search and evacuation - unipi.itsbrinz.di.unipi.it/peppe/mac2017/presentations/5... · searcher as...

MAC, June 2017 1

Search and Evacuation

By

Evangelos Kranakis

School of Computer Science

Carleton University

Evangelos Kranakis, Carleton University, SCS

MAC, June 2017 2

To Start ...

• Concerned with the problem of optimal allocation of effort to

find a target.

– Search complete when first searcher finds the target!

– Evacuation complete when last searcher finds the target!

• Searcher(s) choose(s) how to allocate effort.

– Choice gives rise to an optimization problem.

• Adversary places the target.

– Target cannot evade and is primarily stationary.


MAC, June 2017 3

Outline

• Models

– Robots

– Communication

– Crash and Byzantine Faulty

• Search on a line.

– with non-faulty robots

– with faulty robots

• Evacuation on a cycle.

– with non-faulty robots

– with faulty robots


MAC, June 2017 4

Models


MAC, June 2017 5

Searcher as a Mobile Robot

• It is an autonomous vehicle represented as a point.

• It can move (turn, stop, restart, etc) with a maximum speed.

• It can process information and execute algorithms.

• It can communicate with other robots.

• It is equipped with a pedometer.


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Communication

• Two models of communication between the robots are

considered:

– (F2F) Face-to-Face (or local): robots exchange information

only when simultaneously located at the same point.

– (WiFi) Wireless: robots can communicate with each other

at any time regardless of their distance.

• The more general, CR(r) (communication range r) is not

discussed here.

– CR(0) is F2F.

– CR(∞) is WiFi.


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(Search) Domain

• 1D Search: Infinite Line.

– Search is on the line.

• 1.5D Search: Unit (Radius) Cycle.

– Search is on the perimeter.

– Robot can take short-cuts through interior of cycle.

• 2D Search: Infinite Plane (not discussed here):

– Search is on the plane.


MAC, June 2017 8

Faulty Robots

• Crash Faulty

– Is quiet during the search.

– Looks same as reliable robot.

– Follows its assigned trajectory.

– It may not detect the treasure, and may walk past it.

• Byzantine Faulty

– Is as an active liar during the search.

– May stay silent when it has found the treasure.

– May erroneously claim that it has found the treasure.

– In brief, it is malicious!


MAC, June 2017 9

Why Search and Evacuation?

• Does evacuation provide interesting insights?

– Think of search with faulty robots!

• How does evacuation differ from search?

– Depends on the domain, the communication, robot

behaviour!


MAC, June 2017 10

1D Search(with non-faulty robots)


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Standard Linear Search

• An infinite line.

• A robot is located at the origin O.

• The robot can move with speed 1 in either direction.

O Exit

• An “Exit” placed at an unknown (to robot) location on the

line.

– The exit may be to the left or to the right of the origin.

• What is a good algorithm to find the exit?


MAC, June 2017 12

If the Robot Knew . . .

• . . . in which direction (left or right) the Exit is located

O Exit

– The robot would choose the correct direction and move

towards the Exit (at speed 1).

• Therefore, without knowledge of its actual distance d

– the robot would reach the exit in time d, and this is, of

course, optimal.


MAC, June 2017 13

If the Robot Knew the Distance d (1/2)

• The robot does not know where the exit is (left or right) but

knows its distance d from the exit.O

d

• Consider the following search algorithm.

1. Start from the origin O; choose direction dir;

2. go in dir for distance d;

3. if exit not found;

4. set dir ← opposite dir;

5. walk for distance 2d;

• Since the robot can walk with speed 1 it succeeds in time 3d.


MAC, June 2017 14

If the Robot Knew the Distance d (2/2)

• If the robot knew where the exit were it could find it in time d.

Therefore the competitive ratio of the algorithm is at most3dd = 3.

