e-mail:bengt.nilssonts.mah.se
Abstra t. We studythe approximation omplexity of ertain kineti
variantsoftheTravelingSalesmanProblemintheplanewhere we
on-siderinstan esinwhi hea hpointmoveswithaxed onstantspeedin
axeddire tion.Weprovethefollowingresults.
1. Ifthepointsallmovewiththesamevelo ity,thenthereisaPTAS
fortheKineti TSP.
2. TheKineti TSP annotbeapproximatedbetterthanbyafa torof
twobyapolynomialtimealgorithmunlessP=NP,evenifthereare
onlytwomovingpointsintheinstan e.
3. TheKineti TSP annot beapproximated better thanby afa tor
of
2
Ω
(√n)
byapolynomialtimealgorithmunlessP=NP,evenifthe
maximumvelo ity isbounded.The
n
denotes thesize oftheinput instan e.1 Introdu tion
Considera at in a eld with an ample supply of mi e. The at's obje tive is
to at h all the mi e while exerting the minimum amount of energy. The at
therefore wishes to use the shortest possible path to hase the mi e. A major
di ulty is the fa t that the mi eare moving. This problem is an instan e of
theKineti Traveling SalesmanProblem.
TheTravelingSalesmanProblem,TSPforshort,isprobablythebestknown
intra tableproblem. Itasksfortheshortest losedtourthatvisitsthenodesin
agivenweighted ompletegraphexa tlyon e.Thisde eptivelysimpleproblem
TSP optimization has been shown to be NP-hard, even for restri ted
in-stan es.Spe i ally,ifthegraphismetri orevenembeddableintheEu lidean
plane,theproblemisNP-hard[7,12℄.
EvenamongNP-hardproblems,theTSPis onsideredtobea
omputation-allydi ultproblem.It annotbeapproximatedto withinany onstantfa tor
by apolynomialtime algorithm, unless P=NP[14℄. Formetri graphsthe
sit-uation is better. Christodes [6℄ presents a polynomial time
3
2
-approximation algorithm, but better than a onstant fa tor approximation algorithm is notpossiblein polynomialtimeunlessP=NP[3℄.
Arora[1℄and(independently)Mit hell[11℄,intheirseminalpapers,showed
thatthereexistpolynomialtimeapproximations hemesforTSPwhenthegraph
is embedded in the Eu lidean plane. Polynomialtime approximation s hemes,
PTAS for short,are polynomial time algorithms that, for any
ǫ > 0
, produ ea
(1 + ǫ)
-approximation to a given problem. The running time of a PTAS ispolynomial in the input size, for any xed
ǫ
. Re ently, the running times ofPTAS's for
d
-dimensional Eu lidean TSP has been signi antly improved [2,13℄.
Inthekineti travelingsalesmanproblem,welookatTSPformovingpoints
intheEu lideanplane.We onsiderinstan esinwhi hea hpointmoveswitha
xedvelo ity.Thisisanaturalandboththeoreti allyandpra ti allyimportant
generalizationofTSP(e.g.severals hedulingproblems anberedu edtosolving
variantsofkineti TSP).
PreviousWork Thisresear htopi evolvedoutofaproblemposedbyMorde ai
Golinintheearly1990s.GolinssuggestionwastheTSPformovingpointsonthe
line and it was solvedusing a
O(n
2
)
timedynami programming algorithmin
1998byHelvig,RobinsandZelikovsky[8℄.Theyalsogavea
2 + ǫ
-approximationalgorithm for the Kineti TSP if thenumber of points with non-zero speed is
O(
log log n
log n
)
,andtheyalso onsidereddierentversionsofthek
-deliveryproblem. Previously,in 1996,Chalasaniet al.gavea onstantfa torapproximationalgo-rithmfortheKineti TSPwhenallpointshavethesamevelo ity.Theirresear h
into this areawasjustied by its appli ation in theframework of item
olle -tionon onveyorbelts.Theyalso onsidereddierentversions ofthe
k
-deliveryproblembut amajordisadvantageofalltheirsolutionsisthattheiralgorithms
operateinthe
L
1
-metri ,whi himpliesthatthey annotbenetfromthePTAS of Arora et al.that surfa ed into theresear h ommunityoneyear after theseresultswerepublished.
TheKineti TSP hasalso been addressedby resear hersin the operations
resear h eld using linear programming relaxations [10℄, although their main
optimal,disregardingthesmall
ε
fa tor,andthatP mightbeNP.3. TheKineti TSP annot be approximatedto within a fa tor of
2
Ω(
√
n)
in
polynomialtimeunlessP=NP,evenifthemaximumspeedisbounded.The
n
denotesthesize oftheinputinstan e.Thelastresultinparti ularissurprisinginthelightofexistingpolynomialtime
approximations hemes[1,2,11℄forthestati versionoftheproblem.
Inthenextse tion,westatedenitionsandgivepreliminaryresults
on ern-ing Kineti TSP. Spe i ally, we give an overview of the original redu tion of
Garey et al. [7℄ proving the NP-hardness of the Eu lidean TSP, sin e it plays
an importantrole in ourlater redu tions.InSe tion 3,weprovetheexisten e
of aPTAS for the asewhenall pointsmovewith thesamespeedin the same
dire tion.InSe tions4and5,weprovethestatedinapproximabilityresultsand
we on ludethepresentationwithadis ussionofopenproblems.
2 Preliminaries and Notation
Inthe Eu lidean TravelingSalesman Problem wearegivenaset ofpoints
S =
{s
1
, s
2
, . . . , s
n
}
intheEu lideanplane.Theobje tiveisto omputetheshortest tourthat visits allpoints(the optimal TSP tour). The orrespondingshortestpath that visits all points of
S
, starting ats
1
and ending ats
n
is alled the optimalTSPpath.As in Chalasani et al. [5℄ we distinguish between spa e-points and moving
points.Aspa e-pointisapointina oordinatesystem,whereasamovingpoint
is a point-obje t in spa e, the Eu lidean plane in our ase, that travels with
agivenvelo ity. The oordinates of amoving point
s
anbe des ribed bythefun tion
s(t) = (x + tv cos α, y + tv sin α)
, wherev ≥ 0
isthepoint'sspeedandα
isitsdire tion.Ifv = 0
wesaythatthepointisstati .Thetravelingsalesman isdes ribedbyaspe ialpointthat anmovewithvariablespeedanddire tion.Theinitial position
s
0
of thesalesmanis assumedto be(0, 0)
and itsmaximal speed is assumed to be1
. The path taken by the salesman is denotedP
andthesalesman visits
s
attimet
.P
is alled asalesman pathofS
ifallpointsinS
havebeenvisitedby thesalesman.If thesalesman also returnsto its initial position,thenwe all theresultingtourasalesmantour.We annowdenethekineti travelingsalesmanproblemformovingpoints
in theplane.
