Minimum k-way cut of bounded size is fixed-parameter tractable

0.17 MB·English

by  Ken-ichi Kawarabayashi

k
Minimum -way cut of bounded size is fixed-parameter tractable
Ken-ichiKawarabayashi∗ Mikkel Thorup
NationalInstituteofInformatics AT&TLabs—Research
2-1-2 Hitotsubashi,Chiyoda-ku 180Park Avenue,FlorhamPark, NJ 07932,USA
Tokyo101-8430,Japan [email protected]
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1 k [email protected]
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2 January 27,2011

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2 Abstract

] We considera theminimumk-waycutproblemforunweightedgraphswith a size bounds onthe
M numberof cutedgesallowed. Thuswe seek to removeas few edgesas possible so as to split a graph
D intok components,orreportthatthisrequirescuttingmorethansedges. Weshowthatthisproblemis
fixed-parametertractable(FPT)ins. Moreprecisely,fors=O(1),ouralgorithmrunsinquadratictime
.
s whilewehaveadifferentlineartimealgorithmforplanargraphsandboundedgenusgraphs.
c
Ourtractabilityresultstandsincontrastto knownW[1]hardnessofrelatedproblems. Withoutthe
[
size bound,Downeyetal.[2003]provedthattheminimumk-waycutproblemisW[1]hardink even
2
forsimpleunweightedgraphs.Downeyetal.askedaboutthestatusforplanargraphs.Ourresultimplies
v
tractabilityink fortheplanargraphssincetheminimumk-waycutofaplanargraphisofsizeatmost
9
8 6k (in fact, the size is f(k) foranyboundeddegreegraphsforsome fixedfunctionf of k. Thisclass
6 includesboundedgenusgraphs,andsimplegraphswithanexcludedminor).
4 Asimplereductionshowsthatvertexcutsareatleastashardasedgecuts, sotheminimumk-way
.
1 vertexcutis also W[1] hardin termsof k. Marx [2004]provedthatfindinga minimumk-way vertex
0 cutofsize s isalso W[1]hardins. MarxaskedabouttheFPT status withedgecuts, whichwe prove
1 tractablehere. WearenotawareofanyothercutproblemwherethevertexversionisW[1]hardbutthe
1
edgeversionisFPT.
:
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i
X
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∗Research partly supported by Japan Societyfor the Promotion of Science, Grant-in-Aid for ScientificResearch, by C &C
Foundation,byKayamoriFoundationandbyInoueResearchAwardforYoungScientists.
parametrized byk parametrized bysizes
k-wayvertexcutofsizes W[1]hard[8] W[1]hard[19]
k-wayedgecutofsizes W[1]hard[8] FPT[Thispaper]
Table1: FPTstatusofk-waycutproblems
1 Introduction
Weconsider the minimum k-waycut problem1 ofgraph. Thegoal isto findaminimum set ofcut edges so
astosplit thegraph intoatleast k components. Ifagivengraph isunweighted, minimummeansminimum
cardinality; otherwise it means minimum total weight. Goldscmidt and Hochbaum [10] proved that the
problem is NP-hard when k is part of the input but solvable in polynomial time for any fixed k. Finding
a minimum k-way cut is an extension of the classical minimum cut problem, and it has applications in the
areaofVLSIsystem design, parallel computing systems, clustering, network reliability andfindingcutting
planesforthetravelingsalesmanproblem.
Ourfocusistheminimumk-waycutproblemforanunweighted graph, decidingifthereisak-waycut
of size s. For constant s we solve this problem in quadratic time. In the case of weights, our algorithms
generalize tofindingtheminimumweightk-waycutwithatmostsedges.
Forplanar and, more generally, bounded genus graphs, wepresent adifferent linear time algorithm for
boundedsizeminimumk-waycut. Forsimpleunweightedboundedgenusgraphs,weknowthataminimum
k-waycuthassizeΘ(k),sowegetlineartimewheneverk = O(1).
Our result implies that the k-way cut problem is fixed-parameter tractable when parameterized by the
cutsizes. Recallherethatfixed-parametertractable(FPT)inaparametertmeansthatthereisanalgorithm
with running time O(f(t)nc) for some fixed function f and constant c. In our case we get t = s, c = 2,
andf(t) = ttO(t) forgeneral graphs. Forbounded genus graphs, wegett = s,c = 1andf(t) = 2O(t2). If
theboundedgenusgraphsaresimpleandunweighted, wecanalsouset = k asparameterandgetthesame
asymptotic bounds.
