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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

(√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

(2)

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 not

possiblein 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 e

a

(1 + ǫ)

-approximation to a given problem. The running time of a PTAS is

polynomial in the input size, for any xed

ǫ

. Re ently, the running times of

PTAS'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

-approximation

algorithm for the Kineti TSP if thenumber of points with non-zero speed is

O(

log log n

log n

)

,andtheyalso onsidereddierentversionsofthe

k

-deliveryproblem. Previously,in 1996,Chalasaniet al.gavea onstantfa torapproximation

algo-rithmfortheKineti TSPwhenallpointshavethesamevelo ity.Theirresear h

into this areawasjustied by its appli ation in theframework of item

olle -tionon onveyorbelts.Theyalso onsidereddierentversions ofthe

k

-delivery

problembut amajordisadvantageofalltheirsolutionsisthattheiralgorithms

operateinthe

L

1

-metri ,whi himpliesthatthey annotbenetfromthePTAS of Arora et al.that surfa ed into theresear h ommunityoneyear after these

resultswerepublished.

TheKineti TSP hasalso been addressedby resear hersin the operations

resear h eld using linear programming relaxations [10℄, although their main

(3)

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 orrespondingshortest

path that visits all points of

S

, starting at

s

1

and ending at

s

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 bythe

fun tion

s(t) = (x + tv cos α, y + tv sin α)

, where

v ≥ 0

isthepoint'sspeedand

α

isitsdire tion.If

v = 0

wesaythatthepointisstati .Thetravelingsalesman isdes ribedbyaspe ialpointthat anmovewithvariablespeedanddire tion.

Theinitial position

s

0

of thesalesmanis assumedto be

(0, 0)

and itsmaximal speed is assumed to be

1

. The path taken by the salesman is denoted

P

and

(4)

thesalesman visits

s

attime

t

.

P

is alled asalesman pathof

S

ifallpointsin

S

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 point

s

i

(t)

in

S(t)

moves with the speed

v

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 speedand

dire 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 i

notationisneeded.Givenatourorapath

P

intheEu lideanplane,welet

C(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

}

of

3

-elementsubsets ofaset

U

of

3k

elements (represented by the integers

1, . . . , 3k

), does there exist a subfamily

F

⊆ F

ofpairwisedisjointsetssu h that

S

F

∈F

F = U

?

Wesaythatifthereexistssu hasubfamilyfor

F

then

F ∈ X3C

,otherwisewe

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(−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

of

3

-elements,theobje tiveis to

onstru t asetof points

S

and abound

L

su h that anoptimalTSP tourin

S

haslength at most

L

ifand only if

F

hasan exa t over.

S

is onstru ted

from

F

in stages,startingwith twobasi stru tures: jun tionsand rossovers.

Letus start withthejun tions. Forea h set

F ∈ F

we onstru tonejun tion

in

S

, that is,agadget designedtorepresent

F

inthe TSPinstan e. Fig.1

de-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 tiveregionare

depen-dentonthenumberofsetsin

F

aswellasthenumberofpossibleelements,sin e

r = |F|

and

3k = |U|

;seethea tiveregioninFig.1.Moreover,thelinesegment end point oordinatesoutsidethe a tive regionaredependenton thevalue

K

,

(6)

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 hisdenedas

K = 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 thefamily

F

will berepresented in

S

bya set stru ture onsisting of one jun tion and three rossover sta ks of height

a

i

, b

i

(7)

Setstru turefor

F = {2, 4, 5}

. Crossoversta kofheight

5

.

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. In

this wayatube systemis builtfor ea h set stru ture. These tube systemsare

onne ted intoahugesystemasFig. 6suggests. Theguredes ribesthe nal

TSP onstru tion

S

forthefamily

F = {{1, 2, 3}, {2, 4, 5}, {1, 2, 6}}

.

Let

T

0

denotetheset ofunitlengthedgesbetweenpairsofpointsin

S

.The tubesystem onsistsofedgesfrom

T

0

,andattubeinterse tionpoints rossovers and jun tions are lo ated. Thefollowinglemma servesto justify ourattention

tothetubesegments.

Lemma2. Anoptimal TSP tour

OP T

in

S

ontainsalledges of

T

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

,where

q < 3kr

isthenumber of rossoversin

S

.

