Parallel Searching on m Rays

Parallel Sear hing on m Rays

Mikael Hammar

Departmentof Computer S ien e, Lund University, Box 118, S-22100 Lund,

Sweden

Bengt J. Nilsson

Department of Te hnology andSo iety, MalmöUniversity College,

Citadellsvägen 7, 205 06Malmö, Sweden

Sven S huierer

Institut fürInformatik, Georges-Köhler-Allee, Geb. 051, D-79110 Freiburg,

Germany

Abstra t

Weinvestigateparallelsear hingonm on urrentrays.Weassumethatatargettis

lo atedsomewhere onone of therays; wearegiven agroup ofm point robotsea h

of whi h has to rea h t. Furthermore, we assume that the robots have no way of

ommuni atingoverdistan e.GivenastrategySweareinterestedinthe ompetitive

ratio dened as the ratio of the time needed by the robots to rea h t using S and

thetimeneededto rea h tifthe lo ationof tisknowninadvan e.

If a lower bound on the distan e to the target is known, then there is a simple

strategywhi ha hievesa ompetitiveratioof9independentofm.Weshowthat9

isalowerboundonthe ompetitiveratio fortwolarge lassesofstrategiesifm2.

If the minimum distan eto thetarget is not known inadvan e,we show a lower

boundonthe ompetitiveratio of1+2(k+1) k+1

=k k

where k=dlogmewherelogis

usedtodenotethe base2 logarithm.Wealsogivea strategythatobtainsthis ratio.

Key words: Computational Geometry,On-line Sear h Strategies

?

This resear h is supported by the DFG-Proje t Diskrete Probleme,

Sear hing for a target is an important and well studied problem in roboti s.

In many realisti situations the robot does not possess omplete knowledge

about its environment, for instan e, the robot may not have a map of its

surroundings,orthelo ationofthetargetmaybeunknown[36,8,9,11,12,15

17℄.

Thesear hof therobot an beviewed asanon-line problemsin e therobot's

de isionsabout the sear hare based onlyon the part ofits environmentthat

ithasseensofar.Weusetheframeworkof ompetitiveanalysistomeasurethe

performan e ofanon-linesear hstrategy S [18℄.The ompetitiveratioof S is

dened as the maximum of the ratio of the traveling time needed for arobot

to nd the target using strategy S to the optimal distan e from its starting

pointtothetarget,overallpossiblelo ationsintheenvironmentofthetarget.

Notethat wehavenormalizedthe speed sothat timeequalsdistan e forafull

speed strategy.

A problem with paradigmati status in this framework is sear hing on m

on urrent rays. Here, a point robot oras in our asea group of point

robots is imagined to stand at the origin of m on urrent rays. One of the

rays ontains the target t whose distan e to the origin is unknown. A robot

an dete t t only if it stands on top of it. It an be shown that an optimal

strategy for one robot is to visit the rays in y li order, in reasing the step

lengthea htime by a fa torof m=(m 1).In the beginningthe robotsstarts

with astep lengthof 1 whi his assumed tobea lowerbound onthe distan e

tot [1,7℄.The ompetitive ratio C

m

a hieved by this strategy is given by

1+2 m m (m 1) m 1 :

Thelowerboundforsear hingonmrayshasprovedtobeaveryusefultoolfor

provinglowerboundsfor sear hing ina numberof lassesof simplepolygons,

su h as star-shaped polygons [14℄, generalized streets [6,13℄, HV-streets [5℄,

and -streets[5℄.

In this paper we are interested in obtainingupper and lower bounds for the

ompetitiveratioofparallelsear hingonm on urrentrays.Thisproblemhas

been addressed beforein two ontexts.

The rst ontext is the on-line onstru tion of hybrid algorithmsthe setting

of whi h an be des ribed as follows [10℄: We are given a problem Q and m

approa hestosolvingit.Ea happroa hisimplementedbyanalgorithmwhi h

of its omputation.Onlyasingle basi algorithm an berun by the omputer

at a given time. It is not known in advan e whi h of the algorithms solves

the problemQalthoughwe assumethat thereis atleastoneorhow mu h

time it takes to ompute a solution. In the worst ase only one algorithm

solves Q whereas the othersdo not even halt onQ. One way tosolveQ is to

onstru t a hybrid algorithmthat uses the basi algorithms in the following

way. A basi algorithmis run for some time, and then the omputer swit hes

to another algorithmand so onuntil Q is solved. If k <m, then there is not

enoughmemorytosave allthe intermediateresults.Sosometimesthe urrent

intermediate results have to be dis arded and to be re omputed later from

s rat h.