OurFPTresultstandsincontrasttoknownW[1]hardnessofrelatedproblems(c.f. Table1). Recallthat
ifaproblemisW[1]hardint,thenitisnotFPTintunlessNP=P.Withoutthesizebound,Downeyetal.[8]
provedthattheminimumk-waycutproblemisW[1]hardinkevenforsimpleunweightedgraphs. Downey
et al. [8] asked about the status for planar graphs. Our result implies tractability in k for the planar graphs
since the minimum k-way cut of a planar graph is of size at most 6k. Vertex cuts are at least as hard as
edgecuts, sotheminimumk-wayvertexcutisalsoW[1]hardintermsofk. Marx[19]provedthatfinding
a minimum k-way vertex cut of size s is also W[1] hard in s. Marx [19] asked about the FPT status with
bounded sizeedgecuts,whichweprovetractable here.
The discovered difference in FPT status between the edge and the vertex version of the bounded size
k-way cut problem is unusual for cut problems. As mentioned above, both versions are W[1] hard when
only k is bounded. On the other hand, Marx [19]has proved that the bounded size k-terminal cut problem
is FPT both for vertex and edge cuts. He also proved FPT for a bounded size cut of a bounded number of
pairs. Recentlythiswasstrenthened byMarxandRazgon[20]andBousquetetal. [4],showingthatfinding
abounded sizecutofanunbounded setofpairsisFPTbothforvertexandedgecuts.
1Thereisalotofconfusingterminologyassociatedwithcutproblems,e.g.,intheoriginalconferenceversionof[6],“multiway
cut”referredtotheseparationofgiventerminals,butthattermisfortunatelycorrectedto“multiterminalcut”inthefinaljournal
version. Herewefollow thelattermore explicitterminology: k-waycut for arbitrarysplittingintok pieces, k-terminal cut for
splittingkterminals,k-paircutforsplittingkpairs,andsoforth...
1
Henceforth, unlessotherwise specified,cutsareunderstood tobeedgecuts.
1.1 More history
Generalgraphs. Goldschmidt andHochbaum [10]proved thatfindingaminimumk-waycutisNP-hard
whenkispartoftheinput,butpolynomialtimesolvableforfixedk. Theiralgorithmfindsaminimumk-way
cut in O(n(1/2−o(1))k2)timeand works for weighted graphs. Karger and Stein [16]proposed an extremely
simplerandomizedMontoCarloalgorithmforthek-waycutproblemwhoserunningtimeisO(n(2−o(1))k).
Then Kamidoi et al. [15] presented a deterministic algorithm that runs in O(n(4+o(1))k) time, and finally,
Thorup[24]presented thecurrent fastestdeterministic algorithm witharunningtimeofO˜(mn2k−2).
Theobviousbigtargetwouldbetomovethedependence onk fromtheexponent ofn. However,thisis
impossible duetotheabovementioned W[1]hardness ink byDowneyetal. [8].
As alternative to k, a very natural parameter to look at is the cut size s, i.e., the size of the desired
output. Gettingpolynomial timeforfixedsistrivialsincewecantryallsubsets ofsedgesinO(n2s)time.
Reducing thistoO(ns)timestraightforward using thesparsification from[23]. Thechallenge hereisifwe
can move s from the exponent of n and get FPT in s. As mentioned above, in the case of vertex cuts, the
bounded cut size was considered by Marx [19] who proved the k-way vertex cut to be W[1] hard in the
size s. He asked if the edge version was also hard. Here we show that the edge version is tractable with a
quadratic algorithm for any fixed cut size s. Our algorithm implies that the minimum k-way cut problem
is solvable in polynomial time for bounded degree unweighted graphs. This class includes planar graphs,
bounded genus graphs, and simple graphs withanexcluded minor. Infact, wegive alinear timealgorithm
forplanargraphsandbounded genusgraphs. Letusseemoreprecisely.
Planar graphs. The special case of planar graphs has been quite well-studied. In the case of weighted
planar graphs, Dahlhaus et al. [6] solved the k-way cut problem in O(n3k−1logn) time. The bound was
laterimproved toO˜(n2k−1)timebyHartvigsen [11]. Thiswas,however, matchedbythelaterO˜(mn2k−2)
boundbyThorup[24]forgeneralgraphs.