Denition1. A set of moving points
S(t) = {s
1
(t), s
2
(t), . . . , s
n
(t)}
in the plane with the Eu lidean metri isgiven. A points
i
(t)
inS(t)
moves with the speedv
i
< 1
.Consider atraveling salesman as dened above. The obje tive of theKineti TravelingSalesmanProblem(KTSP)isto omputeasalesmantour,starting andendingatthe initial point
s
0
= (0, 0)
,that minimizesthe traveling time of the salesman. The Translational Traveling Salesman Problem (TTSP)is arestri tedversion of KTSP, where all pointsof
S
havethe same speedanddire tion.
A onvenient way to hara terizekineti TSP instan es is as omplete
di-re ted graphswithtime dependentedge weights. Theweightof an edgeis
de-ned to be equalto the time asalesmanneed to traverse theedge starting at
time
t
.Thefollowinglemmaprovidesanimportantfa t on erningthespeedofthe
salesman. It is a dire t onsequen e of the fa t that the salesman an travel
faster thananyother movingpoint.Note that thesalesman needsto be faster
thanthepointsthatarebeingvisitedsin eotherwisesomemi emayneverbe
aught.
Lemma1. Anoptimal salesmanmoveswith maximalspeed.
Fromnowonweassumethatthesalesmantravelswithmaximalspeedandsin e
themaximalspeedis1,thedistan e traveledequalsthetravelingtime.
Generally,weuse
OP T
todenoteashortesttourorpath,unlessmorespe inotationisneeded.Givenatourorapath
P
intheEu lideanplane,weletC(P )
denotethelengthof
P
.2.1 X3C Redu tion ofGarey etal.
Gareyetal.[7℄provethattheEu lideantravelingsalesmanproblemisNP-hard
byaredu tionfromtheproblemexa t overby3-sets,X3C:
Givenafamily
F = {F
1
, F
2
, . . . , F
r
}
of3
-elementsubsets ofasetU
of3k
elements (represented by the integers1, . . . , 3k
), does there exist a subfamilyF
′
⊆ F
ofpairwisedisjointsetssu h that
S
F
∈F
′
F = U
?Wesaythatifthereexistssu hasubfamilyfor
F
thenF ∈ X3C
,otherwisewe(−9, 45)
(9, 45)
(36, −9)
(9, −45)
(−9, −45)
(−36, −9)
(−36, 9)
(36, 9)
J
(kr)
Fig. 1.TSPjun tions.Wewilldes ribetheinstan eusedintheredu tionperformedbyGareyetal.
Thisse tionfollowsthedes riptionofGareyetal. losely,usingidenti al
nota-tion.Fora ompletepresentationoftheredu tionwereferto thepaper[7℄.
Givenan instan eof X3C,i.e.afamily
F
of3
-elements,theobje tiveis toonstru t asetof points
S
and aboundL
∗
su h that anoptimalTSP tourin
S
haslength at mostL
∗
ifand only if
F
hasan exa t over.S
is onstru tedfrom
F
in stages,startingwith twobasi stru tures: jun tionsand rossovers.Letus start withthejun tions. Forea h set
F ∈ F
we onstru tonejun tionin
S
, that is,agadget designedtorepresentF
inthe TSPinstan e. Fig.1de-s ribesthegeometryofajun tionindetail.Notethatthroughouttheredu tion,
alinesegmentrepresentstheset ofpointswith integer oordinatesit ontains.
Forsimpli ity,weassumethatthejun tiongadgetis enteredaroundtheorigin
(0, 0)
. Theregion within the dotted ir le is alled thea tive region. The end point oordinatesofthelinesegmentsinthejun tion'sa tiveregionaredepen-dentonthenumberofsetsin
F
aswellasthenumberofpossibleelements,sin er = |F|
and3k = |U|
;seethea tiveregioninFig.1.Moreover,thelinesegment end point oordinatesoutsidethe a tive regionaredependenton thevalueK
,WARPED
STANDARD
(1, 2)
(−2, 1)
(2, 1)
(1, −2)
c = 96 − 1/kr
c = 96
(−36, c)
(36, c)
(0, 0)
(36, 15)
(36, −15)
(1, −4)
(4, −1)
(2, −1)
(4, 1)
(−1, 4)
(1, 4)
(−1, 2)
(−4, 1)
(−2, −1)
(−4, −1)
(−1, −2)
(−1, −4)
(−36, −96)
(36, −96)
(−36, −15)
(−36, 15)
(K)
(kr)
Fig. 2.TSP rossovers. whi hisdenedasK = 108kr
2
+ 1008k
2
r
2
+ 108k
2
r,
seeFig.1.These ondstru ture oftheredu tionisthe rossover.There aretwotypes
of rossovers:thestandardtypeandthewarpedtype.The rossoverstru tureis
usedtorepresenttheelementsofasetinthefamily.Fig.2des ribesthestandard
andthewarped rossoverindetail.Notethattheoriginitselfisin ludedinthis
gadget. The dieren e between the standardand the warped rossoverlies in
the oordinatesofthetopmostline segments'lowerend points.
The rossoversareassembledinto verti alsequen es alled rossoversta ks;
seeFig.3.Thetwotopmost endpointsofea h rossover oin ide withthetwo
lowestpointsinthe rossoveraboveit.Thetopmost rossoveriswarpedandthe
other rossoversinthesequen eareofthestandardtype.