• 3d is also a lower bound. Why?O

d−d

• If the robot spent time less than 3d searching then

– it couldn’t visit both endpoints −d, d, and

– if it visited only one of them an adversary would place the

exit in the other

• If the robot knew where the exit were it could find it in time d.

Therefore the competitive ratio of any algorithm is at least (and

hence also exactly) 3dd = 3.


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If the Robot Does not Know the Distance d

• Robot does not know neither the direction nor the distance!

• Previous strategy cannot work!

• New strategy is called zig-zag!O−d d


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What is a Good Way to Search?

• Robot does not know neither the direction nor the distance!O−d d

• Why such a strange zig-zag?

• What is the best zig-zag?


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Doubling Algorithm

• Robot starts at the origin and moves with speed 1 to find a

stationary target at unknown (to the robot) distance d from

the origin.

Zig-Zag Search Algorithm.

1. Start from the origin O; choose direction dir; start r = 1;

2. repeat until target is found;

(a) go in dir for r steps;

(b) return to origin;

(c) set dir ← opposite dir;

(d) set r ← 2r;


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Analysis of the Algorithm (1/2)

• Lemma 1 Robot finds the target in time at most 9d, where d

is the unknown distance of the robot from the target.

• Assume the target is at distance d from the origin.

• Let k be such that 2k < d ≤ 2k+1.

O

d−d

2k

2k+1

...

d

• The robot moves zig-zag and covers distance

2 · 20 + 2 · 21 + · · ·+ 2 · 2k.


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Analysis of the Algorithm (2/2)

• Since 2k < d, the robot could not have found the target.

• Regardless of starting dir the adversary can place target in

direction opposite where robot is when it finishes k + 1 round.

• After additional 2 · 2k+1 it has not found the target.

• Since d ≤ 2k+1 target will be found in additional d steps when

it changes direction. Therefore it takes

2 · 20 + 2 · 21 + · · ·+ 2 · 2k + 2 · 2k+1 + d

≤ 2 · 2k+2 + d ≤ 8 · 2k + d ≤ 9d

• Competitive ratio of this algorithm is 9dd = 9.

• Beck and Newman, 1970, proved that 9 is also a lower bound;

so the doubling strategy has optimal competitive ratio.


MAC, June 2017 20

Previous Work on this Problem

• Originally posed in [Bellman 1963] and Beck [1964]),

• Studied by [Beck and Newman, 1970], [Baeza-Yates, Culberson,

and Rawlins,1993], [Alpern and Gal, 2002] etc.

• Many variations have been studied:

– ray searching.

– cow path.

– searching with turn cost.

– searching with bounds on the distance.

– searching on a linear terrain.


MAC, June 2017 21

Can More Robots Do the Search Faster?

• Sometimes called parallel search as proposed in [Baeza-Yates

and Schott 1995] for the WiFi model.

– Can two robots with WiFi communication solve the

problem faster than 9d?


MAC, June 2017 22

Two Robots: Linear Search (WiFi)

• Divergent Search Algorithm :

1. The two robots start at the origin but go in opposite

direction;

2. The robot that finds the exit shouts: I found it!

O

r1r2

• Search succeeds in time d.


MAC, June 2017 23

1D Search(with faulty robots)


MAC, June 2017 24

Two Robots: Linear Search with a Faulty Robot (WiFi)

• Divergent search fails if one of the two robots is faulty.

• A correct algorithm would be for the non-faulty robot to ignore

the faulty robot!

• Two Robot Search with a Faulty Robot :

1. The non-faulty robot executes the zig-zag algorithm;

2. The non-faulty robot ignores the faulty robot;

• Competitive ratio is 9.


MAC, June 2017 25

Three Robots: Linear Search with a Faulty Robot (WiFi)

• Does it help when you have three robots one of which is faulty?

• Would a form of Divergent search work when one of the three

robots is faulty?