The case of simple unweighted planar graphs has also received attention. Hochbaum and Shmoys [13]
gaveanO(n2)algorithm forsimpleunweighted planargraphs whenk = 3. ThiswasimprovedbyHe[12]
to O(nlogn). The motivation given in [13, 12]is the case of general k, with k = 3 being the special case
forwhichtheyprovideanefficientsolution.
Asmentioned previously, having proved thek-waycutproblem W[1]hard forsimpleunweighted gen-
eralgraphs, Downeyetal. [8]askedabouttheFPTstatusforplanargraphs.
We resolve the question from [8] with an O(2k2n) algorithm for simple unweighted planar graphs.
Even in the special case of k = 3, our result improves on the above mentioned algorithms of Shmoys and
Hochbaum [13]andHe[12].
Ourplanaralgorithm isgeneralized tothebounded genuscase,whereitrunsintimeO(2O(g2k2)n).
Techniques. Ourmain result, the quadratic algorithm for general graphs, is asimple combinatorial algo-
rithm not relying on any previous results. To solve the problem recursively, we will define ”the powercut
problem”, which is much stronger than the minimum k-way cut problem. It also generalizes the muli-pair
cut problem with p pairs for fixed p (this pair problem is, however, known to be FPT both for vertices and
edge cuts [19]). The approach can be seen as a typical example that a stronger inductive hypothesis gives
a much simpler inductive proof. With the powercut problem, it is easy to handle all high degree vertices
2
exceptforone,whichactslikeanapexvertexingraph minortheory. Therestisabounded degreegraphin
whichweidentifyacontractible edge.
For planar graphs, like previous algorithms [13, 12], we exploit that a minimum k-way cut has size
O(k). Otherwise our algorithm for the planar case is based on a decomposition lemma from Klein’s [18]
approximateTSPalgorithm. Infact,thisappearstobethesimplestdirectapplicationofKlein’slemmafora
classic problem. Klein’slemmaisrelatedtoBaker’s layeredapproach [2]toplanar graphs. WhereasBaker
deleteslayerstogetboundedtree-width,Kleincontractslayers(deletioninthedual)andsuchacontraction
does not affect cuts avoiding the contracted layers. We present a linear time version of this approach for
bounded genus graphs. In doing so, we also get a linear time approximate TSP algorithm, improving an
O(nlogn)algorithm basedonworkofCabelloetal.[5]andDemaineetal.[7].
2 FPT algorithm to find minimum k-way cuts of bounded size
We want to find a minimum k-way cut of size at most s. Assuming that a given graph is connected, the
problem can only be feasible if k ≤ s+1. In the spirit of FPT, weare going to O∗ to denote O assuming
thattherelevantparametersareconstant. Wewillsolvetheproblem inO(ssO(s)n2)= O∗(n2)time.
2.1 The powercut problem
Tosolvetheproblemk-waycutproblem inductively, wearegoingtoaddress amoregeneral problem:
Definition2.1 Thepowercut problem takes asinput atriple (G,T,s) where Gisaconnected graph, T ⊆
V(G)asetofterminals, andsasizeboundparameter. Foreveryj ≤ s+1andforeverypartition P ofT
into j sets, some ofwhich maybeempty, wewant aminimalj-way cut C of Gwhose sides partitions T
j,P
according toP but only ifthere issuchafeasible cutofsizeatmosts. Thepowercut isthus acutfamily C
containing thecutsC ,eachofsizeatmosts.
j,P
Itmaybethat wefordifferent partitions P,P′ get thesameedge cutCj,P = Cj,P′. Oftenwewillidentify
apowercutwithitssetofdistinctedgecuts.
Notethatif|T|andsarebounded, thensoisthetotalsizeofthepowercut. Moreprecisely,
Observation 2.2 Thetotalnumberofedgesinapowercutisboundedbys s+1(j|T|/j!) < (s+1)|T|+1.
