Better Approximation Algorithms for Maximum Asymmetric Traveling Salesman and Shortest Superstring

0.36 MB·

by  Katarzyna Paluch

Better Approximation Algorithms for Maximum Asymmetric
Traveling Salesman and Shortest Superstring
KatarzynaPaluch
InstituteofComputerScience,UniversityofWrocław
[email protected]
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0 Abstract
2
Inthemaximumasymmetrictravelingsalesmanproblem(MaxATSP)wearegivenacomplete
n
directedgraphwithnonnegativeweightsontheedgesandwewishtocomputeatravelingsalesman
a
J tourofmaximumweight.Inthispaperwegiveafastcombinatorial 3-approximationalgorithmfor
4
MaxATSP.Itisbasedonanoveluseofhalf-edges,matchingsandanewmethodofedgecoloring.
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1 (Ahalf-edgeofedge(u,v)isinformallyspeaking“eitheraheadoratailof(u,v)”.) Thecurrent
best approximation algorithms for Max ATSP, achieving the approximation guarantee of 2, are
3
] duetoKaplan, Lewenstein, ShafrirandSviridenkoandElbassioni, Paluch, vanZuylen. Usinga
S
recentresultbyMucha,whichstatesthatanα-approximationalgorithmforMaxATSPimpliesa
D
(2+11(1−α))-approximationalgorithmfortheshortestsuperstringproblem(SSP),weobtainalso
. 9−2α
s a (211 2,3667)-approximation algorithm for SSP, beating the previously best known (having
[c appr3o0xim≈ationfactorequalto211 2,4782.)
23 ≈

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1 Introduction
In the maximum asymmetric traveling salesman problem (Max ATSP) we are given a complete di-
rected graph G = (V,E) with nonnegative weights on the edges and we wish to compute a traveling
salesman tour of maximum weight. The problem is known to be APX-hard [19] and the current best
approximation algorithms for it are due to Kaplan, Lewenstein, Shafrir and Sviridenko [9] and El-
bassioni, Paluch, van Zuylen [18]. Both of them achieve the approximation ratio of 2, the former is
3
basedonlinearprogrammingandtheotheriscombinatorialandsimpler. Besidesbeinganinteresting
problem in itself, Max ATSP is also of particular interest because of its applications to a number of
11(1−α)
relatedproblems. Forexample,anα-approximationalgorithmforMaxATSPimpliesa(2+ )-
9−2α
approximationalgorithmforSSP,whichwasrecentlyshownbyMucha[16]. Theshortestsuperstring
(cid:80)
problem is defined as follows. We are given n strings s ,s ,...,s over a given alphabet and we
1 2 n
want to find a shortest string s such that each s for i,1 i n is a substring of s. SSP arises
i
≤ ≤
in DNA sequencing and data compression. Currently the best approximation algorithm for SSP is
duetoMucha[16]andachievesanapproximationfactorof211. Anyα-approximationalgorithmfor
23
Max ATSP implies also an algorithm with the same guarantee for the maximal compression problem
definedbyTarhioandUkkonen[21].
Wedeviseacombinatorial 3-approximationalgorithmforMaxATSP,thusproving
4
Theorem1 Thereexistsa 3-approximationalgorithmforthemaximumtarvelingsalesmanproblem.
4
UsingtheresultofMucha[16],weobtain
Corollary1 Thereexistsa211-approximationalgorithmfortheshortestsuperstringproblem.
30
The approach we have adopted is as follows. We start by computing a maximum weight cycle cover
C ofG,whereacyclecoverC ofgraphGisdefinedasasetofdirectedcyclesofGsuchthateach
max
vertex of G belongs to exactly one cycle of C. A maximum weight cycle cover of G can be found
in polynomial time by a reduction to maximum weight matching. Let OPT denote the weight of a
traveling salesman tour of G of maximum weight. The weight of an edge e will be denoted as w(e)
andforanysubsetE(cid:48)ofedgesE byw(E(cid:48))wewillmean(cid:80) w(e). Sinceatravelingsalesmantour
e∈E(cid:48)
is a cycle cover of G (consisting of just one cycle), we know that w(C ) OPT. By removing
max
≥
the lightest edge from each cycle of C , we obtain a collection of vertex-disjoint paths, which can
max
bearbitrarilypatchedtoformatour. Removingthelightestedgefromcyclecoflengthk resultsina
pathofweightatleast k−1w(c). SinceC maycontaincyclesoflenthtwo(2-cycles),intheworst
k max
case the obtained tour may have weight equal to 1w(C ). If we could find a maximum weight
2 max
cycle cover of G without cycles of length two (2-cycles) or three (3-cycles), then we would achieve
a 3- approximation, but, unfortunately finding a maximum weight cycle cover without 2-cycles is
4
APX-hard[4].
