with Time Zones
BjornBroden 1
,Mikael Hammar 2
, andBengtJ.Nilsson 3
1
DepartmentofComputerS ien e,LundUniversity,
Box118,S-22100Lund,Sweden
2
DipartimentodiInformati aedAppli azioni,UniversitadiSalerno
Baronissi(SA)-84081,Italy
3
S hoolofTe hnologyandSo iety,MalmoUniversityCollege,
SE-20506Malmo,Sweden
Abstra t. Thetimedependenttravelingsalesmanproblemisavariant ofTSPwithtime
dependentedge osts.Westudysomerestri tionsofTDTSPwherethenumberofedge ost
hangesare limited.We nd ompetitiveratiosforonline versionsofTDTSP.Fromthese
we derive polynomial timeapproximation algorithmsfor graphs withedge osts one and
two.Inaddition,wepresentanapproximationalgorithmfortheorienteeringproblemwith
edge ostsoneandtwo.
1 Introdu tion
Transportationands hedulingproblemsmodeledbythetravelingsalesmanproblemareinherently
stati .Tomodeladynami environmentageneralizationisneededthatin orporates hangesinthe
environment.Su hageneralizationisprovidedbythetimedependenttravelingsalesmanproblem
(TDTSP). This problem is a generalizationof TSP in whi h the ost of ea h edge depends on
time, i.e., the ost of an edge depends on the time intervalduring whi h theedge is traversed.
Severalaspe tsofTDTSPhavebeenstudiedanditseemstobediÆ ultto olle tthemallunder
asinglede nition.Themanyvariationsallstemfrom thewaythattimeisbeingmodelled.
In some de nitions time is proportional to the ost of the edges traversed [12℄. This gives
a naturalgeneralization of TSP and is suitable if for example variations in the traÆ load are
important in omputing a TSP tour for a delivery ompany. It also has onne tions to other
generalizationsofTSP, su h asthekineti TSP where movingpointsin theplane [9,10℄orona
line [10℄ are onsidered. We all this formulationthe ost dependent traveling salesman problem
(CDTSP).
Ourfo usisonanotherformulationofTDTSPwherethe ostofanedgedependsonitsposition
inthepath.We allthisproblemthestepdependenttravelingsalesmanproblem,denotedSDTSP.
Thisinterpretationhasbeenstudiedonnumerouso asionsbytheoperationsresear h ommunity,
dueto itsappli ationin s heduling[3,5,6,14℄.WelimitourstudyofTDTSP toinstan eswhere
theedge ostsare restri tedinsize and an hange onlyalimitednumberoftimes.
Dueto ompli ations ausedbytimedependen ies,approximationalgorithmsforTDTSP are
hardtoanalyze.By onsideringonlinealgorithms partofthetime dependen yisremovedanda
new itiesarebeingadded totheset.WearestudyingTSP inadynami environment;although
the itiesareknownfrom thebeginning,thedistan ebetweenapairof ities hangeovertime.
The online algorithms we present here still require a solution to the NP-hard Orienteering
problem [1℄.As weshall see in Se tion 5thisproblem allowsa3=4-approximationalgorithm for
graphswith edge osts oneand two. We onstru ta polynomialtime approximationalgorithm
for SDTSP by hoosing the best answer given by a set of online algorithms. This results in a
2 2=3k-approximationalgorithmfortheSDTSPwheretheedge ostsarerestri tedtothevalues
one and two and the osts an hange at most k 1 times. The restri tions on the edge osts
an be removed given an Orienteering algorithm that handles arbitrary edge osts. Sin e the
inapproximabilityratioofTSPgrowswiththerelativeedge ostswe annotexpe talgorithmsfor
SDTSPwithapproximationratioindependentofthe osts.
InSe tion2westatetheformalde nitionofSDTSP.InSe tion3weanalyzetheonlineversion
oftheproblemand inSe tion 4we onsider polynomialtimeapproximationalgorithms. Wealso
give an approximation algorithm for the Orienteering problem for edge osts one and two. In
Se tion5we onsiderthe ostdependenttravelingsalesmanproblem.Wegiveaninapproximability
resultfortheEu lideanCDTSPandpresentanonlinealgorithmforCDTSPwithtwotimezones
andedge ostsoneandtwo.Thisonlinealgorithmismodi edtogiveanapproximationalgorithm
asin thepreviousse tion.
2 De nition of SDTSP
De nition1. Consider a set of edge ost fun tions f
1 ;:::;
n
g assigned to a omplete graph
G=(V;E)with jVj=n,jEj= n 2 ,and where t
(e)is the ostfun tion for edge e2E.Let e
ij
denotethe edgebetweenv
i andv
j
inV:Thestepdependenttravelingsalesmanproblem(SDTSP )
seeksapermutation ofV thatminimizes
n (e n 1 )+ n 1 X i=1 i (e i i+1 ):
Inorderto omprehendDe nition1ithelpsto onsideraninstan eofSDTSPasann-layered
graph,ea h layer ontaining nverti es.Layeri and layeri+1form a omplete bipartitegraph
wheretheedgesaredire ted,goingfromlayeritolayeri+1,andaregivenweightsa ordingto
weightfun tion
i
. Theobje tiveis to nd apathgoingfrom layeroneto layern thatvisits all
olumns,startingfromand endingatthesame olumn.
We anview
t
asadis retetimedependentedge ostfun tion de nedon t=f1;:::;ng.To
simplify the problemwerestri tourstudy to the asewhere this ost fun tion hanges at most
k 1timesasafun tionoft.Aregionwherethe ostfun tionis onstantis alledatimezone.Let
0=z 0 ;z 1 ;:::;z k 1 ;z k
=n2N denotethetimezonedivisors,i.e.,timezonei=fz
i 1 +1;:::;z i g, and j = j 0 forz i 1 <j;j 0 z i
. We simplify therepresentationby assigningone ostfun tion
to ea h timezone, givingus aninstan e I =f(
1 ;z 1 );:::;( k ;z k )g. An instan eof SDTSP with
ktimezonesisdenotedSD k
TSP. LetM andmdenotethe ostsofthemostandleastexpensive
edgesusinganyofthe ostfun tions
1 ;:::;
k .