Pj=2
We will show how to solve the powercut problem in quadratic time when |T|,k,s = O(1). To solve our
original problem, we solve the powercut problem with an empty set of terminals T = ∅. In this case, for
eachj ≤s+1,weonlyhaveasingletrivialpartitionP ofT consistingofj emptysets,andthenwereturn
j
C .
k,Pk
We can, of course, also use our powercut algorithm to deal with cut problems related to a bounded
numberofterminals,e.g.,thep-paircutproblemwhichforppairs{(s ,t ),...,(s ,t )}askforaminimum
1 1 p p
cut that splits every pair, that is, for each i, the cut separates s from t . If there is such a cut of size at
i i
mosts,wefinditwithapowercutsetting T = {s ,...,s ,t ,...,t }andk = s+1. Intheoutput powercut
1 p 1 p
family{C },weconsiderallpartitionsP splittingeverypair,returningtheminimumofthecorresponding
j,P
cuts. Suchproblems withabounded number ofterminals andabounded cutsize, butnorestrictions onthe
numberofcomponents,areeasiertosolvedirectly,asdoneinmanycasesbyMarx[19]evenforvertexcuts.
However, for vertex cuts, Marx [19] proved that the k-way cut problem is W[1] hard. Hence the hardness
is not in splitting of a bounded set of terminals, but in getting a certain number of components. With our
3
powercutalgorithm,weshowthatgettinganyspecifiednumberofcomponentsisfeasiblewithsizebounded
edgecuts.
Below we will show how to solve the power cut problem recursively in quadratic time, let T be the
0
initialsetofterminals, e.g.,T = ∅forthek-waycutproblem. Wenowfix
0
t = max{2s,|T |}. (1)
0
Inourrecursive problemswewillneverhavemorethantterminals. Theparameterswillnotchange.
Identifying vertices and terminals Our basic strategy will be to look for vertices that can be identified
while preserving some powercut. To make sense of such a statement, we specify a cut as a set of edges,
and view each edge as having its own identity which is preserved even if its end-points are identified with
other vertices. Note that when we identify vertices u and v, then we destroy any cut that would split u
and v. However, the identification cannot create any new cuts. We say that u and v are identifiable if they
are not separated by any cut of some powercut C. It follows that if u and v are identifiably, then C is also
a powercut after their identification, and then every powercut C′ after the identification is also a powercut
before the identification. Since loops are irrelevant for minimal cuts, identifying the end-points of an edge
isthesameascontracting theedge. Therefore, iftheend-points ofanedgeareidentifiably, wesaytheedge
iscontractible.
Wedoallowforthecaseofidentifiable terminalstandt′ thatarenotsplitbyanycutinsomepowercut
C. Bydefinition ofapowercut, this mustimply thatthere isnofeasible cutofsize atmostsbetween tand
t′.
Often we will identify many vertices. Generalizing the above notion, we say a set of vertex pairs is
simultaneously identifiable if there is a powercut that does not separate any of them. This implies that we
canidentifyallthepairswhilepreserving somepowercut.
Notethatwecaneasilyhavecaseswithidentifiablevertexpairsthatarenotsimultaneously identifiable,
e.g.,ifthegraphisapathoftwoedgesbetween twoterminals, theneitheredge iscontractible, yettheyare
notsimultaneously contractible.
Recursing on subgraphs Often we will find identifiable vertices recursing via a subgraph H ⊆ G. If C
is a powercut of G, then C|H denotes C restricted to H in the sense that each cut C ∈ C is replaced by its
edgesC ∩E(H)inH,ignoring cutsthatdonotintersect H.
Lemma2.3 LetH beaconnected subgraph ofG. LetS bethesetofvertices inH withincident edgesnot
inH. DefineT = S∪(T ∩V(H))tobetheterminalsofH. TheneachpowercutC of(H,T ,s)isthe
H H H
restriction toH ofsomepowercut C of(G,T,s). Hence, ifpairs ofvertices aresimultaneously identifiable
in(H,T ,s),thentheyarealsosimultaneously identifiable in(G,T,s).
H
Proof Since the pairs are simultaneously identifiable in(H,T ,s), there isapowercut C of (H,T ,s)
H H H
withnocutseparating anyofthe pairs. Nowconsider apowercut C of(G,T,s), and letC beanycutinC.
ThenH \C has acertain numberj ≤ s+1ofcomponents inducing acertain partition P ofT . InC we
H
now replace C ∩H with C from C , denoting the new cut C′. Since C is a minimal, this can only
j,P H j,P
decreasethesizeofC. ItisalsoclearthatG\C andH\C havethesamenumberofcomponentsinducing
the same partition of T. This way we get a powercut C′ of (G,T,s) such that C′|H = C . In particular it
H
followsthatourpairsfromH aresimultaneously identifiable in(G,T,s).
