Editing to Eulerian graphs

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Contents lists available atScienceDirect

Journal

of

Computer

and

System

Sciences

www.elsevier.com/locate/jcss

Editing

to

Eulerian

graphs

✩

Konrad

K. Dabrowski

a

,

∗

,

Petr

A. Golovach

b

,

Pim van

’t

Hof

b

,

Daniël Paulusma

a

a_School_of_Engineering_and_Computing_Sciences,_Durham_University,_Science_{Laboratories,}_South_Road,_Durham_DH1_3LE,_United_Kingdom b_Department_of_Informatics,_University_of_Bergen,_PB_7803,₅₀₂₀_Bergen,_Norway

a

r

t

i

c

l

e

i

n

f

o

a

b

s

t

r

a

c

t

Articlehistory:

Received25October2014

Receivedinrevisedform 11September 2015

Accepted9October2015 Availableonline11November2015 Keywords:

Euleriangraphs Graphediting Polynomialalgorithm

The Eulerian Editing problem asks, given agraph G and an integer k, whether G can be modiﬁedintoanEulerian graphusingat most k edgeadditions andedgedeletions. Weshowthatthisproblemispolynomial-timesolvableforbothundirectedanddirected graphs.Wegeneralizetheseresultsforproblemswithdegreeparityconstraintsanddegree balanceconstraints,respectively.Wealsoconsiderthevariantswherevertexdeletionsare permitted. Combinedwithknownresults,thisleadstofull complexity classiﬁcationsfor bothundirectedanddirectedgraphsandforeverysubsetofthethreegraphoperations.

1. Introduction

Graphmodificationproblemsplaya centralrole inalgorithmicgraphtheory,partlyduetothefact thatthey naturally ariseinnumerouspracticalapplications.A graphmodificationproblemtakesasinputagraph G andaninteger k,andasks whether G can be modified intoa graphbelongingto a prescribedgraph class

H

,usingat most k operationsofcertain allowedtypes.Themostcommonoperationsthatareconsideredinthiscontextareedgeadditions (

H

-Completion),edge deletions(

H

-EdgeDeletion),vertexdeletions(

H

-VertexDeletion),andacombinationofedgeadditionsandedgedeletions (

H

-Editing).Theintensive studyofgraph modiﬁcationproblemshas produceda plethora ofclassical andparameterized complexityresults(seee.g.[2–8,10,13,15–17,19,21–23,25,26]).

Anundirected(resp.directed)graph G isEulerianifitcontainsawalk thatbeginsandendswiththesamevertexand that uses everyedge (resp.arc) exactly once.As an immediateconsequence,an undirectedgraphis Eulerianifandonly ifitisconnectedandevery vertexhasevendegree.Similarly,a directedgraphisEulerianifitisstronglyconnected1 _and

balanced,i.e.thein-degreeofeveryvertexequalsitsout-degree.Euleriangraphsformawell-knowngraphclassbothwithin algorithmicandstructuralgraphtheory.

Severalgroupsof authorshaveinvestigated theproblemof decidingwhether ornot agiven undirectedgraphcan be madeEulerianusinga smallnumberofoperations.Boesch et al.[2]presentedapolynomial-timealgorithm for Eulerian Completion,andCai andYang [5]showedthat theproblems EulerianVertexDeletion and EulerianEdgeDeletionare NP-complete[5].Whenparameterizedbythenumber k ofallowedoperations,itisknownthat EulerianVertexDeletion

✩ _The_research _leading_to_these_results _has_received_funding_from _the_European_Research _Council_under_the _European_Union’s_Seventh_Framework Programme(FP/2007–2013)/ERCGrantAgreementno. 267959andfromEPSRCGrantEP/K025090/1.Anextendedabstractofthispaperappearedinthe proceedingsofFSTTCS2014[9].

*

Correspondingauthor.

E-mailaddresses:konrad.dabrowski@durham.ac.uk(K.K. Dabrowski),petr.golovach@ii.uib.no(P.A. Golovach),pim.vanthof@ii.uib.no(P. van’tHof), daniel.paulusma@durham.ac.uk(D. Paulusma).

1 _Replacing_“strongly_connected”_by_“weakly_connected”_yields_an_equivalent_deﬁnition_of_Eulerian_digraphs,_as_it_is_well-known_that_a_balanced_digraph isstronglyconnectedifandonlyitisweaklyconnected(seee.g.[8]).

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is W[1]-hard [5],while EulerianEdgeDeletion_is _{ﬁxed-parameter}_tractable _[8]_. _Cygan_{et al.} _[8]_showed_that_the _classical andparameterizedcomplexityresultsfor EulerianVertexDeletion_{and Eulerian}EdgeDeletion_also_hold_for_the_directed variants oftheseproblems.Recently, Goyalet al. [17] improvedtheﬁxed-parametertractability resultsofCygan et al.[8]

forthedirectedandundirectedvariantsof EulerianEdgeDeletionbygivingalgorithmswithrunningtimesthatare single-exponential in k. The sameauthorsalso provedthat the UndirectedConnectedOddEdgeDeletionproblem, whichasks whetherit ispossible toobtain aconnectedgraphin whichall verticeshaveodd degreeby deletingatmost k edges, is ﬁxed-parametertractablewhenparameterizedby k.

AnotherproblemthatcanbeseenasinvolvingeditingtoanEulerianmultigraphisthe ChinesePostmanproblem,also knownasthe RouteInspectionproblem[20].Inthisproblemaconnectedgraph G,togetherwithaninteger k,isgivenand thequestioniswhetherornotthereexistsaclosedwalkin G thatuseseveryedgeof G atleastonceandthathaslengthat most

|

E

(

G

)

|

+

k.Inotherwords,canweaddatotalofatmost k copiesofexistingedgesto G inordertomodify G intoan Eulerianmultigraph?EdmondsandJohnson[12]showedthatboththeundirectedanddirectedvariantofthisproblemcan be solved inpolynomial time. The RuralPostmanproblemgeneralizesthe ChinesePostmanproblem, asitrequires that onlyeveryedgeofsomesubsetof E

(

G

)

needstobeusedatleastonceintheclosedwalk.Dornet al.[10]provedthatthe RuralPostmanisﬁxed-parametertractablefordirectedmultigraphswhenparameterizedbythenumberofarcsthatmay beadded.

Ourcontribution Wegeneralize, extendandcomplementknownresultsongraphmodiﬁcationproblemsdealing with Eu-lerian graphs and digraphs.The main contribution ofthis paper consistsof two non-trivialpolynomial-time algorithms: one forsolving the EulerianEditing _problem, _and_one _for_solving _the_directed _variant_of_this_problem. _Given_the afore-mentioned NP-completenessresultfor EulerianEdgeDeletionandthefactthat

H

-Editing is NP-completeformanygraph classes

H

[3,26],weﬁnditparticularlyinterestingthat EulerianEditing turnsouttobepolynomial-timesolvable.Tothe best ofourknowledge,theonlyother naturalnon-trivialgraphclass

H

forwhich

H

-Editing isknowntobe polynomial-timesolvableistheclassofsplitgraphs[18].