Adierentwaytolookatthisproblemistoassumethatwearegivenkrobots

thathavetosear honmraysforatargettwith k<m.Ea hray orresponds

toabasi algorithm,and a robot orresponds toa memoryarea. At any time

weare allowed tomoveonlyone robot.Dis ardingintermediateresults ofthe

basi algorithmA orresponds to movingthe robot onthe ray orresponding

toA ba k to the origin.

Kao et al. [10,19℄ present an algorithm for the above problem that a hieves

anoptimal ompetitive ratio of

k+2

(m k+1) m k+1

(m k) m k

whi h is,of ourse,alsothe ompetitiveratioofsear hingwith k robotsonm

rays if only one robotis allowed to move at atime.

Inthese ond ontextagroupof mpointrobotssear hesforthe target.Again

neitherthe ray ontainingthe targetnor thedistan etothetargetareknown.

Now allthe robots haveto rea h the targetand the onlyway two robots an

ommuni ate is if they meet, that is, they have no ommuni ation devi e.

We are going touse this modelinour paper. Baeza-Yates and S hott

investi-gate sear hing on the real line, that is, the ase m =2 [2℄. They present two

strategies both of whi h a hieve a ompetitiveratio of 9. They also onsider

sear hing for atarget linein the planewith multiplerobotsand present

sym-metri and asymmetri strategies. However, the question of optimality, that

is, orresponding lowerbounds, is not onsidered.

In this paper we investigate sear h strategies for parallelsear hing onm

on- urrent rays. If a lower bound on the distan e to the target is known, then

thereisasimple strategythat a hievesa ompetitiveratioof 9independent

of m. We show that even in the ase m =2 there is a mat hing lowerbound

implies a lower bound of C for m >2as is to be expe ted. This implies, in

parti ular, that there is no monotone or symmetri strategy for arbitrary m

that has a ompetitive ratio better than 9. For monotone strategies we an

even strengthen this result and show that

C k =1+2(k+1) k+1 =k k ;

where k=dlogme islowerboundon the ompetitiveratio.

Wealso onsiderthe asethattheminimumdistan etothetargetisnotknown

inadvan ewhi hturnsouttobeessentiallyequivalenttorestri tingourselves

tomonotonestrategies.Weagainshowalowerboundonthe ompetitiveratio

ofC

k

wherek =d logme .Wealsopresenta(monotone)strategythata hieves

this ompetitiveratio.

The paper is organized as follows. In the next se tion we present some

def-initions and preliminary results. In parti ular, we present three strategies to

sear h on the line (m = 2), ea h with a ompetitive ratio of 9. In Se tion 3

we show a mat hing lower bound of 9 for two large lasses of strategies. In

Se tion 4 we extend our results to the ase m > 2. Finally, in Se tion 5 we

present an optimal algorithm to sear h on m rays if there is no minimum

distan e to the target.

2 Preliminaries

Inthe followingwe onsider the problemof a groupof m robotssear hing for

atargetof unknown lo ationonm raysinparallel.The robotshave the same

maximal speed whi h we assume without loss of generality to be 1 distan e

unit pertimeunit. Ifthe robots haveunbounded speed, then the timetond

thetarget(botho-lineandon-line) anbemadearbitrarilysmall.Thespeed

ofa robot maybe positive(ifit movesaway fromthe origin)ornegative (ifit

movestowards the origin).

LetS be astrategyforparallelsear hingonm rays andT

S

(D)the maximum

timethe group ofrobots needstond and rea ha targetpla edatadistan e

of D if it uses strategy S. Sin e the maximum speed of a robot is one, the

timeneededtorea hthetargetif thepositionofthetargetisknownisDtime

units.The ompetitiveratioisnowdenedasthe maximumofT

S

(D)=D,over

all D  0. In some appli ations a lower bound D

min

on the distan e to the

target may be known. If su h a lower bound exists, then we assume without

loss of generality that D

min

= 1. It will turn out that the existen e of D

min

Wesayastrategyismonotoneif,atalltimes,alltherobots(thatdonotknow

the lo ationof the target) have non-negativespeed. We say a strategy is full

speed if all the robots travel at a speed of 1 or 1 at all times. We say a

strategy is symmetri if, at all times, all the robots (that do not know the

lo ation of the target)have the same speed.

Weillustratethedierenttypesofstrategiesform =2.Theoptimalmonotone

strategy is for ea h robot to travel at a speed of 1=3 on ea h ray. After one

robot has found the target, it runs ba k to fet h the other. This leads to a

ompetitiveratio of9.This strategyisdes ribed in[2℄. Inthe next se tionwe

showalowerboundof9onthe ompetitiveratioofmonotone strategies.The

optimal(full-speed)symmetri strategyisforea hrobottodoublethedistan e

that has been exploredbeforeand then to returnto the origin.This strategy

an only be applied if a lower bound on the distan e to the target is known.