Since2-and3-cyclesinamaximumweightcyclecoverareanobstacletogettinga 3-approximation,
4
wewouldliketosomehowgetridofthem. Thewaywearegoingtoachievethisisasfollows. IfC
max
containsatleastone2-cycleoratleastone3-cycle(atriangle),wecomputeaacyclecoverofGthat
doesnotcontainany2-cyclethatalreadybelongstoC butmaycontainhalf-edges-ahalf-edgeof
max
edge (u,v) is informally speaking “either a head or a tail of (u,v)”. Such a cycle cover C is going
1
to be called a relaxed cycle cover C improving C . Also we will ensure that a computed C has
1 max 1
weight at least OPT. A similar relaxed cycle cover and half-edges have already been introduced in
[18]. Computing C is done via a reduction to a maximum weight perfect matching. In some cases
1
C will suffice to build a traveling salesman tour of weight at least 3OPT. To (try to) extract such a
1 4
tour from C and C we build a multigraph G consisting of one copy of C and two copies of
1 max 1 max
C . Each occurrence of an edge e in C contributes one copy of e to G and each occurrence of e
1 max 1
2
inC contributestwocopiesofetoG . IfC containsonlyonehalf-edgeofacertainedgee,thenC
1 1 1 1
contributes one copy of e to G . The number of copies of edge e in G may be equal to up to three.
1 1
ThetotalweightofedgesofG isatleast3OPT. WewouldliketodivideedgesofG intofoursets
1 1
Z ,Z ,...,Z in such a way that each Z (1 i 4) is a collection of vertex-disjoint paths. One
1 2 4 i
≤ ≤
ofthesetsZ ,...,Z wouldthenhavetohaveweightatleast 3OPT andbypatchingittoatour,we
1 4 4
wouldobtainthedesiredsolution. DividingedgesofG intofoursetscanbeviewedascoloringthem
1
with four colors. We can immediately see that we are not able to color (the edges of) G in the way
1
describedaboveifC containsanyoneofthefollowing: (i)a2-cyclewhoseoneedgebelongstosome
1
cycleofC ,(ii)atrianglethatalreadybelongstoC oratrianglethatisoppositelyorientedto
max max
sometriangleofC ,(iii)atrianglewhoseoneedgeiscontainedinsome2-cycleofC or(iv)a
max max
cycleoflength4suchthattwoofitsedgesbelongtotwo2-cyclesofC . Inthepaperwewillshow
max
thatifC containsneitheroftheabove,thenweareabletocolorG asrequired.
1 1
Otherwise, if C contains one of the cycles that makes it impossible to color G into four colors,
1 1
we compute another cycle cover C that does not contain: (i) any 2-cycle that already belongs to
2
C , (ii)any2-cyclethatbelongstoC andwhoseoneedgebelongstosomecycleofC , (iii)
max 1 max
anytriangletsuchthatbothC andC containatriangleonthesameverticesast, (iv)atriangle
max 1
thatbelongstoC andwhoseoneedgeiscontainedinsome2-cycleofC or(v)acycleoflength4
1 max
thatbelongstoC andsuchthattwoofitsedgesbelongtotwo2-cyclesofC butmaycontainhalf-
1 max
edges. We call such a cycle cover a relaxed cycle cover improving C and C . We will compute
max 1
C sothatitsweightisatleastOPT. ComputingC isdoneinasimilarspirittocomputingC -via
2 2 1
anovelreductiontoamaximumweightperfectmatching.
OncewehavecyclecoversC ,C andC describedabove,webuildamultigraphG consisting
max 1 2 2
oftwocopiesofeachofthesecyclecovers. ThetotalweightofedgesofG isatleast6OPT. From
2
G we will want to extract a tour of weight at least 3OPT. To this end we will divide edges of G
2 4 2
intoeightsetsZ ,Z ,...,Z insuchawaythateachZ (1 i 8)isacollectionofvertex-disjoint
1 2 8 i
≤ ≤
paths. OneofthesetsZ ,...,Z willthenhavetohaveweightatleast 3OPT.
1 8 4
The task of dividing edges of G or G into appropriately four or eight sets can equivalently be
1 2
phrasedasfollows: wewanttocoloreachedgeofG orG withoneoffourorcorrespondinglyoneof
1 2
eightcolorsinsuchawaythateachcolorclassconsistsofvertex-disjointpaths.ForcoloringbothG
1
andG wepresentanewmethod,whichwethinkisinterestinginitsownright. Oneofthesurprisingly
2
simple ideas on which this method is based is as follows: let S be a subset of V and e = (u,v) an
edgegoingintoS (i.e. u / S andv S),whichiscoloredwithacolork. Thenifthereexistsnoedge
∈ ∈
e(cid:48) = (u(cid:48),v(cid:48)) outgoing from S (i.e. such that u(cid:48) S and v(cid:48) / S) which is colored k, then e does not
∈ ∈
belong to any cycle, whose all edges are colored k. Another ingredient of the method are alternating
cycles(anotionfrommatchingtheory)withcertainproperties.
PreviousandrelatedresultsThehistoryofapproximatingtheproblemsofmaximumasymmetric
travelingsalesmanandshortestsuperstringisquitelongasprovethefollowinglistsofpapers[14],[5],
[22],[7],[10],[1],[1],[6],[20],[9],[18],[16]and[8],[10][3],[13],[9],[18].