• Attempt 1:

Three Robot Divergent Search with a Faulty Robot :

1. Two of the robots go left one robot goes right;

• Attempt 2:

Three Robot Divergent Search with a Faulty Robot :

1. Robot1 goes left, Robot2 goes right; Robot3 stays still;

• It seems you must make all three robots move at the same time!


MAC, June 2017 26

Crash Faults

• Assume f of the n robots are faulty, and the remaining reliable.

• Search time

– Time the first reliable robot reaches the treasure.

– Since the first f robots to visit could be faulty, this is the

time the (f + 1)st robot reaches the treasure.


MAC, June 2017 27

Time-Space Representation of Trajectory

0(x ,t )

0

2(x ,t )

2

3(x ,t )

3

(x ,t )1 1

L

4(x ,t )

4

t

0

Figure 1: Time Space Representa-

tion

• The robot trajectory

is represented in time space.

• Robot trajectory:

Sequence of turning points

x0, x1, . . . of a robot is infinite.

• It is called zig-zag strategy.


MAC, June 2017 28

Zig-Zag strategy defined by Cβ

x-axis

t-axis

Cone Cβ

β

1

x = −βt x = βt

OL

(x0, t0)

(x1, t1)

(x2, t2)

Figure 2: Zig-Zag strategy

• We consider

zig-zag strategies where

all turning points lie on a

cone Cβ delimited by the lines

t = xβ for x ≥ 0 and

t = −xβ for x ≤ 0,

and where all the robots

(once they start moving)

travel at maximum speed 1.

• Given a value β and

a starting point (x0, t0) on Cβ ,

the trajectory of the unit-speed

robot is completely specified.


MAC, June 2017 29

Proportional schedule Sβ(n) within Cone CβL 0 x-axis

t-axis

β

1Cone CβA0

A1

A2

An

An+1

d

dr

An−1

drn−1

drn

B0

B1

B2

τ0

τ1

τ2

X

Y

Z

t0

t1

tn

tn+1

τn

B2


MAC, June 2017 30

Expansion Factor

• A zigzag trajectory defined by a cone Cβ has a fixed expansion

factor κ, that is: ∀i xi+1 = κxi.

– The doubling algorithm is defined by C3 and has expansion

factor 2.

• Lemma 2 Let (x0, t0) be the starting point of a robot on the

boundary of the cone Cβ. Then the turning points of the robot

are given by

xi = x0

Åβ + 1

β − 1

ãi(−1)i

with corresponding expansion factor (β + 1)/(β − 1).


MAC, June 2017 31

Multiple robots use the same cone

• In algorithms, all robots follow zigzag strategies defined by the

same cone Cβ .

Cβ

L

tCβ

L

tCβ

L

t

They all have the same expansion factor, but they have

different turning points.


MAC, June 2017 32

Proportional schedules

• Let λ0 < λ1 < λ2 . . . be the infinite sequence of positive turning

points of the (different) robots. We say that these constitute a

proportional schedule if for all i,

λi+1 − λiλi − λi−1

= r

We call r the proportionality ratio of the schedule.

• Lemma 3 Consider any β > 1 and n robots moving on the

line according to the proportional schedule Sβ(n). The

proportionality ratio of schedule Sβ(n) equals

r =

Åβ + 1

β − 1

ã2/n.


MAC, June 2017 33

Competitive ratio of proportional schedule Sβ(n)

• Definition 1 Denote by Tf+1(x) the time of the visit of point

x by the (f + 1)-st robot according to the schedule Sβ(n). Let

CRn,fβ = supx∈R

Tf+1(x)

|x|denote the competitive ratio of the proportional schedule Sβ(n)

when there are f faulty robots.

• Lemma 4 Consider n robots executing the proportional

schedule Sβ(n). If up to f robots may be faulty where

f < n < 2f + 2 then

CRn,fβ = (β + 1)2f+2

n (β − 1)1− 2f+2

n + 1.

• To obtain the main theorem. we choose β = 4f+4n − 1 to

minimize CRn,fβ .