Ita hieves a ompetitiveratio of9.Againthisstrategy isdes ribed in[2℄and

we show a lower bound of 9 onthe ompetitive ratio of symmetri strategies

inthenext se tion.Finally,anasymmetri strategyisforbothrobotstowalk

together and to use the optimalstrategy for one robotto sear h ontwo rays.

This again yieldsa ompetitive ratio of 9.

3 Sear hing on Two Rays

In this se tion we onsider the problem of two robots sear hing for a target

ofunknown lo ationonthe reallineinparallel;therobotsare initiallypla ed

atthe origin.We assumein the following that a lowerbound onthe distan e

fromthe originto the target of D

min

=1 isknown.

Formonotone strategies wehave the followinglowerbound.

Theorem 1 There isnomonotone strategy thata hievesa better ompetitive

ratio than 9 to sear h on two rays in parallel.

PROOF. The proof uses anadversary to pla e the target point in order to

maximizethe ompetitive ratio.

Let us enumerate the robots and dene v

i

(T) as the average speed of robot

i 2 f1;2g at time T, i.e., the distan e of the robot to the origin at this time

divided by the time.It is learthat amonotone sear hstrategy is ompletely

spe ied by the twoaverage speed fun tions.

First of all we realize that the two robots willnot both go in the positive or

Hen e, the two robots go in dierent dire tions. As time passes the robots

move ontinuously and monotoni ally with some speed along the line until

one ofthem ndsthe target.This robotnowtravels atfullspeed tothe other

robotand ommuni ates toitthe lo ationof thetarget and they both return

tothis target point.Consider the value v

min spe ied by v min = inf tT min minfv 1 (t);v 2 (t)g; where T min

is the rst point in time su h that both robots are at least a

distan e D

min

from the origin. From the denition of v

min , we know that v 1 (T)v min and v 2 (T)v min ,for allT >T min .

For any " > 0, there is a time T

"  T

min

su h that either v

1 (T " )  v min +" orv 2 (T " )v min

+". Assume withoutlossof generality that for somespe i

" it holds for v

1 (T

"

). By the denition of monotoni ity T

"  T min and hen e v 1 (T " )T "  v 1 (T min )T min = D min

, so the adversary pla es the target at the

pointD " =v 1 (T " )T "

.Alowerboundonthe ompetitiveratioCofthestrategy

an now be expressed as C  T " +2T R D " ; where T R

denotes the time for the robot that found the target to rea h the

se ond robot at full speed. Sin e they both have to return to the target this

adds an extra term T

R

. Be ause of the monotoni ity of the strategy we an

express the time T

R by T R =D " +v 2 (T " +T R )(T " +T R ):

This is be ause the robot that nds the target goes towards the other robot

at full speed traveling rst the distan e D

"

and then the distan e v

2 (T " + T R )(T " +T R

) untilthey meet. We have that

T R = D " +v 2 (T " +T R )T " 1 v 2 (T " +T R )

where we an assume that v

2 (T

" +T

R

) < 1 sin e otherwise robot 1 never

rea hes robot2. Hen e,

C T " +2(v 1 (T " )T " +v 2 (T " +T R )T " )=(1 v 2 (T " +T R )) v 1 (T " )T "

= v 1 (T " ) + 1 v 2 (T " +T R ) + v 1 (T " ) 2 " R (1 v 2 (T " +T R ))  1 v min +" + 2 1 v min + 2 (v min +") v min (1 v min ) 9 18" 1+3" ;

sin e the above expression is minimized for v

min = 1 3 and it tends to 9 as " tends to0. 2

Next we look at symmetri strategies. The following lemma shows that only

full speed strategies need to be onsidered.

Lemma 2 For all ">0 and all strategies S, there is a full speed strategy S 0

su h that the ompetitive ratio of S 0

is at most (1+") times the ompetitive

ratio of S.

PROOF. Divide the time intosmall intervals of length Æ. If strategy S has

average speed v in an interval I, then let S 0

be the full-speed strategy that

rst travels ba k for (1 v)Æ=2 time units and then forward for (1+v)Æ=2

time units in I. At the end of I the robot is atthe same position asif ithad

used S. Byletting Æ goto 0,the laimfollows. 2

The previous lemma tells us that we an simulate any strategy with a full

speed strategy and therefore we only have to onsider full speed strategies if

we are given alowerboundon the distan e to the target.

In the following we show a lower bound for symmetri full speed strategies.