L

is goingto bethe lowerbound ofaTSP tourin

S

ifno

(8)

a)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

)

(9)

J

J

J

Fig. 6.Final

S

when

F = {{1, 2, 3}, {2, 4, 5}, {1, 2, 6}}

.

Lemma3. There isanoptimal TSP tour

OP T

oflengthlessor equalto

L

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 of

S

, i.e. an optimal

TSP 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 hthat

C(AP X) < C(OP T ) + 1

unless

P = 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

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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 movingpointsisgiventogetherwithastartingpoint

s

0

= (0, 0)

.Allpointsmove in thesamedire tion

α

and withthesamespeed

v

. We anassume,w.l.o.g.,

that

α =

π

2

.Thusapoint

s

i

(t)

isdened as

s

i

(t) = (x + tv cos

π

2

, y + tv sin

π

2

) = (x, y + tv).

The traveling distan e

c

ij

between two points

s

i

(t)

and

s

j

(t)

is the time neededbyasalesman,movingwithspeed

1

,totravelfrom

s

i

to

s

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 tion

d(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)

fromamovingpoint

s = (x, y)

toastati point in theEu lideanplane:

f

v

(s) =



x

1 − v

2

,

y

1 − v

2



Let

P = (s

i

1

, . . . , s

i

n

)

beapathinthetranslationalinstan e,then

C(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 let

S

E

bethe orresponding Eu lidean instan eafter the transformation using

f

v

. A salesman path in

S

T

is optimal if and only if the orrespondingEu lidean Hamiltonianpath in

S

E

isoptimal.

(11)

using a modied PTAS for the Eu lidean TSP [1,2,11℄. Here

OP T

E

denotes anoptimalTSP pathin theEu lidean instan e.Let

AP X

T

and

OP T

T

denote the orrespondingpathsinthetranslationalinstan e.Observethat

OP T

T

isan optimalsalesmanpathbyLemma6.Wehave

C(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 between

s

1

and

s

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 followsthat

AP X

T

is a

(1 +

ǫ

1−v

)

-approximationof theoptimalpath.Thus, wehaveaPTAS forthetranslationalTSPpathproblem.Notethatthe

approx-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 that

OP T = (s

0

, s

1

(t), . . . , s

n

(t), s

0

)

is an optimal salesman tour. It follows easily that the optimalsalesman path startingfrom

s

0

and endingat

s

n

(t)

isa part of that tour. For ea h possible su h ending point

s

i

(t)

,

i ≥ 1

, we ompute a

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Theorem1. 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 from

s

0

and ending at

s

i

(t)

. The length of the optimal salesman tour

OP T

is

C(OP T ) = OP T

H

+ l

,assuming

s

i

isthelastmovingpointvisitedbythetour

OP T

and

l

isthelengthofthelastsegmentinthetour(betweenthepoints

s

i

(t)

and

s

0

). There is atour

AP X

among thetours our algorithm omputes, that alsohas

s

i

(t)

asthelastunvisitedpoint.Thelengthof

AP X

is

C(AP X) = (1 +

ǫ

2(1 − v)

)OP T

H

+ l

,

where

l

isthelengthbetween

s

i

(t)

and

s

0

.Asalesmanthattravelsalong

AP X

visits

s

i

(t)

at time

t = (1 +

ǫ

2(1−v)

)OP T

H

, whereastheoptimalsalesmanvisits

s

i

(t)

at time

t = OP T

H

. Thetime dieren eis

ǫ

2(1−v)

OP T

H

and in that time period,

s

i

(t)

movesthelength

2(1−v)

OP T

H

. Thetriangle inequalityassures us that

l

doesnot ex eed

l +

2(1−v)

OP T

H

. The ost of our approximate tour is thus

C(AP X) ≤ (1 +

ǫ

2(1 − v)

)OP T

H

+ l +

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

.Theratiobetween

the ostsof

AP X

and

OP T

be omes

C(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.Ifthespeedsarebounded

bya onstant

c < 1

thenthealgorithmisatruePTAS.

4 Two Redu tions for KTSP

The kineti traveling salesman problem is in general not as easily

(13)

O

p

2

Fig. 7.Kineti TSP-instan ewithtwomovingpoints.

