Unknown direction of movement but known speed of the bus

4.3.2 Unknown direction of movement but known speed of the bus

In this subsection we assume that the robots do not know thedirection ofmovement but know the speed s ofthe bus.

Theorem 8. If the robots know the speed s of the bus but do not know its direction ofmovement then the optimal search time is exactly

1. 2n/ks ifs > 1.

2. 2n/k ifs < 1 and k is even

3. Furthermore, ifs < 1 and k is odd, then the lower bound is 2ix/{k— 1) and the upper bound is

(a) % forse (^,1) (b) £ fors < j±ï

Proof. (Theorem 8) We prove separately the upper and lower bounds for ail cases in the statement ofthe theorem.

Upper bounds. Firstconsider Statement 1. Assume s > 1. Consider thefollowing algorithm.

Search Algorithm (Direction Unknown, Speed Known s > 1).

1. The robots are initiallyplaced on the perimeter ofthe cycle at distance

~ from each otherand wait motionless for the bus to arrive.

It is clear that the bus will meet one ofthe robots in time at most 2tr

ks'

This upper bound is valid regardless ofthe parity of k.

Next consider Statement 2. Recall that in this case k is even. Assume s < 1.

Consider the following algorithm.

Search Algorithm (Direction Unknown, Speed Known s < 1: k even).

1. The robots are initially placed in pairs on the perimeter of the cycle at distance y fromeach other;

2. The robots in each pair move in opposite directions.

For k even, theresulting distance between pairsis exactly = y- Observe that the bus is located between two robots moving against each other. Since these two robots will meet no later than in time ^

algorithm will be

Y, the resulting running time for this

2tr k

Finally, consider the case of k odd. We evaluate two algorithms and choose the best, depending on s. The first option is to use the algorithms for speed larger than 1, i.e. spread the agents evenly and wait until the bus meets you. The meeting time is 2n/ks in such case, The second option is to use the following algorithm:

Figure 4.7: The algorithm for odd numberofagents. Here k = 5.

Search Algorithm (Direction Unknown, Speed Known s < 1: k odd) 1. Let X = _k—s and Y= _k—1

2. k+1 robots are initially placed in pairs on the perimeter ofthe cycle at distance Y from each other; One robot at, a node u neighbouring a segment oflength X is then removed to bringdown the number ofrobots used to k (see Figure 4.7)

3. The robots in each pair move in oppositedirection, the lone robot moves awav from the X segment.

Observethatifthe busstartedina Fsegment,tworobotswillbetravelingtowards each other from the opposite ends of this segment and will meet it at time at most Y/2. Ifthe bus started in the Xsegment, the lone robot Crossingthe X segment will catch it in time at most X was selected so that these two times are equal:

2ir(k-s)~27t(1—s)

Y 2ir-X _k—s

T— —

k - 1

2 k-l

2ir(k — 1)

(k — s)(k— 1) k —s

2z X

1 — s

For s > y5 thebound 2ir/kofthewaiting algorithm isbetter, while for s <

this algorithm yields a better bound of

S+Vi(t) '—s

Vi(t)+s -s

*^>i(t) Vi{t)-S 's

/ // /

t=0

S~Vi(t)

t=0

Figure4.8: Sizesofexcluded régions. Left: case 5 > 1. Right: case s < 1.

Lower bounds. Firstconsider thelowerbound inStatement 1. Assumethat s > 1.

Let Vi(t) dénoté the speed ofthe z-th robot at time t as it is searching for the bus.

R,ecall that alwaysu,(i) < 1 andhence also Uj(f) < s. Further, consider themovement ofthe robot at an arbitrary time dt. Let us express how much ofthe possible initial bus positions can the robot exclude in time dt, i.e. if the robot does not meet the bus at time interval dt, then the bus could not hâve started at those positions (see Figure4.8). Summingup thesizeofexluded régions forboth busdirectionswe obt.ain

(s+Vi(t))dt+(s —Vi(t))dt = 2sdt.

Let T be the time it takesfor at least one ofthe robots to find the bus according to the execution ofan optimal search algorithm. Therefore in time T, the i-th robot can coverat most, (ifat everv time moment it excluded different régions) length

J

f

Thus, ail k robots taken together can cover at most a length of 2Tks, and only ifail of them cover different areas. However, this last quantity must be at least 4zr (2n forclockwise and another 2tt forcounterclockwise busdirections, otherwisethere isa trajectoryofthe buswhichwillescapetherobots’search). Itfollows that 2Tks > 4ir, which yields T > 2ir/ks. This proves the lower bound in Statement 1.

Now consider the lower bound in Statement 2. Assume that s < 1. Let Vi(t) dénoté the speed of the i-th robot at time t as it is searching for the bus. Using a similar argument we observe that the z-th robot in time dt covers a length equal to

(s+Vi(t))dt4-(vi(t) —s)dt =2Vi(t)dt.

Therefore in time T, the i-th robot covers a length

/ 2Vi{t)dt < 2T,

•/o

where the last inequality is valid since the speed of the robot never exceeds 1, It follows that k robots can cover a length ofat most 2Tk. However, this last quantity must be at least 4zr (otherwise there is a trajectory ofthe bus which will escape the robots’ search). It follows that 2Tk > 47T, which yields T > 2ir/k. This proves the lower bound in Statement 2.

The lower bound for the case of .s < 1 and odd k follows directly from the lower bound of Statement 2 byignoring the last robot.

The proofofTheorem 8is now complété.

□

Remark 1. We note the open problem arisingfrom thefact that the lower bound in Theorem 8 is not tightfor k odd.

4.3.3 Robots know neither the direction nor the speed of the