We start with the ase m=2and onsider the general ase later.

Lemma 3 LetS beasymmetri strategy.Then,thereisasequen eofpositive

numbers (y 0 ;y 1 ;y 2

;:::) su h thatthe ompetitive ratio C

S of S satises C S  sup k1 1+2 P k+1 i=0 y i y k :

PROOF. Sin e the strategy is symmetri , the two robots willuse the same

lo alstrategytosear h itsownray.Furthermore,by Lemma2we an assume

that itisafull speed strategy. We anmodelafull speed strategyfor arobot

by sayingthat itrstmovesadistan e x

0

forwardalong therayatfullspeed,

thenitmovesadistan ey

0

ba kwardsatfullspeed,thenadistan ex

1

forward,

a distan e y

1

s x k y k k+1 L k L k+1 U k

Fig.1. Thedenitionof U

k and L

k .

itruns ba k atfull speed untilitmeetsthe other robot,and they both run to

the target atfull speed.

The proof uses an adversary to pla e the target point in order to maximize

the ompetitiveratio.

We say that a robot is in step k when it moves forward and ba kward the

k +1 st

time. Let L

k

denote the distan e to the origin of the turning point

where the robot begins step k and let U

k

denote the distan e to the origin

of the turning point where the robotstarts to moveba kwards during step k

(see Figure1). We have that L 0 =0; L k =L k 1 +x k 1 y k 1 = k 1 X i=0 x i y i ; U k =L k +x k = x k + k 1 X i=0 x i y i = y k + k X i=0 x i y i :

The total time that the robot has traveled whenit ompletes step k is

T k = k X i=0 x i +y i :

We an assumethat U

k 1 <U

k

, sin eotherwise, the strategy willnot explore

any new part of the ray during step k, and we an ex hange the strategy for

another equivalent one, where the assumption holds. In parti ular, we an

assume that y

k <x

k+1

, for allk 0.

Furthermorewe anassumethat y

k U

k

forallk 0,thatis,arobotalways

staysonthesameray,sin eifthisdoesnothold,we anex hangethestrategy

foranequivalentonewhere,ifthetworobotsmeetattheorigin,theyex hange

pla es and ontinue ontheir own ray instead of the other robot's ray.

Assume that the target is pla ed atdistan e D with 1=D

min

from U k to L k+1 , say at point q 2 [L k+1 ;U k

℄. The ompetitive ratio for this

pla ementis given by C k (D)= T k 1 +x k +(U k q)+(q+D) D = 1+ T k 1 +x k +U k D =1+2 P k 1 i=0 y i +U k D ; (1) sin e T k 1 +x k +(U k

q) is the time needed to rea h point q and q+D is

the time needed for both robots to return to the target after they have met.

It is interesting to note that, for a given D, the above analysis is ompletely

independent of the stepinwhi hthe targetis found and onlydependsonthe

step k inwhi hthe robotsmeet.

Toea hstepk wewillasso iateapla ementofthetargetpoint,su hthatthe

robot that nds the target nds it during a step no earlier than step k +1.

The orresponding ompetitive ratio of the pla ementis denoted C

k .

Considerpla ingthetargetatdistan eDaftery

k

.ByLemma4we anassume

that y

k  U

k 1

, so the robot that nds the target, say robot 1, dis overs it

earliestduringstep k.We laimthattherobots donot meetbeforestepk+1.

The total distan e between the robots is 2D at the time when the target is

found. The distan e is only redu ed during the times when robot 2 travels

ba k towards the origin. During su h a phase the distan e is redu ed by two

distan e units per time unit sin e the robots travel towards ea h other. Now,

if the target is found in step k,then the robotsmeet in the rst step k  su h that P k  i=k y i  D. Sin e D > y k

, this implies that k 

 k+1. If the target

is not found in step k, then the robots meet earliest in step k+1 anyway.

Hen e, using Equation 1 the ompetitive ratio C

k

of pla ing the target after

y k satises C k  sup D2℄y k ;1[ ( 1+2 P k i=0 y i +U k+1 D ) = 1+2 P k i=0 y i +U k+1 y k 1+2 P k+1 i=0 y i y k ; sin e y k+1 U k+1 .

Asaresultthe ompetitiveratioforanysymmetri strategyisbounded below

by C=sup k0 C k sup k0 1+2 P k+1 i=0 y i y k

Lemma 4 Usingthe terminologyof theproof of Lemma3thereisan optimal

symmetri strategy with y

k U

k 1

, for allk 1.