In fact, ourpolynomial-timealgorithms are implications oftwo more generalresults.In orderto formally state these results, we need to introduce some terminology. Let ea, ed and vd denote the operations edge addition, edge deletion and vertexdeletion, respectively. Forany set S

⊆ {

ea

,

ed

,

vd

}

andnon-negative integer k, we say that a graph G canbe

(

S

,

k

)

-modiﬁed intoagraph H if H canbeobtainedfrom G byusingatmost k operationsfrom S.Wedeﬁnethefollowing problemforevery S

⊆ {

ea

,

ed

,

vd

}

:

CDPE

₍

S

₎

: Connected Degree Parity Editing

₍

S

₎

Instance: A (simple) graph G, an integer k and a function

δ

: V

(

G

)

→ {

0

,

1

}

.

Question: Can G be

(

S

,

k

)

-modiﬁed into a connected graph H with dH

(

v

)

≡ δ(

v

) (

mod 2

)

for each v

∈

V

(

H

)

?

Inspired by thework of Cygan et al. [8]on directed Euleriangraphs, we also studya naturaldirected variant of the CDBE

₍

S

₎

problem. Denoting the in- and out-degreeof a vertex v ina digraph G by din

G

(

v

)

and doutG

(

v

)

, respectively, we

deﬁnethefollowingproblemforevery S

⊆ {

ea

,

ed

,

vd

}

:

CDBE

₍

_S

₎

: Connected Degree Balance Editing

₍

_S

₎

Instance: A (simple) digraph G, an integer k and a function

δ

: V

(

G

)

→ Z

.

Question: Can G be

(

S

,

k

)

-modiﬁed into a weakly connected digraph H with dout_H

(

v

)

−

din_H

(

v

)

= δ(

v

)

for each

v

∈

V

(

H

)

?

In Section 3, we prove that CDPE

₍

S

₎

can be solved in polynomial time when S

= {

ea

}

and when S

= {

ea

,

ed

}

. The ﬁrst ofthesetwo results extendsthe resultby Boesch et al.[2]on EulerianCompletion andthesecond yields the ﬁrst polynomial-time algorithm for EulerianEditing_, _as_these_problems _are _equivalent _to CDPE

₍

{

_ea

})

_andCDPE

₍

{

_ea

_,

_ed

})

_, re-spectively, whenweset

δ

≡

0 (i.e. when

δ(

v

)

=

0 forevery v

∈

V

(

G

)

).Thecomplexity oftheproblemdrasticallychanges when vertex deletion is allowed: we prove that forevery subset S

⊆ {

ea

,

ed

,

vd

}

with vd

∈

S, the CDPE

(

S

)

problem is NP-completeand W[1]-hardwithparameter k,evenwhen

δ

≡

0.ThiscomplementsresultsbyCaiandYang[5]statingthat CDPE

(

S

)

is NP-completeand W[1]-hardwithparameter k when S

= {

vd

}

and

δ

≡

0 or

δ

≡

1.Ourresults,together withthe

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Table 1

AsummaryoftheresultsforCDPE(S)andCDBE(S).All results arenewexceptthosefor which areferenceis given.The numberofallowedoperations k isthe pa-rameterintheparameterizedresults,andifa parame-terizedresultisstated,thenthecorrespondingproblem is NP-complete. S CDPE(S) CDBE(S) ea,ed P P ea P P ed FPT[8] FPT[8] vd W[1]-hard[5] W[1]-hard[8] ea,vd W[1]-hard W[1]-hard ed,vd W[1]-hard W[1]-hard ea,ed,vd W[1]-hard W[1]-hard

aforementionedresults duetoCygan et al.[8]2 _and_Cai _and_Yang_[5]_,_yield _a_complete _{classiﬁcation}_of_both _the_classical

andtheparameterizedcomplexityofCDPE

₍

_S

₎

_for_all _S

⊆ {

_ea

_,

_ed

_,

_vd

}

_;_see_the_middle_column_of_{Table 1}_.

InSection4,weusedifferentandmoreinvolvedargumentstoclassifytheclassicalandparameterizedcomplexityofthe CDBE

(

S

)

problemforall S

⊆ {

ea

,

ed

,

vd

}

.Interestingly,theclassiﬁcationweobtainforCDBE

(

S

)

turnsouttobeidenticalto theoneweobtainedforCDPE

₍

S

₎

.Inparticular,ourproofofthefactthatCDBE

₍

S

₎

ispolynomial-timesolvablewhenS

= {

ea

}

andS

= {

ea

,

ed

}

impliesthatthedirectedvariantsof EulerianCompletionand EulerianEditingarenotsigniﬁcantlyharder thantheirundirectedcounterparts.AllresultsonCDBE

₍

_S

₎

_are_summarized_in_the_right_column_of_{Table 1}_.

We would like to emphasize that there are no obvious hardness reductions between the different problemvariants. Theparameter k in theproblemdeﬁnitionsrepresentsthebudgetforalloperationsintotal;addinganewoperationto S maycompletelychangethe problem,asthereis nowayofforbiddingitsuse.Hence, ourpolynomial-timealgorithmsfor CDPE

(

{

ea

,

ed

})

andCDBE

(

{

ea

,

ed

})

do notgeneralize the polynomial-timealgorithmsfor CDPE

(

{

ea

})

andCDBE

(

{

ea

})

,and assuchrequiresigniﬁcantlydifferentarguments.Inparticular,ourmainresult,statingthat EulerianEditingis polynomial-time solvable, is nota generalization ofthe fact that EulerianCompletionispolynomial-timesolvable andstands inno relationtothe FPT-resultbyCyganet al.[8]for EulerianEdgeDeletion_.

Weendthissectionbymentioningtwosimilargraphmodiﬁcationframeworksintheliteraturethatformedadirect mo-tivationfortheframeworkdeﬁnedinthispaper.MathiesonandSzeider[23]consideredthe DegreeConstraintEditing(S) problem, whichisthatoftestingwhetheragraph G canbe

(

S

,

k

)

-modiﬁedintoa graph H inwhich thedegreeofevery vertexbelongstosomelistassociatedwiththatvertex;recentlysomenewresultsforthisproblemwereobtainedbyFroese et al.[13]andGolovach[16].Golovach[15]performedasimilarstudytothatofMathiesonandSzeider [23],butwiththe additionalconditionthattheresultinggraphmustbeconnected.

2. Preliminaries

We considerﬁnitegraphs G

= (

V

,

E

)

that maybe undirectedor directed;in thelattercasewe will always callthem digraphs.Allourundirectedgraphswillbesimple,thatis,withoutloopsormultipleedges;inparticular,thisisthecasefor both theinputandtheoutput graphinevery undirectedproblemweconsider. Similarly,forevery directedproblemthat we consider,we onlyconsider simpledigraphs,that is,we donot allowthe input oroutputdigraph to containmultiple arcs(butwedoallow oppositearcs).Inourproofswewillalsomakeuseofdirectedmultigraphs,whicharedigraphsthat arepermittedtohavemultiplearcs.

The completegraph on n verticesis denoted Kn and thecomplete bipartite graphwithclasses ofsize s and t is

de-noted Ks,t.Thelength ofapathorcycleisequaltothenumberofedgesthatitcontains.