Ea h set
F
i
= {a
i
, b
i
, c
i
}
in thefamilyF
will berepresented inS
bya set stru ture onsisting of one jun tion and three rossover sta ks of heighta
i
, b
i
Setstru turefor
F = {2, 4, 5}
. Crossoversta kofheight5
.Fig. 3.Compoundgadgetsofthe onstru tion
S
.and
c
i
respe tively.These arejoined by onne ting thethree topmost pairs of points in the jun tion to the lowest pair of points in the rossoversta ks. Inthis wayatube systemis builtfor ea h set stru ture. These tube systemsare
onne ted intoahugesystemasFig. 6suggests. Theguredes ribesthe nal
TSP onstru tion
S
forthefamilyF = {{1, 2, 3}, {2, 4, 5}, {1, 2, 6}}
.Let
T
0
denotetheset ofunitlengthedgesbetweenpairsofpointsinS
.The tubesystem onsistsofedgesfromT
0
,andattubeinterse tionpoints rossovers and jun tions are lo ated. Thefollowinglemma servesto justify ourattentiontothetubesegments.
Lemma2. Anoptimal TSP tour
OP T
inS
ontainsalledges ofT
0
.To omplete aHamiltonian y lein this graph,followingtheline segments
of thetubes, someof thesetubesmust be onne tedto ea hother. These
on-ne tionsarerealizedinthea tiveregionsofthegadgets.Forjun tionsthereare
twotypesof onne tions:either theyare onne tedortheyarenot;seeFig.4.
For rossovers(standardand warped),there are threepossible onne tions:no
onne tion, the upward onne tion and the downward onne tion; see Fig. 5.
Notethatthenon- onne tionalternativeisthe heapestonesoanoptimalTSP
tourneedstominimizethenumberofother onne tions.Wesaythatagadget
isa tiveifitperformsa onne tion.
Let
L
∗
= |T
0
| + 72kr
2
+ 312krq + 108k
2
r − 6k
,whereq < 3kr
isthenumber of rossoversinS
.L
∗
is goingto bethe lowerbound ofaTSP tourin
S
ifnoa)NOCONNECTION
Length:
72kr
b)CONNECTION
Length:
180kr
Fig. 4.Possible onne tionsina tiveregionsofjun tions.
a)NO
Length:
312kr
Length:252kr
)UPWARD b)DOWNWARD
CONNECTION CONNECTION CONNECTION
Length:
312kr(−2
ifWARPED)
J
J
J
Fig. 6.Final
S
whenF = {{1, 2, 3}, {2, 4, 5}, {1, 2, 6}}
.Lemma3. There isanoptimal TSP tour
OP T
oflengthlessor equaltoL
∗
if
andonly ifthereisan exa t over ofthe family
F
.Gareyetal.provethistheorembyshowingthatonly onne tions
orrespond-ing toanexa t over angiveatouroflengthlessorequalto
L
∗
. Thismeans
thatonlythejun tions orrespondingtothesetsinvolvedintheexa t overare
a tive,and in the sameway,only the rossoversin the rossoversta ks
repre-sentingtheelementsin theexa t overarea tive.Fortheproofofthistheorem
werefertoGareyet al.[7℄.
Weendthisse tionwithalemmathatappearedasthenalremarkofGarey
et al.in theirpaper.Thislemmais ru ialforourredu tionsinlaterse tions.
Lemma4. Let
OP T
denote the minimum length ir uit ofS
, i.e. an optimalTSP tour.Then
OP T
has integral length.Thisfollowsfrom Lemma 2and thewayin whi h thea tiveregionsof the
gadgetswere onstru ted.Outofthisfa t nextlemma followsdire tly.
Lemma5. There isnopolynomialtimeapproximation algorithm forTSP
pro-du ing atour
AP X
su hthatC(AP X) < C(OP T ) + 1
unlessP = N P
.ThelemmaholdsalsofortheoptimalTSPpathproblem,sin etheredu tion
ofGareyetal.alsoworksforthatproblem.Inthis ase,theredu tionfromX3C
results in instan es where wehaveopened thebottom tube.Observe that the
length of theoptimal TSP path for instan es thus onstru ted isequal to the
X3Credu tionwillfromnowonbe alledGGJ-instan es.Werefertotheoriginal
instan es astypeawhereastypeb instan esare thosewith anopenedbottom
tube.
3 A PTAS for TTSP
To nd a PTAS for the Translational TSP we are going to nd a bije tive
mappingbetweenthis problemand thestati Eu lideanTSP. Letus startthis
se tion by analyzing the translationalTSP. At this point we are interestedin
nding the optimal translationalTSP path. A set
S(t) = {s
1
(t), . . . , s
n
(t)}
of movingpointsisgiventogetherwithastartingpoints
0
= (0, 0)
.Allpointsmove in thesamedire tionα
and withthesamespeedv
. We anassume,w.l.o.g.,that
α =
π
2
.Thusapoints
i
(t)
isdened ass
i
(t) = (x + tv cos
π
2
, y + tv sin
π
2
) = (x, y + tv).
The traveling distan e
c
ij
between two pointss
i
(t)
ands
j
(t)
is the time neededbyasalesman,movingwithspeed1
,totravelfroms
i
tos
j
,i.e.c
ij
=
v
1 − v
2
(y
j
− y
i
) +
s
(x
j
− x
i
)
2
1 − v
2
+
(y
j
− y
i
)
2
(1 − v
2
)
2
Notethatthetravelingdistan eisindependentoftimeandthatthe ostfun tion
c
ij
isasymmetri .Considerthefun tiond(s
i
, s
j
) =
c
ij
+ c
ji
2
=
s
(x
j
− x
i
)
2
1 − v
2
+
(y
j
− y
i
)
2
(1 − v
2
)
2
andthebije tivemapping
f
v
(s)
fromamovingpoints = (x, y)
toastati point in theEu lideanplane:f
v
(s) =
x
√
1 − v
2
,
y
1 − v
2
LetP = (s
i
1
, . . . , s
i
n
)
beapathinthetranslationalinstan e,thenC(P ) =
n
−1
X
k=1
c
i
k
,i
k+1
=
v(y
i
n
− y
i
1
)
1 − v
2
+
n
−1
X
k=1
d(s
i
k
, s
i
k+1
).