WeprimarilystudytheonlineversionofSD k
TSPforwhi hanalgorithmre eivesinformation
regarding the time zones and the ostfun tions online. Let I
j =(
j ;z
j
) representthe instan e
restri tedtotimezonej,i.e.,I=fI
1 ;:::;I
k
g.Thealgorithmre eivestheinputinstan eonetime
unvisitedverti esandwithedgeweightsgivenby
j
.WhenP
j
hasbeen omputed,thealgorithm
re eivesthenextpartoftheinputinstan e,i.e.,I
j+1
.Afterktimezonesthealgorithmhasre eived
theentireinputinstan eandaHamilton y lein Ghasbeenbuilt.
De nition2. Given an arbitrary algorithm A for SD k
TSP wewrite A[I℄ for the verti esof the
y le that the algorithm hooses on the instan e I and A(I) for the ost of the resulting y le
produ edon I. Thiswill alsobeusedfor spe i timezones,for instan eA(I
1
) isthe ostof the
pathin timezoneone.
LetA j [I℄=[ j i=1 A[I i
℄,i.e.,the verti esthat A hooses fromtimezoneone totimezonej.
De nition3. Let R (A) denote the ompetitive ratio for algorithm A and R the smallest
om-petitive ratio for any online algorithm, i.e., R = min
A
fR (A)g. We use ALG to denote an
arbi-traryonlinealgorithm,OFFanarbitraryoinealgorithmandOPTthe optimaloinealgorithm
for SD k TSP. 3 Online SD k TSP
Herewepresentalowerbound onthe ompetitiveratioforSD k
TSP. Wealsopresentastrategy
with a ompetitive ratio mat hing the lower bound. To ompute the lower bound we use an
adversaryargument.Theadversarybuildsahardinstan eforthestrategy.Thisisdoneonlinein
responsetothede isionsmadebythestrategy.LetI z
denotetheinstan e reatedbytheadversary.
Inadditiontothisinstan ewede neanadversaryoinealgorithmZIG.Thisalgorithmisadapted
bothto I z
and to the de isionsthat were made by theonline strategy beinganalyzed. We an
think of the adversary algorithm as being run in hindsight after the ompletion of the online
strategy.NotethatZIGis designedto beoptimalforI z
.
Nodesvisitedbyboth
algorithmspriortotimezonej. neitherALGnorZIG.
Nodesvisitedsofarby
ZIGpriortotimezonej. Nodesvisitedonlyby ALGpriortotimezonej. Nodesvisitedonlyby Allotheredgeshave ostM. Edges with ostm.
Fig.1.Timezonejofatypi alinstan e reatedbytheadversary.
The adversary onstru ts theinstan e I z
asfollows:alledge osts in I z 1 areset tom. In I z j ,
for2jk,all ostsofedgesadja enttoverti esinALG j 1
[I z
℄aresettomandallotheredge
isdesigned tomaximize thenumberof ostm edgesused. Itis des ribedbelowandits strategy
fortimezonej isillustratedinFigure2.
Algorithm ZIG
Input: Thetimezoneinformationz
0 ;:::;z
k
andtheHamiltonpathP
A
produ edbyALG.
Output: AHamiltonpathPZ.
PZ=;;bottom:=1;top:=n for i:=1 to z 1 do P Z [i℄:=P A
[top℄;top:=top 1
endfor
for j:=2 to k do
/*forea htimezonej>1*/
for i:=z j 1 +1 to z j do if bottom<z j 1
andodd(i) then
PZ[i℄:=PA[bottom℄;bottom:=bottom+1
else
P
Z [i℄:=P
A
[top℄;top:=top 1
endif endfor endfor End ZIG 3 2 1 ZIG ALG n n z 1 2 n z1 1 n z 1
Fig.2.Des riptionofthestrategyusedbyZIG.
Todeterminethe ompetitiveratioweneedto measurethe numberof ostmedges that are
usedbyZIG.Espe ially,wewouldliketo ountthenumberof ostmedgesthatareusedbyZIG
butnotbyALG.Theusagefun tionisageneral on eptthat ountsthenumberof ostmedges
usedbyanarbitraryoinealgorithmOFFuptoaspe i edtimezonej.A ountededgeeshould
omply withthe following onditions: (1)If theonline algorithm usese in timezone l<j then
l
(e)=M.(2)Theonlinealgorithm annotuseeinatimezonelj.
Tomakesurethat the se ond onditionholds weonly ountedges that ontainat least one
De nition4. The bla kverti esinzone j are de nedas V b j (OFF )=ALG j 1 [I℄nOFF j 1 [I℄:
WeassumethatanonlinestrategyALGhasbeenrunonaninstan eI resultingina y leP
A .
We label the verti es with uniquenumbers from oneup to n a ordingto theirposition in the
y leP
A .
De nition5. Theusagefun tionU
j
(P)forapathPinaSD k
TSPinstan e,isde nedasfollows:
U 1 (P)=0 U j (P)=U j 1 (P)+jfe2P j j j
(e)=m and ehas an endpointin V b
j
(OFF)gj
HereP
j
denotesthesubpathof P restri tedtotimezonej.Wewillviolatethenotationanduse
U
j
(A)todenotetheusagefun tionfora y leprodu edbythealgorithmA.WeletU
j =U j (P Z ), P Z
beingthe y leprodu edbyZIGrunonI z
,andweuseV b
j
todenotethe oloringoftheverti es
in time zonej, using ZIG astheoine algorithm onI z
. Wewantto nd thevalueof U
j .Note
thattheusagefun tiondoesnot ountedges hosenin timezoneone.
Lemma1. If ALG visitsa vertexin timezonej >1 thathas beenvisitedbyZIG in aprevious
timezonethen jALG j
[I℄[ZIG j
[I℄j=n,where I isanarbitrary inputinstan e.