Wedenoteanedgebetweentwovertices u and v inagraphby uv.Wedenoteanarcbetweentwovertices u and v by

(

u

,

v

)

,where u isthetail of

(

u

,

v

)

and v isthehead.Wesaythataset A ofarcsoredgeshassize

|

A

|

.Thedisjointunion oftwographs G1 and G2isdenotedG1

+

G2.

Let G

= (

V

,

E

)

be a graphor a digraph. Throughout thepaper we assume that n

= |

V

|

andm

= |

E

|

.For U

⊆

V ,we let G

[

U

]

bethegraph(digraph)withvertexset U andanedge(arc)betweentwovertices u and v ifandonlyifthisisthe casein G; wesaythat G

[

U

]

isinducedby U .Wewrite G

−

U

=

G

[

V

\

U

]

.ForE

⊆

E,we let G

(

E

)

be thegraph(digraph) withedge(arc)set E whosevertexsetconsistsoftheend-vertices ofthe edgesin E; wesaythat G

(

E

)

isedge-induced by E.Let S beasetof(ordered)pairsofverticesof G.WeletG

−

S bethegraph(digraph)obtainedbydeletingalledges (arcs)of S

∩

E from G,andweletG

+

S bethegraph(digraph)obtainedbyaddingalledges(arcs)of S

\

E to G.Wemay write G

−

e or G

+

e if S

= {

e

}

.

LetG

= (

V

,

E

)

beagraph.A component of G isamaximalconnectedsubgraphof G.Thecomplement of G isthegraph

G

= (

V

,

E

)

withvertexset V andanedgebetweentwodistinctvertices u and v ifandonlyif uv

∈

/

E.Foravertexv

∈

V ,

2 _{The FPT-results}_by_Cygan_{et al.}_[8]_only_cover_CDPE₍_{_ed_})_and_CDBE₍_{_ed_})_when_δ_≡_0,_but_it_can_easily_be_seen_that_their_results_carry_over_to CDPE({ed})andCDBE({ed})foranyfunctionδ.

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we let NG

(

v

)

= {

u

|

uv

∈

E

}

denoteits (open)neighbourhood.The degree of v isdenoteddG

(

v

)

= |

NG

(

v

)

|

.The graph G is even ifall itsverticeshaveevendegree,anditisEulerian ifit isevenandconnected.Wesaythat aset D

⊆

E isan edge cut-set in G ifG

−

D hasmorecomponentsthan G.Anedgecut-setofsize 1 issaidtobeabridge.

Amatching ofa graph G isaset ofedges,inwhichnotwo edgeshaveacommonend-vertex; itiscalledamaximum

matching ifits numberof edges ismaximum overall matchings of G.We need the followinglemma dueto Micaliand Vazirani.

Lemma1.([24]) Amaximummatchingofann-vertexgraphcanbefoundin O

(

n5/2

₎

_time.

Let G

= (

V

,

E

)

be adigraph. If

(

u

,

v

)

isan arc, then

(

v

,

u

)

isthe reverse ofthisarc. Fora subset F

⊆

E, we let FR

₌

{(

u

,

v

)

|(

v

,

u

)

∈

F

}

denote the setof arcswhose reverse isin F . The underlying graph of G is theundirected graphwith vertexset V wheretwoverticesu

,

v

∈

V areadjacentifandonlyif

(

u

,

v

)

or

(

v

,

u

)

isanarcin G.Wesaythat G is(weakly) connected if its underlying graphis connected.A component of G is a connected componentof its underlyinggraph. An arc a

∈

E is a bridge in G ifit is a bridge inthe underlying graphof G. A vertex u isan in-neighbour orout-neighbour

of avertex v if

(

u

,

v

)

∈

E or

(

v

,

u

)

∈

E,respectively. LetNin_G

(

v

)

= {

u

| (

u

,

v

)

∈

E

}

andNout_G

(

v

)

= {

u

| (

v

,

u

)

∈

E

}

,where we call din_G

(

v

)

= |

Nin_G

(

v

)

|

and dout_G

(

v

)

= |

Nout_G

(

v

)

|

the in-degree and out-degree of v, respectively. A vertex v

∈

V is balanced

if dout_G

(

v

)

=

d_Gin

(

v

)

,or equivalently,its degreebalance dout_G

(

v

)

−

din_G

(

v

)

=

0.Recall that G is Eulerian if it is connectedand

balanced,thatis,theout-degreeofeveryvertexisequaltoitsin-degree.

Let G

= (

V

,

E

)

be a graph and let T

⊆

V . A subset J

⊆

E is a T -join if the set of odd-degree vertices in G

(

J

)

is precisely T . If G is connectedand

|

T

|

is eventhen G has atleast one T -join. InSection 3 we need to ﬁnda minimum T -join,thatis,oneofminimumsize.WeusethefollowingresultofEdmondsandJohnson [12]todoso.

Lemma2.([12]) LetG

= (

V

,

E

)

beagraph,andletT

⊆

V .ThenaminimumT -join(ifoneexists)canbefoundin O

(

n3

)

time. Lemma 2wasusedbyCyganet al.[8]tosolve

H

-EdgeDeletion_in_polynomial_time_when

H

_is_the_class_of_even_graphs. Itwouldimmediatelyyieldapolynomial-timealgorithmforCDPE

₍

{

ed

})

ifwedroppedtheconnectivitycondition.

WeneedavariantofLemma 2fordigraphsinSection4.LetG

= (

V

,

E

)

beadirectedmultigraphandlet f

:

T

→ Z

bea functionforsomeT

⊆

V .A multisetE

⊆

E withT

⊆

V

(

G

(

E

))

isadirectedf -join in G ifthefollowingtwoconditionshold:

dout_G₍_E)

(

v

)

−

d

in

G(E)

(

v

)

=

f

(

v

)

forevery v

∈

T and d

out

G(E)

(

v

)

−

d

in

G(E)

(

v

)

=

0 forevery v

∈

V

(

G

(

E

))

\

T .A directed f -joinis

minimum ifithasminimumsize.ThenextlemmawasusedbyCyganet al.[8]tosolve

H

-EdgeDeletioninpolynomialtime when

H

istheclassofbalanced digraphs;itwouldalsoyieldapolynomial-timealgorithmforCDBE

(

{

ed

})

ifwedropped theconnectivitycondition.

Lemma3.([8]) LetG

= (

V

,

E

)

beadirectedmultigraphand f

:

T

→ Z

beafunctionforsomeT

⊆

V .A minimumdirected f -join F (ifoneexists)canbefoundinO

(

nmlog nlog log m

)

time.Moreover,F consistsofmutuallyarc-disjointdirectedpathsfromvertices u with f

(

u

)

>

0 tovertices v with f

(

v

)

<

0.

3. Connecteddegreeparityediting

LetS

⊆ {

ea

,

ed

,

vd

}

.InSection3.1wewillshowthatCDPE

₍

_S

₎

_is_{polynomial-time}_solvable_if_S

= {

_ea

}

_or_S

= {

_ea

_,

_ed

}

_and inSection3.2wewillshowthatitis NP-completeand W[1]-hardwithparameter k ifvd

∈

S.