PROOF. Let S be any strategy su h that y

k < U

k 1

for some k  1. We

showthat we an ex hange S for anotherstrategy S 

su h that itsspe ifying

parameters y  i and U  i

have the property y  i = y i , for 0i <k, U  i = U i , for 0ik and y  k U  k 1 . Let k 

be the smallest index with P k  i=k y i  U k 1 . The index k  must exist

and be nite sin e otherwise we an pla e the target at distan e D > U

k 1

and get an innite ompetitiveratio for the strategy.

Wenowdenetwonewspe ifyingsequen esy  i andU  i ofS 

whi hwillrepla e

the sequen es y

i

and U

i .

For0i<k, nothing hanges,that is,we set y  i =y i and U  i =U i .Fork we deneU  k =U k and y  k = P k  l =k y l

. Finally,wedrop allindi esfromk+1 tok  and dene U  k+j =U k  +j and y  k+j =y k  +j , for j 1. Let C n (D) and C  n

(D) be the orresponding ompetitive ratios for strategies

Sand S 

respe tively,whenthetargetispla edatdistan eDfromtheorigin,

and the robotsmeet at step n. Re all that

C n (D)=1+2 P n 1 i=0 y i +U n D ;

wherenisthestepinwhi hthetworobotsmeet.ForC  n (D)wehaveasimilar formula. We will ompare C n (D) and C  n

(D) for all possible meeting steps from one

onwards.

If the robots meet instep i where 0ik, then obviously C

i

(D)=C 

i (D).

Ifin strategyS the robotsmeet instep k 

+j,for j 1,then they willmeet

in step k+j in strategy S 

and it holds that C

k  +j (D)= C  k+j (D).Hen e it

onlyremains to onsider the steps k+1;:::;k 

of strategy S.

Iftherobotsmeetinstepl ofstrategyS wherek+1lk 

,thenthey meet

instep k in strategyS 

. Tosee this, we provethat if robot2(whi hdoesnot

knowabout the target)rea hes the pointU 

k =U

k

, thenthe distan e between

the two robots is at most 2y 

k

and they meet when robot 2 travels ba k in

l

most a distan e 2y

l

from it (sin e they meet in step l). Similarly, if robot 2

rea hes pointU

l 1

, then the distan e is atmost 2(y

l +y

l 1

).By indu tion we

have that when robot 2 is at point U

k

, then the distan e between them is at

most 2 P l i=k y i 2 P k  i=k y i =2y  k as laimed.

The ompetitiveratio of strategy S if the robots meet during step l>k is

C l (D)=1+2 P l 1 i=0 y i +U l D 1+2 P k 1 i=0 y  i +U  k D = C  k (D) sin e y  i = y i , for 0  i  k 1, and U  k = U k < U l

. Thus, the ompetitive

ratio of strategy S 

isat most aslarge as the ompetitiveratio of S.

Repeatingthe pro ess forall k in S su h that y

k <U

k 1

we get a strategy S 0

with spe ifyingparameters y 0 i and U 0 i su h that y 0 i U 0 i 1 for alli1.

IfS wastaken asanoptimalstrategythenobviouslyS 0

mustalsobeoptimal,

thus on ludingthe proof. 2

Sin e for any sequen e of positivenumbers (y

0 ;y 1 ;y 2 ;:::) the value of sup k0 1+2 P k+1 i=0 y i y k

is bounded from belowby 9[1,7℄, we have shown the followingtheorem.

Theorem 5 Thereisnosymmetri strategythata hievesabetter ompetitive

ratio than 9 to sear h on two rays in parallel.

4 Sear hing on m Rays

We now turn to sear hing on m rays in parallel. We assume that a group of

m robots is lo ated at the origin of the rays in the beginning and that again

alowerbound onthe distan e to the targetis known. Werst show thatany

lower bound forsear hing ontwo rays with two robotsimpliesa lowerbound

for sear hing on m rays with m robots.

Theorem 6 If C is a lower bound for sear hingon two raysin parallel,then

beginning about the lo ation of the target, then the ompetitive ratio for

a strategy that exploits this information does not in rease. Assume that the

robotsknowthatthetargetisononeofthersttworays.They anallexplore

these rays in ommon using strategy S. We now dene a new strategy S 0

for

sear hing inparallel with two robots on two rays whi h depends onS. At all

times the two robots of S 0

follow the robots in S that are furthest from the

originonea hofthetworays.Clearly,itispossiblefortworobotstomaintain

the position of the furthest of the m robots on both rays sin e this position

hanges ontinuously.

Now assume that one robot (of the m robots), say robot i, nds the target.

Clearly, robot i is a robot whi h is the furthest from the origin on its ray.