Proof. AssumethatALGvisitsthevertexlabellediduringtimezonej,andthatZIGvisitsvertex
i during time zone j 0
, where j 0
<j. Consider algorithm ZIG at the pointwhen it visits vertex
number i. By onstru tion, bottom z
j
0. Hen e, P
A
[bottom℄ hasalready been visited by ALG
whi h impliesthat i 6=bottom.Thus, i =top. This means that allverti esP
A
[l℄,for li have
alreadybeen visitedby ZIG. After time zonej all verti es P
A
[l℄, for li havebeenvisited by
ALG,bytheassumptionabove.Wetherefore on ludethat jALG j
[I℄[ZIG j
[I℄j=n. ut
Thenumberof ostmedgesthat ZIGusein timezonej is limitedbythetime zone'slength
and2jV b
j
jasfollows.
Lemma2. Thenumber of ostm edges ountedby U
j
(OFF )intimezone j>1is atmost
min f2jV b j (OFF)j;z j z j 1 g:
Proof. LetP bethe y legenerated byOFF andP
j
thesubpath ofP in time zonej. All edges
ountedbyU
j
(OFF)intimezonejbelongtoP
j byde nition,andjP j j=z j z j 1 .Se ondly,all edges ountedby U j
(OFF) in time zonej haveat least oneendpointin V b
j
(OFF). This implies
that ea h vertexin V b
j
(OFF)is adja entto atmost2 ountededges.It followsthat thenumber
of ostmedges ountedisatmost2jV b
j
(OFF)j. ut
Lemma3. Thenumber of ostm edges ountedby U
j
in timezonej>1isexa tly
minf2jV b j j;z j z j 1 g:
Proof. FromLemma2itfollowsthatZIGusesatmostminf2jV b j j;z j z j 1
g ostmedgesintime
zonej.Toseethat ZIGusesatleastminf2jV b j j;z j z j 1 g ostmedgesofI z weneedtoexamine
IfjALG j 1 [I z ℄[ZIG j [I z
℄j=n thenalledgestakenbyZIG in timezone j haveendpointsin
ALG j 1
[I z
℄ and by de nition, these edges have ost m in I z . Hen e, exa tlyz j z j 1 ost m
edgesare ountedbyU
j in timezonej. IfjALG j 1 [I z ℄[ZIG j [I z
℄j<nthen thereare verti esneithervisitedbyALG inatime zone
priortoj norbyZIGinatimezonepriortoj+1.Lett
j
denotethevalueoftopandb
j
thevalue
ofbottomatthebeginningoftimezonej andlett 0
j andb
0
j
denotethe orrespondingvaluesatthe
endofthetimezone.Now,z
j 1
istheindexofthelastvertexvistedbyALG intimezonej 1,
andsin ethere areunvisitedverti esleft, t 0
j >z
j 1
.Thebla kverti esthereforehavethelabels
b j ;:::;z j 1 .Thus, z j 1 b j +1=jV b j
j. Furthermore,anyvertextakenfrom the topof P
A has
notbeenvisitedyet byALG.
Consideralgorithm ZIG at the beginning of time zone j. Everytime bottom is in reased, a
bla k vertexis in orporated into ZIG's path. Everytime top is de reasedavertex unvisited by
ALG is inserted into the path. Thus, everyse ond vertex pi ked is bla k saveperhaps astring
of un oloredverti esat theend. Ifexa tlyeveryse ondvertexin thepath onstru tedfor time
zonej isbla kthenallz
j z
j 1
edgesare ountedbyU
j
.Iflessthaneveryse ondvertexisbla k
thentheif-statementinsidethefor-loophasbeenviolatedmorethan(z
j z j 1 )=2timesandwe inferthatb 0 j =z j 1
.Thus,bottomhasbeenin reasedjV b
j
jtimesandallbla kverti eslieonthe
subpath of P
Z
in time zonej. Foreverybla k vertex twoedgesare ounted byU
j
. Inthis ase
2jV b
j
jedgesare ounted. ut
Lemma2givesthefollowingupperbound onU
j (OFF).
Lemma4. U
j
(OFF)isgiven by the following re urren erelation:
U 1 (OFF)0 U j (OFF)U j 1 (OFF)+min 2jV b j (OFF )j;z j z j 1 ,if 2jk;
Lemma3givesus are urren erelationforU
j
similartotheupperbound onU
j (OFF).
Lemma5. U
j
isgiven bythe following re urren erelation:
U 1 =0 U j =U j 1 +min 2jV b j j;z j z j 1 ,if2jk;
Toevaluatethe re urren ewewould liketo express jV b j j in termsof U j 1 and z j 1 . Tothis
endweusethefollowinglemma.
Lemma6. If2jV b j j<z j z j 1 then ALG j 1 [I z ℄[ZIG j 1 [I z ℄ <n:
Proof. Atthebeginningoftimezonej,thenumberofverti esvisitedbyALGbutnotbyZIGis
jV b
j
j.Thenumberofverti esvisitedbyZIG inprevioustimezonesisz
j 1 .Therefore, ALG j 1 [I z ℄[ZIG j 1 [I z ℄ = ALG j 1 [I z ℄nZIG j 1 [I z ℄ + ZIG j 1 [I z ℄ =jV b j j+z j 1 2jV b j j+z j 1 <z j z j 1 +z j 1 =z j n:
Lemma7. If2jV b j j<z j z j 1 thenU j 1 =2 ALG j 1 [I z ℄\ZIG j 1 [I z ℄ . Proof. If 2jV b j j <z j z j 1 then ALG j 1 [I z ℄[ZIG j 1 [I z ℄
< n, a ording to Lemma 6. From
Lemma1itfollowsthatpriortotimezonej,ALGhasnevervisitedavertexalreadyvisitedbyZIG.
Weinfertwo onsequen esfromthisfa t,the rstbeingthatallverti esinALG j 1 [I z ℄ \ ZIG j 1 [I z ℄
were bla k when ZIG visited them. The se ond onsequen e is that top > z
j 1
at the end of
time zone j 1. This implies that no edge in the path produ ed by ZIG has two endpoints in
ALG j 1 [I z ℄\ZIG j 1 [I z
℄. Thenumber ofedges in identto verti esin ALG j 1 [I z ℄\ZIG j 1 [I z ℄ is therefore 2 ALG j 1 [I z ℄\ZIG j 1 [I z ℄
and all of them are ounted by U
j 1
, sin e U
j 1 only
ountsedgesthatALG hasalreadyvisited.Hen e,theresultfollows. ut
Now we an nd a simpler expression for U using the following transformation. If 2jV b j j < z j z j 1 thenjV b j j= ALG j 1 [I z ℄ ALG j 1 [I z ℄\ZIG j 1 [I z ℄ =z j 1 U j 1 =2, byLemma 7.
