evacuaiton problem: group search on the line leszek gąsieniec, thomas gorry, russell martin, marek...

Evacuaiton Problem: Group Search on the LineLeszek Gąsieniec, Thomas Gorry, Russell Martin, Marek Chrobak

[email protected] HTTP://WWW.CSC.LIV.AC.UK/~TGORRY/

The General Evacuation Problem

2 /24 EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK HTTP://WWW.CSC.LIV.AC.UK/~TGORRY/

• There are K Mobile Entities located at some point of origin.

• They are tasked with locating an Evacuation Point.

• Once located all K Mobile Entities must occupy the Evacuation Point simultaneously.

Possible Models• Environment settings

• Fixed or mobile target

• Randomised or Deterministic strategies

• Variance of communication

• Number of Mobile Entities

• Variance on speed of Mobile Entities


Group Search on the Line• K Mobile Entities all initially located at a point of origin on a line.

• Evacuation Point is fixed at an unknown direction and distance d from the origin point.

• Communication is limited to when two or more Mobile Entities occupy a location simultaneously.

AIM: All Mobile Entities must simultaneously occupy the Evacuation Point.


OriginsThe cow-path problem was introduced by Baeza-Yates, et al. in 1988 [1] .

1. The cow does not know the value of d2. Does not know which of the w paths leads to the goal

3. The cow’s eyesight is not very good, so it will not know it has found the goal until it is standing on it.


“A single cow stands at a crossroads (defined as the origin) with w paths leading off into unknown territory. Traveling with unit speed, the goal of the cow is to locate a

destination that is at distance d from the origin in as little time as possible.”

[1] R.A. Baeza-Yates , J.C. Culberson , and G.J.E. Rawlins, Searching with uncertainty, Proc. SWAT 88: 1st Scandinavian workshop on algorithm theory, no. 318 pp. 176–189, 1988.

OriginsBaeza-Yates, et al. [1, 2] studied the cow-path problem, and proposed a deterministic algorithm as a solution.

In the case that w = 2 (two paths), their algorithm will find the goal in time at most 9d and that this is optimal up to lower order terms.


In the same work, the authors considered the case of w > 2 paths, showing they could find the destination with an optimal (up to lower order terms) result of

[2] R.A. Baeza-Yates , J.C. Culberson , and G.J.E. Rawlins, Searching in the plane, Information and Computation, vol. 106, no. 2, pp. 234–252, 1993.

One Mobile Entity with Uniform Speed

7/ 24 EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK HTTP://WWW.CSC.LIV.AC.UK/~TGORRY/

This deterministic search strategy for a single Mobile Entity yields the search time of 9d, which is optimal up to lower order terms [2, Theorem 2.1].

Multiple Mobile Entities with Uniform Speeds


Strategy 1: Ignore everyone else!Strategy 2: Teamwork


d1 = The distance from the origin to the destination (= d).

t1 = Time to discover the destination by one Mobile Entity.

t2 = Additional time for this Mobile Entity to inform the other Mobile Entity.

d1 + d2 = The distance from the origin where the two Mobile Entities will meet in this scenario.

α = = = Speed used during initial exploration, and by the second Mobile Entity until it is informed of the location of the destination.

Total Evacuation Time is t1 + t2 + 2d1 + d29 /24 EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK

HTTP://WWW.CSC.LIV.AC.UK/~TGORRY/


d1 = The distance from the origin to the destination (= d).

t1 = Time to discover the destination by one Mobile Entity.

t2 = Additional time for this Mobile Entity to inform the other Mobile Entity.

d1 + d2 = The distance from the origin where the two Mobile Entities will meet in this scenario.


Total Evacuation Time is t1 + t2 + 2d1 + d210 /24

Theorem 1: Algorithm 2, with α = gives an evacuation procedure with time bound 9d, where d is the distance from the origin to the destination.

EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK HTTP://WWW.CSC.LIV.AC.UK/~TGORRY/


d1 = The distance from the origin to the destination (= d). t1 = Time to discover the destination by one Mobile Entity. t2 = Additional time for this Mobile Entity to inform the

other Mobile Entity. d1 + d2 = The distance from the origin where the two

Mobile Entities will meet in this scenario.


Only satisfied when α = So using an “exploration speed” of α = gives an evacuation

procedure for 2 Mobile Entities that works in 9d.

11 /24

Proof 1:

t1 + t2 + 2d1 + d2 ≤ 9d1

EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK HTTP://WWW.CSC.LIV.AC.UK/~TGORRY/

Two Mobile Entities with Different Speeds• K = 2• Different maximum speeds. (maximum speed of the faster Mobile Entity is 1, and the speed of the second Mobile Entity is some value 0 < s < 1).