Hen e, one of the tworobots of S 0

nds the target atthe same time. Nowall

of the otherrobotshavetobenotied. It iseasy toshowby simple indu tion

that we an assume that roboti notiesall the other robotsitself sin e

(1) noother robot an travel faster to the opposite ray than roboti and

(2) arobot an onlystart hasingother robotson eit meetsarobotthat at

some point afterthe dis overy of the target met roboti.

Obviously, the last robot that is going to be notied is, at the moment of

noti ation,furthest from the originon the opposite ray.But this robotis at

the same position as the se ond robot of S 0

. Hen e, both strategies need the

same time for all robots to rea h the target and the ompetitive ratio is the

samethis impliesthe laim. 2

We have the following orollary.

Corollary 7 There is no monotone or symmetri strategy that a hieves a

better ompetitive ratio than 9 to sear h on m rays in parallel.

Now, there is a symmetri strategy that a hieves a ompetitive ratio of 9 to

sear honmraysinparallel.Thestrategyisknownasthedoublingstrategyand

goesasfollows[1,2,7℄.Ea h robotstartsby goingone unit atfullspeed onits

ray andthen goesba k tothe origin.Thenthey ea hgotwounits, fourunits,

and so on, on their orresponding ray, always doubling the distan e traveled

andrepeatedly goingba ktothe origin.On earobotndsthe target,itgoes

ba katfullspeed tothe originandwaitsthereuntilthe otherrobotsrea hit.

It then ommuni ates the lo ation of the target tothe otherrobots and they

allmove atfull speed to that lo ation. The ompetitiveratio of the doubling

Csup k1 sup D2℄2 k 1 ;2 k ℄ 2 P k i=0 2 i +D D = 1+sup k1 sup D2℄2 k 1 ;2 k ℄ 2 P k i=0 2 i D =1+sup k1 2 P k i=0 2 i 2 k 1 = 1+2sup k1 2 k+1 1 2 k 1 = 9:

We have proved the followingtheorem.

Theorem 8 The doubling strategy a hieves a ompetitiveratio of 9 to sear h

on m rays in parallel givena lower bound on the distan e to the target.

Corollary 7 an be strengthened onsiderably for monotone strategies. This

also illustrates ni ely that there is a dieren e between monotone and

sym-metri strategies for m>2.

Consider a monotone strategy. As time passes the robots move ontinuously

andmonotoni allywithsomespeedalongtheraysuntiloneofthemhasfound

the target.Thisrobotnowtravelsatfullspeed tooneof theother robotsand

ommuni ates to him the lo ation of the target, they both travel to other

robotsto ommuni atethe lo ationof the target,and soon. When allrobots

know where the target is they allgo tothis target point. We use this idea to

showthe following lower bound.

Theorem 9 There isnomonotone strategy thata hievesa better ompetitive

ratio than C k =1+2 (k+1) k+1 k k ;

where k =dlogme,to sear hon m raysin parallel givena lowerbound on the

distan e to the target.

PROOF. Westart similartothe proofof the asem =2.The m robotswill

ea hgo ona dierent ray sin e otherwise the ompetitive ratio isinnite.

Letv

i

(T)be the average speed of robot 1im attime T and

v min = inf tTmin minfv 1 (t);v 2 (t);:::;v m (t)g; whereT min

istherstpointintimesu hthatallrobotsareatleastatdistan e

D

min

from the origin.

For every " > 0, there is a T " > T min , su h that v i (T " )  v min +", for one

1im.The adversary pla esthe target atapointD

" =v i (T " )T  onray i.

Robots of the rst type are alled the hunters,i.e., the robots that urrently

know the positionof the target and are hasing afterother robotsinorder to

onvey this information.The remainingrobots are alled the prey, and these

ontinue on their respe tive ray using their monotone strategy. Initially at

time T

"

, onlyroboti is a hunter, allthe othersare prey.

Denote by T

n

the rst time whenat least n of the m robots are hunters.

We use indu tion to prove the following inequality

T n  T "  1+2 v min 1 v min  dlogne :

Note that we an assume that v

min

< 1 sin e otherwise the robot that nds

the target willnever rea h any other robot.

The base ase n =1 follows dire tlysin e weknowthat T

1 =T

" .

