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 1/55

2/55

3/55

4/55

5/55

6/55

7/55

8/55

9/55

Outline Model Definition 1 Problem Statement Time/Relay Model What to measure Manhattan Hopper Strategy 2 Strategy Description Static Scenario Performance Dynamic Scenario Performance Conclusion 3 10/55

Problem Statement 11/55

Problem Statement v n + 1 v 0 v 1 . . . Grid size: 0 . 5 Transmission distance: 1 11/55

Problem Statement Grid size: 0 . 5 Transmission distance: 1 11/55

Problem Statement Grid size: 0 . 5 Transmission distance: 1 11/55

Time/Relay Model 12/55

Time Model Synchronized Look – Compute – Move 12/55

Relay Model - Sensory Input α d 1 d 2 Sees its chain neighbors Memoryless No communication 13/55

Relay Model - Sensory Input α d 1 d 2 Sees its chain neighbors Memoryless No communication 13/55

Relay Model - Sensory Input α d 1 d 2 Sees its chain neighbors Memoryless No communication . . . must sense when predecessor has stepped 13/55

Relay Model - Movement Moves with constant speed 14/55

Relay Model - Movement Moves with constant speed Can be removed everywhere Inserted only at home 14/55

Chain Attributes Valid condition Optimal condition 15/55

What to measure 16/55

Static Scenario Explorer fixed Quality measurement: Time to optimal chain 16/55

Dynamic Scenario Chain in optimal condition Explorer moving Quality measurement: Possible speed of explorer Maximal chain length 17/55

What can we expect? Dynamic Scenario Explorer can move as fast as a relay constant 18/55

What can we expect? Dynamic Scenario Explorer can move as fast as a relay constant Chain length? O ( minimal length ) 18/55

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. 18/55

Strategy Description 19/55

Manhattan Hopper All stations move on a grid Chain remains valid Relays move at most constant distance Uses Manhattan distance 19/55

Manhattan Hopper All stations move on a grid Chain remains valid Relays move at most constant distance Uses Manhattan distance d = ∆ x + ∆ y 19/55

Manhattan Hopper Description v n + 1 v 0 v 1 . . . Executed sequentally. v i + 1 moves after v i One sequence is called a run 20/55

Manhattan Hopper Description Neighbors not in line → move Neighbors in line → stay 21/55

Manhattan Hopper Description v i + 2 v i + 1 v i If v i moves to v i + 2 . v i + 1 and v i + 2 are removed. 22/55

Manhattan Hopper Description v i + 2 v i + 1 v i If v i moves to v i + 2 . v i + 1 and v i + 2 are removed. v i + 1 and v i + 2 are removed. 22/55

Manhattan Hopper Description v i If v i moves to v i + 2 . v i + 1 and v i + 2 are removed. v i + 1 and v i + 2 are removed. A remove operation ends the run. 22/55

A little example 23/55

Static Scenario Performance 24/55

Static Scenario Theorem 1 After n runs, the chain has optimal length 24/55

Configuration � u i = position ( v i + 1 ) − position ( v i ) 25/55

Configuration � u i = position ( v i + 1 ) − position ( v i ) C = ( ⇒ , ⇑ , ⇑ , ⇒ , ⇒ , . . . , ⇒ , ⇑ , ⇑ ) = ( � u 0 , � u 1 , . . . , � u k ) 25/55

Configuration � u i = position ( v i + 1 ) − position ( v i ) C = ( ⇒ , ⇑ , ⇑ , ⇒ , ⇒ , . . . , ⇒ , ⇑ , ⇑ ) = ( � u 0 , � u 1 , . . . , � u k ) u i and � � u j are oppositional ↔ � u i = − � u j 25/55

Configuration � u i = position ( v i + 1 ) − position ( v i ) C = ( ⇒ , ⇑ , ⇑ , ⇒ , ⇒ , . . . , ⇒ , ⇑ , ⇑ ) = ( � u 0 , � u 1 , . . . , � u k ) u i and � � u j are oppositional ↔ � u i = − � u j Optimal (Manhattan) length configuration? 25/55

Static Scenario - Strategy Effects On Configuration Lemma 2 Let C = ( � u 0 , � u 1 , � u 2 . . . , � u k ) . Assume a run finishes without removing any relay. C ′ = ( � u 1 , � u 2 , . . . , � u k , � u 0 ) is the configuration after the run. Also afterwards � u 0 is not oppositional to any other. 26/55

Static Scenario - Strategy Effects On Configuration C = ( ⇒ , ⇑ � �� � , ⇑ , ⇒ , ⇒ , . . . , ⇒ , ⇑ , ⇑ ) 27/55

Static Scenario - Strategy Effects On Configuration C = ( ⇑ , ⇒ , ⇑ � �� � , ⇒ , ⇒ , . . . , ⇒ , ⇑ , ⇑ ) 28/55