3.1. Thepolynomial-timesolvablecases

First,let

{ea}

⊆

S

⊆ {ea

,

ed}.Let

(

G

,

δ,

k

)

bean instanceofCDPE

₍

S

₎

withG

= (

V

_,

E

₎

.Let A beasetofedgesnot in G, andlet D beasetofedgesin G,withD

= ∅

ifS

= {

ea

}

.Wesaythat

(

A

,

D

)

isasolution for

(

G

,

δ,

k

)

ifitssize

|

A

|

+ |

D

|

≤

k,

the congruence dH

(

u

)

≡ δ(

u

) (

mod 2

)

holdsfor every vertex u and thegraph H

=

G

+

A

−

D is connected; ifwe drop

thecondition that H isconnectedthen

(

A

,

D

)

isasemi-solution for

(

G

,

δ,

k

)

.IfS

= {

ea

}

we maydenotethesolutionby A ratherthan

(

A

,

D

)

(sinceD

= ∅

).WeconsidertheoptimizationversionforCDPE

₍

_S

₎

_._The_input_is_a_pair

₍

_G

_,

_δ)

_,_and_we_aim toﬁndtheminimum k suchthat

(

G

,

δ,

k

)

hasasolution(ifoneexists).Wecallsuchasolutionoptimal anddenoteitssize by opt_S

(

G

,

δ)

.We saythata(semi)-solutionfor

(

G

,

δ,

k

)

isalsoa(semi)-solutionfor

(

G

,

δ)

.If

(

G

,

δ,

k

)

hasnosolutionfor anyvalueof k,then

(

G

,

δ)

isano-instance ofCDPE

(

S

)

andopt_S

(

G

,

δ)

= +∞

.

Let T

= {

v

∈

V

|

dG

(

v

)

≡ δ(

v

) (

mod 2

)

}

.Deﬁne GS

=

Kn if S

= {

ea

,

ed

}

and GS

=

G if S

= {

ea

}

. Notethat if S

= {

ea

}

then GS containsnoedgesof G,sointhiscaseanyT -joinin GS canonlycontainedgesin E

(

G

)

.Thefollowingkeylemma

isaneasyobservation.

Lemma4.Let

{

ea

}

⊆

S

⊆ {

ea

,

ed

}

.Let

(

G

,

δ)

beaninstanceofCDPE

₍

S

₎

and A

⊆

E

₍

G

₎

,D

⊆

E

₍

G

₎

.Then

₍

A

_,

D

₎

isasemi-solutionof CDPE

(

S

)

ifandonlyifA

∪

D isaT -joinin G_S.

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WerecallthatBoesch et al.[2]provedthat EulerianCompletion_can_be_solved _in_polynomial_time, _that_is,CDPE

₍

_S

₎

_is polynomialtimesolvableifS

= {

ea

}

and

δ

≡

0.Weextendthisresulttoarbitrary

δ

.Ourproofisbasedaroundsimilarideas butwe alsohadtodosome furtheranalysis. Themaindifferenceinthetwoproofsisthefollowing.If

δ

≡

0 thennoneof theaddededgesinasolutionwillbeabridge inthemodiﬁed graph(asthenumberofverticesofodddegreeinagraph isalwayseven).Howeverthisisnolongertrueforarbitrary

δ

andextraargumentsareneeded.Wethereforepresentafull proofofourresult.

Theorem1.LetS

= {

ea

}

.ThenCDPE

₍

_S

₎

_can_be_solved_{in O}

₍

_n3

₎

_time.

Proof. Let S

= {

ea

}

andlet

(

G

,

δ)

bean instanceof CDPE

₍

_S

₎

_._We _ﬁrst_use_{Lemma 2}_to _check _{in O}

₍

_n3

₎

_time _{whether G}_S hasa T -join.Ifnot then

(

G

,

δ)

hasnosemi-solutionby Lemma 4,andthus nosolutioneither.Wemaythereforeassume that

|

T

|

isevenand F isaminimumT -joinin GS.(RecallthatLemma 2statesthatwecanﬁnd F in O

(

n3

)

timeifitexists.)

Wealsoassumethateither T

= ∅

or G isnotconnected,otherwisethetrivialsolutionA

= ∅

isclearlyoptimal.Let p bethe numberofcomponentsof G thatdonotcontainanyvertexof T andlet q bethenumberofcomponentsof G thatcontain atleastonevertexof T .

Wewillprove thefollowingseriesofstatements.Undertheassumptions madeinthe previousparagraph,these state-mentsgivenecessaryandsuﬃcientconditionsfor

(

G

,

δ)

tobeayes-instance,andif

(

G

,

δ)

isayes-instancetellustheexact sizeofanoptimalsolutionfor

(

G

,

δ)

.WerecallthatG1

+

G2denotesthedisjointunionoftwographs G1 and G2.

–

(

G

,

δ)

isano-instanceifp

=

2

,

q

=

0 andG

=

K1

+

Kt fort

≥

1.

– opt_S

(

G

,

δ)

=

4 if p

=

2

,

q

=

0 andG

=

Ks

+

Kt fors

,

t

≥

2.

– optS

(

G

,

δ)

=

3 if p

=

2

,

q

=

0 and G hasacomponentthatisnotcomplete.

– optS

(

G

,

δ)

=

p if p

≥

3

,

q

=

0.

– optS

(

G

,

δ)

=

max

{|

F

|,

p

+

q

−

1

,

p

+

12

|

T

|}

ifq

>

0.

Wesplitourproofintotwopartsdependingonthevalueof q.

Case1: q

=

0.

In this case T

= ∅

, so by Lemma 4 for any semi-solution A, every vertex in GS

(

A

)

must have even degree in GS

(

A

)

.

In other words, every vertexof G must be incident to an even number ofedges in A. Since T

= ∅

,we assumedabove that G wasdisconnected,sop

≥

2 andanysolution A mustbenon-empty.Thismeansthat GS

(

A

)

mustcontainacycle,so optS

(

G

,

δ)

≥

3.Recallthat GS

(

A

)

isasubgraphof G.

Suppose p

=

2.IfG

=

K1

+

Kt fort

≥

2 then G

=

K1,t,whichdoesnotcontainacycle.Therefore

(

G

,

δ)

isano-instance

in this case. If G

=

Ks

+

Kt for s

,

t

≥

2 then G

=

Ks,t, which contains no cycles of length 3. Therefore optS

(

G

,

δ)

≥

4

in this case. Indeed, if u

,

v are vertices in the Ks component of G and u

,

v are vertices in the Kt component, then A

= {

uu

,

uv

,

v v

,

vu

}

is a solution of size 4 and this solution must therefore be optimal. Finally, suppose G contains exactlytwocomponents,atleastoneofwhichisnotaclique.Let x

,

y benon-adjacentverticesinthiscomponentandlet z beavertexintheothercomponent.Then A

= {

xy

,

yz

,

zx

}

isasolutionofsize 3,whichmustthereforebeoptimal.

Finally,supposethat p

≥

3.Since G

+

A mustbeconnectedforanysolution A,everycomponentin G mustcontainat leastonevertexincidenttoanedgeof A.ByLemma 4,thisvertexmustbeincidenttoanevennumberofedgesof A, mean-ingthat itmustbeincidenttoatleasttwosuchedges.Therefore optS

(

G

,

δ)

≥

p.Indeed,ifwechoosevertices v1

,

. . . ,

vp,

onefromeachcomponentof G then A

= {

v1v2

,

v2v3

,

. . . ,

vp−1vp

,

vpv1

}

isasolutionofsize p,whichisthereforeoptimal. Thisconcludestheq

=

0 case.