For n > 1, we prove the laim as follows. The time T

n

is of ourse

non-de reasing in n. Consider now the point in time T

n

, for some spe i n. At

some time prior to T

n

, at least dn=2e robots are hunters. Assume otherwise,

i.e.,thattherearen 0

<dn=2ehunters atanytimepriortoT

n

.Thenthelargest

number of robots that an be ome hunters at time T

n is 2n

0

< n, sin e ea h

hunter an produ eatmostonenewhunterattimeT

n

.Thisisa ontradi tion

sin e at time T

n

we have at least n hunters. For the same reason, there is a

hunter and a prey that meet at a time T no earlier than T

dn=2e

and then one

of them, say robot l, hases some other prey, say robot j, and at hes itat a

time T 0 nolater than T n .Let T R =T 0 T.We have that T n  T dn=2e +T R :

The distan e of the robot onray l tothe originis now v

l

(T)T and itrea hes

the other roboton ray j at adistan e of v

j (T +T R )(T +T R ). Hen e, T R  min T dn=2e TTn fv l (T)T +v j (T +T R )(T +T R )g  min TT dn=2e fv l (T)gT dn=2e + min TT dn=2e fv j (T +T R )g(T dn=2e +T R ): Therefore, T R T dn=2e min TT dn=2e fv l (T)g+min TT dn=2e fv j (T +T R )g 1 min TT dn=2e fv j (T +T R )g

2T dn=2e min 1 v min :

From above we have that

T n T dn=2e +2T dn=2e v min 1 v min  T dn=2e  1+2 v min 1 v min  T "  1+2 v min 1 v min  dlogne ;

by our indu tionhypothesis.

So,afterT

m

time,allrobotsarehunters.Assumethat(oneof)thelastrobot(s)

tobe omeahunterisrobotl.The ompetitiveratioofanymonotonestrategy

an now be expressed by C T m +v l (T m )T m +D D = 1+ T m +v l (T m )T m v i (T " )T " 1+  1+2 v min 1 v min  dlogme 1+v min v min +" (k =dlogme)=1+  1+v min 1 v min  k 1+v min v min +" =1+ (1+v min ) k+1 v min (1 v min ) k (") 1+ (1+ 1 2k+1 ) k+1 1 2k+1 (1 1 2k+1 ) k (") =1+2 (k+1) k+1 k k (");

sin e the expression is minimizedfor v

min

=1=(2k+1) and ittends to C

k as

" tends to0, whi h on ludes the proof. 2

In the next se tion we show that there is a monotone strategy that a hieves

this lower bound.

5 Without Lower Bound on the Minimum Distan e

In this se tion we onsider the problem of a group of m robots sear hing for

a target of unknown lo ation onm rays in parallel where no lower bound on

then itis not ne essary toknowthe minimum distan e to the target.

However,usingthestrongerdenitionof ompetitiveratiothisspe ial asehas

been onsidered beforeforsear hing onthe lineeven with onlyone sear her

whi h requires somewhat arti ial assumptions on the strategies [7℄. These

assumptions are unne essary in our ase. Moreover, there are problems in

whi h no lower bound on the distan e to the target is known, for instan e,

when one wants to he k whether apolygon isstar-shaped ornot [14℄.

We begin by presenting a strategy that a hieves the ompetitiveratio

C k =1+2 (k+1) k+1 k k ;

wherek=dlogme.Wethenshowthat,infa t,nostrategy andobetterthan

this.

5.1 The Strategy

Theoptimalstrategyisamonotonestrategywherealltherobotsmove,oneon

ea hray,witha onstantspeed v.When onerobotndsthe targetitsear hes

for arobotatfull speed totellitwhere the targetis lo ated.Then they both

go at full speed to sear h for two more robots and tell them the lo ation of

the target, and so on. After ea h step the number of robots that know the

lo ation of the target is doubled. On e all robotsknow the lo ation,they all

movetothe target.Supposethe targetisonsomeray andatdistan e Dfrom

the origin. The strategy onsists of steps. Step i starts when 2 i

robots know

the lo ationtothetarget andends when2 i+1

robotsknowthe lo ationtothe

target;that is,in step i the 2 i

robots that urrently know the positionof the

target hase 2 i

of those robotsthat do not. In the last step k =dlogme only

m 2

k 1

robots sear hfortheremainingrobotswhereasthe restmovestothe

target.

Let T

i

denote the time it takes to omplete step i.It takes any of the robots

D=v time to nd the target, and when all robots know the lo ation of the

target,ittakesthemtimeT

F

togotothe target.Hen e, the ompetitiveratio

of the strategy is C = D=v+ P k 1 i=0 T i +T F D ;

Thetimethathaspassedinordertoinform2 i robotsisD=v+ i 1 j=0 T j .Hen e,

atthe end of step i 1 allthe robotshavea distan e of D+v P i 1 j=0 T j tothe origin. The 2 i

informed robots now start hasing 2 i

uninformed robots. In

order torea hthem, they have totravel adistan e of D+v P i 1 j=0 T j torea h

the origin, a distan e of D +v P

i 1

j=0 T

j

to rea h the lo ation of the hased

robot at the beginning of step i, and a distan e of vT

i

until the uninformed

robotsare nally rea hed. Hen e, T

i

isgiven by the equation

T i =2(D+v i 1 X j=0 T j )+vT i and T i = 2D+2v P i 1 j=0 T j 1 v = 2D+2v P i 2 j=0 T j 1 v + 2v 1 v T i 1 T i 1 1+v 1 v :