Othervariantsofthemaximumtravelingsalesmanproblemthathavebeenconsideredareamong
others: the maximum symmetric traveling salesman problem (MAX TSP), in which the underlying
graph is undirected - currently the best known approximation ratio is 7 [?], the maximum metric
9
symmetric traveling salesman problem, in which the edge weights satisfy the triangle inequality -
the best approximation factor is 7 [11], the maximum asymmetric traveling salesman problem with
8
triangleinequality-thebestapproximationratiois 35 [12].
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3
2 New upper bounds - relaxed cycle covers improving C
max
SupposewehavecomputedamaximumweightcyclecoverC ofagivencompletedirectedgraph
max
G = (V,E). Wewillsaythatacyclec C ishardifeachedgeeofcsatisfiesw(e) > 1w(c). We
∈ max 4
aregoingtocallcyclesoflengthi,i.e. consistingofiedges,i-cycles. Alsosometimes3-cycleswillbe
calledtriangles. Letusnoticethatonly2-cyclesandtrianglescanbehard. IfC doesnotcontaina
max
hardcycle,thenwecaneasilybuildatravelingsalesmantourofweightatleast 3w(C ) 3OPT.
4 max ≥ 4
If C contains at least one hard cycle, we would like to obtain two other cycle covers C and C
max 1 2
which may contain half-edges - a half-edge of edge (u,v) is informally speaking “either a head or a
tailof(u,v)”.
Below we give the precise definitions of a relaxed cycle cover C improving C and a relaxed
1 max
cyclecoverimprovingC andC .
max 1
Definition1 Given a complete directed graph G = (V,E) with edge weights w(u,v) 0 for every
≥
(u,v) E andamaximumweightcyclecoverC ofG,letG˜ = (V˜,E˜)bethegraphobtainedfrom
max
∈
Gbyreplacingeach(u,v) E byavertexx andtwoedges(u,x )and(x ,v),eachwith
(i,j) (u,v) (u,v)
∈
weight 1w(u,v). Edges (u,x ),(x ,v) will be called half-edges (of (u,v)). A relaxed cycle
2 (u,v) (u,v)
coverimprovingC isasubsetC˜ E˜ suchthat
max
⊆
(i) eachvertexinV hasexactlyoneoutgoingandoneincominghalf-edgeinC˜;
(ii) foreach2-cyclecofC onverticesu,v C˜ containsatmosttwohalf-edgesfrom
max
(u,x ),(x ,v),(v,x ),(x ,u) . Moreover if C˜ contains one half-edge of (u,v)
(u,v) (u,v) (v,u) (v,u)
{ }
andonehalf-edgeof(v,u),thenoneofthemisincidentwithuandtheotherwithv.
LetC beanyrelaxedcyclecoverimprovingC .
1 max
ArelaxedcyclecoverimprovingC andC isasubsetC˜ E˜ suchthat
max 1
⊆
(i) eachvertexinV hasexactlyoneoutgoingandoneincominghalf-edgeinC˜;
(ii) foreach2-cycleconverticesu,vsuchthatceitherbelongstoC orcbelongstoC andone
max 1
ofitsedgesbelongstosomecycleofC C˜ containsatmosttwohalf-edgesfrom
max
(u,x ),(x ,v),(v,x ),(x ,u) . Moreover if C˜ contains one half-edge of (u,v)
(u,v) (u,v) (v,u) (v,u)
{ }
andonehalf-edgeof(v,u),thenoneofthemisincidentwithuandtheotherwithv;
(iii) for each triangle t on vertices p,q,r containing edges (p,q),(q,r),(r,p) such that either (1)
t C C or (2) t C and C contains a 2-cycle on two vertices of t or (3) t C
max 1 max 1 1
∈ ∩ ∈ ∈
andC containsa2-cycleontwoverticesoft,
max
C˜ containsatmostfourandtheevennumberofhalf-edgesofedgesoft t(cid:48),wheret(cid:48) denotesa
∪
triangleoppositelyorientedtot. Alsoforanyedge(v ,v ),where(v ,v ) (q,p),(r,q),(p,r) ,
1 2 1 2
∈ { }
C˜ satisfiesthefollowing: ifw(v ,v ) > 1(w(t) w(v ,v )),thenC˜ doesnotcontainallhalf-
1 2 2 − 2 1
edges from the set (v ,x ),(x ,v ),(v ,x ),(x ,v ) , where v belongs
{ 1 (v1,v2) (v1,v2) 2 3 (v3,v2) (v2,v3) 3 } 3
to p,q,r v ,v ;
1 2
{ }\{ }
(iv) let c be a cycle of length 4 containing edges (p,q),(q,r),(r,s),(s,p) and such that c either
belongstoC andC hastwo2-cyclesthatshareanedgewithcorviceversa. Thenforeach
max 1
such c C˜ contains at most six and the even number of half-edges of edges of c c(cid:48), where c(cid:48)
∪
denotesa4-cycleoppositelyorientedtoc.
4
In graph G˜ each original edge of G is replaced by two half-edges. Since this time on, by saying
an edge, we will mean an edge of G and by saying a half-edge we will mean an edge of G˜. Also by
sayingthatanedge(u,v)ofGbelongstoC˜,wewillmeanthatbothhalf-edgesof(u,v)belongtoC˜.
