Optimal strategies for maintaining a chain of relays between an - - PowerPoint PPT Presentation

Optimal strategies for maintaining a chain of relays between an explorer and a base camp

Lukas Humbel

• 2. Mai 2012

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Outline

1

Model Definition Problem Statement Time/Relay Model What to measure

2

Manhattan Hopper Strategy Strategy Description Static Scenario Performance Dynamic Scenario Performance

3

Conclusion

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Problem Statement

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Problem Statement

v0 v1. . . vn+1 Grid size: 0.5 Transmission distance: 1

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Problem Statement

Grid size: 0.5 Transmission distance: 1

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Problem Statement

Grid size: 0.5 Transmission distance: 1

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Time/Relay Model

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Time Model

Synchronized Look – Compute – Move

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Relay Model - Sensory Input

α d1 d2 Sees its chain neighbors Memoryless No communication

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Relay Model - Sensory Input

α d1 d2 Sees its chain neighbors Memoryless No communication

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Relay Model - Sensory Input

α d1 d2 Sees its chain neighbors Memoryless No communication . . . must sense when predecessor has stepped

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Relay Model - Movement

Moves with constant speed

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Relay Model - Movement

Moves with constant speed Can be removed everywhere Inserted only at home

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Chain Attributes

Valid condition Optimal condition

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What to measure

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Static Scenario

Explorer fixed Quality measurement: Time to optimal chain

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Dynamic Scenario

Chain in optimal condition Explorer moving Quality measurement:

Possible speed of explorer Maximal chain length

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What can we expect?

Dynamic Scenario

Explorer can move as fast as a relay constant

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What can we expect?

Dynamic Scenario

Explorer can move as fast as a relay constant Chain length? O( minimal length )

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What can we expect?

Dynamic Scenario

Explorer can move as fast as a relay constant Chain length? O( minimal length )

Static Scenario

There are cases where a (constant speed moving) relay needs n timesteps to get close to the direct line.

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Strategy Description

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Manhattan Hopper

All stations move on a grid Chain remains valid Relays move at most constant distance Uses Manhattan distance

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Manhattan Hopper

All stations move on a grid Chain remains valid Relays move at most constant distance Uses Manhattan distance d = ∆x + ∆y

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Manhattan Hopper Description

v0 v1. . . vn+1 Executed sequentally. vi+1 moves after vi One sequence is called a run

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Manhattan Hopper Description

Neighbors not in line → move Neighbors in line → stay

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Manhattan Hopper Description

vi vi+1 vi+2 If vi moves to vi+2. vi+1 and vi+2 are removed.

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Manhattan Hopper Description

vi vi+1 vi+2 If vi moves to vi+2. vi+1 and vi+2 are removed. vi+1 and vi+2 are removed.

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Manhattan Hopper Description

vi If vi moves to vi+2. vi+1 and vi+2 are removed. vi+1 and vi+2 are removed. A remove operation ends the run.

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A little example

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Static Scenario Performance

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Static Scenario

Theorem 1 After n runs, the chain has optimal length

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Configuration

• ui = position(vi+1) − position(vi)

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Configuration

• ui = position(vi+1) − position(vi)

C = (⇒, ⇑, ⇑, ⇒, ⇒, . . . , ⇒, ⇑, ⇑) = ( u0, u1, . . . , uk)

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Configuration

• ui = position(vi+1) − position(vi)

C = (⇒, ⇑, ⇑, ⇒, ⇒, . . . , ⇒, ⇑, ⇑) = ( u0, u1, . . . , uk)

• ui and

uj are oppositional ↔ ui = − uj

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Configuration

• ui = position(vi+1) − position(vi)

C = (⇒, ⇑, ⇑, ⇒, ⇒, . . . , ⇒, ⇑, ⇑) = ( u0, u1, . . . , uk)

• ui and

uj are oppositional ↔ ui = − uj Optimal (Manhattan) length configuration?

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Static Scenario - Strategy Effects On Configuration

Lemma 2 Let C = ( u0 , u1, u2 . . . , uk). Assume a run finishes without removing any relay. C′ = ( u1, u2, . . . , uk, u0 ) is the configuration after the run. Also afterwards u0 is not oppositional to any other.