MAC, June 2017 34

Search with Crash Faults [CKK+16](PODC 2016)

• Theorem 1 (Upper Bound) Consider a collection of n

robots up to f of which are faulty. Then the competitive ratio

of algorithm A(n, f) is at mostÅ4f + 4

n

ã 2f+2nÅ

4f + 4

n− 2

ã1− 2f+2n

+ 1.

• Theorem 2 (Lower Bound) The competitive ratio of any

algorithm for search by n < 2f + 2 robots, of which f are

faulty, is at least α for any α > 3 satisfying

(α− 1)n(α− 3) ≤ 2n+1

• Corollary 1 The optimal competitive ratio of a proportional

schedule for n = 2f + 1 robots, exactly f of which are faulty, is

at most 3 + 4 lnnn + O(1)

n and at least 3 + 2 lnnn − 2 ln lnn

n .


MAC, June 2017 35

Upper and lower bounds for specific values of n and f

n f competitive ratio lower bound on expansion factor

of A(n, f) competitive ratio of A(n, f)

2 1 9 9 2

3 1 5.24 3.76 4

3 2 9 9 2

4 1 1 1

4 2 6.2 3.649 3

4 3 9 9 2

5 1 1 1

5 2 4.43 3.57 6

5 3 6.76 3.57 2.67

5 4 9 9 2

11 5 3.73 3.345 12

41 20 3.24 3.12 42


MAC, June 2017 36

Additional Research: Byzantine Fauts

• Small number of robots

n, f Byz. UB Byz. LB Crash-UB Crash-LB

3, 1 9d 3.93d 5.24d 3.76d

4, 1 3d 3d d d

5, 1 2d 2d d d

6, 1 d d d d

5, 2 9d 3.57d 4.43d 3.57d

6, 2 4d 3d d d

• For larger values of n and f we express our results in terms of

the density β = fn and show how to extend our algorithms from

small values of n and f to this setting.

• Additional Information [CGK+16](ISAAC 2016)


MAC, June 2017 37

Additional Research

• Search in linear terrains.

O x-axis

• Search in the F2F Model

• Search with robots of different speed.


MAC, June 2017 38

1.5D Search(with non-faulty robots)


MAC, June 2017 39

Searching for an Exit

• A robot is located at some point in the plane.

• Exit is at distance 1 from the robot.

– Robot’s max speed is 1.

– Exact location of the exit is unknown to the robot.

• What is the best (time optimal) algorithm to find the exit?


MAC, June 2017 40

Evacuation from a Circle

• k ≥ 2 robots are initially located (anywhere) inside a disk.

exit

k robots

• Exit is located on the perimeter of the disk.

• To simplify the geometry: we assume all k robots start from

the center of a disk.


MAC, June 2017 41

Evacuation Problem

• k mobile robots located inside a disk of unit radius.

– The robots have the same (unit) maximal speed and start

at the centre of the disk.

– An exit is located at an unknown (to the robots) position

on the perimeter.

– The robots may communicate and share information about

the presence or absence of an exit and broadcast its position.

• The robots are required to evacuate the disk through this

unknown exit.

• The goal is to minimize the evacuation time (defined as the

time it takes the last robot to exit).


MAC, June 2017 42

Evacuation Time

• Robots do not necessarily evacuate at the same time.

– Robots can try to find the exit on their own!

– As soon as a robot finds the exit it tries to inform the rest!

• Measuring the complexity of an algorithm.

– Minimum time required so that all robots evacuate.


MAC, June 2017 43

Important Things to Know!

• Arc, Chord, Radius

exit

α

2 sin(α/2)

arc

radius

chord


MAC, June 2017 44

Two Algorithms: F2F vs WiFi [CGG+14](DISC 2014)

• Evacuation Algorithm for 2 robots.

x x

yy

αK

B

A

C

D

x x

A

B

c(x)

• NB: In both algorithms the robots take shortcuts!