Condition(ii)intheabovedefinitionsensuresthatC˜ doesnotcontaincertain2-cycles. Thefirstpart
ofcondition(iii)saysthatC˜ doesnotcontainalltheedgesofagiventriangleoratriangleoppositely
oriented. Thereasonforhavingthesecondpartofcondition(iii),inwhichweforbidthegivensubsets
ofhalf-edges,isillustratedinFigure1.
We will say that a cycle c of C is problematic if it is a 2-cycle that shares an edge with some
1
cycleofC oritisatriangleora4-cyclehavingpropertiesdescribedinconditions(iii)and(iv)of
max
Definition1. InananalogouswaywedefineaproblematiccycleofC .
2
TocomputearelaxedcyclecoverC improvingC andarelaxedcyclecoverimprovingC
1 max max
andC weconstructthefollowingundirectedgraphsG(cid:48) = (V(cid:48),E(cid:48))andG(cid:48)(cid:48) = (V(cid:48)(cid:48),E(cid:48)(cid:48)).
1
First we describe graph G(cid:48). For each vertex v of G we add two vertices v ,v to V(cid:48). For each
in out
edge(u,v) Eweaddverticese1 ,e2 ,anedge(e1 ,e2 )ofweight0andedges(u ,e1 ),(v ,e2 ),
uv uv uv uv out uv in uv
∈
eachofweight 1w((u,v)). Nextwebuildso-calledgadgets.
2
Foreach2-cycleofC onverticesuandvweaddverticesa ,b andedges(a ,e1 ),
max {u,v} {u,v} {u,v} uv
(a ,e2 ,(b ,e1 ),(b ,e2 )havingweight0.
{u,v} vu {u,v} vu {u,v} uv
GraphG(cid:48)(cid:48) isanextensionofgraphG(cid:48). LetC beanyrelaxedcyclecoverimprovingC . Addi-
1 max
tionalgadgetscontainedinG(cid:48)(cid:48) areasfollows.
Foreachtriangletonverticesp,q,r containingedges(p,q),
(q,r),(r,p) such that either (1) t C C or (2) t C and C contains a 2-cycle on two
max 1 max 1
∈ ∩ ∈
vertices of t, we add vertices a ,b and connect them appropriately to vertices e2 ,e1 ,e1
{p,q,r} {p,q,r} pq rq rp
and e1 ,e2 ,e2 via edges of weight 0. The gadget is depicted in Figure 2. Let us notice that it
pq rq rp
cannot happen that w((p,r)) > 1(w((p,q)) + w((q,r)) and w((q,p)) > 1(w((q,r)) + w((r,p))
2 2
and w((r,q)) > 1(w((p,q)) + w((r,p)) because it would mean that triangle consisting of edges
2
(q,p),(r,q),(p,r)hasgreaterweightthantandthusthatC isnotamaximumweightcyclecover
max
ofG. Thereforew.l.o.g. wecanassumethatw((p,r)) 1(w((p,q))+w((q,r)).
≤ 2
For each 4-cycle c such that c C and C has two 2-cycles that share an edge with c or
1 max
∈
c C and C has two 2-cycles that share an edge with c, we add a similar gadget which will
max 1
∈
enforcethatatmostsixhalf-edgesofedgesc c(cid:48) arepresentinC˜.
∪
Theorem2 AnyperfectmatchingofG(cid:48) yieldsarelaxedcyclecoverC improvingC . Anyperfect
1 max
matching of G(cid:48)(cid:48) yields a relaxed cycle cover improving C and C . A maximum weight perfect
max 1
matchingofG(cid:48)yieldsarelaxedcyclecoverC improvingC suchthatw(C ) OPT. Amaximum
1 max 1
≥
weight perfect matching of G(cid:48)(cid:48) yields a relaxed cycle cover C improving C and C such that
2 max 1
w(C ) OPT.
2
≥
Proof. First we will show that any perfect matching of G(cid:48)(cid:48) yields a relaxed cycle cover improving
C and C . Let M be any perfect matching of G(cid:48)(cid:48). M defines a set of half-edges C˜ E˜ of the
max 1
⊆
same weight that satisfies property (i) of Definition 1. Let us now verify that the thus defined set C˜
satifiesalsotheotherpropertiesofDefinition1.
Let c be a 2-cycle of G on vertices u,v. Then G(cid:48) contains vertices a ,b and the fact
{u,v} {u,v}
that these vertices have to be matched in a perfect matching ensures that C˜ does not contain all four
half-edges corresponding to (u,v) and (v,u). If vertices a ,b are matched so that a is
{u,v} {u,v} {u,v}
matchedwithe1 andb ismatchedwithe2 ,ora ismatchedwithe2 andb ismatched
uv {u,v} uv {u,v} vu {u,v}
withe1 ,thenbothhalf-edgesofoneoftheedges(u,v),(v,u)areexcludedfromC˜ andfortheother
vu
edge,eitherbothofitshalf-edgesbelongtoC˜ orbothdonotbelongtoC˜. Ifa ,b arematched
{u,v} {u,v}
intheotherway,thenmatchingM doesnotcontainappropriatelyeither(1)(u ,e1 )and(e1 ,v )
out uv vu out
or(2)(u ,e2 )and(e2 ,v ). Thereforecondition(ii)isindeedsatisfied.