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2.2 Goodseparation
For our recursion, we are going to look for good separations as defined below. A separation of the graph
G is defined via an edge partition into two connected subgraphs A and B, that is, each edge of G is in
exactly one of Aand B. Werefer to Aand B asthe sides of the separation. LetS bethe set of vertices in
both A and B. Then S separates V(A) from V(B) in the sense that any path between them will intersect
S. Contrasting vertex cut terminology, we include S in what is separated by S. In order to define a good
separation, wefix
p = (s+1)t+1 andq = 2(p+1). (2)
Thenpistheupper bound fromObservation 2.2onthetotal numberofedgesinapowercutwithatmostt
terminals. Theseparation isgood if|S| ≤ sandbothAandB haveatleastq vertices.
Supposewehavefoundagoodseparation. ThenoneofAandB willcontainatmosthalftheterminals
from T because |T| ≤ 2s. Suppose it is A. Recursively we will find a powercut C of A with terminal
A
set T = S ∪(T ∩V(A)) as in Lemma2.3. Finally in G wecontract all edges from Athat are not in the
A
powercutC .
A
For the validity of the recursive call, we note that |T | ≤ s + t/2 ≤ t. The last inequality follows
A
because t ≥ 2s. For the positive effect of the contraction, recall that A has at least q = 2(p+1) vertices
where p bounds the number of edges in C . We know that A is connected, and it will remain so when we
A
contracttheedgesfromAthatarenotinC . Intheend,Ahasatmostpnon-contracted edges,andtheycan
A
span at most p+1 distinct vertices. The contractions thus reduce the number of vertices in G by at least
|A|−p−1.
Below,splitting intoafewcases,wewilllookforgoodseparations torecurseover.
2.3 Multiplehigh degree vertices
Avertexhashighdegreeifithasatleast
d= q+s−1 (3)
neighbors. Here q was the lower bound from (2) on the number of vertices in a side of a good separation.
Suppose that the current graph G has two high degree vertices uand v. In that case, check if there isa cut
D betweenu-v-cutofsizeatmosts. Ifnot,wecantrivially identifyuandv andrecurse.
IfthereisacutDofsizeatmostsbetweenuandv,letAbethecomponentcontaining uinG\D,and
letB bethesubgraph withalledgesnotinA.
Lemma2.4 Thesubgraphs AandB formagoodseparation.
Proof The set S of vertices in both A and B are exactly the end-points on the u-side of the edges in D.
Therefore|S| ≤ |D| ≤ s. WenowneedtoshowthateachofAandB spanatleastq vertices. Thisistrivial
for B since B contains v plus allthe d = q +s−1neighbors of v. Forthecase ofA, wenote that D can
separateufromatmostsofitsneighbors. Thismeansthatuisconnected toatleastd−s = q−1vertices
inG\D,soAcontains atleastq nodes.
Thus, if we have two high degree vertices, depending on the edge connectivity between u and v, we can
eitherjustidentifyuandv,orrecurseviaagoodseparation. Belowwemaythereforeassumethatthegraph
hasatmostonehighdegreevertex.
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2.4 No highdegree vertex
Belowweassumethatnovertexwithhighdegree≥ d—c.f.(3). Thecaseofonehighdegree“apex”vertex
willlaterbeaddedastraightforward extension.
Akernelwithsurroundinglayers Westart bypicking astart vertexv andgrow anarbitrary connected
0
subgraph H , called the kernel from v such that H contains all edges leaving v and H spans h ≥ d
0 0 0 0 0
vertices where h = O∗(1) is a parameter to be fixed later. Next we pick edge disjoint minimal layers H ,
i
i = 1,...,p,subjecttothefollowingconstraints:
(i) ThelayerH containsnoedgesfromH = H ,butH containsallotheredgesfromGincident
i <i Sj<i j i
totheverticesinH .
<i
(ii) Eachcomponent ofH iseitherabigcomponentwithatleastq vertices, oralimitedcomponent with
i
noedgefromG\H leaving it—ifacomponentisboth,weviewitasbig.
