Evacuaiton Problem: Group Search on the Line
Leszek Gąsieniec, Thomas Gorry, Russell Martin, Marek Chrobak
[email protected]
HTTP://WWW.CSC.LIV.AC.UK/~TGORRY/
The General Evacuation Problem
• 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.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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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
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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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.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Origins
The cow-path problem was introduced by Baeza-Yates, et al. in 1988 [1] .
“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.
The cow does not know the value of d
2.
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.
[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.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Origins
Baeza-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.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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One Mobile Entity with Uniform Speed
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].
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Multiple Mobile Entities with Uniform Speeds
Strategy 1: Ignore everyone else!
Strategy 2: Teamwork
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Multiple Mobile Entities with Uniform Speeds
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=
𝐝𝟏
𝐝𝟐
=
=
𝐭𝟏
𝐭𝟐
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 + d2
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Multiple Mobile Entities with Uniform Speeds
𝟏
α = 𝟑 ,gives
Theorem 1: Algorithm 2, with
an
evacuation procedure with time bound 9d,
where d is the distance from the origin to the
destination.
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.
𝒅𝟏
𝒅𝟐
α = 𝒕𝟏 = 𝒕𝟐 = 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 + d2
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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10 /24
Multiple Mobile Entities with Uniform Speeds
Proof 1:
t1 + t2 + 2d1 + d2 ≤ 9d1
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.
𝒅𝟏
𝒅𝟐
α = 𝒕𝟏 = 𝒕𝟐 = Speed used during initial exploration, and by the
second Mobile Entity until it is informed of the location of the
destination.
𝟏
Only satisfied when α = 𝟑
𝟏
So using an “exploration speed” of α = 𝟑 gives an evacuation
procedure for 2 Mobile Entities that works in 9d.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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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.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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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.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Two Mobile Entities with Different Speeds
Theorem 2: The strategy outlined for the SME
and FME gives a 9d bound for the evacuation
problem.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Two Mobile Entities with Different Speeds
• 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
𝟏
Entities walked together at s = to the
𝟑
evacuation point.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Two Mobile Entities with Different Speeds
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 some
0 < ε ≤ 3·2k−2.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Two Mobile Entities with Different Speeds
Proof 2:
Discovery phase time:
Inform phase time:
4
2
ε
÷
3
3
=2ε
Evacuation phase time: 2 ε ÷
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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1
=
3
6ε
17 /24
Two Mobile Entities with Different Speeds
Proof 2:
Therefore, the entire evacuation procedure (in the
𝟏
worst-case, with a speed for the SME) will take
𝟑
time at most 9d.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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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.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Three Mobile Entities with Different Speeds
• FME_1 = red
• FME_2 = green
• SME = blue
• Evacuation Point = Yellow
• FME_1 & FME_2 = orange
𝟏
1. The two FMEs explore the line with speed = 𝟑 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.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Three Mobile Entities with Different Speeds
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
𝟏
speed ≥ .
𝟓
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Three Mobile Entities with Different Speeds
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.
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Three Mobile Entities with Different Speeds
Proof 3:
Time for FME_1 to find d:
𝑑 ÷
1
3
Time to inform SME:
𝑑
Time for FME_1 to inform FME_2 and return to the Evacuation
Point:
5𝑑
Time for SME to get from the origin to the Evacuation Point:
𝑑 ÷ 𝛼
Therefore the speed 𝛼 of the SME must satisfy the following:
𝑑 ÷ 𝛼 ≤ 5d
𝟏
𝟓
This means that 𝜶 must be at least .
EVACUATION PROBLEM: GROUP SEARCH ON THE LINE - LESZEK GĄSIENIEC, THOMAS GORRY, RUSSELL MARTIN,, MAREK CHROBAK
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Summary
•
Thank You
𝟏
Using an exploration speed of α = 𝟑 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
𝟏
procedure (in the worst-case, with a 𝟑 speed for the SME) will take time at
most 9d.
•
𝟏
Therefore, the entire evacuation procedure (in the worst-case, with a 𝟓 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 Problem
J. 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
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