With thisitiseasytoprovethefollowingresult.
Lemma6. Let
S
T
beaninstan eofthe translational TSPwithspe ied start-ing andendpoint, and letS
E
bethe orresponding Eu lidean instan eafter the transformation usingf
v
. A salesman path inS
T
is optimal if and only if the orrespondingEu lidean Hamiltonianpath inS
E
isoptimal.using a modied PTAS for the Eu lidean TSP [1,2,11℄. Here
OP T
E
denotes anoptimalTSP pathin theEu lidean instan e.LetAP X
T
andOP T
T
denote the orrespondingpathsinthetranslationalinstan e.ObservethatOP T
T
isan optimalsalesmanpathbyLemma6.WehaveC(AP X
T
) =
v(y
n
− y
1
)
1 − v
2
+ C(AP X
E
)
≤
v(y
n
− y
1
)
1 − v
2
+ (1 + ǫ)C(OP T
E
)
= C(OP T
T
) + ǫC(OP T
E
)
≤ (1 +
ǫ
1 − v
)C(OP T
T
).
The last inequality holds sin e
OP T
T
is at least as longas the shortest path betweens
1
ands
n
,i.e.C(OP T
T
) ≥
|y
n
−y
1
|
1+v
.Hen e,C(OP T
E
) = C(OP T
T
) −
v(y
n
− y
1
)
1 − v
2
≤ C(OP T
T
) +
v|y
n
− y
1
|
(1 − v)(1 + v)
≤ C(OP T
T
) +
vC(OP T
T
)
1 − v
=
C(OP T
T
)
1 − v
.
It followsthatAP X
T
is a(1 +
ǫ
1−v
)
-approximationof theoptimalpath.Thus, wehaveaPTAS forthetranslationalTSPpathproblem.Notethattheapprox-imationfa tordoesdependonthemaximalspeed.Thus,ifwewantto ompute
a onstantfa torapproximationforaninstan ein whi hthepointsaremoving
withthespeed
1 −
1
n
thenouralgorithm mightneedexponentialtime.With this approximation s heme we an now give a PTAS for the
trans-lational TSP. We use the same approa h as Chalasani et al. [4,5℄. The
dif- ulty of the translational TSP is that the initial point
s
0
, is not moving, whi h reates an asymmetry that we must be able to handle. Assume thatOP T = (s
0
, s
1
(t), . . . , s
n
(t), s
0
)
is an optimal salesman tour. It follows easily that the optimalsalesman path startingfroms
0
and endingats
n
(t)
isa part of that tour. For ea h possible su h ending points
i
(t)
,i ≥ 1
, we ompute aTheorem1. The algorithm des ribed above is a PTAS for the translational
TSP.
Proof. Let
OP T
H
denote the length of the optimal salesman path, starting froms
0
and ending ats
i
(t)
. The length of the optimal salesman tourOP T
isC(OP T ) = OP T
H
+ l
,assumings
i
isthelastmovingpointvisitedbythetourOP T
andl
isthelengthofthelastsegmentinthetour(betweenthepointss
i
(t)
ands
0
). There is atourAP X
among thetours our algorithm omputes, that alsohass
i
(t)
asthelastunvisitedpoint.ThelengthofAP X
isC(AP X) = (1 +
ǫ
2(1 − v)
)OP T
H
+ l
′
,
where
l
′
isthelengthbetween
s
i
(t)
ands
0
.AsalesmanthattravelsalongAP X
visitss
i
(t)
at timet = (1 +
ǫ
2(1−v)
)OP T
H
, whereastheoptimalsalesmanvisitss
i
(t)
at timet = OP T
H
. Thetime dieren eisǫ
2(1−v)
OP T
H
and in that time period,s
i
(t)
movesthelengthvǫ
2(1−v)
OP T
H
. Thetriangle inequalityassures us thatl
′
doesnot ex eed
l +
vǫ
2(1−v)
OP T
H
. The ost of our approximate tour is thusC(AP X) ≤ (1 +
ǫ
2(1 − v)
)OP T
H
+ l +
vǫ
2(1 − v)
OP T
H
= (1 +
(1 + v)ǫ
2(1 − v)
)OP T
H
+ l
< (1 +
ǫ
1 − v
)OP T
H
+ l.
Thelast inequalityholds sin ethespeeddoesnotex eed
1
.Theratiobetweenthe ostsof
AP X
andOP T
be omesC(AP X)
C(OP T )
<
(1 +
ǫ
1−v
)OP T
H
+ l
OP T
H
+ l
≤ (1 +
ǫ
1 − v
).
The proof is ompleted, sin ethe ost ofthe returnedsalesman tour doesnot
ex eedthatof
AP X
.⊓
⊔
Observethat ourte hniquediersfrom thatof Chalasaniet al. [5℄onlyin the
bije tivemapping
f
v
withwhi h thetransformationwasperformed. Thanksto thismappingwewereabletogeneralizetheirresult.Ifthespeedsareboundedbya onstant
c < 1
thenthealgorithmisatruePTAS.4 Two Redu tions for KTSP
The kineti traveling salesman problem is in general not as easily
O
p
2
Fig. 7.Kineti TSP-instan ewithtwomovingpoints.
Thegure ontainsaGGJ-instan eoftypea,atdistan e
D
fromtheoriginand twomovingpoints
p
1
andp
2
,bothwithspeedv < 1
.Weassumethat the salesmanstartsattheoriginattimet = 0
.Thepointp
1
moveshorizontallyand liesat thesamealtitudeasthebottomlineof theGGJ-instan eand interse tstheline
l
attimeD
.Pointp
2
movesparalleltop
1
andrea hestheoriginattime2D + L
∗
. Re all that aGGJ-instan e is onstru tedso that if
F ∈ X3C
, thenan optimal salesmantouris at most
L
∗
long. Otherwisethe length is at least
L
∗
+ 1
.