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Static Scenario - Strategy Effects On Configuration

C = ( ⇒ , ⇑ , ⇑, ⇒, ⇒, . . . , ⇒, ⇑, ⇑)

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Static Scenario - Strategy Effects On Configuration

C = (⇑, ⇒ , ⇑ , ⇒, ⇒, . . . , ⇒, ⇑, ⇑)

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Static Scenario - Strategy Effects On Configuration

C = (⇑, ⇑, ⇒ , ⇒ , ⇒, . . . , ⇒, ⇑, ⇑)

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Static Scenario - Strategy Effects On Configuration

C = (⇑, ⇑, ⇒, ⇒ , ⇒ , . . . , ⇒, ⇑, ⇑)

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Static Scenario - Strategy Effects On Configuration

If u0 is oppositional to any other ui, u0 will meet it at some point C = (. . . , ⇒ , ⇐ , . . .) triggers a removal

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Static Scenario - Strategy Effects On Configuration

Lemma 3 Let C = ( u0 , u1, u2 . . . , uk). The run finishes with removing vi and vi+1 if and only if

• ui+1 is the first vector oppositional to

u0. C′ = ( u1, u2, . . . , ui, ui+2, . . . uk) is the configuration after the run.

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Static Scenario - Strategy Effects On Configuration

C = ( ⇑ , ⇑ , ⇒, ⇒, ⇒, ⇓, . . . , ⇑, ⇑, ⇒)

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Static Scenario - Strategy Effects On Configuration

C = (⇑, ⇑ , ⇒ , ⇒, ⇒, ⇓, . . . , ⇑, ⇑, ⇒)

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Static Scenario - Strategy Effects On Configuration

C = (⇑, ⇒, ⇑ , ⇒ , ⇒, ⇓, . . . , ⇑, ⇑, ⇒)

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Static Scenario - Strategy Effects On Configuration

C = (⇑, ⇒, ⇒, ⇑ , ⇒ , ⇓, . . . , ⇑, ⇑, ⇒)

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Static Scenario - Strategy Effects On Configuration

C = (⇑, ⇒, ⇒, ⇒, ⇑ , ⇒ , ⇓, . . . , ⇑, ⇑, ⇒) C′ = (⇑, ⇒, ⇒, ⇒, ⇒, ⇑ , ⇓ , . . . , ⇑, ⇑, ⇒) C′′ = (⇑, ⇒, ⇒, ⇒, ⇒, . . . , ⇑, ⇑, ⇒)

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Static Scenario - Strategy Effects On Configuration

Lemma 3 Let C = ( u0 , u1, u2 . . . , uk). The run finishes with removing vi and vi+1 if and only if

• ui+1 is the first vector oppositional to

u0. C′ = ( u1, u2, . . . , ui, ui+2, . . . uk) is the configuration after the run.

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Static Scenario - Some Observations

Vectors are never created, label them uniquely C1 = ( a0, a1, . . . , ak)

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Static Scenario - Some Observations

Vectors are never created, label them uniquely C1 = ( a0, a1, . . . , ak) In every run ui (i = 0) reduces its position at least by one

Case 1: No removal Case 2: Removal happens and ui is before the removal Case 3: Removal happens and ui is after the removal

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Static Scenario

Assume after n runs, there is an oppositional pair up and

• uq with p < q.

C = (. . . , up, . . . , un

• Distance: n−p

) At most n − p + 1 runs earlier, up was at position 0 and hence would have been removed.

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Static Scenario

Assume after n runs, there is an oppositional pair up and

• uq with p < q.

C = (. . . , up, . . . , un

• Distance: n−p

) At most n − p + 1 runs earlier, up was at position 0 and hence would have been removed. After n rounds, there are no more oppositional pairs.

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Static Scenario

It takes n rounds to reach minimal length. Timesteps?

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Static Scenario

It takes n rounds to reach minimal length. Timesteps? Pipeline! Start new run every 3 time steps.

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Static Scenario

It takes n rounds to reach minimal length. Timesteps? Pipeline! Start new run every 3 time steps. After 3n + n = 4n time steps the chain is optimal

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Dynamic Scenario Performance

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Dynamic Scenario

Must handle explorer moves

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Dynamic Scenario

Must handle explorer moves Perform Follow run Then perform Hopper run

The Hopper run is what we have seen before

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Follow Run

Explorer moves

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Follow Run

Relays follow

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Follow Run

Base inserts new relay

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Follow Run

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Dynamic Scenario Performance

Lemma 4 Let the chain have optimal length prior to the explorer’s

• movement. Then after the explorer’s movement, the Hopper

and Follow run bring the chain to an optimal length.

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Dynamic Scenario Performance

Lemma 4 Let the chain have optimal length prior to the explorer’s

• movement. Then after the explorer’s movement, the Hopper

and Follow run bring the chain to an optimal length. Proof. Let C be the configuration before the movement and C′ after the follow run.