Thegure ontainsaGGJ-instan eoftypea,atdistan e

D

fromtheorigin

and twomovingpoints

p

1

and

p

2

,bothwithspeed

v < 1

.Weassumethat the salesmanstartsattheoriginattime

t = 0

.Thepoint

p

1

moveshorizontallyand liesat thesamealtitudeasthebottomlineof theGGJ-instan eand interse ts

theline

l

attime

D

.Point

p

2

movesparallelto

p

1

andrea hestheoriginattime

2D + L

. Re all that aGGJ-instan e is onstru tedso that if

F ∈ X3C

, then

an optimal salesmantouris at most

L

long. Otherwisethe length is at least

L

+ 1

.

Be auseofthelargespeedofthemovingpoints,itfollowsthattherstpoint

to bevisitedis

p

1

. Furthermore,thepoint

p

2

must be visitedbeforeit rea hes theorigin.Otherwise,thetourhasunboundedlength.

If

F ∈ X3C

,thenanoptimalsalesman an at h

p

1

and ontinuealongthe GGJ-instan e'soptimalsalesmantour.AfterallpointsintheGGJ-instan ehave

beenvisited,thesalesmanstillhastimeto at h

p

2

beforeitrea hestheorigin. If

F 6∈ X3C

, thenanoptimal salesmandoesnothavetimeto at h allthe

pointsintheGGJ-instan ebeforehehastoturnba kand at h

p

2

.Thismeans that hemusttraversethedistan e

D

at leastfourtimes.Fromthisweget:

Theorem 2. It is NP-Hard to get an approximation ratio less than

2

for the

kineti TSP,evenif thereareonly twomoving pointsin the instan e.

InTheorem 2weneed the speed

v

to be arbitrarily loseto

1

. With more

movingpointswe anrestri tthespeedto

v = 1/2

andstill getalowerbound

of

2

ontheapproximationratio.

Theorem3. It is NP-Hard to get an approximation ratio less than

2

for the

(14)

Proof. To prove the theorem we use two pairs of sets with

k = m

points in

ea 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 pair

1

and

2

in pair

2

. The moving points

performahorizontalmotionfromleftto 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

. At

the same time, the lowest moving point of pair

1

interse ts the originand at

time

D + L

,thetopmostmovingpointinpair

2

ollideswiththe orresponding

stati 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 pairs

of 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

Pair

2

Pair

1

D

W

Fig. 8.KTSP-instan ewithmovingpointsofspeed

1/2

.

Ifthereexistsanexa t overfortheX3C-instan e,thenanoptimalsalesman

hastimetovisitallpointsinbothpair

1

and theGGJ-instan ebeforetherst

twopointsin pair

2

ollide.Thetime that anoptimalsalesman needsin order

to 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 all

(15)

3D/k

v = 1/2

ostofmovingbetweenthetwosetsba kandforthdoublesea htime.The ost

ofthisapproa histhus

4D + L

− 3Dz/k + 2

z

−1

,where

z

isthenumberof

zig-zagmovements.Minimizingthis expression,using theknowledgethat

L

≥ m

,

D = m

2

and

k = m

,provesthatthelengthofthetourisatleast

4m

2

−cm log m

,

forsome onstant

c

.

Sin ethesalesmanmoveswithspeed

1

,lengthequalstime.Thelengthofan

optimalsalesmantouristhereforeat least

4m

2

− cm log m

if

F 6∈ X3C

and at

most

2(m

2

+ L

)

if

F ∈ X3C

.Thelimitoftheratiogivesalowerbound onthe

approximationratioand 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)

.Thisimpliesthatattime

t = 0

allpointsarelo ated attheorigin,andthat therelativedistan eswithin theinstan edonot hange

overtime. Weassume that the salesman begins his pursuit at time

t

0

> 0

(if he starts at time

t = 0

, he visits all pointsat on e withoutmoving). Observe

that thelengthofasalesmantour

P

in auniformlyexpandingTSP-instan e is

dire tion independent. That is, ifwereverse theorder in whi h thepoints are

visited,thelengthofthenewtourisequaltothelengthof

P

.Asgadgetsinthe

KTSP-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

(16)

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 reate

auniformlyexpandingGGJ-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.

km

2

v

i

2π/k

Fig. 9.One ir leoftheinapproximableKTSP-instan e.