≤i
RecallthatapowercutCcanhaveatmostpedges. Thismeansthattheremustbeatleastoneofthep+1edge
disjointsubgraphs H whichhasnoedgesinC. SupposingwehaveguessedthisH ,wewillfindasetF of
i i i
simultaneously contractible edgesfromH (notethatwemeanH ,notH ). Bydefinition, F = E(H ). If
0 0 i 0 0
i > 0,condition (i)impliesthatH isacutbetweenH andtherestofthegraph, andwewillusethisfact
i <i
tofindthesetF . Sinceoneoftheguesses mustbecorrect, theintersection F = F = p F mustbe
i Ti i Ti=1 i
simultaneously contractible. Analternativeoutcomewillbethatwefindagoodseparationwhichrequiresq
vertices on either side. Thisis wherecondition (ii) comes in, saying that wehave to grow each component
of H until either it becomes a big component with q vertices, or it cannot be grown that big because no
i
moreedgesareleaving it.
Beforeelaborating ontheabovestrategy, wenotethatthegraphsH areoflimitedsize:
i
Lemma2.5 ThegraphH hasatmosthdi vertices.
≤i
Proof We prove the lemma by induction on i. By definition |V(H )| = h. For the inductive step with
0
i > 0,weprovethemoreprecisestatementthatlayerH hasatmostdtimesmoreverticesthanthevertices
i
itcontains fromlayerH .
i−1
Since each layer is a cut, the edges leaving H must all be incident to H . We will argue that each
<i i−1
component of H has at most d vertices. This is trivially satisfied when we start, since each vertex from
i
H comeswithitsatmostd−1neighbors. Now,ifwegrowacomponent along anedge, itisbecause it
i−1
has less than q < d vertices, including at least one from H . Either the edge brings us to a new vertex,
i−1
increasingthesizeofthecomponentby1,whichisfine,ortheedgeconnectstosomeothercomponentfrom
H whichbyinduction hadatmostdverticespervertexinH .
i i−1
IftheV(G) = V(H ),thenGhasonlyhdi = O∗(1)vertices,andthenwecansolvethepowercutproblem
≤q
exhaustively. Belowweassumethisisnotthecase.
Pruning layers checking for good separations Consider a layer H , i > 0, and let H− be the union
i i
of the big components of H . Moreover let H+ be H combined with all the limited components from
i <i <i
H . WecallH− thepruned layer. Sincethelimitedcomponents have noincident edges from G\H ,we
i i ≤i
note that H− is a cut between H+ and the rest of the graph. The lemma below summarizes the important
i <i
properties obtained:
6
Lemma2.6 Fori = 1,...,p:
(i) PrunedlayerH− ⊆H isacutseparatingH fromG\V(H ). Inparticular,wegetanarticulation
i i 0 ≤q
pointifweidentifyallofH− inasinglevertex.
i
(ii) Eachcomponent ofH− isofsizeatleastq.
i
Now,wetakeeachpruned layer H− separately, andorderthecomponents arbitrarily. ForeverypairAand
i
B of consecutive components (of order at least q), wecheck if their edge connectivity is atleast s in G. If
not, there is a cut D in G with at most s edges which separates A and B. We claim this leads to a good
separation. Ontheonesideoftheseparation, wehavethecomponent AofG\D containing A,andonthe
other wehave the reminder B of Gwhichincludes B and cut edges from D. ThenAandB intersect inat
most|D|≤ svertices, andbothAandB haveatleastq vertices. Thuswegetagoodseparation.
Belowweassumethatforeachpruned layer, theedgeconnectivity betweenconsecutive components is
atleasts.
Articulationpointsfromprunedlayers
Lemma2.7 If there is a powercut of (G,T,s) that does not use any edge from H , then all vertices in the
i
prunedlayerH− canbeidentified inasinglevertexv .
i i
Proof We are claiming that no cut D from C separates any vertices from H−. Otherwise, since D does
i
notcontainanyedgesfromH ,thecutwouldhavetogobetweencomponentsfromH−. Inparticular, there
i i
would be two consecutive components of H− separated by D. However, D has at most s edges, and we
i
alreadychecked thattherewasnosuchsmallcutbetweenanycomponents ofH−.
i
BelowweassumewehaveguessedalayersH thatisnotusedinsomepowercutof(G,T,s). Let[H− 7→ v ]
i i i
denote that all vertices from H− are identified in a single vertex v , which we call the articulation point.
i i
FromLemma2.7itfollowsthatsomepowercutispreserved in(G[H− 7→ v ],T[H− 7→ v ],s).
i i i i
Next,fromLemma2.6(i)wegetthatH [H− 7→ v ]isablockofG[H− 7→ v ]separatedfromtherest
≤i i i i i
bythearticulation pointv . AsinLemma2.3,wenowfindapowercutC of
i i
H [H− 7→ v ], {v }∪(T ∩V(H ))[H− 7→ v ], s .