Be auseofthelargespeedofthemovingpoints,itfollowsthattherstpoint
to bevisitedis
p
1
. Furthermore,thepointp
2
must be visitedbeforeit rea hes theorigin.Otherwise,thetourhasunboundedlength.If
F ∈ X3C
,thenanoptimalsalesman an at hp
1
and ontinuealongthe GGJ-instan e'soptimalsalesmantour.AfterallpointsintheGGJ-instan ehavebeenvisited,thesalesmanstillhastimeto at h
p
2
beforeitrea hestheorigin. IfF 6∈ X3C
, thenanoptimal salesmandoesnothavetimeto at h allthepointsintheGGJ-instan ebeforehehastoturnba kand at h
p
2
.Thismeans that hemusttraversethedistan eD
at leastfourtimes.Fromthisweget:Theorem 2. It is NP-Hard to get an approximation ratio less than
2
for thekineti TSP,evenif thereareonly twomoving pointsin the instan e.
InTheorem 2weneed the speed
v
to be arbitrarily loseto1
. With moremovingpointswe anrestri tthespeedto
v = 1/2
andstill getalowerboundof
2
ontheapproximationratio.Theorem3. It is NP-Hard to get an approximation ratio less than
2
for theProof. To prove the theorem we use two pairs of sets with
k = m
points inea h set evenly distributed on a line, and a GGJ-instan e of type b with
m
stati points.Ea hpair ontainsonesetwithstati pointsandonewithmoving
points. The stati points lie on averti alline of length
D = m
2
, whereasthe
movingpointslieonalinewiththesameheightbutwiththelength
√
5
2
D
.The slope of the se ond line is−2
in pair1
and2
in pair2
. The moving pointsperformahorizontalmotionfromleftto rightwiththespeed
v = 1/2
.Given these building blo ks, we onstru t an instan e of the Kineti TSP
asdes ribedin Figure 8.The salesmanbeginsat the originat time
t = 0
. Atthe same time, the lowest moving point of pair
1
interse ts the originand attime
D + L
∗
,thetopmostmovingpointinpair
2
ollideswiththe orrespondingstati point in that pair. The pairs are onstru ted so that the salesman an
move (with speed
1
) along the verti al line of the stati points, visiting pairsof points from the twosets at their ollision point ifand onlyif the salesman
visitsthe rsttwo ollidingpointsat thetime ofimpa t. Itthus takestime
D
to visit allpointsin apair, assuming the onditionsabove,sin ethe lengthof
theverti allineis
D
.instan e
GGJ-O
Pair2
Pair1
D
W
Fig. 8.KTSP-instan ewithmovingpointsofspeed
1/2
.Ifthereexistsanexa t overfortheX3C-instan e,thenanoptimalsalesman
hastimetovisitallpointsinbothpair
1
and theGGJ-instan ebeforethersttwopointsin pair
2
ollide.Thetime that anoptimalsalesman needsin orderto visit all points in the whole instan e is thus at most
2D + L
∗
+ W
, where
W ≤ L
∗
isthelengthofthebottomlinesegmentoftheGGJ-instan e.
Ifthereisnoexa t overfortheX3C-instan e,thenanoptimalsalesmandoes
nothaveenoughtimetovisitallpointsintheGGJ-instan ebeforethersttwo
points in pair
2
ollide. Thus, he is left with two options. Either he visits all3D/k
v = 1/2
ostofmovingbetweenthetwosetsba kandforthdoublesea htime.The ost
ofthisapproa histhus
4D + L
∗
− 3Dz/k + 2
z
−1
,where
z
isthenumberofzig-zagmovements.Minimizingthis expression,using theknowledgethat
L
∗
≥ m
,
D = m
2
and
k = m
,provesthatthelengthofthetourisatleast4m
2
−cm log m
,
forsome onstant
c
.Sin ethesalesmanmoveswithspeed
1
,lengthequalstime.Thelengthofanoptimalsalesmantouristhereforeat least
4m
2
− cm log m
if
F 6∈ X3C
and atmost
2(m
2
+ L
∗
)
if
F ∈ X3C
.Thelimitoftheratiogivesalowerbound ontheapproximationratioand on ludestheproofofTheorem3:
lim
m
→∞
4m
2
− cm log m
2(m
2
+ L
∗
)
= 2.
⊓
⊔
5 An Exponential Lower Bound for the general KTSP
In this se tion, we present a gap produ ing redu tion from X3C to the
ki-neti TSP.Theinstan e usedintheredu tionisauniformlyexpanding
KTSP-instan e. A uniformly expanding instan e ontainsmoving points of the form
s
i
(t) = (v
i
t cos α, v
i
t sin α)
.Thisimpliesthatattimet = 0
allpointsarelo ated attheorigin,andthat therelativedistan eswithin theinstan edonot hangeovertime. Weassume that the salesman begins his pursuit at time
t
0
> 0
(if he starts at timet = 0
, he visits all pointsat on e withoutmoving). Observethat thelengthofasalesmantour
P
in auniformlyexpandingTSP-instan e isdire tion independent. That is, ifwereverse theorder in whi h thepoints are
visited,thelengthofthenewtourisequaltothelengthof
P
.AsgadgetsintheKTSP-instan eweuseaspe ialkindofGGJ-instan essuitablefortranslation.
We therefore begin with a des ription of these, and ontinue with a detailed
des riptionoftheKTSP-instan e.
Considerastati typebGGJ-instan e.Fromthisinstan ewewantto reate
asthe stati instan e. This is a hieved by applying theinverseof thefun tion
f
v
des ribed in Se tion 3 to all points in the stati instan e. Let us all su h instan estranslational inverseGGJ-instan es.Givensu haninstan ewe reateauniformlyexpandingGGJ-instan e in whi h therelativedistan es mat hthe
relative distan es in the translational instan e. We all these small expanding
GGJ-instan es,orsimplyGGJ-instan es.
2π
km
2
v
i
2π/k
Fig. 9.One ir leoftheinapproximableKTSP-instan e.
The KTSP-instan e onsists of
l
on entri ir lesC
1
, . . . , C
l
, with radiir
1
, r
2
, . . . , r
l
, enteredaroundtheorigin.Ea h ir le onsistsofk = ⌈m
2
l⌉
identi- alsmallexpandingGGJ-instan es.TheGGJ-instan esarepla edonthe ir les
asshowninFigure9.
