv, thenG R[ uv ∪C2]is a hole ofG. 3. IfRxy is an xy‐path of^G, then   C3 ≤ Rxy .

4. IfRxy is an xy‐path of^G with   C3 = Rxy , thenG R[ xy ∪C2]is a hole ofG.

Proof of Lemma4. LetGbe connected without loss of generality. For anyU⊆V G( ), let N UG[ ]denote the vertex subset ofGconsisting of the vertices in U and their neighbors in G. For any verticesuandv, letPuvbe an arbitrary shortestuv‐path of^G. For any vertex‐ disjoint edgesux andvy ofG with   Puv = Pxy − 1such thatuandv are connected in

H u v x y( , , , ) =G V G[( ( )⧹N V PG[ (uv∪Pxy) { , }])⧹ u v ∪{ , }],u v

letQ u v x y( , , , ) be a shortestuv‐path of^{H u v x y}^{( , , , )}. If there are edgesux and vy ofG such that Q u v x y( , , , ) exists andG P[ xy∪Q u v x y( , , , )] is an even hole, then report a

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shortest such even hole; otherwise, just report the empty graph. The procedure takes overallO n( )⁶ time.

The rest of the proof shows that the subgraph reported by the above procedure is a shortest even hole ofGas long asGis not antishallow, that is, auv‐path^Pis aC‐shallow worstC‐shortcut for a bad shortest even hole C of^G. LetC1andC2be theuv‐paths of C with   C1 ≤ C2 .

We first show thatG P[uv ∪C2]is a hole ofG. By Lemma6(2),G P[ ∪C2]is a hole. By Equations (1) and (2),     P ≤ C1 ≤ P + 1. We have

P = C1 − 1

    (3)

or else   P = C1 would imply thatG P[ ∪C2]is a shortest even hole ofG, contradicting that Pis aC‐bad C‐shortcut. By Lemma6(1) and the definition of Puv,

Puv = P ,

    (4)

implying thatG P[ uv∪C2]is a hole ofG by Lemma6(2).

Let x (respectively, y) be the neighbor ofu (respectively,v) inC1. By Lemma6(3), Pxy = C3 = C1 − 2.

      (5)

By Lemma 6(4), G P[ xy ∪C2] is a shortest even hole of G. Since G P[ uv ∪C2] and G P[ xy∪C2] are holes ofG, the interior ofC2 is disjoint from N V PG[ ( uv∪Pxy) { , }]⧹ u v , implyingC2 ⊆H u v x y( , , , ) and thatu and v are connected in H u v x y( , , , ). Therefore, Q u v x y( , , , ) exists with

Q u v x y( , , , ) C2 .

   ≤ (6)

By Equations (3)–(5), we have    Puv = Pxy − 1, implying that exactly one of G P[ uv∪Q u v x y( , , , )] and G P[ xy∪Q u v x y( , , , )] is an even hole of G. By Equations (3)–(6), we have

G P Q u v x y C G P Q u v x y C

[ ( , , , )] − 1,

[ ( , , , )] .

uv xy

   

∪ ≤

Therefore,G P[ xy∪Q u v x y( , , , )]is a shortest even hole ofG. □

2.2 | Proving Lemma 5

Proof of Lemma 5. Let the long graphG be connected without loss of generality. For eachi∈[0, 7], leti⁺= ( + 1) mod 8i . For each of theO n( )⁸ choices of distinct vertices v₀, …, v₇, let

C v( , …,0 v7) =P0 ∪ ⋯ ∪P7,

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where eachPiwithi∈[0, 7]is an arbitrary shortestv v_{i i}⁺‐path of^G. If one of theseO n( )⁸ subgraphsC v( , …,0 v7)is an even hole ofGwith  ≥Pi 3for eachi∈[0, 7], then report a shortest such one; otherwise, just report the empty graph. A naive implementation of the algorithm takesO n( ¹⁰) time. The algorithm can be implemented to run inO n( )⁸ time:

Spend overallO n( )⁴ time to obtaind u vG( , )and a shortestuv‐path ^{P u v}^{( , )}ofG for any verticesu andv. Spend overallO n( )⁶ time to obtain a data structure from which it takes O (1)time to determine (1) whetherG P u v[ ( , )∪P v w( , )]is a path for any three vertices u,v, andw and (2) whetherG P u v[ ( , )∪P x y( , )]is disconnected for any four verticesu, v,x, andy. It then takesO (1)time for any given verticesv₀, …, v₇to obtainC v( , …,0 v7) and whetherC v( , …,0 v7)is an even hole with  ≥Pi 3for eachi∈[0, 7].