Wehaveprovedthefollowingtheorem.
Theorem1. Theusage fun tion of ZIGisequal to
U 1 =0; U j =U j 1 +minf2z j 1 U j 1 ;z j z j 1 g,if2j k.
GivenTheorem1we omputeamaximumvalueofU
k .
Theorem2. Themaximum valueof the fun tionU
k isU k =n z 1 ando urs whenz i =(2 i 1)n=(2 k 1). Proof. AssumingU 0 =0wehavethat U j =U j 1 +minf2z j 1 U j 1 ;z j z j 1 g: U j is maximizedif2z j 1 U j 1 =z j z j 1 . Thatis, U j =2z j 1 ; (3.0.1) U j =U j 1 +z j z j 1 : (3.0.2) Substituting U j with2z j 1 in (3.0.2)yields z j 3z j 1 +2z j 2 =0:
Wesolvetheequationgiventhat z
0 =0andz k =n: z j = 2 j 1 2 k 1 n: Aftersubstitutingz j
a ordinglyin(3.0.1) wehavethat
U k =n z 1 ; sin ez 1 =n=(2 k 1). ut
Westatea nal lemma on erningtheusagefun tion foranyoinealgorithm OFF. Weuse
Lemma8. Let 0=z 0 ;z 1 ;:::;z j
n2 N. Forany instan e I =f(
1 ;z 1 );:::;( j ;z j )g of SD k TSP
andany oinealgorithm OFF,
U
j
(OFF )U
j
Proof. Byde nition U
1
(OFF )=U
1 =0.
For j > 1, assume that U
j 1
(OFF) U
j 1
. By de nition, U
j 1
ounts the number of low
ostedgesvisited byOFFin time zonej 0
<j havingat least oneend pointin ALG j
0
1
[I℄. The
numberofsu hverti esisat least Uj 1 2 ,andhen e, U j 1 2 ALG j 0 1 [I℄\OFF j 0 [I℄ ALG j 1 [I℄\OFF j 1 [I℄ =j ALG j 1 [I℄ ALG j 1 [I℄nOFF j 1 [I℄ =z j 1 V b j (OFF) :
FromLemma4and ourindu tionhypothesiswehavethat
U j (OFF)U j 1 (OFF)+min 2jV b j (OFF)j;z j z j 1 U j 1 (OFF)+minf2z j 1 U j 1 (OFF);z j z j 1 g =minf2z j 1 ;z j z j 1 +U j 1 (OFF)g minf2z j 1 ;z j z j 1 +U j 1 g =U j 1 +minf2z j 1 U j 1 ;z j z j 1 g =U j u t
3.1 Lower Boundon the CompetitiveRatio
Thisse tionusestheresultswehavearrivedatsofartostatethe rstofourmainresults.
Theorem3. The ompetitiveratio, R ,of anyonlinealgorithm for SD k TSP is R1+ (M m)U k ZIG(I z )
Proof. Consideran arbitraryonline strategyALG. Toanalyze the ompetitive ratioweuse our
adversary argument, in ludingthe instan e I z
together with the adversaryalgorithm ZIG. The
onlinealgorithm hastopayM onalledgesex eptthoseused intimezone one,sin eedges with
ost set to m in other zones anonly be used by ZIG. The ost for ZIG on the instan e I z is ZIG(I z )=mU k +M(n (U k +z 1 ))=Mn (M m)(U k +z 1
);givingusthe ompetitiveratio
R= min ALG max I ALG(I) OPT(I) min ALG ALG (I z ) ZIG(I z ) M(n z 1 )+mz 1 ZIG(I z ) :
Doingthe al ulationsba kwardswegetthat
R M(n z 1 )+mz 1 Mn+(M m)(U k +z 1 )+ZIG(I z ) ZIG(I z ) =1+ (M m)U k ZIG(I z )
ApplyingTheorem2to Theorem3produ esthefollowing orollary.
Corollary 1. The worst ase ompetitive ratio R of any strategy ALG on any instan e I =
f( 1 ;z 1 );:::;( k ;z k )g isatleast R M m 2 k 2 2 k 1 + 1 2 k 1 :
3.2 Upper Boundon the CompetitiveRatio
Thereisasimplestrategythata hievestheupperboundfoundinthelastse tion.Thisalgorithm,
presentedbelow,usesagreedyapproa h.
Algorithm GREEDYforSD k
TSP
Input: Asequen eofk instan esI
1 ;:::;I
k .
Output: AHamilton y leP.
1 Givenaninstan eIj,produ ethe heapestpathofsize zj zj
1
ontheunvisitedverti esinV.
End GREEDY forSD k
TSP
Theorem4. The ompetitiveratioof GREEDYasafun tion of U
k isatmost 1+max I (M m)U k OPT(I)
Proof. LetI beanarbitrarySD k
TSPinstan e, andP
I
theoptimalTSP touronI.Intime zone
oneGREEDY(I
1
)=OPT(I
1
).Assume thatthefollowingholdsforj>1:
GREEDY(I j )OPT(I j )+(M m)(U j (P I ) U j 1 (P I )):
Summinguptheinequalities(in ludingtimezoneone)yieldsthat
k X i=1 GREEDY(I i )OPT(I)+(M m)(U k (P I )) OPT(I)+(M m)U k ;
a ordingto Lemma8.Thisgivesa ompetitiveratioof
R (GREEDY)max I OPT(I)+(M m)U k OPT(I) =1+max I (M m)U k OPT(I) :
It remains to prove our assumed inequality. This is equivalent to showing that there is a path
usable byGREEDY in time zone j >1 that ostsat mostOPT(I
j )+(M m)(U(OPT j [I℄) U(OPT j 1
[I℄)). First observethat the number of edges ounted by the usage fun tion in time
zonej isU j (P I ) U j 1 (P I
).OPTpaysatleastm(U
j (P I ) U j 1 (P I
))forthese edges.This ost
is in luded in OPT j
[I℄. Edgesnot ounted by the usage fun tion an be used by ALG for the
same ost asOPT. These edges are onne ted with possibly expensive edges for atotal ostof
OPT(I j )+(M m)(U j (P I ) U j 1 (P I
m m ALG OPT M M M m M m m m m
Fig.3.Pat hingtogetherthepie esoftheoptimalalgorithm'spath.