Ea hGGJ-instan e ontains
m
points.Weletl = m
a
, forsome
a > 1
.Thetotalnumberofpointsintheinstan e,denoted
n
,isthereforemkl = m
2a+3
.The
rightmostpointofaGGJ-instan eispla edonthe ir leandthebottomlineof
theGGJ-instan efollowsthe ir le'stangentlineatthatpoint.Ea h ir le
C
i
is expandingwithaspeedv
i
,i.e.therightmost pointof ea hGGJ-instan e onC
i
hasaspeedv
i
,dire tedawayfrom the enterandtheradiusofC
i
isr
i
= v
i
t
.Welet
v
1
=
1
2e
andv
i
= (1 +
m
2
k
)v
i
−1
.This impliesthatthelargest ir le's speed,v
l
,is bounded byv
l
= (1 +
m
2
k
)
l
v
1
= (1 +
1
l
)
l
v
1
< ev
1
=
1
2
.
Notealsothat
r
1
≤ r
2
≤ . . . ≤ r
l
.ObservethattheGGJ-instan esareuniformly expanding, i.e. the distan e between two pointss
i
ands
j
isd(s
i
, s
j
) = d
ij
t
, whered
ij
is onstant.Theexpansionrateissu hthatthedistan e betweenthe leftmostandtherightmostpointofaGGJ-instan e islessthantv
i
tan
2π
m
2
k
.Weu
i
= v
i
tan
m
2
k
theanglebetweentheleftmostpoint
L
,theorigin,andtherightmostpointR
inaGGJ-instan e is
∠LOR =
2π
m
2
k
and theanglebetweentwo onse utive GGJ-instan es is2π
k
; see Fig.9.Thefastest point ofaGGJ-instan e is theleftmost pointandthispointhasthespeedr
v
2
i
+ (v
i
tan
2π
m
2
k
)
2
≤ v
i
(1 +
2π
m
2
k
).
Thus, no point belonging to ir le
C
i
has aspeedex eedingv
i
(1 +
2π
m
2
k
)
, and thisimpliesthatthemaximalspeedisbounded.Fromnowonwesimplydenoteournewly onstru tedinstan etheKTSP-instan e.
Wewillshowthat the onstru tionaboveguaranteesthat anoptimal
sales-manvisitstheGGJ-instan es ir lewise,oneatatime.Furthermore,thetimeit
takestogofromanymovingpoint
s
i
toanyothermovingpoints
j
inthe KTSP-instan edependslinearlyonthestartingtime.Assumingthatasalesmanstartsfrom
s
i
attimet
itwouldtakehimtimeκ
ij
t
togotos
j
,wherewe onsiderκ
ij
tobe onstantsin eitdoesnotdependont
.Asa onsequen e,travelingtimeismultipli ativeintheedgeweight onstants
κ
ij
.Thus,ifthesalesmanbeginshis pursuitat timet
0
andnishestherst GGJ-instan eat timeKt
0
, thenit will takehimatleasttimeCK
kl
t
0
tovisitallpointsintheKTSP-instan e,whereC
denotesa onstantandK
isindependentoft
.Notethatanon-optimalsalesmanmust doworseonea hGGJ-instan e and sin etheerrorisalsomultipli ative,
theinapproximabilityratioisgoingtobe omelarge.Therestofthepresentation
isdevotedtoprovingthefollowingtheorem.
Theorem 4. For any
γ > 0
, there exists no polynomial-time algorithm thata hieves anapproximation ratioof
2
Ω(n
1/2−γ
)
for the kineti travelingsalesman
problem, unlessP=NP.
Considerasalesmanmovingbetweenthepointsin theKTSP-instan e. Let
P
bethepathtaken by thesalesman.WedeneC
P
(t)
tobethelengthof this path (whi h issynonymouswith thetime ittakesfor thesalesmanto traversethepath
P
,giventhestartingtimet
,i.e.,T
P
(t) = t + C
P
(t)
.A simpleproofof indu tion yieldsthatT
P
(t) = t
Y
[s
i
,s
j
]∈P
(1 + κ
ij
) = tK
P
andtherefore,C
P
(t) = t
Y
[s
i
,s
j
]∈P
(1 + κ
ij
) − t,
where
tκ
ij
is the time it takes for the salesman to travel betweens
i
ands
j
, starting froms
i
at timet
. Let us examine a small expanding GGJ-instan e produ edbyanX3C-instan e.Weassumethattheinstan eislo atedsomewhereon ir le
C
i
. The optimal salesmanpath for the GGJ-instan e starting at the rightmost point and ending at the leftmost point is denotedopt
ifF ∈ X3C
and
opt
′
if
F 6∈ X3C
.Wewouldliketondanupperboundonopt
andalowerbound on
opt
′
. Todo this, we need some denitions. Consider a translational
inverseGGJ-instan eofasatisfyingX3Cinstan ewiththesamesizeasthesmall
expanding GGJ-instan e at time
t
. Letall points in the translationalinstan emove with the same velo ity as the rightmost point in the expanding
GGJ-instan eattime
t
.WedeneD(t)
asthelengthoftheoptimalpath,startingattherightmostpointandendingattheleftmostpoint.Let
D
′
(t)
denotethelength
oftheoptimalpathinasimilarinstan e onstru tedfromanX3C-instan ewith
noexa t over.Clearly,
D(t) ≤ C
opt
(t) ≤ D(t + C
opt
(t)),
and (1)D
′
(t) ≤ C
opt
′
(t) ≤ D
′
(t + C
opt
′
(t))
(2)Furthermore, let
f (t)
denote the distan e between the two losest pointsin the expanding instan e at time
t
and lett
i
bethe point in time su h thatf (t
i
) = 1 − v
i
2
.Notethatt
i
isthetimewhentheexpandingGGJ-instan esrea h the size of theoriginal GGJ-instan es.That is, ifweapply the fun tionf
v
on anexpandinginstan eattimet
i
, thenthestati instan ethat resultswillhave thesamesize astheoriginalGGJ-instan esofSe tion2.Thus,D(t
i
) ≤ L
∗
and
D
′
(t
i
) ≥ L
∗
+ 1
. Observealsothatt
i
depends onthe ir le where theinstan e islo ated.D(t)
,D
′
(t)
and
f (t)
arelinearmappings,whi himpliesthatD(t)
D(˜
t)
=
D
′
(t)
D
′
(˜
t)
=
f (t)
f (˜
t)
=
t
˜
t
.