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Dynamic Scenario Performance

Lemma 4 Let the chain have optimal length prior to the explorer’s

• movement. Then after the explorer’s movement, the Hopper

and Follow run bring the chain to an optimal length. Proof. Let C be the configuration before the movement and C′ after the follow run. No pair of oppositional vectors in C

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Dynamic Scenario Performance

Lemma 4 Let the chain have optimal length prior to the explorer’s

• movement. Then after the explorer’s movement, the Hopper

and Follow run bring the chain to an optimal length. Proof. Let C be the configuration before the movement and C′ after the follow run. No pair of oppositional vectors in C At most one pair of oppositional in C′

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Dynamic Scenario Performance

Lemma 4 Let the chain have optimal length prior to the explorer’s

• movement. Then after the explorer’s movement, the Hopper

and Follow run bring the chain to an optimal length. Proof. Let C be the configuration before the movement and C′ after the follow run. No pair of oppositional vectors in C At most one pair of oppositional in C′ One Hopper removes the first pair of oppositional vectors

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Dynamic Scenario Performance

Lemma 4 Let the chain have optimal length prior to the explorer’s

• movement. Then after the explorer’s movement, the Hopper

and Follow run bring the chain to an optimal length. Proof. Let C be the configuration before the movement and C′ after the follow run. No pair of oppositional vectors in C At most one pair of oppositional in C′ One Hopper removes the first pair of oppositional vectors Hence there is no pair at the end and hence the chain has

• ptimal length

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Dynamic Scenario Performance

dr := (Manhattan) distance between explorer and home at beginning of round r.

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Dynamic Scenario Performance

dr := (Manhattan) distance between explorer and home at beginning of round r. dr = 4.5 Number of relays = 9 Optimal chain: Number of relays = 2dr

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Dynamic Scenario Performance

Explorer speed?

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Dynamic Scenario Performance

Explorer speed? Must pipeline Time Explorer moves t Follow run started, v1 moves t + 1 Hopper run started, v1 moves t + 4 Explorer moves t + 6 One round

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Dynamic Scenario Performance

Theorem 5 Assume we start with an optimal chain. Then, the chain maintained by the strategy has the following properties before each round r.

1

The chain remains connected

2

The explorer may move a distance of 1

2 every round, i.e.

every 6th time step

3

Relays move at most constant distance per round

4

The number of relays used in the chain is at most 3dr + 2

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Dynamic Scenario Performance - Number Of Relays

Each Hopper run operates on an optimal chain.

Chain has 2dr relays. Run takes at most 2dr + 2 time steps.

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Dynamic Scenario Performance - Number Of Relays

Each Hopper run operates on an optimal chain.

Chain has 2dr relays. Run takes at most 2dr + 2 time steps.

Fix round r Number of relays ≤ 2dr + 2 (number of unfinished Hopper runs)

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Dynamic Scenario Performance - Number Of Relays

Run of round r Run of round r − 1 Run of round r − 2

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How Many Unfinished Hopper runs Are There?

Lemma 6 There are at most dr+1

2

unfinished runs in round r. ↔ The run started in round r − dr+1

2

is finished at round r

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How Many Unfinished Hopper Runs Are There?

r := current round z := earlier round Rounds r z Run started in round z

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How Many Unfinished Hopper Runs Are There?

r := current round z := earlier round Rounds r z Run started in round z

1

z < r − dr+1

2

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How Many Unfinished Hopper Runs Are There?

r := current round z := earlier round Rounds r z Run started in round z

1

z < r − dr+1

2

2

Run of round z needs < 2dz + 2 timesteps to finish

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How Many Unfinished Hopper Runs Are There?

r := current round z := earlier round Rounds r z Run started in round z

1

z < r − dr+1

2

2

Run of round z needs < 2dz + 2 timesteps to finish

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• Max. distance of explorer between z and r = r−z

2

→ dz ≤ dr + r−z

2

Run of round z ends in which round?

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Dynamic Scenario Performance

z < r − dr+1

2

Rounds r z Unfinished runs at r ? At most r − z = dr+1

2

many Number of relays ≤ 2dr + 2 (number of unfinished Hopper runs) Number of relays ≤ 2dr + 2dr+1

2

= 3dr + 1

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Dynamic Scenario Performance

z < r − dr+1

2

Rounds r z Unfinished runs at r ? At most r − z = dr+1

2

many Number of relays ≤ 2dr + 2 (number of unfinished Hopper runs) Number of relays ≤ 2dr + 2dr+1

2

= 3dr + 1 The strategy keeps chain length in O(dr)

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Outlook

Can be generalized (drop grid requirement) Keeps optimal characteristics

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Conclusion

The oscillation of the strategy and its sequential nature improve the Go-to-the-Middle strategy It converts a chain into an optimal in O(n) timesteps (n = number of relays)

Which is optimal

It allows the explorer to move with constant speed.

Which is optimal

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Questions?

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