Static Scenario - Strategy Effects On Configuration C = ( ⇑ , ⇑ , ⇒ , ⇒ � �� � , ⇒ , . . . , ⇒ , ⇑ , ⇑ ) 29/55

Static Scenario - Strategy Effects On Configuration C = ( ⇑ , ⇑ , ⇒ , ⇒ , ⇒ � �� � , . . . , ⇒ , ⇑ , ⇑ ) 30/55

Static Scenario - Strategy Effects On Configuration If � u 0 is oppositional to any other � u i , � u 0 will meet it at some point C = ( . . . , ⇒ , ⇐ � �� � , . . . ) triggers a removal 31/55

Static Scenario - Strategy Effects On Configuration Lemma 3 Let C = ( � u 0 , � u 1 , � u 2 . . . , � u k ) . The run finishes with removing v i and v i + 1 if and only if u i + 1 is the first vector oppositional to � � u 0 . C ′ = ( � u 1 , � u 2 , . . . , � u i , � u i + 2 , . . . � u k ) is the configuration after the run. 32/55

Static Scenario - Strategy Effects On Configuration C = ( ⇑ , ⇑ , ⇒ , ⇒ , ⇒ , ⇓ , . . . , ⇑ , ⇑ , ⇒ ) � �� � 33/55

Static Scenario - Strategy Effects On Configuration C = ( ⇑ , ⇑ , ⇒ , ⇒ , ⇒ , ⇓ , . . . , ⇑ , ⇑ , ⇒ ) � �� � 34/55

Static Scenario - Strategy Effects On Configuration C = ( ⇑ , ⇒ , ⇑ , ⇒ , ⇒ , ⇓ , . . . , ⇑ , ⇑ , ⇒ ) � �� � 35/55

Static Scenario - Strategy Effects On Configuration C = ( ⇑ , ⇒ , ⇒ , ⇑ , ⇒ , ⇓ , . . . , ⇑ , ⇑ , ⇒ ) � �� � 36/55

Static Scenario - Strategy Effects On Configuration C = ( ⇑ , ⇒ , ⇒ , ⇒ , ⇑ , ⇒ , ⇓ , . . . , ⇑ , ⇑ , ⇒ ) � �� � C ′ = ( ⇑ , ⇒ , ⇒ , ⇒ , ⇒ , ⇑ , ⇓ , . . . , ⇑ , ⇑ , ⇒ ) � �� � C ′′ = ( ⇑ , ⇒ , ⇒ , ⇒ , ⇒ , . . . , ⇑ , ⇑ , ⇒ ) 37/55

Static Scenario - Strategy Effects On Configuration Lemma 3 Let C = ( � u 0 , � u 1 , � u 2 . . . , � u k ) . The run finishes with removing v i and v i + 1 if and only if u i + 1 is the first vector oppositional to � � u 0 . C ′ = ( � u 1 , � u 2 , . . . , � u i , � u i + 2 , . . . � u k ) is the configuration after the run. 38/55

Static Scenario - Some Observations Vectors are never created, label them uniquely C 1 = ( � a 0 , � a 1 , . . . , � a k ) 39/55

Static Scenario - Some Observations Vectors are never created, label them uniquely C 1 = ( � a 0 , � a 1 , . . . , � a k ) In every run � u i ( i � = 0) reduces its position at least by one Case 1: No removal Case 2: Removal happens and � u i is before the removal Case 3: Removal happens and � u i is after the removal 39/55

Static Scenario Assume after n runs, there is an oppositional pair � u p and � u q with p < q . C = ( . . . , u p , . . . , u n ) � �� � Distance: n − p At most n − p + 1 runs earlier, � u p was at position 0 and hence would have been removed. 40/55

Static Scenario Assume after n runs, there is an oppositional pair � u p and � u q with p < q . C = ( . . . , u p , . . . , u n ) � �� � Distance: n − p At most n − p + 1 runs earlier, � u p was at position 0 and hence would have been removed. After n rounds, there are no more oppositional pairs. 40/55

Static Scenario It takes n rounds to reach minimal length. Timesteps? 41/55

Static Scenario It takes n rounds to reach minimal length. Timesteps? Pipeline! Start new run every 3 time steps. 41/55

Static Scenario It takes n rounds to reach minimal length. Timesteps? Pipeline! Start new run every 3 time steps. After 3 n + n = 4 n time steps the chain is optimal 41/55

Dynamic Scenario Performance 42/55

Dynamic Scenario Must handle explorer moves 42/55

Dynamic Scenario Must handle explorer moves Perform Follow run Then perform Hopper run The Hopper run is what we have seen before 42/55

Follow Run Explorer moves 43/55

Follow Run Relays follow 43/55

Follow Run Base inserts new relay 43/55

Follow Run 43/55

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. 44/55