Case2: q

>

0.

In this case T

= ∅

.We ﬁrst show that optS

(

G

,

δ)

≥

max

{|

F

|,

p

+

q

−

1

,

p

+

12

|

T

|}

. Since F is a minimum T -join in GS, Lemma 4impliesthatoptS

(

G

,

δ)

≥ |

F

|

.Since G has p

+

q components,anysolution A mustcontainatleastp

+

q

−

1 edges

toensure that G

+

A isconnected,so optS

(

G

,

δ)

≥

p

+

q

−

1.Finally, let G1

,

. . . ,

Gp be thecomponentsof G that donot

containanyverticesof T .If A isasolutiontheneverycomponent Gi mustcontainavertexincidenttosomeedgein A.By Lemma 4,thisvertexmustbeincidenttoanevennumberofedgesof A,meaningthatitmustbeincidenttoatleasttwo suchedges.ByLemma 4,everyvertexof T mustbeincidenttosomeedgein A.Therefore A mustcontainatleastp

+

1₂

|

T

|

edges,sooptS

(

G

,

δ)

≥

p

+

12

|

T

|

.

Nextweshowthatwecanalwaysconstructasolutionofsizemax

{|

F

|,

p

+

q

−

1

,

p

+

1₂

|

T

|}

.Todothis,wetrytoreplace edges of F insuch awaythat F remainsa minimum T -joinin GS,butthe numberofcomponentsin G

+

F isreduced.

Afterwehaveﬁnishedthisprocess,ifG

+

F isconnectedthensettingA

=

F givesasolutionofsize

|

F

|

,whichistherefore optimal.Otherwise,wewillbeabletousethestructureof F toconstructasolutionofsizeeitherp

+

q

−

1 or p

+

1₂

|

T

|

.

Consider thegraph GS

(

F

)

.Since F isaminimum T -join, GS

(

F

)

cannotcontain anycycles(otherwise theedgesinthe

cyclecouldberemovedfrom F togiveasmallerT -join).Weprovethefollowingclaim.

(6)

Suppose,forcontradiction,thatthereissuchapathwithedgeset P andend-vertices u and v.Notethat u and v arein the samecomponentofG

+

F .Since G

+

F isnot connected(otherwise A

=

F wouldbe anoptimalsolutionofsize

|

F

|

), theremustbeavertexx

∈

V

(

G

)

whichisinadifferentcomponentofG

+

F fromtheonecontaining u and v.Inthiscase

ux

,

xv

∈

E

(

GS

)

.Let F

= (

F

\

P

)

∪ {

ux

,

xv

}

.Then F isalsoa T -join in GS,since thedegreeparityofanyvertexin G

+

F

isthesameasitsdegreeparityinG

+

F .However,

|

F

|

<

|

F

|

,whichcontradictsthefactthat F isaminimumT -join.This provesClaim 1.

Nowsupposethatu

,

v

,

u

,

varefourdistinctverticesin F withuv

,

uv

∈

F ,suchthat uv isnotabridgeinG

+

F and

thevertices u and uareindifferentcomponentsofG

+

F .Let F

= (

F

\ {

uv

,

uv

})

∪ {

uv

,

uv

}

.Then Fisalsoaminimum

T -joinin GS.However,G

+

F hasonecomponentlessthanG

+

F .Indeed,since uv isnotabridgeinG

+

F ,thevertices u

,

u

,

v

,

v mustall be inthesamecomponentof G

+

F.Therefore,ifsuchedges uv

,

uv

∈

F exist,we replace F by F. Wedothisexhaustivelyuntilnofurthersuchpairsofedgesexist.Atthispointeithereveryedgein F mustbeabridgeor everyedgein F isinthesamecomponentofG

+

F .Weconsiderthesepossibilitiesseparately.

First suppose that every edge in F is a bridge. Chooseuv

∈

F andlet G1

,

. . . ,

Gk be the componentsof G

+

F , with u

,

v

∈

V

(

G1

)

.Notethatsinceeveryedgein F isabridge,k

=

p

+

q

− |

F

|.

Nowlet vi

∈

V

(

Gi

)

fori

∈ {2

,

. . . ,

k

}.

LetA

=

F if k

=

1 and A

= (

F

\ {

uv

})

∪ {

uv2

,

v2v3

,

. . . ,

vk−1vk

,

vkv

}

otherwise.Now everyvertexin G

+

A hasthesamedegreeparity

asinG

+

F ,so A isaT -joinin GS.ThegraphG

+

A isconnected,so A isasolution.However,

|

A

|

= |

F

|

−

1

+

p

+

q

− |

F

|

=

p

+

q

−

1.Therefore A isanoptimalsolution.

We maynowassume thateveryedgein F is inthesamecomponentofG

+

F .IfG

+

F isconnected,then A

=

F isa solutionofsize

|

F

|

andisthereforeoptimal,sowemayassumethatG

+

F isnotconnected.

Supposeuv

,

v w

∈

F .Thenu w

∈

E

(

G

)

,asotherwisewecouldreplace uv

,

v w in F by u w togetasmallerT -join in GS.

Suppose that uv

,

v w do not forman edge cut-set in G

+

F .In other words, we suppose that u and v are inthe same componentofG

+ (

F

\ {

uv

,

v w

})

.Let x beavertexinadifferentcomponentofG

+

F fromtheonecontainingu

,

v

,

w.Then

ux

,

xw

∈

E

(

GS

)

.Let F

= (

F

\ {

uv

,

v w

})

∪ {

ux

,

xw

}

.Then F must alsobe a minimumT -joinin GS.However, G

+

F has

one lesscomponentthan G

+

F .Indeed,x is inthe samecomponentofG

+

F asu

,

v

,

w.Inthiscasewemayreplace F by F. Again, we apply this replacement exhaustively until it can no longer be applied. This process ends when either

G

+

F becomesconnected(inwhichcase A

=

F isan optimalsolutionofsize

|

F

|

)or,foreverypairofedgesoftheform

uv

,

v w

∈

F ,weﬁndthat

{

uv

,

v w

}

isanedgecut-setinG

+

F .Wemayassumethelatteristhecase. Wewillprovethefollowingclaim.

Claim 2.Let uv

,

v w

∈

F .Thenthecomponent C of G

+ (

F

\ {

uv

,

v w

})

thatcontains v containsnovertices of T .Moreover, dGS(F)

(

v

)

=

2 and v istheuniquevertexof GS

(

F

)

in C .

WeproveClaim 2asfollows.Weﬁrstshowthat C containsnoverticesof T .Suppose,forcontradiction,thatx

∈

T

∩

V

(

C

)

(x is notnecessarily distinctfrom v). Then byLemma 4, x mustbe theend-vertex ofsomeedge in F

\ {

uv

,

v w

}

,say xy (again y isnotnecessarilydistinctfrom v).Notethat x and y areinthesamecomponentofG

+ (

F

\ {

uv

,

v w

})

,whichis differentfromthecomponentcontaining u and w. Let F

= (

F

\ {

xy

,

uv

,

v w

})

∪ {

ux

,

w y

}

.Then F isalsoa T -join in GS,

but

|

F

|

= |

F

|

−

1,contradictingtheminimalityof F .Hence,C containsnoverticesof T .