in vu uv in
5
Lettbeatrianglecontainingedges(p,q),(q,r),(r,p)satisfyingthepropertiesgiveninDefinition
1. Suppose that w((r,p)) 1(w((p,q))+w((q,r)). Then the gadget for t in G(cid:48)(cid:48) looks as in Figure
≤ 2
2. Eachofthepairofverticesa ,b anda ,b ,anda ,b ensuresthatexactlytwo
{p,q} {p,q} {r,q} {r,q} {p,r} {p,r}
half-edges of edges belonging to corresponding 2-cycles are excluded from C˜. Vertices a and
{p,q,r}
b gurantee that two more half-edges connecting two vertices from the set V are excluded from
{p,q,r} t
C˜. ThereforeC˜ containsatmostfourhalf-edgesconnectingtwoverticesfromthesetV .
t
WewillprovenowthatC˜containsneitherallthehalf-edgesfromthesetS = (r,q),(p,x ),(x ,p)
1 (p,q) (r,p)
{ }
norallthehalf-edgesfromthesetS = (q,p),(r,x ),(x ,r) . LetussupposethatC˜ contains
2 (r,p) (q,r)
{ }
all the half-edges from the set S . Since edge (r,q) belongs to C˜ vertex b cannot be matched
1 {p,q,r}
with e2 . Since both (p,x ) and (x ,p) belong to C˜ vertex b cannot be matched with
(r,q) (p,q) (r,p) {p,q,r}
e1 orwithe2 . Butb isincidentonlywithverticese2 ,e1 ,e2 andthesearetheonly
(p,q) (r,p) {p,q,r} (r,q) (p,q) (r,p)
verticesitcanbematchedwith-acontradiction.
Let us now suppose that C˜ contains all the half-edges from the set S . Since edge (q,p) belongs
2
to C˜, vertex b must be matched with e2 . Since (x ,r) C˜ vertex b must be matched
{p,q} (p,q) (q,r) ∈ {q,r}
withe1 . Since(r,x ) C˜ vertexa cannotbematchedwithe1 . Buta isincident
(r,q) (r,p) ∈ {p,q,r} (r,p) {p,q,r}
onlywithe2 ,e1 ,e1 -acontradiction.
(p,q) (r,q) (r,p)
It is fairly straightforward to show that each traveling salesman tour T of G corresponds to some
perfect matching of G(cid:48)(cid:48). It is so because a tour cannot contain a 2-cycle or triangle if the graph
G contains at least four vertices. Therefore any maximum weight perfect matching of G(cid:48)(cid:48) yields a
relaxedcyclecoverC improvingC andC suchthatw(C ) OPT.
2 max 1 2
≥
TheproofforarelaxedcyclecoverC improvingC isanalogous. (cid:50)
1 max
p
in p p out
b out p
a in
1 {p,q}
xab x 1
2 bc b{p,q,r}
a c b
1 q {p,q}
out
d i q in a{p,q,r} r in
r
e h qin rout out
a{q,r} b {q,r}
f g q r
out in
Figure 1: Cmax consists of triangle abc and 6-cycle Figure 2: Agadgetcorrespondingtoatriangletonver-
defghi.C˜consistsoftwo4-cyclesacid,efghandtwohalf- ticesp,q,r suchthattconsistsofedges(p,q),(q,r),(r,p)
edges (x(ab),b) and (b,x(bc)). The only edges with posi- andw((p,r))≤ 21(w((p,q))+w((q,r)). Tomakethefig-
tiveweightare: (ab),(bc),(ca),(a,c). Thusw(Cmax) = uremorereadableverticespin,pout,qin,qout,rin,routare
w(C˜) = 3. Also edge (a,c) is such that w((a,c)) > drawntwiceeach. Verticesa ,b areconnected
{p,q,r} {p,q,r}
1(3−1). Amaximumtravelingsalesmantourinthisgraph correspondingly to vertices e2 ,e1 ,e1 and e1 ,e2 ,e2 .
2 pq rq rp pq rq rp
hasweight2,whichmeansthatC˜isuselessingettinganap- Notice that a ,b are connected to vertices on
{p,q,r} {p,q,r}
proximationbetterthan 2. edge(r,q)whichdoesnotbelongtot.
3
LetusnoticethatarelaxedcyclecoverC andarelaxedcyclecoverC improvingC andC
1 2 1 max
consist either of (1) directed cycles or (2) directed paths such that each one of them begins and ends
withahalf-edgeorboth.
WewillsaythatC (C )isintegralifforeachedgee E C (C )containseitherbothhalf-edges
1 2 1 2
∈
6
e1uv e 2uv
uout a b vin
{u,v} {u,v}
u v
in out
2 1
e e
vu vu
Figure3: Agadgetcorrespondingtoa2-cycleonverticesu,v.
of e or none. If C or C is integral, then it corresponds to a cycle cover of G i.e. consists only of
1 2
directedcycles.