This is are urren e relationwith the solution

T i  T 0  1+v 1 v  i = 2D 1 v  1+v 1 v  i : We obtain k 1 X i=0 T i  2D 1 v k 1 X i=0  1+v 1 v  i = D v  1+v 1 v  k 1 ! :

By the above onsiderations the time T

F

it takes for the robots to get to

the target is the time of the last robot that is informed to go to the origin

D+v P i 1 j=0 T j

plus the distan e D tothe target

T F =D+v k 1 X i=0 T i +D  D  1+v 1 v  k +D:

So, the ompetitive ratioof the strategy is

C 1 v + 1 v  1+v 1 v  k 1 v +  1+v 1 v  k +1=1+  1 v +1  1+v 1 v  k =1+ 1 1 2k+1 +1 ! 1+ 1 2k+1 1 1 2k+1 ! k =1+2 (k+1) k+1 k k ;

if we set the speed v = 1

2k+1 .

Theorem 10 There is a monotone strategy that a hieves a ompetitive ratio of C k =1+2(k+1) k+1 =k k

, where k =dlogme,to sear honm raysin parallel

if no lower bound on the distan e to the target is known.

Sin eTheorem9remainsvalidif nolowerboundonthedistan e tothe target

is known, this strategy is an optimal monotone strategy (with or without

minimum distan e to the target). Butwe an showan even stronger resultif

nominimum distan e to the target isknown.

5.2 A Lower Bound

Theorem 11 There is no strategy at all that a hieves a better ompetitive

ratio than C k = 1+2(k+1) k+1 =k k

, where k =dlogme, to sear h on m rays

in parallel if no minimum distan e to the target is known.

PROOF. Anystrategy anbespe iedbytheaveragespeedfun tionsv

1 (T);

:::;v

m

(T)of them robots. (In onjun tionwithinformationaboutwhether a

robotswit hesray,atsomepointintime.)Giventheseaveragespeedfun tions,

anadversary anextra tinformationaboutthetimelengththatarobotmoves

monotoni ally along a ray. (This in ludes also the time that a robot stands

stillat the origin.)Let T

i

, for 1i m denotethe time that robot i moves

monotoni ally along a ray. Ea h T

i

is greater than 0 sin e either the robot

stands stillormoves alongsome ray. LetT

min =min 1im fT i g.

Now onsider the speeds of the robotsin the beginning. Assume that robot1

is (one of) the robot(s) that starts with the least speed v

min , that is, v min = lim T!0 v 1 (T). 1 For" >0, letT "

be the time su h that,for all 0<T T

" , (1) v 1 (T)v min +" and (2) v i (T)v min ", for all1im.

Theadversarynowpla esthetargetonray1atdistan eD=v

1 (T D )T D where T D = minfT " ;T min g C k ;

fork =dlogme. Ifthe strategy uses morethan C

k T

D

time,then the

ompeti-tive ratio is triviallybounded frombelow by C

k .

1

We assumethat this limit exists, for all 1 im.Hen e we disallow fun tions

k D

is monotone inthe interesting time intervaland we an apply a proof similar

tothe proof of Lemma 9. Onlynow T

n satises T n  T D  1+2 v min " 1 v min +"  dlogne :

This provesour laim. 2

6 Con lusions

We onsider sear h strategies for parallel sear h on m on urrent rays. We

showthat astraightforward generalizationofthe so alled doublingstrategy,

fromsear hingonthe linetosear hingonm on urrentrays,yieldsa

ompet-itive ratio of 9 if a minimum distan e fromthe origin to the target is known

inadvan e.Furthermore,weprovethat 9isalowerboundonthe ompetitive

ratio for both monotone and symmetri strategies inthis ase.

We alsoprove alowerbound of

1+2

(k+1) k+1

k k

onthe ompetitiveratio,wherek =dlogme,whi happliestomonotone

strate-gies and strategies for the ase when the minimum distan e from the origin

tothe target isnot known in advan e. Finally,we give a sear h strategythat

a hieves this ratio regardless of whether su h a minimum distan e is known

ornot, giving usanoptimal sear h strategy in the latter ase.

The question that remains unanswered is whether the lower bound of 9 an

be generalized tohold for any strategy.

A knowledgments

We would like to thank three anonymous referees for the areful reading of

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