ByClaim 1andthedeﬁnitionofaT -join,everyvertexof GS

(

F

)

thatisnotin T musthaveaneighbourin T .Inparticular,

this meansthat inthe graph G

+ (

F

\ {

uv

,

v w

})

,no vertexof V

(

C

)

\ {

v

}

hasaneighbour in T andthe vertex v has no neighboursin T

\ {

u

,

w

}

,otherwiseinbothcasessuchaneighbourwouldbein C ,acontradiction.Thiscompletestheproof ofClaim 2.

Recall,thatbyClaim 1,GS

(

F

)

isaforestinwhicheachcomponentisastar.Then,byClaim 2,eachcomponentof GS

(

F

)

is in fact a path of length 1 or 2. Hence, GS

(

F

)

consists of 1₂

|

T

|

vertex-disjoint paths of length at most 2 with their

ends in T .Since GS

(

F

)

has

|

F

|

edges, GS

(

F

)

consistsof

|

T

|

− |

F

|

paths oflength 1 and

|

F

|

−

1₂

|

T

|

paths oflength 2.By Claim 2,themiddlevertexofeverypathoflength 2 liesinadifferentcomponentofthose p componentsof G thatdonot contain anyverticesof T .LetG0

,

G1

,

. . . ,

Gk be thecomponentsofG

+

F suchthat G0 istheonly componentcontaining vertices of T . Note that k

=

p

− (|

F

|

−

1₂

|

T

|)

. Let vi

∈

V

(

Gi

)

for i

∈ {

1

,

. . . ,

k

}

. Choose uv

∈

F and let A

= (

F

\ {

uv

})

∪

{

uv1

,

v1v2

,

. . . ,

vk−1vk

,

vkv

}

. Then every vertexin G

+

A hasthe samedegree parity asin G

+

F and the graph G

+

A

is connected, so A isa solution. Furthermore,

|

A

|

= |

F

|

+

p

− (|

F

|

−

1₂

|

T

|)

=

p

+

1₂

|

T

|

,so A is an optimalsolution. This concludestheproofofCase 2.

Notethat p and q canbecomputedand T canbefoundin O

(

n

+

m

)

time.RecallthataminimumT -joinin GS canbe

found in O

(

n3

)

time by Lemma 2,sothe valueof optS

(

G

,

δ)

canbe computedin O

(

n3

)

time. Notethat theconstructive

proofsforCases 1and 2canbeturnedintoalgorithms.ForCase 1,ifp

=

2,thenwecancheckin O

(

n

+

m

)

timewhether ornot G isthedisjointunionoftwocliquesandeitherreturnano-answerorﬁndasolution.If p

≥

3,thenwecanﬁnda solution in O

(

n

+

m

)

time.ForCase 2,observethat fora given F ,we canﬁndthecomponentsandthebridges ofG

+

F

in O

(

n2

₎

_time._Hence,_as_G

₊

_{F has}_at_{most n components,}_glueing_these_components_of_G

₊

_{F via}_non-bridge_edges_uv

_∈

_F can be done in O

(

n3

)

time. Ifevery edge of F is a bridgeof G

+

F , then therest ofthe construction ofa solution can be done in O

(

n

)

time.Otherwise wedo asfollows.We partition theedges of GS

(

F

)

into pathsoflength 1 andpaths of

length 2 with their endsin T . This canbe done in O

(

n

)

time. The replacements ofpaths oflength 2 consisting oftwo edges uv

,

v w thatdonotformanedgecut-setofG

+

F bypairsofedgesux

,

xw thatdocanbedonein O

(

n3

₎

_time._After

(7)

this, wecan ﬁndanoptimalsolution in O

(

n

)

time.We concludethat an optimalsolution A canbe foundin O

(

n3

)

total

time.

2

Wearenowreadytopresentthemainresultofthissection.Provingthisresultrequiressigniﬁcantlydifferentarguments thantheonesusedintheproofofTheorem 1.LetS

= {

ea

,

ed

}

andlet

(

G

,

δ)

beaninstanceofCDPE

(

S

)

.If F isaT -join in

GS

=

Kn,let D

=

F

∩

E

(

G

)

and A

=

F

\

D.ThenbyLemma 4,

(

A

,

D

)

isasemi-solution.Notethatif F isaminimumT -join

in GS thenitisamatchinginwhicheveryvertexof T isincidenttopreciselyoneedgeof F ,so

|

F

|

=

1₂

|

T

|

.Wewillshow

howthisallowsustocalculateoptS

(

G

,

δ)

directlyfromthestructureof G,withouthavingtoﬁndaT -join.Notethatthere

isnoconnectedgraphonexactlytwoverticesinwhichboth verticeshaveodddegree.Similarly, nographcancontain an oddnumberofverticesofodddegree.Thismeansthatnot everyinstanceofCDPE

(

S

)

isayes-instance. However,wewill showthatallno-instancesaretrivial,thatis,theyoccurwhen

|

T

|

isoddor G containsonlytwovertices.

Theorem2.LetS

= {

ea

,

ed

}

.ThenCDPE

₍

S

₎

canbesolvedinO

₍

n

+

m

₎

timeandanoptimalsolution(ifoneexists)canbefound

in O

(

n3

₎

_time.

Proof. Let S

= {

ea

,

ed

}

andlet

(

G

,

δ)

be aninstanceofCDPE

₍

_S

₎

_._By_{Lemma 4}_,_we_may_assume_that

|

_T

|

_is_even,_otherwise

(

G

,

δ)

is a no-instance.IfG

=

K2 and T

=

V

(

G

)

, or G

=

K1

+

K1 and T

= ∅

, then

(

G

,

δ)

isa no-instance.If G

=

K2 and

T

= ∅

then,trivially,optS

(

G

,

δ)

=

0,andifG

=

K1

+

K1 andT

=

V

(

G

)

thenoptS

(

G

,

δ)

=

1.Toavoidthesetrivialinstances,

wethereforeassumethat G containsatleastthreevertices.UndertheseassumptionswewillshowthatoptS

(

G

,

δ)

isalways

ﬁniteandgiveexactformulasforthevalueofoptS

(

G

,

δ)

.Let p bethenumberofcomponentsof G thatdonotcontainany

vertexof T and let q be thenumber ofcomponents of G thatcontainat leastone vertexof T . Weprove thefollowing seriesofstatements.

– optS

(

G

,

δ)

=

0 if p

=

1

,

q

=

0,

– optS

(

G

,

δ)

=

max

{

3

,

p

}

ifp

≥

2

,

q

=

0,

– optS

(

G

,

δ)

=

12

|

T

|

+

1 if p

=

0

,

q

=

1,G

[

T

]

=

K1,r,forsomer

≥

1,andeachedgeof G

[

T

]

isabridgeof G,

– opt_S

(

G

,

δ)

=

max

{

p

+

q

−

1

,

p

+

1₂

|

T

|}

inallothercases.