3 3 - approximation algorithm for Max ATSP
4
The algorithm starts with computing a maximum weight cycle cover C of a given graph G. If
max
C does not contain any hard cycles, we can easily build a tour of weight 3OPT. Otherwise
max 4
we compute a relaxed cycle cover C improving C and, if necessary, a relaxed cycle cover C
1 max 2
improving C and C . Both of them are such that w(C ),w(C ) OPT . We compute C and
max 1 1 2 1
≥
C in the way described in the previous section. We create two multigraphs G and G as described
2 1 2
in Introduction. The total weight of all edges of G (corr. G ) is equal to w(C +2w(C ) (corr.
1 2 max 1
2(w(C ) + w(C ) + w(C ))), which is at leas t3OPT (corr. 6OPT.). Next we would like to
max 1 2
divideedgesofG intofoursetsZ ,Z ,...,Z insuchawaythateachZ (1 i 4)isacollection
1 1 2 4 i
≤ ≤
of vertex-disjoint paths. If this turns out to be impossible, because C contains a problematic cycle,
1
we will show how to divide edges of G into eight sets Z ,Z ,...,Z in such a way that each Z
2 1 2 8 i
(1 i 8) is a collection of vertex-disjoint paths.The task of dividing edges of G or G into
1 2
≤ ≤
appropriatelyfouroreightsetscanequivalentlybeviewedascoloringeachedgeofG orG withone
1 2
of four or correspondingly one of eight colors. Let denote 1,2,3,4,5,6,8 , = 1,2,3,4
8 4
K { } K { }
and = 1,2,3 .
3
K { }
FurtheronwewilltreatG andG assimplegraphsbutwillrequirethateachedgeiscoloredwith
1 2
anappropriatenumberofcolorsequaltothenumberofitscopiesinthegraph.
We will saythat two edgese ande arein-coincidentif there existsa vertexv such that bothof
1 2
them are going into v. ( We will also say that e is in-coincident with e .) We will say that they are
1 2
out-coincident if there exists a vertex v such that both of them are outgoing from v. Two edges are
coincidentiftheyarein-coincidentorout-coincident. Wewillalsosaythatanedgeeisin-coincident
withacyclecofC ifeisin-coincidentwithsomeedgeofc.
1
Cyclecoversareaspecialcaseofb-matchingsandthereforemanynotionsandfactsfrommatching
theory will prove useful. Below we recall some of them. A sequence P of edges (e ,e ,...,e ) is
1 2 k
said to be an alternating path (with respect to a given cycle cover C) if its edges alternately belong
and do not belong to C, every two consecutive edges of P are coincident and for each 1 < i < k
edgee isin-coincidentwithe andout-coincidentwithe orviceversa. AnalternatingpathP =
i i−1 i+1
(e ,e ,...,e ) such that exactly one of the edges e ,e belongs to C and either e is in-coincident
1 2 k 1 k k
with e and out-coincident with e or the other way round is called an alternating cycle (with
k−1 1
respecttoC). ItcanbeobservedthatasymmetricdifferenceC C = (C C ) (C C )oftwo
1 2 1 2 2 1
⊕ \ ∪ \
cyclecoversC ,C ofGconsistsofedge-disjointalternatingcycles(wrt. bothC andC ). (Seealso
1 2 1 2
[15]forexample.)
SinceC isnotalwaysintegral,weneedtoextendthedefinitionofanalternatingcycle. Wedoit
rel
asfollows. LetC beanycyclecoverofG. AsequenceP = (e ,e ,...,e )isgoingtobecalledan
1 2 k
7
alternatingh-cycle(wrt. C)ife ande arehalf-edgesandotherwiseP satisfiesanalogousconditions
1 k
asanalternatingpath.
Wel also introduce the term of an alternating weight. Suppose that P is an alternating path or
(cid:80)
cycle (wrt. C), then the alternating weight (wrt. C) of P is defined as w˜ (P) = w(e)
(cid:80) C e∈P\C −
w(e). If C is a cycle cover of G and A an alternating cycle wrt. C, then by applying A to
e∈P∩C
C wewillmeantheoperation,whoseresultisC A. C AisalsoacyclecoverofGandwehave
⊕ ⊕
w(C A) = w(C)+w˜ (A). Wecaneasilyobservethefollowingpropertyofanyalternatingcycle
C
⊕
withrespecttoC .
max
Fact1 LetAbeanyalternatingcyclewithrespecttoC . Thenw˜ (A) 0.
max Cmax ≤
Proof. C is a maximum weight cycle cover of graph G and C A is a cycle cover of G.
max max
⊕
Thereforew(C A) w(C ). Sincew(C A) = w(C)+w˜(A),wegetthatw˜(A) 0. (cid:50)
max max
⊕ ≤ ⊕ ≤
BothC C andC C disintegrateintoalternatingcyclesand h-cycles, i.e. eachedge
max 1 max 2
⊕ ⊕
ofcorrespondinglyC C andC C belongstoexactlyonealternatingcycleorh-cycle.
max 1 max 2
⊕ ⊕
WewillsaythatanalternatingcycleAofC C isnecessaryifC Acontainsa2-cyclewhich
max 1 1
⊕ ⊕
alsobelongstoC .
max
WewillsaythatanalternatingcycleAofC C isnecessaryifC Acontainsanyofthecycles
max 2 2
⊕ ⊕
describedinDefinition1.