For the correctness, the rest of the proof shows that ifG is not bad, that is, there is a good shortest even holeC ofG, then one of theO n( )⁸ iterations yields a shortest even hole ofG. SinceG is long,  ≥C 24, implying integersa≥3 andb ∈[0, 7]with

C = 8 + .a b

 

Let v₀, …, v₇be vertices ofC such that the shortestv v_{i i}⁺‐paths^CiofC withi∈[0, 7]

are edge‐disjoint and satisfy

C a a C C a b

{ , + 1} and + 2 +

4 .

i i i⁺

      









∈ ≤ 

1. We first ensure for eachi∈[0, 7]that Pi = Ci .

   

Byd v vC( ,i i⁺)≥a≥3and the fact thatC is a hole, we have  ≥Pi 2. By Pi =d v vG( ,i i⁺)≤ d v vC( ,i i) = Ci a+ 1 < 2a C

+  ≤ ≤ ^{ }4 ,Pi is aC‐shortcut. Since C is a good shortest even hole ofG, theC‐shortcut^Pⁱ isC‐good, implying ^Pⁱ ⁼^{d v v}^C^{( ,}ⁱ ⁱ⁺^{) =} ^Cⁱ .

2. Assume for contradiction that aG P[ i∪Pi⁺]withi ∈[0, 7]is not a path. Thus, P <

Ci + Ci⁺

    holds for a shortest v v_{i i}⁺⁺‐path ^P ofG P[ i∪Pi⁺]. SinceC is good and the union ofPand the longerv v_{i i}⁺⁺‐path of C is a hole of^G,Pcannot be aC‐shortcut. Hence,

a b C

P C C a b

2 + 4 =

4 < + 2 +

4 ,

i i⁺

        









≤ ≤ 

contradicting that P is an integer. Therefore, eachG P[ i∪Pi⁺]withi∈[0, 7]is a path.

3. To see thatG P[ 0 ∪ ⋯ ∪P7]is a hole, assume for contradiction integersi∈[0, 7]

andd∈[1, 3]such thatG P[ i∪Pj]with j= (i⁺+ ) mod 8d is connected. There arev vi k‐ pathQandv vi⁺ ℓ‐path R of^{G P}^[ i∪^Pj^]with{ , ℓ} = { ,k j j⁺}andV Q( ) ∩V R( )≤2. Thus,

Q + R Pi + Pj + 2 = Ci + Cj + 2 2 + 4.a

       ≤     ≤ (7)

We have

Q a + 3

  ≥ (8)

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or else ^Q ≤^a^{+ 2 < 2}^a≤^min

{

^{d v v}^C^{( ,}ⁱ ^k^), ^{ }^C4

}

and the fact thatC is good would imply thatQis aC‐good C‐shortcut, contradicting ^Q ^<^{d v v}C^{( ,}i k⁾. Similarly, we have

R a + 3

  ≥ (9)

or else^{ }^R ^≤^a^{+ 2 < 2}^a^≤^min

{

^{d v}^C⁽ ⁱ⁺^{, ),}^v^ℓ ^{ }^C⁴

}

and the fact thatC is good would imply that R is a C‐good C‐shortcut, contradicting  ^R ^<^{d v}C⁽ i⁺^{, )}^vℓ. Combining Equations (7)–(9), we have2 + 6a ≤   Q + R ≤2 + 4a , contradiction. □

3 | C O N C L U D I N G R E M A R K S

We resolve the open problem on the tractability of reporting a shortest even hole in ann‐vertex graph by presenting anO n( ³¹)‐time algorithm. The complexity is much higher than that of the currentO n( ¹⁰)time for reporting an arbitrary even hole, implied by theO n( )⁹ ‐time algorithm of Lai et al. [39, Theorem 1.6] for detecting even holes. The current time for reporting an arbitrary odd hole isO n( )⁹ , implied by theO n( )⁸ ‐time algorithm of Lai et al. [39, Theorem 1.4] for detecting odd holes. The shortest‐odd‐hole algorithm of Chudnovsky et al. [17] runs inO n( ¹⁴) time. The O n( )⁵ gap for odd holes is much smaller than theO n( ²¹)gap for even holes. It would be of interest to reduce either one of the gaps [2,12].

A C K N O W L E D G M E N T S

We thank Professor Paul Seymour for his supportive comments. Research of Hsueh‐I Lu is supported by MOST grants 110–2221–E–002–075–MY3 and 107–2221–E–002–032–MY3.

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How to cite this article: H.‐T. Cheong and H.‐I. Lu, Finding a shortest even hole in polynomial time, J. Graph Theory. (2021), 1–10.https://doi.org/10.1002/jgt.22748