Notethatifp

=

1

,

q

=

0,thentheﬁrststatementappliesandthetrivialsolution

(

A

,

D

)

= (∅,

∅)

isoptimal.Wenowconsider theremainingthreecasesseparately.

Case1: p

≥

2 andq

=

0.

Then T

= ∅

,sobyLemma 4foranysemi-solution

(

A

,

D

)

,everyvertexinGS

(

A

∪

D

)

musthaveevendegreeinGS

(

A

∪

D

)

.

In other words, every vertexof G must be incident to an even numberof edges in A

∪

D. Since p

≥

2, the graph G is disconnected,soanysolution

(

A

,

D

)

isnon-empty.Thismeansthat GS

(

A

∪

D

)

mustcontainacycle,sooptS

(

G

,

δ)

≥

3 ifa

solutionexits.Suppose p

=

2.As G hasatleastthreevertices,itcontains acomponentcontaining an edge xy.Let z bea vertexinits other component. Weset A

= {

xz

,

yz

}

and D

= {

xy

}

to obtaina solutionfor

(

G

,

δ)

.Since

|

A

|

+ |

D

|

=

3, this solutionisoptimal.Supposep

≥

3.SinceG

+

A

−

D mustbeconnectedforanysolution

(

A

,

D

)

,everycomponentin G must containatleastonevertexincidenttoanedgeof A.ByLemma 4,thisvertexmustbeincidenttoanevennumberofedges of A

∪

D,meaningthatitmustbeincidenttoatleasttwosuchedges.Thereforeopt_S

(

G

,

δ)

≥

p.Indeed,ifwechoosevertices

v1

,

. . . ,

vp,onefromeachcomponentof G,thensettingA

= {

v1v2

,

v2v3

,

. . . ,

vp−1vp

,

vpv1

}

andD

= ∅

givesasolutionof size p,whichisthereforeoptimal.ThisconcludesCase 1.

Case2: p

=

0

,

q

=

1,G

[

T

]

=

K1,rforsomer

≥

1 andeachedgeof G

[

T

]

isabridgeof G.

Then G isconnected.Let v0 bethe centralvertexof thestar G

[

T

]

andlet v1

,

. . . ,

vr be theleaves. ByLemma 4, inany

semi-solution

(

A

,

D

)

,everyvertexof T mustbeincidenttoanoddnumberofedgesin A

∪

D,sooptS

(

G

,

δ)

≥

12

|

T

|

. Weclaimthat optS

(

G

,

δ)

≥

12

|

T

|

+

1.Suppose,forcontradiction, that

(

A

,

D

)

isasemi-solutionofsize

|

A

|

+ |

D

|

=

1 2

|

T

|

. Then A

∪

D mustbeamatchingwitheachedgejoiningapairofverticesof T .However,thenv0vi

∈

A

∪

D forsome i.Since v0vi

∈

E

(

G

)

,wemusthavev0vi

∈

D.However,since v0viisabridgeof G, v0and vimustthenbeindifferentcomponents

ofG

+

A

−

D,soG

+

A

−

D isnotconnectedand

(

A

,

D

)

isnotasolution,acontradiction.ThereforeoptS

(

G

,

δ)

≥

12

|

T

|

+

1. Nextweshowhowtoﬁndasolutionofsize 1₂

|

T

|

+

1.Since

|

T

|

iseven,r mustbeodd.Firstsupposethatr

=

1.Since G isconnectedand v0v1 isabridge,G

\ {

v0v1

}

hasexactlytwocomponents.Since G containsatleastthreevertices,one of thesecomponentscontainsanothervertex x.Withoutlossofgeneralityassumexv0

∈

E

(

G

)

,inwhichcasexv1

∈

/

E

(

G

)

.Then settingA

= {

xv1

}

andD

= {

xv0

}

givesasemi-solution.Sincex

,

v0

,

v1areallinthesamecomponentofG

+

A

−

D,thegraph

G

+

A

−

D mustbeconnected,so

(

A

,

D

)

isasolution.Since

|

A

|

+ |

D

|

=

2

=

1₂

|

T

|

+

1,thissolutionisoptimal.Nowsuppose

r

≥

3.LetA

= {

v1v2

,

v2v3

}

∪{

v2iv2i+1

|

2

≤

i

≤

1₂

(

r

−

1

)

}

andD

= {

v0v2

}

.Then

(

A

,

D

)

isasemi-solutionandsincev0

,

. . . ,

vr

areallinthesamecomponentofG

+

A

−

D,weﬁndthat

(

A

,

D

)

isasolution.Since

|

A

|

+|

D

|

=

2

+

1₂

(

r

−

1

)

−

1

+

1

=

1₂

|

T

|

+

1, thissolutionisoptimal.ThisconcludesCase 2.

(8)

Case3: q

≥

1 andCase 2doesnothold.

Then T

= ∅

.LetG1

,

. . . ,

Gp bethecomponentsof G withoutverticesof T andletG

=

G

− (

V

(

G1

)

∪ · · · ∪

V

(

Gp

))

.Notethat G

=

G if p

=

0 andthat G isnottheemptygraph,asq

>

0.Choosevi

∈

V

(

Gi

)

fori

∈ {

1

,

. . . ,

p

}

.

WeﬁrstshowthatoptS

(

G

,

δ)

≥

max

{

p

+

q

−

1

,

p

+

12

|

T

|}

.Since G hasp

+

q components,anysolution

(

A

,

D

)

mustcontain atleastp

+

q

−

1 edgesin A toensurethat G

+

A

−

D isconnected,sooptS

(

G

,

δ)

≥

p

+

q

−

1.If

(

A

,

D

)

isasolutionthen

every component Gi mustcontain avertexincident tosome edgein A. ByLemma 4, thisvertexmust beincident toan

evennumberofedgesofA

∪

D,meaningthatitmustbeincidenttoatleasttwosuchedges.ByLemma 4,everyvertexof T mustbeincidenttosomeedgein A

∪

D.Therefore A

∪

D mustcontainatleastp

+

1₂

|

T

|

edges,sooptS

(

G

,

δ)

≥

p

+

12

|

T

|

.

We now show howto ﬁnda solutionof sizemax

{

p

+

q

−

1

,

p

+

1₂

|

T

|}

.We start by ﬁndinga maximummatching M in G

[

T

]

.Let U bethesetofverticesin T thatarenot incidenttoanyedgein M.Wedividetheargumentintotwocases, dependingonthesizeof U .

Case3a: U

= ∅

.