ItwillbeconvenientforustodealwithrelaxedcyclecoversC andC suchthatbothC C
1 2 max 1
⊕
andC C containonlynecessaryalternatingcyclesandpossiblyalternatingh-cycles.
max 2
⊕
Lemma1 Let C be a relaxed cycle cover improving C such that w(C ) OPT and let C
1 max 1 2
≥
be a relaxed cycle cover improving C and C such that w(C ) OPT. Then there exists a
max 1 2
≥
relaxed cycle cover C(cid:48) improving C such that w(C(cid:48)) w(C ) and every alternating cycle from
1 max 1 1
≥
C C(cid:48) isnecessary. Similarly,thereexistsarelaxedcyclecoverC(cid:48) improvingC andC(cid:48) such
max 1 2 max 1
⊕
thatw(C(cid:48)) w(C )andeveryalternatingcyclefromC C(cid:48) isnecessary.
2 2 max 2
≥ ⊕
Because of the above lemma since this time on we will assume that every alternating cycle from
C C orC C isnecessary.
max 1 max 2
⊕ ⊕
3.1 ColoringofG
1
WewillnowshowhowtocolortheedgesofG withcolorsfrom insuchawaythateachcolorclass
1 4
K
consistsofvertex-disjointpaths. (IfitisnotpossibletocolorthewholeG ,thenweleaveproblematic
1
cyclesuncolored.)
LetusfirstdemonstatehowwecolorthegraphG whenC isintegral.
1 1
Formally,thecoloringofG isanassignmentofcolorsof toedgesofG . EachedgeofC
1 4 1 max
K
is assigned one color and each edge ofC is assigned two colors. (Ifan edge belongs to C C ,
1 max 1
∩
then,asaconsequence,itisassignedthreecolors.) WesaythatacoloringofG ispartialifsomeof
1
itsedgeshavecolorsassignedtothem(theseedgesarecolored)andsomeofthemdonot(theseedges
are uncolored.) Still, in a partial coloring of G an edge of C either has two colors assigned to it or
1 1
none. A(partial)coloringofG iscalledgoodifeachcolorclassconsistsofvertex-disjointpaths
1
TheideabehindcoloringG isthatforeachcycleofC wemarkatleastoneofitsedges,that
1 max
is going to be colored with one of the colors of and color the unmarked edges of C with 4.
3 max
K
Whenever possible we want to color the edges of C only with colors of . Each edge e = (u,v)
1 3
K
of C has to be colored twice. Let e ,e be two edges of C that are coincident with e, i.e. they
1 1 2 max
havetheforme = (u,v(cid:48)),e = (u(cid:48),v)(theymaydenotethesameedge). Then,ofcourseecannotbe
1 2
8
coloredwiththesamecolorase ore . Ife ismarkedandwewanttocolorit1ande isunmarked,
1 2 1 2
thenehastobecoloredwith2and3. Ifbothe ande aremarkedandwewanttocolore with1and
1 2 1
e with2,thenwehavetouse4forcoloringeandcoloritwith3and4. Ontheotherhand,ifbothe
2 1
ande aremarkedandwewanttocolorthemwiththesamecolor,say1forexample,thenwehavea
2
choiceandcancolorewith2and3or2and4or3and4.
In a (partially) colored G a cycle c is called monochromatic if there exists a color i of such
1 4
K
that each edge of c is colored i. Of course, if a (partial) coloring of G is good, then G does not
1 1
containanymonochromaticcycles. Wewillsaythatanedgeeissafeifnomatterhowwecolortheso
far uncolored edges, it is guranteed not to belong to any monochromatic cycle. For example suppose
that edge e = (u,v) C is colored with color 1 and that edges (z,u) C ,(z(cid:48),u) C are
max max 1
∈ ∈ ∈
colored appropriately 2 and 3,4. Then clearly e is safe. An edge e = (u,v) C will be called
max
∈
externalifuandv belongtotwodifferentcyclesofC . Otherwiseitwillbecalledinternal.
1
We state now the observation that will prove very useful in guaranteeing that we do not create
monochromaticcycles.
Observation1 Supposethatedgee = (u,v)colored1goesintocyclecofC (i.e. udoesnotbelong
1
to c and v does). If there exists no edge e(cid:48) = (u(cid:48),v(cid:48)) going out of c (u(cid:48) is on c and v(cid:48) is not) that is
coloredk,thenedoesnotbelongtoanycycle,whosealledgesarecoloredk. Inotherwords,eissafe.