In certain situations, an evacuation time of 9d is still achievable.


Two Mobile Entities with Different Speeds

• FME = red

• SME = blue

• Yellow lines = turning points in the SME’s trajectory are the turning points of the FME from its previously completed stage in its own trajectory.

• During the kth stage of the FME, it is exploring up to a distance of 2k (on one side of the origin).

• FME and SME will meet at distance 2k−2 from the origin,

• After which the FME is exploring virgin territory up to a distance of 2k from the origin.

13 / 24EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK HTTP://WWW.CSC.LIV.AC.UK/~TGORRY/


Theorem 2: The strategy outlined for the SME and FME gives a 9d bound for the evacuation problem.



• Figure 3 shows in detail the paths taken by the two Mobile Entities once the evacuation point has been found.

• Evacuation point = orange line.

• The path the FME took = red

• The path that the SME took = blue

• Originally intended paths shown as a dashed lines.

• The purple line = where the two Mobile Entities walked together at s = to the evacuation point.



Proof 2:

Think of the evacuation procedure as a three-step process:

1. The FME locates the evacuation point

2. The FME informs the SME of that location

3. The two entities proceed (back) to the evacuation point.

Assume that d ≥ 2. (The 9d bound for small d is easy to verify.)k = The integer such that 2k−2 < d ≤ 2k. In particular, we can write d = 2k−2 + ε for some0 < ε ≤ 3·2k−2.



Proof 2:

Discovery phase time:

Inform phase time: = 2 ε Evacuation phase time: 2 ε = 6 ε



Proof 2:

Therefore, the entire evacuation procedure (in the worst-case, with a speed for the SME) will take

time at most 9d.


Three Mobile Entities with Different Speeds• K = 3• Different maximum speeds. (maximum speed of two Mobile Entities is 1, and the speed of the third Mobile Entity is some value 0 < s < 1.

In certain situations, an evacuation time of 9d is still achievable.


Three Mobile Entities with Different Speeds

• FME_1 = red

• FME_2 = green

• SME = blue

• Evacuation Point = Yellow

• FME_1 & FME_2 = orange

1. The two FMEs explore the line with speed = while the SME remains stationary at the origin.

2. Once a FME finds the Evacuation Point they sprint with speed = 1 to tell the other Mobile Entities.

3. The SME learns of the location of the Evacuation Point and proceeds there with its maximum speed.

4. As before once FME_2 is notified by FME_1 they sprint with their maximum speed of 1 to the Evacuation Point.



Theorem 3: The strategy outlined for the SME, FME_1 and FME_2 gives a 9d bound for the evacuation problem if the SME travels with a speed .



Proof 3:

Again think of the evacuation procedure as a three step process:

1. FME_1 locates the evacuation point

2. FME_1 informs the SME of that location as it passes on its way to inform FME_2.

3. FME_1 and FME_2 then proceed (back) to the evacuation point.

Also from Proof 1 we know that this strategy for FME_1 and FME_2 gives 9d.

Therefore, the SME must simply move 1d at least as fast as the FME_1 can get the FME_2 and return to the Evacuation Point.



Proof 3:

Time for FME_1 to find d:

Time to inform SME:

Time for FME_1 to inform FME_2 and return to the Evacuation Point:

Time for SME to get from the origin to the Evacuation Point:

Therefore the speed of the SME must satisfy the following: 5dThis means that must be at least .23 / 24EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK

HTTP://WWW.CSC.LIV.AC.UK/~TGORRY/

Thank You

Summary• Using an exploration speed of α = gives an evacuation procedure for 2

Mobile Entities that share the same maximum speed of 1 that works in 9d.

• For two Mobile Entities with different speeds the entire evacuation procedure (in the worst-case, with a speed for the SME) will take time at

most 9d.

• Therefore, the entire evacuation procedure (in the worst-case, with a speed for the SME) will take time at most 9d when there are three Mobile

Entities, two with the maximum speed of 1 and one with a slower speed.

Other work on the Evacuation ProblemJ. Czyzowicz, L. Gąsieniec, T. Gorry, E. Kranakis, R. Martin, D. Pająk, Evacuating

Robots from an Unknown Exit in a Disk.

EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK HTTP://WWW.CSC.LIV.AC.UK/~TGORRY/ 24 / 24

evacuaiton problem: group search on the line leszek gąsieniec, thomas gorry, russell martin, marek...

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