MAC, June 2017 45

Upper bounds

• Theorem 3 (F2F Model) There is an algorithm for

evacuating the robots from an unknown exit located on the

perimeter of the disk which takes time

1 + α/2 + 3 sin(α/2)

where the angle α satisfies the equation cos(α/2) = −1/3. It

follows that the evacuation algorithm takes time ∼ 5.74.

• Theorem 4 (WiFi Model) There is an algorithm for

evacuating two robots from an unknown exit located on the

perimeter of the disk which takes time at most

1 + 2π3 +√

3 ≈ 4.826.


MAC, June 2017 46

Competitive Ratio of F2F Algorithm (1/2)

x x

yy

αK

B

A

C

D

• We calculate the time required until both robots reach the exit.

• Denote x = |BA| = |AC|, y = |BD| = |CD| and α = |BD|.• The total time required is f(α) = 1 + x+ 2y.

• Observe that α = 2x+ y, and y = 2 sin(α/2), because y is a

chord of the angle α.


MAC, June 2017 47

Competitive Ratio of F2F Algorithm (2/2)

• Substituting x and y in the function f we can express the

evacuation time as a function of the angle α as follows

f(α) = 1 +α− y

2+ 2y

= 1 +α

2+

3y

2

= 1 +α

2+ 3 sin(α/2).

• Now we differentiate with respect to α and we obtain:

df(α)

dα=

1

2+

3

2cos(α/2).

• Set the derivative equal to 0 to find the maximum of the

function f(α), which yields as value for α the solution of

cos(α/2) = −1/3.

• This proves Theorem 3.


MAC, June 2017 48

Two Lower bounds

• Proving lower bounds for two robots: F2F vs WiFi

A

K

B

CD

A

A1 A2

√3

√3

B

C


MAC, June 2017 49

Lower bounds

• Theorem 5 (F2F) It takes at least 3 + π4 +√

2 (∼ 5.199)

time units for two robots to evacuate from an unknown exit

located in the perimeter of the disk.

• Theorem 6 (WiFi) For any algorithm it takes at least

1 + 2π3 +√

3 time in the worst case for two robots to evacuate

from an unknown exit located in the perimeter of the disk.


MAC, June 2017 50

Not yet the Full Story!

Table 1: Recent F2F evacuation algorithms for 2 Robots

Paper Year Upper Bound Lower Bound

[CGG+14](DISC 2014) 5.74 5.199

[CGK+15](CIAC 2015) 5.628 5.255

[BLL+17](CIAC 2017) 5.625

More than 2 Robots

Table 2: Evacuation algorithms for more robots.

# Robots Lower/Upper (F2F) Lower/Upper (WiFi)

3 4.519/5.09 4.159/4.22

k →∞ ≈ 3 + 2πk ≈ 3 + π

k


MAC, June 2017 51

1.5D Search(with faulty robots)


MAC, June 2017 52

Three Robots One of Which is Faulty

• Three robots are located at the center of a unit cycle.

• WiFi communication model.

• One of the robots is (Byzantine or Crash) Faulty.

• Give an (optimal) evacuation algorithm to find the exit?


MAC, June 2017 53

Evacuation Algorithm (1/2)

• Robots go to the perimeter and walk for time 2π3

r1

r2

r3

2π3

2π3

2π3

A

B

C

• Several events may occur during this time . . . but an additional2π3 walk in same direction is needed.


MAC, June 2017 54

Evacuation Algorithm (2/2)

• After the initial searching for time 2π3 (left) robots r1, r2 and r3

r1

r2

r3

2π3

2π3

2π3

A

B

C

r1

r2

r3

x

y

• The robots continue searching for additional time 2π3 ;

• After that the robots perform the final evacuation movements

(right).