ThistheoremtogetherwithTheorem3givesasharpworst ase ompetitiveratio.
Corollary2. The ompetitiveratioof GREEDY is
R= M m 2 k 2 2 k 1 + 1 2 k 1 :
Note that we have no implementation of GREEDY that runs in polynomial time, sin e it
ontainsanNP- ompletesubproblem.
4 Polynomial Time Approximation Algorithms for SD k
TSP
Next wedes ribe apolynomialtime algorithm for SD k
TSP with edge osts from the set f1;2g.
Weusetheexponentialtimegreedyalgorithmdesignedfortheonline ase.Thisgreedyalgorithm
isexponentialsin eit ndsanoptimalk-TSPpath.Wegiveanapproximationalgorithmforthe
longestpath problem in agraphwith edge ostsin the set f1;2g. This problemis also referred
toastheOrienteeringproblemintheliterature[1℄.Thenewalgorithmgivesanapproximationto
k-TSP,whi h anbeusedtomakethegreedyalgorithmpolynomial.
4.1 Orienteeringwith Edge Costs One and Two
De nition6. GivenavalueTandagraphGwithedgeweightsoneandtwo,theOrienteering(1;2)
problemisto omputethelongestpathinGwith ostatmostT.Thelengthofapathisthenumber
of edges it onsistsof.
Let us start with a straightforwardalgorithm that a hieves anapproximationfa tor of 2=3.
Thenwedes ribeasimpleenhan ement,improvingtheapproximationfa torto3=4.
The algorithm is based on mat hing. We simply perform amaximummat hing on the ost
oneedgesintheinputgraphG.Observethatthemat hing onsistsofatleasthalfthenumberof
ostoneedgesintheoptimalpath.Hen e,we onstru tapathwith ostT inwhi heveryse ond
edge omesfromthemat hingaslongasthereareedgesin themat hing leftto hoose.Thereby
weguaranteethatthelengthofourpathisatleasttwothirds oftheoptimallength.
bothmat hingsindu eapossiblydis onne tedsubgraphofG ontainingpathsoflengthsbetween
oneand nand y lesof lengthsbetweenfour and n.If we breakupall the y lesintopaths by
removing oneedgewetransformthe indu edgraphinto aforest of paths.A y le oflength iis
thereby transformed into a path of length i 1. Denote the resulting forest G 0
. Our algorithm
buildsalongpathwith ostT by on atenatingthepathsofG 0
in de reasinglengthorder,using
arbitraryedgesas\glue".
Algorithm Enhan edOrienteering(1;2)
Input: AvalueT anda ompletegraphG=(V;E)withedge ostseitheroneortwo.
Output: Apathwith ostT.
1 E1 Amaximummat hingonthe ostoneedgesinG.
2 E
2
Amaximummat hingontheremaining ostoneedgesinG.
3 G 0 (V;E 1 [E 2 )
4 Break upthe y lesinG 0
asdes ribedabove.
5 Constru taHamiltonpathHofGby on atenatingthepathsinG 0
inde reasinglengthorder,
appendingtheremaining osttwoedgestotheend.
6 return thelongest(initial)partofH with ostlessorequaltoT.
End Enhan edOrienteering(1;2)
The time omplexity of this algorithm is dominated by the maximum mat hing pro edure,
whi hispolynomial[7℄.
To on rm the approximation ratio we ompare the path built by the enhan ed algorithm
(APX) withthe path onstru tedbyan optimalalgorithm(OPT).Let p
i
denotethe numberof
pathswithlengthiinG 0
.LetthelengthoftheshortestpathfromG 0
usedinAPXbejandletx
j
denotethenumberoflengthj pathsusedinAPX.Also,letp
0
denote thenumberof onse utive
ost two edges in APX. The optimal path ontains a orresponding set of paths built up by
onse utive ost oneedges. We divide these paths into aset K
2
ontainingpairs of onse utive
edgesand aset K
1
ontainingtheremaininglooseedgesasdes ribedin Figure 4.Letk
2 denote
the numberof edge pairsin K
2 , k
1
thenumber of edgesin K
1 , and k
0
thenumber of osttwo
edgesinOPT.ThelengthsofAPXandOPTare
K2 K1 K1 K1 K2 K 2 1 2 2 2 1 1 1 2 1 1 1 1 1
Fig.4. Dividethe paths of ostone edges withinOPT into the setsK
1
OPT=2k 2 +k 1 +k 0 ; (4.1.1) APX= X i>j (i+1)p i +(j+1)x j +p 0 ; (4.1.2)
andthevalueT an beexpressedas
T =OPT+k 0 ; (4.1.3) T X i>j (i+2)p i +(j+2)x j +2p 0 : (4.1.4)
Let A denote the number of ost one edges in G 0
that originate from the rst maximum
mat hing.LetBdenotethenumberof ostoneedgesfromthese ondmat hinginG 0
.Thefollowing
Lemma des ribes ru ial relationsregardingthe numberof edgesmat hed bythealgorithm, the
numberofpathsin G 0
and thenumberof ostoneedgesinOPT.
Lemma9. Thefollowing four onditionshold.