The KTSP-instan e onsists of

l

on entri ir les

C

1

, . . . , C

l

, with radii

r

1

, r

2

, . . . , r

l

, enteredaroundtheorigin.Ea h ir le onsistsof

k = ⌈m

2

l⌉

identi- alsmallexpandingGGJ-instan es.TheGGJ-instan esarepla edonthe ir les

asshowninFigure9.

Ea hGGJ-instan e ontains

m

points.Welet

l = m

a

, forsome

a > 1

.The

totalnumberofpointsintheinstan e,denoted

n

,istherefore

mkl = m

2a+3

.The

rightmostpointofaGGJ-instan eispla edonthe ir leandthebottomlineof

theGGJ-instan efollowsthe ir le'stangentlineatthatpoint.Ea h ir le

C

i

is expandingwithaspeed

v

i

,i.e.therightmost pointof ea hGGJ-instan e on

C

i

hasaspeed

v

i

,dire tedawayfrom the enterandtheradiusof

C

i

is

r

i

= v

i

t

.

Welet

v

1

=

1

2e

and

v

i

= (1 +

m

2

k

)v

i

−1

.This impliesthatthelargest ir le's speed,

v

l

,is bounded by

v

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 points

s

i

and

s

j

is

d(s

i

, s

j

) = d

ij

t

, where

d

ij

is onstant.Theexpansionrateissu hthatthedistan e betweenthe leftmostandtherightmostpointofaGGJ-instan e islessthan

tv

i

tan

m

2

k

.We

(17)

u

i

= v

i

tan

m

2

k

theanglebetweentheleftmostpoint

L

,theorigin,andtherightmostpoint

R

in

aGGJ-instan e is

∠LOR =

m

2

k

and theanglebetweentwo onse utive GGJ-instan es is

k

; see Fig.9.Thefastest point ofaGGJ-instan e is theleftmost pointandthispointhasthespeed

r

v

2

i

+ (v

i

tan

m

2

k

)

2

≤ v

i

(1 +

m

2

k

).

Thus, no point belonging to ir le

C

i

has aspeedex eeding

v

i

(1 +

m

2

k

)

, and thisimpliesthatthemaximalspeedisbounded.Fromnowonwesimplydenote

ournewly onstru tedinstan etheKTSP-instan e.

Wewillshowthat the onstru tionaboveguaranteesthat anoptimal

sales-manvisitstheGGJ-instan es ir lewise,oneatatime.Furthermore,thetimeit

takestogofromanymovingpoint

s

i

toanyothermovingpoint

s

j

inthe KTSP-instan edependslinearlyonthestartingtime.Assumingthatasalesmanstarts

from

s

i

attime

t

itwouldtakehimtime

κ

ij

t

togoto

s

j

,wherewe onsider

κ

ij

tobe onstantsin eitdoesnotdependon

t

.Asa onsequen e,travelingtimeis

multipli ativeintheedgeweight onstants

κ

ij

.Thus,ifthesalesmanbeginshis pursuitat time

t

0

andnishestherst GGJ-instan eat time

Kt

0

, thenit will takehimatleasttime

CK

kl

t

0

tovisitallpointsintheKTSP-instan e,where

C

denotesa onstantand

K

isindependentof

t

.Notethatanon-optimalsalesman

must doworseonea hGGJ-instan e and sin etheerrorisalsomultipli ative,

theinapproximabilityratioisgoingtobe omelarge.Therestofthepresentation

isdevotedtoprovingthefollowingtheorem.

Theorem 4. For any

γ > 0

, there exists no polynomial-time algorithm that

a 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.Wedene

C

P

(t)

tobethelengthof this path (whi h issynonymouswith thetime ittakesfor thesalesmanto traverse

(18)

thepath

P

,giventhestartingtime

t

,i.e.,

T

P

(t) = t + C

P

(t)