(cid:0) ≤i i i 0 ≤i i i (cid:1)
ThepowercutC canbefoundexhaustively sinceH hasatmosthdi = O∗(1)vertices. Sincethisisnota
i ≤i
recursivecall,soitisOKif{v }∪(T ∩V(H ))[H− 7→ v ]involves t+1terminals.
0 ≤i i i
Lemma2.8 If there is a powercut of (G,T,s) that does not use any edge from H , then there is such a
i
powercutwhichagreeswithC onH ,andonH inparticular.
i ≤i 0
Proof From Lemma 2.3 we get that C is the restriction to H [H− 7→ v ] of some powercut C′ of
i ≤i i i i
(G[H− 7→ v ],T[H− 7→ v ],s). From Lemma2.7 it follows that C′ is also a powercut of (G,T,s). Since
i i i i i
C′ doesnotcontain anyedgescontracted inH−,weconclude thatC istherestriction ofC′ toH .
i i i i ≤i
WithourassumptionthatH isnotusedinsomepowercut,wegetthatalledgesinF = E(H )\E(C )are
i i 0 i
identifiable. Disregarding theassumption, wearenowreadytoprove
Lemma2.9 Let F = p F be the set of edges from H that are not used in any C , i = 1,...,p. The
Ti=1 i 0 i
edgesfromF aresimultaneously contractible.
7
Proof GivenanypowercutCof(G,T,s),sinceithasatmostpedges,weknowthereiscomei∈ {0,...,p}
suchthatC doesnotuseanyedgefromH . Ifiis0,thismeansalledgesfromH arecontractible. Forany
i 0
otheri,theclaimfollowsfromLemma2.8.
With Lemma 2.9 we contract all edges from H that are not in some C . From Observation 2.2, we know
0 i
thateachC involvesatmost(s+1)t+2 edges,socombinedtheyinvolveatmostp(s+1)t+2 un-contracted
i
edges,spanning atmostp(s+1)t+2 +1distinct vertices. AstheinitialsizeforH ,westartwith
0
h = 2(p(s+1)t+2 +1) (4)
vertices. Therefore, when we contracting all edges from H that are not in some C , we get rid of half the
0 i
verticesfromH .
0
2.5 A singlehigh degree “apex” vertex
All that remains is to consider the case where there is a single high degree vertex r with degree ≥ d—c.f.
(3). Wearebasically going torun the reduction forno high degree from Section 2.4on thegraph G\{r},
butwithsomesubtleextensions described below.
Starting from an arbitrary vertex that isv neighbor tor, weconstruct the layers H in G\{r}. If this
0 i
includes allverticesofG\{r},thenGhasO∗(1)vertices, andthenwefindthepowercutexhaustively.
Nextweaddthevertexr toeachlayerH ,including alledgesbetweenrandH \H . Wedenote this
i i <i
graphHr. NotethatHr isconnected sincerisaneighborofv . AlsonotethatalltheHr areedgedisjoint
i 0 0 i
liketheH .
i
Aftertheadditionofr,fori> 0,weturnanylimitedcomponentinvolvingrbig. Moreprecisely,inHr
i
wesaythat acomponent isbig ifitishasq ormorevertices orifitcontains r. Theremaining components
are limited. Removing all other limited components from Hr we get the pruned layer Hr−. Similarly,
i i
we have the graph Hr+ which is Hr expanded with the limited components from Hr. Corresponding to
<i <i i
Lemma2.6,weget
Lemma2.10 Fori= 1,...,p:
(i) The vertices from the pruned layer Hr− form a vertex separator in Gbetween Hr and G\Hr . In
i 0 ≤q
particular, wegetanarticulation pointifweidentify Hr inasinglevertex.
i
(ii) Eachcomponent ofHr− whichdoesnotcontainr hasatleastq vertices.
i
Proof Above (ii) is trivial. Concerning (i), we already have from Lemma 2.6 that the edges from H−
i
provideacutofG\{r}betweenH andG\{r}\H . TheverticesinH− provideacorresponding vertex
0 ≤q i
separationinG\{r}. Whenaddingrtothegraphandtotheseparation,wegetavertexseparationV(Hr−)
i
betweenH andG\Hr .