B X i>1 (i 1)p i ; (4.1.5) A> 3k 1 +6k 2 8 ; (4.1.6) B k 2 ; (4.1.7) A+B = X i1 ip i : (4.1.8) Proof. (4.1.5)B = P i>1 bi=2 p i
,sin eeveryse ondedgeinapathfromG 0
omesfromthese ond
mat hing.Furthermore,sin ebi=2 i 1ifi1,
X i>1 bi=2 p i X i>1 (i 1)p i :
(4.1.6)The rstmat hingwillin ludeatleasthalfthenumberof ostoneedgesintheoptimal
path,i.e.(2k
2 +k
1
)=2.Butaswestatedinthealgorithm,someoftheedgesfromthe rstmat hing
anbelost as webreak upthe y les.Butea h y lehasat least four edges,whi h meansthat
we anlooseat mostaquarteroftheedgesfrom themat hing. Thus,
A 3 4 2k 2 +k 1 2 = 6k 2 +3k 1 8
(4.1.7)The rstmat hingwillin ludeatmostoneoftheedgesinea hpairinK
2 .Therestof thek 2 edgesinK 2
areleftforthese ondmat hing,andallofthem anbeusedforthemat hing.
Thus,thenumberofedgesusedin these ondmat hingisatleastk
2 .
The orre tnessof(4.1.8)followsdire tly fromthede nitionofp
i
,AandB. ut
Toprovetheapproximationratioweneedto onsidertwo ases:p
0 =0andp 0 6=0. Case1:p 0 =0.
From(4.1.3) and(4.1.4)itfollowsthat
x j OPT+k 0 P i>j (i+2)p i :
Thus, APX X i>j (i+1)p i +(j+1) OPT+k 0 P i>j (i+2)p i j+2 = P i>j p i ((j+2)(i+1) (j+1)(i+2))+(j+1)(k 0 +OPT) j+2 : Ifj=1then X i>j p i ((j+2)(i+1) (j+1)(i+2))= X i>1 (i 1)p i B k 2 :
InsertingthisintotheexpressionforAPXandtakingtheratiobetweenAPXandOPTyieldsthat
APX OPT k 2 +2(k 0 +OPT) 3OPT : Sin ek 0 k 1 ,andOPT=2k 2 +k 1 +k 0 itfollowsthat APX OPT 4k 2 +4k 1 +4k 0 +8OPT 12OPT 10OPT 12OPT =5=6:
Ifontheotherhandj2then(j+2)(i+1) (j+1)(i+2)0,sin eji>0.Therefore
APX
(j+1)OPT
j+2 :
Theapproximationratioisthus
APX OPT j+1 j+2 3=4;
sin e,on eagain,j2.
Case2:p
0 6=0.
Observethatp
0
6=0impliesthatalledgesinthemat hingareusedinAPX,i.ej=1andx
1 =p
1 .
Againweuse(4.1.3) and(4.1.4)togetthat
p 0 OPT+k 0 P i1 (i+2)p i 2 :
Insertingtheexpressionforp
0 into(4.1.2)yields APX X i1 (i+1)p i + OPT+k 0 P i1 (i+2)p i 2 (4.1.9) = P i1 ip i +OPT+k 0 2 (4.1.10) = A+B+OPT+k 0 2 (4.1.11) 6k2+3k1 8 +k 2 +OPT+k 0 2 (4.1.12) = 14k 2 +3k 1 +8k 0 +8OPT 16 (4.1.13) 14k 2 +5k 1 +6k 0 +8OPT 1 6 (4.1.14) = 4k 2 +k 0 +13OPT 1 6 (4.1.15)
In(4.1.11)and(4.1.12)weuseLemma9,andin(4.1.14)weuseon eagainthefa tthatk
0 k
1 .
With this,Case2hasbeenprovedtohold,andTheorem 5follows.
Theorem5. Theenhan edalgorithm hasthe approximation ratio3=4.
From the analysis we inferthat the worst ase appears when p
0
=0 and j = 2. Thefollowing
example shows that the analysis is tight. The example ontainsa ost oneskeleton of a graph,
i.e., all ostoneedges in thegraphareshownin the gure. Itis leftfor thereaderto apply the
enhan edalgorithmonthis example.
Fig.5.Costoneskeletonofaworst ase instan efortheenhan ed
orien-teeringalgorithm.
4.2 Polynomial Time GreedyAlgorithmsfor SD k
TSP
Letusnowde nethepolynomialtimegreedyalgorithm(PGforshort).
Algorithm PGforSD k
TSP(1,2)
Input: Asequen eofkinstan esI
1 ;:::;I
k .
Output: AHamilton y leP.
1 GivenanelementIj,useapolynomialtimealgorithmfor(zj zj
1 )-TSP(1,2)ontheunvisited verti esinV. End PGfor SD k TSP(1,2)
Tosimplifytheanalysis wede nethe on eptof ostoneratio.
De nition7. Let#
1
(ALG [I℄)bethenumberof ostoneedgesusedbyalgorithmALGoninstan e
I and#
1
(OPT[I℄)thenumberof ostoneedgesusedbytheoptimalalgorithmonthesameinstan e.
The ostoneratio ofALG is
q 1 (ALG )=min I # 1 (ALG [I℄) # 1 (OPT[I℄) : With q 1
wedenotethelargestknown ostoneratioforanypolynomialtimeSD k
TSPalgorithm.
Theorem6. PGhas a ompetitive ratioatmost
q 1 U k +(2 q 1 )n
Proof. Let P
I
denote the y le produ ed by the newgreedy algorithm. In everytime zone, the
numberof ostoneedgesused byPG isat least q
1
timesthe numberof ost oneedgesused by
OPTthatwerenot ountedbytheusagefun tion.
Thisaddsupto 2U k (P I )+q 1 (n U k (P I ))+2(1 q 1 )(n U k (P I )) n = q 1 U k (P I )+(2 q 1 )n n q 1 U k +(2 q 1 )n n : u t
We an adopt theenhan ed orienteering algorithm in thepreviousse tion to get ak-TSP path
algorithmwithq
1 =
2
3
byreturningapathoflengthT insteadofapathwith ostT.(Toseethat
thealgorithm attainsthis ost oneratio itissuÆ ient to onsider itsworst ase,in whi h every
thirdedgein thepathhas osttwowhenalledgesintheoptimalpathhave ostone.) UsingPG
withthisnewalgorithmasasubroutinewegetthe ompetitiveratio
2 3 U k + 4 3 n n :
Usingtheresultsin Theorem2wegetthefollowinglargest ompetitiveratio:
2(2 k 2) 3(2 k 1) + 4 3 :
Wepresentthe ompetitiveratiosforsomevaluesofkin Table1.