WesaythatanedgeebelongingtoacycleofC iswingedifallexternaledgescoincidentwithit
1
are marked. (An edge that has no external edges coincident with it is thus wingy as well.) For each
cycle c of C we will require that at least c 1 of its edges are winged. ( c denotes the length of
1
| |− | |
c.) An external marked edge e coincident with a cycle c is called a tail (of c) if all external edges
coincident with c are marked or if e is coincident with an edge of c that is not winged. An external
markededgeecoincidentwithacyclecthatisnotitstailisawing(ofc). Wealsosaythatacyclecis
tailyifchasexactly2 c externalmarkededgescoincidentwithit.
| |
Now,wearereadytogivethedefinitionofanicesetofmarkededges.
Definition2 WesaythatasetofmarkededgesofC isniceif
max
foreachcyclecofC atleast c 1ofitsedgesarewinged,
1
• | |−
noedgeofC isatailoftwodifferentcyclesofC ,
max 1
•
for each2-cyclecofC , ifc has exactlyone tail, then italso hasa wing whichis alsoa wing
1 1
•
ofanothercycleofC .
1
Lemma2 There exists an algorithm of marking the external edges of C in such a way that the
max
resultingsetisnice. TherunningtimeofthealgorithmisO(n).
NextwearegoingtoshowthealgorithmofcoloringG whichconsistsofthreephases. InPhase1
1
wecolorallmarkedexternaledgesinsuchawaythatallofthemaresafe. Forclarity,wefirstpresent
Phase1andPhase2forthecasewhentherearenotailycycles.
Phase1ofAlgorithmColorG1
whilethereexistsacyclecofC1thathasanuncoloredtailorwingdo
coloralltheuncol.markedexternaledgesgoingintocwith1andalltheuncol.markedexternaledgesgoingoutofcwith2or3;
Lemma3 InPhase1ofAlgorithmColorG ,aftergettingcolored,amarkedexternaledgeeissafe.
1
9
Proof. Let e denote any external marked edge. At some point, e gets colored together with all other
so far uncolored external marked edges coincident with some cycle c of C . If at this point, c has no
1
previously colored marked external edges coincident with it, then by Observation 1, e becomes safe,
becausewecolorallmarkededgesgoingintocwith1andallmarkededgesgoingoutofcwith2or3.
If,ontheotherhand,chassomepreviouslycoloredmarkedexternaledgescoincidentwithit,then,by
thisproof,theyarealreadysafe. Thus,ifeisgoingintoc,thenwecoloritwith1andifthereexistsan
edgeegoingoutofcandcolored1,thenitmusthavebeencoloredpreviouslyandissafe. Therefore,
eissafeaswell. Thecasewheneisgoingoutofcisanalogous. (cid:50)
After Phase 1 all colored edges are safe but it may happen that there is no way of completing
this coloring so that it is good. For example if a cycle c has 2c external edges coincident with it and
all edges going into c are colored 1 and all edges going out of c are colored 2, then the only way of
completing the current coloring is to color each edge of c with 3 and 4. This way we would create a
cyclewithintheclassofedgescolored3andwithintheclassofedgescolored4. Wesaythatacycle
c(cid:48) of G is a cycle within a cycle c if every vertex lying on c(cid:48) lies also on c. We say that a cycle c of
1
C isblockedifthereisnowayofcompletingthecurrentpartialcoloringsothatitdoesnotcontaina
1
monochromaticcyclewithinc. Aninternaledgee = (u,v)willbecalledquasiexternalifthereexist
externaledgese ,e suchthate isincidentwithuande isincidentwithv.
1 2 1 2
Lemma4 AcyclecofC canbeblockedonlyifcsatisfiesoneofthefollowing:
1
c has exactly 2 c external edges coincident with it and the number of edges of c whose wings
• | |
are colored the same plus the number of colors of that occur on the external edges of c is
4
K
lessthan4,
allexternaledgesofcarecoloredwiththesamecolor,
•
allexternaledgescoincidentwithcaremarked,allinternaledgesofcarequasiexternalandfor
•
eachquasiexternaledgee = (u,v)ofcthereexisttwodifferentcolorsi ,i of suchthatan
1 2 3
K
externaledgeofcincidentwithuiscoloredi ,anexternaledgeofcincidentwithv iscolored
1
i andanedgeofcthatiscoincidentwithehasanexternalmarkededgecoincidentwithitthat
2
iscoloredwithacolori belongingto i ,i ,
3 3 1 2
K \{ }
cisproblematic.
•
Otherwise the current partial coloring can be completed in such a way that it does not contain a
monochromaticcyclewithinc.
LetusnoticethatifacyclecofC isblockedanditisnottaily, thenwecanchangethecoloring
1
ofitstailttok,wherek issuchacolorof thatnoexternaledgeincidentwithciscoloredk,and,
3
K
asaresult,cwillnotbeblockedanymoreandtwillbesafe. Roughly,thisiswhatwearegoingtodo
inPhase2withblockedcyclesthatarenottaily.
InPhase2wearegoingtochangethecoloringofsomeoftheexternalmarkededgesinsuchaway
thateachcolorededgeissafeandthereisnoblockedcycleofC .
1
Phase2ofAlgorithmColorG2
whilethereexistsablockedcycleofC1thatisnottaily
c←anyblockedcycleofC1thatisnottaily
whilecisblockeddo
changethecoloringofthetailtofcsothatcisnotblockedanymore
10