.A simpleproofof indu tion yieldsthat

T

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

ij

is the time it takes for the salesman to travel between

s

i

and

s

j

, starting from

s

i

at time

t

. Let us examine a small expanding GGJ-instan e produ edbyanX3C-instan e.Weassumethattheinstan eislo atedsomewhere

on ir le

C

i

. The optimal salesmanpath for the GGJ-instan e starting at the rightmost point and ending at the leftmost point is denoted

opt

if

F ∈ X3C

and

opt

if

F 6∈ X3C

.Wewouldliketondanupperboundon

opt

andalower

bound 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 e

move with the same velo ity as the rightmost point in the expanding

GGJ-instan eattime

t

.Wedene

D(t)

asthelengthoftheoptimalpath,startingat

therightmostpointandendingattheleftmostpoint.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 points

in the expanding instan e at time

t

and let

t

i

bethe point in time su h that

f (t

i

) = 1 − v

i

2

.Notethat

t

i

isthetimewhentheexpandingGGJ-instan esrea h the size of theoriginal GGJ-instan es.That is, ifweapply the fun tion

f

v

on anexpandinginstan eattime

t

i

, thenthestati instan ethat resultswillhave thesamesize astheoriginalGGJ-instan esofSe tion2.Thus,

D(t

i

) ≤ L

and

D

(t

i

) ≥ L

+ 1

. Observealsothat

t

i

depends onthe ir le where theinstan e islo ated.

D(t)

,

D

(t)

and

f (t)

arelinearmappings,whi himpliesthat

D(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 onstants

c

1

and

c

2

)

,

(3)

D

(t) > D(t) + f (t)

(19)

u

i

≥ v

f

u

i

m

.

Now,

t

i

=

1−v

2

i

v

f

,so

1 − v

2

i

u

i

≤ t

i

m(1 − v

2

i

)

u

i

.

Re allthat

u

i

= v

i

tan

m

2

k

andsobyTaylorexpansion,wegetthat

2π +

1

3

m

2

k

≥ tan

m

2

k

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

yieldsthat

c

1

km

2

≤ t

i

≤ c

2

km

3

, for some on-stants

c

1

and

c

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)

wegetthat

D(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 e

t

i

≥ c

1

km

2

> D(t

i

)

.

(20)

(6) Bythelinearityof

D(t)

and

f (t)

wehavethat

C

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 expanding

GGJ-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 7

T

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

and

c

1

km

2

≤ t

i

≤ c

2

km

3

. Expressed in terms of

k

and

m

, the quotientisatleast

T

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

,and

c = 2c

2

.

LetusreturntotheKTSP-instan ethat ontainsthesmallexpanding

GGJ-instan es. Weassumethat thesalesman startsattheoriginattime

t

0

> 0

.We willprovethattheKTSP-instan eisinapproximableusingtheinapproximability

resultinLemma8forthesmallexpandingGGJ-instan es.

Considerasalesmantour

P

fortheKTSP-instan e.Asubpathof

P

starting

atapoint

s

i

andendingatthepoint

s

j

is alledanedgeof

P

ifnootherpoints are visitedalongthe subpath.Edgesthat onne ttwodierent GGJ-instan es

are alled leaps.A leapthatexits the

j

thGGJ-instan eon ir le

C

i

isdenoted

J

ij

. Theshortest leapthat onne ts ir le

C

i

with ir le

C

i+1

is theleap that onne tstheleftmost pointofthe topmostGGJ-instan e on ir le

C

i

withthe rightmost point of the orresponding GGJ-instan e on ir le

C

i+1

. This leap will be denoted

E

i

(exterior leap exiting ir le

C

i

);

T

E

i

(t) = K

E

i

t

, for some

(21)

C

opt

i

(t

i

) = Θ(m),

(7)

C

I

i

(t

i

) = Ω(m

2

),

and

C

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 e

t

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 ethereare

m

pointsin theinstan eandtheminimum distan ebetweenpairsof pointsis

1

at time

t

i

.It followsthat

C

opt

(t

i

)

and thereforealso

C

opt

(t

i

)

is

Θ(m)

.We on ludethat

C

opt

i

(t

i

) = Θ(m)

.

(8) Thedistan e,andthusthelengthofthetravelingpath,tothe losest

GGJ-instan eonthesame ir leisshowntobe

Θ(t

i

/k)

inFigure11,and

c

1

km

2

t

i

≤ c

2

km

3

.