Notsurprisinglythesevaluesarerather loseto2.Notethattheonly ostoneedgeswe anhope
forintheworst aselieintimezoneone.Thissituationdoesnot hangewithabetterorienteering
algorithm.Thus,toimprovetheapproximationratioweneedtomodifythepolynomialtimegreedy
algorithm.We allthenewalgorithmIPG(improvedgreedy).Tomeasuretheperforman eofthis
Algorithm IPGforSD k
TSP (1,2)
Input: Asequen eofk instan esI1;:::;Ik.
Output: AHamilton y leP.
1 UsePGoneverypermutationofthesetfI
1 ;:::;I
k
g,andreturnthe heapestsolution.
End IPGforSD k
TSP(1,2)
algorithmwenotethatnooinealgorithm ansimultaneouslyfor eavalueonthe orresponding
usagefun tionU 0 k higherthan U 0 k =min fU(A )g;
where is a permutation of f1;:::;kg, and A
is the polynomial time greedy algorithm using
the inputsequen e I
1 ;I
2 ;:::;I
k
. A trivialupperbound on U 0 k is if z j = jn k , for whi h U 0 k max z i = k 1 k
n.UsingthesameanalysisasforPGwegetthatIPGhasanapproximationrationo
higherthan q 1 U 0 k +(2 q 1 )n q 1 k 1 +2 q 1 =2 q 1 :
With ourbest k-TSP(1,2) algorithmthis simpli esto 2 2
3k
. Weprintthese resultsfor smallk
inTable1.
In ontrasttothevaluesoftheGREEDYalgorithm,thataresharp,thevaluesfortheIPGare
onlyupperbounds. It ispossibletomakeasmallimprovementonthese resultsif wein thelast
time zoneuse adedi ated TSP(1,2) algorithm[13℄ instead ofthe themoregeneralOrienteering
algorithm. kGREEDY(PGwithq 1 =1)PGwithq 1 =2=3IPGwithq 1 =1IPGwithq 1 =2=3 1 1.00 1.33 1.00 1.33 2 1.67 1.78 1.50 1.67 3 1.86 1.91 1.67 1.78 4 1.93 1.96 1.75 1.83 5 1.97 1.98 1.80 1.87 6 1.98 1.99 1.83 1.89
Table1.CompetitiveratioforSDTSPalgorithms.
5 The Cost Dependent Traveling Salesman Problem
Letusendourexpositionwiththese ondvariantofTDTSP:the ostdependenttravelingsalesman
problem (CDTSP).InCDTSP theobje tiveis tominimizethetravelingtime,whi h isidenti al
tothetotal ostoftheedgestraversed.
De nition8. Consider a set of edge ost fun tions f
1 ;:::;
tn
g assigned to a omplete graph
G=(V;E)with jVj=n,jEj= n 2 ,and where t
(e)is the ostfun tion for edge e2E.Let e
ij
denotetheedgebetweenv
i andv
j
inV.The ostdependenttravelingsalesmanproblem(CDTSP )
seeksapermutation ofV thatminimizes
t 1 (e 1 ; 2 )+ t n (e n ; 1 )+ n 1 X i=2 t i (e i ; i+1 ); wheret i = i 1 X j=1 t j (e j ; j+1 ):
The di eren e between SDTSP and CDTSP is that the ost fun tion now depends on the
weightsumoftheedgesalreadytraversedbythesalesman.Alsointhis aseone anvisualizethe
instan easadire tedlayeredgraph.However,anedgeof ost goingfromavertexinlayeriwill
onne twithavertexinlayeri+ .Note,though,that thenumberoflayersmightbeverylarge,
e.g.,equaltothelengthofthelongestsalesmanpath.
As for SDTSP we restri t the number of di erenttime zones to k and denote the resulting
problemCD k
TSP.Webeginwithadis ouragingresultfortheEu lidean ase.
LetS =fs 0 ;s 1 ;:::;s n
gbeaset of pointsin theplane, withs
0
beingthe startingpointand
z 0 ;z 1 ;:::;z k 1
beingthetimezonedivisors.Furthermore,letC
1 ;:::;C
k
bekpositivevaluessu h
thatC
1
<<C
k
.Wede nethefun tion
C(t)=C i ; if i 1 t< i ; for1ik.
ConsiderasalesmanthatvisitsthepointsinS.Wede nethedeparturetimet
i
ofpoints
i asthe
TheRestri tedEu lidean CDTSP istheproblemof ndingatourthatvisitsallthepointsin
S, minimizing thetotal travelingtime. The time needed to go between two points s
i and s j is givenbyd(s i ;s j )C(t i ),whered(s i ;s j
)isthedistan ebetweenthetwopoints.Weassumethatthe
departuretimeofthestartingpointist
0 =z
0 =0.
Garey,Graham and Johnson[8℄ provethat theEu lidean travelingsalesmanproblem is
NP-hardby aredu tionfrom theNP- ompletede isionproblem exa t overby 3-sets,X3C. Weuse
their redu tionto prove that theRestri ted Eu lidean CDTSP is inapproximable. If F is a yes
instan eofX3C,thenwesaythatF 2X3C,otherwisewesaythat F62X3C.
LetS be an instan e of TSP produ ed by Garey et al.'s redu tion, su h that jSj = n. The
pointsofSlieonaunitgridGofsizelessthatnnandanaivetourvisitingallpointsofS has
a ostl<2n.These resultsfollowdire tlyfromGareyetal.'s onstru tion.
Gareyet al.provethat the ost ofan optimaltouris lessorequalto somespe i valueL
ifan exa t overexists forthe X3C-instan e.If there is noexa t over,then it hasat least the
ostL
+1.
Theorem7. TheEu lidean CD 2
TSP annotbeapproximatedbyany onstant fa tor.
Proof. We onstru taRestri tedEu lideanCDTSP-instan ebasedupontheinstan eusedinthe
redu tionofGareyetal.Letrbeanarbitrary onstant.WetakeS fromtheredu tionofGarey
et al. asthe set of pointsin theRestri ted Eu lidean CDTSP-instan e,with the lowerleftmost
pointasdepot. Weletz
0 =0,z 1 =L ,C 1 =1,andC 2 =(r 1)L .