0 ≤q
Now, as in Section 2.4, we order the components of Hr− arbitrarily, and check if the edge connectivity
i
between pairsofconsecutive components isatleastsinG. Ifnot, weclaim thereisagoodseparation. Let
D be a cut in G of size at most s between two components A and B of Hr−. If A and B both have at
i
least q vertices, then we have the same good separation as in Section 2.4. Otherwise, one of them, say B
involves r. In this case we have an argument similar to that used for two high degree vertices in Section
2.3. Ononeside ofthegood separation, wehave thecomponent AofG\D including A. Clearlyithasat
least |V(A)| ≥ q vertices. The other side B is the rest of G including B and the cut edges from D. Then
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B includes the neighborhood of all vertices in B including all neighbors of r, so B has at least d+1 > q
vertices. Belowweassumethatwedidnotfindsuchagoodseparation.
Wenow continue exactly asin Section 2.4. Fori = 1,...,q weidentify thevertices ofHr− ina vertex
i
v whichbecomesanarticulation point,andthenwefindapowercutC of
i i
Hr [Hr− 7→ v ], {v }∪(T ∩V(Hr ))[Hr− 7→ v ]), s .
(cid:0) ≤i i i 0 ≤i i i (cid:1)
Corresponding toLemma2.9,weget
Lemma2.11 TheedgesfromHr thatarenotusedinanyC ,i = 1,...,paresimultaneously contractible.
0 i
As in Section 2.4, we conclude that we get at most p(s + 1)t+2 un-contracted edges from Lemma 2.11
and they span at most p(s+1)t+2 +1 distinct vertices. As the initial size for Hr, we start with h+1 =
0
2(p(s+1)t+2 +1)vertices. Thenthecontractions ofLemma2.11allowsustogetridofatleasthalftheh
verticesinHr.
0
2.6 Analysisandimplementation
Wearenowgoingtoanalyzetherunningtimeincludingsomeimplementationdetailsoftheaboverecursive
algorithm, provingatimeboundof
T(n) = O stO(t)n2 . (5)
(cid:16) (cid:17)
First,wearguethatwecanassumesparsity withatmostO(sn)edges. Moreprecisely, ifthegraphatsome
point has m ≥ 2sn edges, as in [23] we find s edge disjoint maximal spanning forests If an edge (v,w)
is not in one of these spanning forests, then v and w are s edge connected. We can therefore contract all
suchoutsideedges, leavinguswithatmostsn ≤ m/2edges. Thismayalsoreduce thenumberofvertices,
whichisonlypositive. TheoverallcostofthisprocessiseasilyboundedbyO(sn2).
In ouranalysis, for simplicity, wejust focus onthe case withno high degree vertices from Section 2.4.
Whenwelookforgood separations, wecheck iftheedge connectivity betweentwovertexsetsiss. Aswe
sawabove,thegraphcanbeassumedtohaveatmost2nsedges,sothistakesonlyO(s2n)time[9]including
identifying acutwithsedgesifitexists. Thenumberofsuchgoodseparation checksislimitedbythetotal
number of components in all the layers H , and for each layer, this is limited by the number of vertices.
i
Thus, by Lemma 2.5, we have at most p hdi < 2hdp good separation checks, each of which takes
Pi=1
O(s2n)time. Witht ≥ 2s,p = (s+1)t+1,q = 2(p+1),d= q+s−1,andh = 2(p(s+1)t+2+1)—c.f.
(1),(2),(3),and(4)—wegetthatthetotaltimeforgoodseparation checksisboundedby
O(2hdps2n)= O(stO(t)n).
If we do find a good separation, we recurse on one of the sides A, which we know has at least q vertices.
Including the separating vertices, we know that A has at most n−q +s vertices. After the recursion, we
canidentifyallbutq/2verticesinA. Allthisleadstoatherecurrence
T(n)≤ max O stO(t)n +T(ℓ)+T(n−ℓ+q/2).
(cid:16) (cid:17)
q≤ℓ≤n−q+s
Inductively thisrecurrence satisfies(5),theworst-case beingwhenℓattainsoneofitsextremevalues.
Ifwedonot findagood separation, for i = 1,...,p, weexhaustively findapowercut ofagraph withat
most hdi vertices and shdi edges. Wesimply consider all the (shdi)s potential cuts withs edges, and that
9