(9) Thelengthofanyleap

J

, onne ting ir le

C

i

withanother ir leattime

t

i

isatleasttheminimumof

min

1≤j<i

 v

i

− (1 + 2π/m

2

k)v

j

1 + v

j

t

i



and

min

i<j≤n

 v

j

− (1 + 2π/m

2

k)v

i

1 − v

j

t

i



.

Inserting the bounds of

t

i

and

v

i

and observing that

v

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)

and

C

P

(t)

.

(22)

r = v

i

t

i

= Θ(t

i

)

y

x

r

2π/k

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 attime

t

0

isnishedattime

T

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. Let

K

tsp

ij

denote the total ost spentin the

j

th GGJ-instan eon ir le

C

i

.Be ause ofthe salesman tourpropertyweknowthat theremustbeat leastone leapexiting ea h

GGJ-instan e and a leap

J

0

from the origin to the rst GGJ-instan e in the tour. Thuswe an hara terizethenishingtimeof

P

as

T

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

,

where

K

J

ij

des ribesthe ostofexitingthe

j

thGGJ-instan eon ir le

C

i

,and

H

ij

is the numberof extraleaps exiting this GGJ-instan e. The ost of these extraleapsaredes ribedby

K

J

(23)

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

,

where

J

i

isaleapexiting ir le

C

i

.Thisa tuallyprovesthattheGGJ-instan es mustbetaken ir lewise.Thequestionthatremainsisinwhat orderthe ir les

aretaken.

Note that the ost of the shortestpath between theorigin and ir le

C

l

is exa tly

K

E

0

K

E

1

· · · K

E

l−1

.Furthermore,theshortestpathbetweenthefurthest pointon ir le

C

l

andtheoriginhasthe ost

K

E

l

.Now,anysalesmanpathneeds tovisitthese points.Itturnsoutthat

K

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 e

C

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

(24)

= (1 +

1

ckm

3

)

kl

∈ 2

Ω(n

1/2−γ

)

,

for

γ = 3/a

,and

a

anbe hosenarbitrarilylarge.Rememberthat

n = 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 esperform

onstanttranslationalmovementintime.

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.

Referen es

1. S.Arora. Polynomialtime approximations hemesfor Eu lideanTSP andother

geometri problems.InPro .37thAnnualSymposiumonFoundationsofComputer

S ien e (FOCS),pages211,1996.

2. S.Arora. Nearlylineartimeapproximations hemesforEu lideanTSPandother

geometri problems.InPro .38thAnnualSymposiumonFoundationsofComputer

S ien e (FOCS),pages554563,1997.

3. S.Arora,C.Lund,R.Motwani,M.Sudan,andM.Szegedy.Proofveri ationand

intra tabilityofapproximationproblems.InPro .33rdFOCS,pages1322,1992.

4. P.Chalasani and R. Motwani. Approximating apa itated routingand delivery

problems. Te hni al Report STAN-CS-TN-95-24,Department of Computer

S i-en e,StanfordUniversity,1995.

5. P.Chalasani,R.Motwani,andA.Rao.Approximationalgorithmsforrobotgrasp

and delivery. InPro eedings of the 2nd International Workshop on Algorithmi

Foundations ofRoboti s,pages347362,1996.

6. N.Christodes. Worst- aseanalysisofanewheuristi forthetravelingsalesman

problem. Te hni alReport388, GradS hoolofIndustrialAdministration,CMU,

1976.

7. M. R.Garey, R.L. Graham, and D. S.Johnson. Some NP- omplete geometri

(25)

andrelatedproblems. ToappearinSIAMJ.Computing.

12. C.H.Papadimitriou. Eu lideanTSPisNP- omplete. TCS,4:237244,1977.

13. S.B.RaoandW.D.Smith.Approximatinggeometri algraphsviaspannersand

banyans. InSTOC,pages540550,1998.

14. S.Sahniand T.Gonzales. P- ompleteapproximation problems. JACM, 23:555

Figure

Fig. 5. Possible 
onne
tions in a
tive regions of 
rossovers.
Fig. 6. Final S when F = {{1, 2, 3}, {2, 4, 5}, {1, 2, 6}} .
Fig. 7. Kineti
 TSP-instan
e with two moving points.
Fig. 8. KTSP-instan
e with moving points of speed 1/2 .
+3

References

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