Lemma7. Thefollowing boundshold:
c
1
km
2
≤ t
i
≤ c
2
km
3
(for some onstantsc
1
andc
2
),
(3)D
′
(t) > D(t) + f (t)
u
i
≥ v
f
≥
u
i
m
.
Now,t
i
=
1−v
2
i
v
f
,so1 − v
2
i
u
i
≤ t
i
≤
m(1 − v
2
i
)
u
i
.
Re allthat
u
i
= v
i
tan
2π
m
2
k
andsobyTaylorexpansion,wegetthat2π +
1
3
m
2
k
≥ tan
2π
m
2
k
≥
2π
m
2
k
.
Usingthiswe anbound
t
i
asfollows:m
2
k(1 − v
2
i
)
(2π +
1
3
)v
i
≤ t
i
≤
m
3
k(1 − v
2
i
)
2πv
i
.
Insertingthe bounds of
v
i
yieldsthatc
1
km
2
≤ t
i
≤ c
2
km
3
, for some on-stantsc
1
andc
2
.(4) Itfollowsfromthelinearityofthefun tions:
D
′
(t) = D
′
(t
i
)·
t
t
i
> (D(t
i
)+f (t
i
))·
t
t
i
=
D(t
i
)D(t)
D(t
i
)
+
f (t
i
)f (t)
f (t
i
)
= D(t)+f (t)
(5) Westartwiththeinequality
C
opt
(t) ≤ D(t + C
opt
(t))
takenfrom(1). Usingthelinearityof
D(t)
wegetthatD(t + C
opt
(t)) = D(t
i
)(t + C
opt
(t))/t
i
.
Thisimpliesthat
C
opt
(t) ≤
D(t
i
)
t
i
− D(t
i
)
t,
sin et
i
≥ c
1
km
2
> D(t
i
)
.(6) Bythelinearityof
D(t)
andf (t)
wehavethatC
opt
′
(t) ≥ D
′
(t) > D(t) + f (t) =
D(t
i
) + 1 − v
i
2
t
i
· t.
⊓
⊔
With theseboundswe anprovealowerbound ontheapproximationratio
for the small expanding GGJ-instan es. To simplify the analysis later on, we
omputetheapproximationratioin termsofthearrivaltime.Weassume that
thesalesmanvisitshisrstpointoftheinstan eattime
t
.Lemma8. ItisNP-hardtond asalesman path
apx
,for an expandingGGJ-instan e, with
T
apx
(t) ≤ (1 +
1
ckm
3
)T
opt
(t).
Proof. The following ratio measures the gap between the arrival times of
apx
and
opt
.A ordingtoLemma 7T
apx
(t)
T
opt
(t)
≥
T
opt
′
(t)
T
opt
(t)
=
C
opt
′
(t) + t
C
opt
(t) + t
≥
D(t
i
)+1−v
2
i
t
i
t + t
D(t
i
)
t
i
−D(t
i
)
t + t
≥ 1 +
1 − v
2
i
t
i
−
2D(t
i
)
2
t
2
i
D(t
i
) < 12m
andc
1
km
2
≤ t
i
≤ c
2
km
3
. Expressed in terms ofk
andm
, the quotientisatleastT
apx
(t)
T
opt
(t)
> 1 +
1 − v
2
i
c
2
km
3
−
2m
2
(c
1
km
2
)
2
> 1 +
1
ckm
3
forsu ientlylarge
k
,andc = 2c
2
.⊓
⊔
LetusreturntotheKTSP-instan ethat ontainsthesmallexpanding
GGJ-instan es. Weassumethat thesalesman startsattheoriginattime
t
0
> 0
.We willprovethattheKTSP-instan eisinapproximableusingtheinapproximabilityresultinLemma8forthesmallexpandingGGJ-instan es.
Considerasalesmantour
P
fortheKTSP-instan e.AsubpathofP
startingatapoint
s
i
andendingatthepoints
j
is alledanedgeofP
ifnootherpoints are visitedalongthe subpath.Edgesthat onne ttwodierent GGJ-instan esare alled leaps.A leapthatexits the
j
thGGJ-instan eon ir leC
i
isdenotedJ
ij
. Theshortest leapthat onne ts ir leC
i
with ir leC
i+1
is theleap that onne tstheleftmost pointofthe topmostGGJ-instan e on ir leC
i
withthe rightmost point of the orresponding GGJ-instan e on ir leC
i+1
. This leap will be denotedE
i
(exterior leap exiting ir leC
i
);T
E
i
(t) = K
E
i
t
, for someC
opt
i
(t
i
) = Θ(m),
(7)C
I
i
(t
i
) = Ω(m
2
),
andC
I
i
(t
i
) = O(m
3
),
(8)C
E
i
(t
i
) = Ω(m
4
).
(9) Proof. (7)D(t
i
) < D
′
(t
i
) < 2m
,andsin et
i
≥ c
1
km
2
,wehavethat:C
opt
(t
i
) < C
opt
′
(t
i
) ≤
D
′
(t
i
)t
i
t
i
− D
′
(t
i
)
= D
′
(t
i
) +
D
′
(t
i
)
2
t
i
− D
′
(t
i
)
= O(m).