If F 2 X3C, then the time needed by an optimal salesman to visit all points and go ba k
to thedepot is L
asin theTSP-instan eof Gareyet al. Onthe other hand,ifF 62X3C then
the ostof the optimal tourin theTSP-instan e is at least L
+1. The shortest Hamilton path
startingfrom thelowerleftmostpointisthereforeat leastL
long.FortheRestri tedEu lidean
CDTSP-instan e this implies that the departure time of the last point s
i
to be visited by the
salesmanisat least L
. The ostof thelast edge istherefore (r 1)L d(s i ;s 0 )(r 1)L , sin e d(s i ;s 0
)1. Thetotaltravelingtime isthus at leastrL
andthe approximationratiobe omes
atleast rL
L
=r.Sin ewe an hooserarbitrarilylarge,thetheoremfollows. ut
Let us again restri tthe edge osts to either oneor two and onsider the online versionof
CD 2
TSP.On eagainthegreedyalgorithmisoptimal.
Theorem8. The ompetitive ratio of online CD 2
TSP(1;2) is 5=3 and the greedy algorithm is
optimal.
Wegivetwolemmasthat togethergiveustheproofofTheorem8.
Lemma10. The ompetitiveratioofanyonlinealgorithmforCD 2
TSP(1;2)withtwotimezones
isatleast5=3.
Proof. Letz
1
=n=3andassumethatea hedgeintimezoneonehas ostone.Takeanarbitrary
on-linealgorithmALGforthisproblemand onsideritsperforman e.Thealgorithmmustprodu e
aninitialpathof ostandlengthn=3beforea quiringthe ostoftheedgesintimezonetwo.The
adversary makessure that ZIG[1℄ and ALG [1℄ are disjointand that alledges in ident to bla k
verti esget ostoneintimezonetwo.Allotheredgesget osttwo.Sin etheonlyedgeswith ost
onein timezonetwoarethosein identto verti esalreadyvisitedbyALG, the ostoftime zone
Below,wepresentanoptimalexponential-timeonlinealgorithmforCD 2
TSP(1;2)withtwotime
zones.Theinputtothealgorithmisthe ostmatrixfortimezoneoneandattimez
1
thealgorithm
isgiventhe ostfun tionoftimezonetwo.
Algorithm A
exp
1 Computethelongestpathof ostz
1
2 Getnewinput(new ostfun tion)
3 Computethemin ostpathfromthe urrentpositiontothestartingpoint.
End Aexp
Lemma11. A
exp
has ompetitiveratio5=3.
Proof. Letm
1
bethenumberofedgesinthetourprodu edbyA
exp
,restri tedtotimezoneone,
andA
exp (I
2
)the ostofthetourrestri tedtotimezonetwo.Furthermore,letm
2
andOPT(I
2 )be
the orrespondingvariablesfortheoptimaltour.The ompetitiveratioofA
exp anbeexpressed as R (A exp )= z 1 +A exp (I 2 ) z 1 +OPT(I 2 ) :
The ostoneedgesthat anbe usedbyan oinealgorithm but notby theonlinealgorithm
arethe edgesin time zonetwothat arein ident tobla kverti es. Themaximalnumberof su h
edgesin ludedin atourisminf2m
1 ;n m
2
g.Therestoftheedgesusedbytheoinealgorithm
inthese ond timezonemayalsobeusedbyA
exp
.Theseedges onne tanumberofverti esthat
bothtoursvisitinthistimezone.IfweletoptdenotetheshortestTSP-pathamongtheseverti es,
wegetthatOPT(I
2 )minf2m 1 ;n m 2 g+opt.
Letusapplythesameargumentforthe osttwoedgesin timezonetwo.The osttwoedges
that A
exp
's tourmaybefor edtovisitbut that theoptimaltourmayes apearetheedgesthat
areadja enttoverti esalreadyvisitedbytheoptimaltour.Themaximalnumberofsu hedgesin
A exp 'stourisminf2m 2 ;n m 1 g,andsin eA exp
omputesanoptimalTSP-pathintimezonetwo,
itfollowsthatA exp (I 2 )2minf2m 2 ;n m 1 g+opt.Sin em 2 m 1
itfollowsthatthe ompetitive
ratiois R (A exp ) z 1 +2minf2m 1 ;n m 1 g+opt z 1 +minf2m 1 ;n m 1 g+opt =1+ min f2m 1 ;n m 1 g z 1 +min f2m 1 ;n m 1 g+opt :
Thisratioismaximizedform
1
=n=3for whi hopt=0,andsin ez
1 m 1 ,wegetthat R (A exp )1+ 2n=3 n=3+2n=3 = 5 3 : u t
Using the improved greedy algorithm for CD 2
TSP we an a hieve the same approximation
ratioinpolynomialtime.
Corollary3. Thereisapolynomialtimeoine5=3-approximationalgorithm for CD 2
TSP(1;2).
Note that wea hievethesameresult asfor thetwotime zoneSDTSP. However,be auseof
6 Con lusions
We study online strategies for two versions of the time dependent traveling salesman problem
(SD k
TSPandCD k
TSP).
FortheonlineversionofSD k
TSPwea hieveanoptimalexponentialtimestrategywith
om-petitiveratio( M m 1) 2 k 2 2 k 1
+1;whereM isthelargestandmthesmallestedge ost,andkisthe
numberoftimezones.
Inorder tomaketheonline strategy timeeÆ ientwestudy theorienteering problem, whi h
appears as a subproblem in the online strategy. We nd an approximation algorithm for the
orienteeringproblemwithapproximationratio3=4iftheedge ostsarerestri tedtooneandtwo.
Using the online result for SD k
TSP together with the approximationalgorithm for the
ori-enteeringproblem weareabletoprodu eagreedyapproximationalgorithm withapproximation
fa tor2 2
3k
forgraphswithedge ostsoneortwo.WealsogivesimilarresultsforCD 2
TSP(1;2),
mat hingthosefoundforSD 2
TSP(1;2).
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