Inthiscase,byLemma 4,setting A

=

M andD

= ∅

givesasemi-solution.Nowsupposethatuv

,

uv

∈

M,suchthat uv is notabridgeinG

+

M andthevertices u and uareindifferentcomponentsofG

+

M.Let M

= (

M

\ {

uv

,

uv

})

∪ {

uv

,

uv

}

. Then M isalso amaximum matchingin G

[

T

]

.However, G

+

M has onecomponent lessthan G

+

M. Indeed,since uv is nota bridgein G

+

M,the verticesu

,

u

,

v

,

v mustall beinthe samecomponentofG

+

M.Therefore,ifsuchedges

uv

,

uv

∈

M exist,we replace M by M.We dothisexhaustivelyuntil nofurther suchpairs ofedges exist.At thispoint eitherevery edgein M isabridge inG

+

M orevery edgein M isinthesamecomponentofG

+

M. Weconsiderthese possibilitiesseparately.

Firstsupposethateveryedgein M isabridgeinG

+

M.Chooseuv

∈

M andletQ1

,

. . . ,

QkbethecomponentsofG

+

M,

withu

,

v

∈

V

(

Q1

)

.Notethat sinceeveryedge in M isabridge,k

=

p

+

q

− |

M

|

.Now let xi

∈

V

(

Qi

)

fori

∈ {

2

,

. . . ,

k

}

.Let D

= ∅

andletA

=

M ifk

=

1 andA

= (

M

\ {

uv

})

∪ {

ux2

,

x2x3

,

. . . ,

xk−1xk

,

xkv

}

otherwise.NoweveryvertexinG

+

A

−

D has

thesamedegreeparityasinG

+

M,so

(

A

,

D

)

isasemi-solutionbyLemma 4.ThegraphG

+

A

−

D isconnected,so

(

A

,

D

)

isasolution.As

|

A

|

+ |

D

|

= |

M

|

−

1

+

p

+

q

− |

M

|

+

0

=

p

+

q

−

1,weﬁndthat

(

A

,

D

)

isanoptimalsolution.

Now supposethatevery edgein M isinthesamecomponentofG

+

M.Notethat G1

,

. . . ,

Gp aretheremaining

com-ponents of G

+

M.Chooseuv

∈

M. Let D

= ∅

andlet A

=

M if p

=

0 and A

= (

M

\ {

uv

})

∪ {

uv1

,

v1v2

,

. . . ,

vp−1vp

,

vpv

}

otherwise.Then everyvertexinG

+

A

−

D hasthesameparityasinG

+

M andG

+

A

−

D isconnected,sobyLemma 4

(

A

,

D

)

isasolution.Since

|

A

|

+ |

D

|

=

1₂

|

T

|

−

1

+

p

+

1

=

p

+

1₂

|

T

|

,thissolutionisoptimal.ThisconcludesCase 3a.

Case3b: U

= ∅

.

Notethat z

= |

U

|

mustbe evensince

|

T

|

iseven.Everypairofverticesin U mustbenon-adjacent in G,asotherwise M wouldnotbemaximum.Therefore G

[

U

]

isaclique.LetU

= {

u1

,

. . . ,

uz

}

.

RecallthatG

=

G

− (

V

(

G1

)

∪ · · · ∪

V

(

Gp

))

.Weclaimthat Q

=

G

+

M isconnected.Clearlyeveryvertexoftheclique U

must be inthe samecomponentof Q

=

G

+

M. Suppose forcontradictionthat Q1 isa componentof Q thatdoesnot contain U . Then Q1 mustcontain some edge w1w2

∈

M. However, inthiscase M

= (

M

\ {

w1w2

})

∪ {

u1w1

,

u2w2

}

is a largermatchinginG

[

T

]

than M,whichcontradictsthemaximalityof M.Therefore Q isconnected.

First suppose that z

≥

4 (recallthat z is even). Let M

= {

u1u2

,

u3u4

, . . . ,

uz−1uz

}

. Since U isa clique, G

+

M

−

M is

connected.Ifp

=

0 setA

=

M andD

=

M.Ifp

>

0 set A

=

M

∪ {

u1v1

,

v1v2

,

. . . ,

vp−1vp

,

vpu2

}

andD

=

M

\ {

u1u2

}

.Then

G

+

A

−

D isconnected,so

(

A

,

D

)

isasolutionbyLemma 4.Thissolutionhassize

|

A

|

+ |

D

|

=

p

+

1₂

|

T

|

,soitisoptimal. Now supposethat z

≤

3.Then z

=

2.Ifp

>

0,let A

=

M

∪ {

u1v1

,

v1v2

,

. . . ,

vp−1vp

,

vpu2

}

and D

= ∅

.Then G

+

A

−

D isconnected,so

(

A

,

D

)

isasolutionbyLemma 4.Thissolutionhassize

|

A

|

+ |

D

|

=

p

+

1₂

|

T

|

,soitisoptimal.Assumethat

p

=

0,soG

+

M containsonlyonecomponent.If u1u2isnotabridgeinG

+

M,let A

=

M andD

= {

u1u2

}

.Then G

+

M is connected,so

(

A

,

D

)

isasolution.Thissolutionhassize

|

A

|

+ |

D

|

=

p

+

1₂

|

T

|

,soitisoptimal.

Nowassumethat u1u2isabridgeinQ

=

G

+

M.Let Q1and Q2denotethecomponentsofQ

− {

u1u2

}

withu1

∈

V

(

Q1

)

andu2

∈

V

(

Q2

)

.Notethat u1u2isalsoabridgein G.Weclaimthattheedgesof M areeitherallin Q1orallin Q2.Suppose forcontradictionthat y1z1

∈

E

(

Q1

)

∩

M and y2z2

∈

E

(

Q2

)

∩

M.Then M

= (

M

\ {

y1z1

,

y2z2

})

∪ {

u1y2

,

u2y1

,

z1z2

}

wouldbe alargermatchingin G

[

T

]

than M,contradictingthemaximalityof M.Withoutlossofgenerality,wemaythereforeassume thatalledgesof M arein Q1.

LetM

= {

x1y1

,

. . . ,

xryr

}

,wherer

=

1₂

|

T

|

−

1.Weclaimthat u1 mustbeadjacentin G toallverticesofT

\ {

u1

}

.Suppose forcontradictionthat u1 isnon-adjacentin G tosome vertexofT

\ {

u1

}

.Sinceu1u2

∈

E

(

G

)

,thisvertexwouldhavetobe incident tosome edgein M.Withoutlossofgenerality, assumeu1x1

∈

/

E

(

G

)

.Then M

= (

M

\ {

x1y1

})

∪ {

u1x1

,

u2y1

}

would be alargermatchingin G

[

T

]

than M,contradicting themaximalityof M.Therefore u1 isadjacentin G toevery vertexof

T

\ {

u1

}

.Inparticular,since p

=

0,itfollowsthatq

=

1 and G isconnected.

Suppose that everyedge between u1 and T

\ {

u1

}

isa bridge in G.Then no twoverticesof T

\ {

u1

}

canbe adjacent, and G

[

T

]

=

K1,r. However,then Case 2applies, whichwe assumedwas not thecase. Withoutloss ofgenerality,we may

thereforeassumethat u1x1isnotabridgein G.LetA

= (

M

\ {

x1y1

})

∪ {

y1u2

}

andD

= {

u1x1

}

.ThenG

+

A

−

D isconnected, so

(

A

,

D

)

isasolution.Since

|

A

|

+ |

D

|

=

1₂

|

T

|

−

1

−

1

+

1

+

1

=

p

+

1₂

|

T

|

,thissolutionisoptimal.ThisconcludesCase 3b andthereforealsoconcludesCase 3.

It isclear that opt_S

(

G

,

δ)

can be computed in O

(

n

+

m

)

time. We also observe that the above proof isconstructive. Hence,wenotonlysolvethedecisionvariantofCDPE

(

ea

,

ed

)

butwecanalsoﬁndanoptimalsolutioninpolynomialtime.