Now,
D(t
i
) ≥ m
sin etherearem
pointsin theinstan eandtheminimum distan ebetweenpairsof pointsis1
at timet
i
.It followsthatC
opt
(t
i
)
and thereforealsoC
opt
′
(t
i
)
isΘ(m)
.We on ludethatC
opt
i
(t
i
) = Θ(m)
.(8) Thedistan e,andthusthelengthofthetravelingpath,tothe losest
GGJ-instan eonthesame ir leisshowntobe
Θ(t
i
/k)
inFigure11,andc
1
km
2
≤
t
i
≤ c
2
km
3
.(9) Thelengthofanyleap
J
, onne ting ir leC
i
withanother ir leattimet
i
isatleasttheminimumofmin
1≤j<i
v
i
− (1 + 2π/m
2
k)v
j
1 + v
j
t
i
andmin
i<j≤n
v
j
− (1 + 2π/m
2
k)v
i
1 − v
j
t
i
.
Inserting the bounds of
t
i
andv
i
and observing thatv
j
= (1 −
m
2
k
)
j−i
v
i
yieldsthatthelengthofanysu hleapis
Ω(m
4
)
.
Corollary 1. Forsu iently large
m
K
E
i
> K
I
i
> K
opt
i
.
Proof. Itfollowsfrom Lemma9andthedenitionof
T
P
(t)
andC
P
(t)
.r = v
i
t
i
= Θ(t
i
)
y
x
r
2π/k
2π
m
2
k
2π(m
2
−1)
m
2
k
⇒ y = Θ(t
i
/k)
.y ≈ x = 2r sin(
2π(m
2
−1)
2km
2
)
Fig. 11.Shortestdistan ebetweentwoinstan esattime
t
i
.Lemma10. An optimal salesmantour,
OP T
,startingattheorigin attimet
0
isnishedattimeT
OP T
(t
0
) = K
E
0
l
Y
i=1
(K
opt
i
K
I
i
)
k
K
E
i
K
I
i
t
0
,
where
E
0
is theleap betweenthe origin andthe rst ir le.Proof. Firstofallletusonly onsidertoursthatvisitstheGGJ-instan es
oun-ter lo kwiseoneatatime.Thelengthoftheshortestsu htourisequaltothe
givenbound.Toprovethatthisisalsoalowerbound wedoasfollows.
Let
P
be any salesman tour of the KTSP-instan e. LetK
tsp
ij
denote the total ost spentin thej
th GGJ-instan eon ir leC
i
.Be ause ofthe salesman tourpropertyweknowthat theremustbeat leastone leapexiting ea hGGJ-instan e and a leap
J
′
0
from the origin to the rst GGJ-instan e in the tour. Thuswe an hara terizethenishingtimeofP
asT
P
(t
0
) = K
J
′
0
l
Y
i=1
k
Y
j=1
K
tsp
ij
K
J
ij
H
ij
Y
h=1
K
J
h
t
0
,
whereK
J
ij
des ribesthe ostofexitingthej
thGGJ-instan eon ir leC
i
,andH
ij
is the numberof extraleaps exiting this GGJ-instan e. The ost of these extraleapsaredes ribedbyK
J
T
P
(t
0
) = K
J
′
0
Y
i=1
Y
j=1
K
opt
i
K
I
i
K
J
′
i
K
I
i
t
0
,
whereJ
′
i
isaleapexiting ir leC
i
.Thisa tuallyprovesthattheGGJ-instan es mustbetaken ir lewise.Thequestionthatremainsisinwhat orderthe ir lesaretaken.
Note that the ost of the shortestpath between theorigin and ir le
C
l
is exa tlyK
E
0
K
E
1
· · · K
E
l−1
.Furthermore,theshortestpathbetweenthefurthest pointon ir leC
l
andtheoriginhasthe ostK
E
l
.Now,anysalesmanpathneeds tovisitthese points.ItturnsoutthatK
E
0
K
E
1
· · · K
E
l
≥ K
J
0
′
K
J
1
′
· · · K
J
l
′
.
Wegetthefollowinglowerbound onanysalesmanpath
P
:T
P
(t
0
) > K
E
0
l
Y
i=1
(K
opt
i
K
I
i
)
k
K
E
i
K
I
i
t
0
.
⊓
⊔
Corollary 2. Thereexistsnopolynomial-timealgorithmprodu ing atourAPX
for ourKTSP-instan esu hthat,unless P=NP,
T
AP X
(t
0
) < K
E
0
l
Y
i=1
((1 +
1
ckm
3
)K
opt
i
K
I
i
)
k
K
E
i
K
I
i
t
0
.
Proof. ItfollowsbyLemma8andtheproofofLemma10.
⊓
⊔
We an now al ulatealowerbound ontheapproximationratio, assuming
that
P 6= NP
.Proof. (Theorem 4) We are a tually interestedin theratio
C
AP X
(t
0
)
C
OP T
(t
0
)
but sin eC
AP X
(t
0
)
C
OP T
(t
0
)
>
T
AP X
(t
0
)
T
OP T
(t
0
)
wemightaswellusethelatterratio.FromLemma 10and
theresulting orollarywehavethat
T
AP X
(t
0
)
T
OP T
(t
0
)
≥
K
E
0
Q
l
i=1
((1 +
1
ckm
3
)K
opt
i
K
I
i
)
k K
Ei
K
Ii
t
0
K
E
0
Q
l
i=1
(K
opt
i
K
I
i
)
k K
K
Ei
Ii
t
0
= (1 +
1
ckm
3
)
kl
∈ 2
Ω(n
1/2−γ
)
,
for
γ = 3/a
,anda
anbe hosenarbitrarilylarge.Rememberthatn = mkl = m
2a+3
.
⊓
⊔
6 Con lusions
Wehaveinvestigatedkineti variantsofTSP.Ourmajorresultprovesan
expo-nentiallowerboundontheapproximationfa torforsu hproblemsunlessP=NP
evenwhenthevelo itiesarebounded.Evenso,wefeelthattheboundis oarse
and anprobablybeimproved.Also,thequestionofgoodupperbounds onthe
approximationratio omes tomind.
Theuseofthemapping
f
v
des ribedinSe tion3isa tuallyageneri method that an beusedtosolvealarge lassoftourproblemswheninstan esperformonstanttranslationalmovementintime.
The translational TSP PTAS that we present an be used in an
O(1)
-approximationalgorithm for KTSP, assuming that thenumberof dierent
ve-lo itiesoftheinstan esisboundedbya onstantandthatthemaximumvelo ity
oftheinstan es onsideredisbounded.
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