MAC, June 2017 55

Algorithm: Three Robots One Byzantine Faulty

1. Robot k goes to the circumference, at position 2πk3 , 1 ≤ k ≤ 3;

2. For time 2π3 follow the circumference CW;

3. If a robot claims to have found more than one exit, continue

execution of algorithm as though robot remained silent end;

4. If no information about exit then for additional time 2π3

follow perimeter CW and Investigate exit at unexplored

location. finish.

5. If one robot claims to have found exit go to closest part of

segment containing exit and explore entire segment. finish

6. If two robots claim to have found the exit, investigate both

exits. finish.

7. Inform all robots of the location of the exit.


MAC, June 2017 56

3 Robots, One Faulty: WiFi Communication

• Theorem 7 (Byzantine Faulty) 1 + 4π3 +√

3 ≈ 6.921 is an

upper bound for the evacuation problem of three robots one of

which is Byzantine faulty.

Theorem 8 (Crash Faulty) 6.309 is an upper bound for the

evacuation problem of three robots one of which is Crash faulty.

• Comparisons of evacuation costs ([CGG+17][SIROCCO 2017])

Table 3: Crash vs Byzantine Faults

Fault Lower Bound Upper Bound

Crash ≈ 5.082 ≈ 6.309

Byzantine ≈ 5.948 ≈ 6.921


MAC, June 2017 57

Additional Research

• Evacuation with n robots.

• Evacuation with n robots, f faulty.

• Search-and-Fetch [GKK16][Algosensors 2016], Treasure

Evacuation [GKK17](ICORES 2017).

• Evacuation with robots of different speed.

• Evacuation in squares and rectangles [CKK+15](ADHOCNOW

2015)


MAC, June 2017 58

References[BLL+17] S. Brandt, F. Laufenberg, Y. Lv, D. Stolz, and R. Wattenhofer. Collaboration without

communication: Evacuating two robots from a disk. In CIAC 2017. Springer, Athens,

Greece, 2017.

[CGG+14] J. Czyzowicz, L. Gasieniec, T. Gorry, E. Kranakis, R. Martin, and D. Pajak.

Evacuating robots from an unknown exit located on the perimeter of a disc. In DISC

2014. Springer, Austin, Texas, 2014.

[CGG+17] J. Czyzowicz, K. Georgiou, M. Godon, E. Kranakis, D. Krizanc, W. Rytter, and

M. Wlodarczyk. Evacuation from a disc in the presence of a faulty robot. In SIROCCO

2017, Porquerolles, France, 19-22 June 2017.

[CGK+15] J. Czyzowicz, K. Georgiou, E. Kranakis, L. Narayanan, J. Opatrny, and

B. Vogtenhuber. Evacuating robots from a disc using face to face communication. In

CIAC 2015. Springer, Paris, France, 2015.

[CGK+16] J. Czyzowicz, K. Georgiou, E. Kranakis, D. Krizanc, L. Narayanan, J. Opatrny, and

S. Shende. Search on a line with byzantine robots. In ISAAC. LIPCS, 2016.

[CKK+15] J. Czyzowicz, E. Kranakis, K. Krizanc, L. Narayanan, J. Opatrny, and S. Shende.

Wireless autonomous robot evacuation from equilateral triangles and squares. In

ADHOCNOW 2015. Springer, Athens, Greece, 2015.

[CKK+16] J. Czyzowicz, E. Kranakis, D. Krizanc, L. Narayanan, and Opatrny J. Search on a

line with faulty robots. In PODC. to appear, 2016.

[GKK16] K. Georgiou, G. Karakostas, and E. Kranakis. Search-and-fetch with one robot on a

disk. In Algorithms for Sensor Systems, ALGOSENSORS 2016, Aarhus, Denmark, August 25-26,

2016, pages 80–94, 2016.

[GKK17] K. Georgiou, G. Karakostas, and E. Kranakis. Search-and-fetch with 2 robots on a

disk: Wireless and face-to-face communication models. In ICORES 2017, Porto, Portugal,

volume abs/1611.10208, February 23-25, 2017.


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