The angular blowing-a-kiss 001 010 000 problem 011 111 Kevin - - PowerPoint PPT Presentation

001 010 011 100 101 110 111 000

The angular blowing-a-kiss

and Martijn Struijs

problem

Kevin Buchin, Irina Kostitsyna, Roel Lambers,

The angular blowing-a-kiss problem

• n oriented agents in the plane

The angular blowing-a-kiss problem

• n oriented agents in the plane
• stationary, but can rotate

The angular blowing-a-kiss problem

• n oriented agents in the plane
• stationary, but can rotate
• a pair of agents oriented towards eachother can scan

The angular blowing-a-kiss problem

• n oriented agents in the plane
• stationary, but can rotate
• a pair of agents oriented towards eachother can scan
• how long does it take to scan all pairs?

The angular blowing-a-kiss problem

• n oriented agents in the plane
• stationary, but can rotate
• a pair of agents oriented towards eachother can scan
• how long does it take to scan all pairs?

The angular blowing-a-kiss problem

• n oriented agents in the plane
• stationary, but can rotate
• a pair of agents oriented towards eachother can scan
• how long does it take to scan all pairs?

Related work

• kissing problem [Bender et al. 2014]
• angular freeze-tag [Fekete, Krupke 2018]
• scan cover for general graphs [Fekete, Kleist, Krupke 2020]

Problem description

• rotations take time proportional to their angle

π/2 time π time

Problem description

• rotations take time proportional to their angle

π/2 time π time

• goal: find a schedule that minimizes makespan

π/3 π/3 makespan: 2π/3

Problem description

• rotations take time proportional to their angle

π/2 time π time

• goal: find a schedule that minimizes makespan

π/3 π/3 makespan: 2π/3

• synchronous schedule: ⌊n/2⌋ simultaneous scans per round

synchronous asynchronous

• synchronous schedule: ⌊n/2⌋ simultaneous scans per round

synchronous asynchronous

• synchronous schedule: ⌊n/2⌋ simultaneous scans per round

synchronous asynchronous

• synchronous schedule: ⌊n/2⌋ simultaneous scans per round

synchronous asynchronous

Results

line/1D uniform circle general 2D async. π(⌈log n⌉ − 1) ∼ π log n π( 3

2⌈log n⌉ − 1 2)

sync. π(⌈log n⌉ − 1) 2π log n a – LB π(⌈log n⌉ − 1) ∼ π log n –

awhen n is a power of 2

Results

line/1D uniform circle general 2D async. π(⌈log n⌉ − 1) ∼ π log n π( 3

2⌈log n⌉ − 1 2)

sync. π(⌈log n⌉ − 1) 2π log n a – LB π(⌈log n⌉ − 1) ∼ π log n –

awhen n is a power of 2

Synchronous on a circle with n = 2k

for phase ℓ = 1, . . . , k: scan pairs that differ in ℓ-th bit 001 010 011 100 101 110 111 000 1-bit agents rotate clockwise 0-bit agents rotate counter-clockwise

Synchronous on a circle with n = 2k

for phase ℓ = 1, . . . , k: scan pairs that differ in ℓ-th bit 001 010 011 100 101 110 111 000 1-bit agents rotate clockwise 0-bit agents rotate counter-clockwise phase 1:

Synchronous on a circle with n = 2k

for phase ℓ = 1, . . . , k: scan pairs that differ in ℓ-th bit 001 010 011 100 101 110 111 000 1-bit agents rotate clockwise 0-bit agents rotate counter-clockwise phase 1:

Synchronous on a circle with n = 2k

for phase ℓ = 1, . . . , k: scan pairs that differ in ℓ-th bit 001 010 011 100 101 110 111 000 1-bit agents rotate clockwise 0-bit agents rotate counter-clockwise phase 1:

Synchronous on a circle with n = 2k

for phase ℓ = 1, . . . , k: scan pairs that differ in ℓ-th bit 001 010 011 100 101 110 111 000 1-bit agents rotate clockwise 0-bit agents rotate counter-clockwise phase 1:

Synchronous on a circle with n = 2k

for phase ℓ = 1, . . . , k: scan pairs that differ in ℓ-th bit 001 010 011 100 101 110 111 000 1-bit agents rotate clockwise 0-bit agents rotate counter-clockwise phase 1:

Synchronous on a circle with n = 2k

for phase ℓ = 1, . . . , k: scan pairs that differ in ℓ-th bit 001 010 011 100 101 110 111 000 1-bit agents rotate clockwise 0-bit agents rotate counter-clockwise phase 2:

Synchronous on a circle with n = 2k

for phase ℓ = 1, . . . , k: scan pairs that differ in ℓ-th bit 001 010 011 100 101 110 111 000 1-bit agents rotate clockwise 0-bit agents rotate counter-clockwise phase 2:

Synchronous on a circle with n = 2k

for phase ℓ = 1, . . . , k: scan pairs that differ in ℓ-th bit 001 010 011 100 101 110 111 000 1-bit agents rotate clockwise 0-bit agents rotate counter-clockwise phase 3:

Synchronous on a circle with n = 2k

for phase ℓ = 1, . . . , k: scan pairs that differ in ℓ-th bit 001 010 011 100 101 110 111 000 1-bit agents rotate clockwise 0-bit agents rotate counter-clockwise

• phase ℓ:

scan pairs at distance 2ℓ−1(2i + 1) for i = 0, . . . , 2k−ℓ − 1 phase 3:

Synchronous on a circle with n = 2k

for phase ℓ = 1, . . . , k: scan pairs that differ in ℓ-th bit 001 010 011 100 101 110 111 000 1-bit agents rotate clockwise 0-bit agents rotate counter-clockwise

• phase ℓ:

scan pairs at distance 2ℓ−1(2i + 1) for i = 0, . . . , 2k−ℓ − 1

• costs:

π per phase to scan pairs π between phases 2π log n in total phase 3:

Synchronous on a line

p1 p2 pn · · · pn−1 · · ·

Synchronous on a line

Algorithm: iteratively construct schedule Sn with rules:

• 1. Sk → Sk−1 if k even, with the same time
• 2. Sk → S2k, adding π time

base case: S2, with 0 time p1 p2 p1 p2 pn · · · pn−1 · · ·

Synchronous on a line

Algorithm: iteratively construct schedule Sn with rules:

• 1. Sk → Sk−1 if k even, with the same time
• 2. Sk → S2k, adding π time

base case: S2, with 0 time p1 p2 S2 Sn with ⌈log(n/2)⌉ applications of rule 2, so π(⌈log n⌉ − 1) time in total Time: p1 p2 pn · · · pn−1 · · ·

Synchronous on a line

• 1. Sk → Sk−1 if k even, in the same time

Synchronous on a line

• 1. Sk → Sk−1 if k even, in the same time

p1 p2 pk−1 · · · pk · · · pj round i of Sk:

Synchronous on a line

• 1. Sk → Sk−1 if k even, in the same time

p1 p2 pk−1 · · · pk · · · pj round i of Sk: p1 p2 pk−1 · · · · · · pj round i of Sk−1: bye: pj pk

Synchronous on a line

• 2. Sk → S2k, adding π time

Synchronous on a line

• 2. Sk → S2k, adding π time

p1 p2 · · · pk pj · · · Sk . . . . . . . . . . . .

Synchronous on a line

• 2. Sk → S2k, adding π time

p1 p2 · · · pk pj p′

1

p′

2

p′

k

p′

j

· · · · · · · · · Sk . . . . . . . . . . . . mirror(Sk) . . . . . . . . . . . .

Synchronous on a line

• 2. Sk → S2k, adding π time

p1 p2 · · · pk p′

1

p′

2

p′

k

· · · · · · · · · Sk . . . . . . . . . . . . mirror(Sk) . . . . . . . . . . . . pj p′

j

Synchronous on a line

• 2. Sk → S2k, adding π time

p1 p2 · · · pk p′

1

p′

2

p′

k

· · · · · · · · · Sk . . . . . . . . . . . . mirror(Sk) . . . . . . . . . . . . pj p′

j

Synchronous on a line

• 2. Sk → S2k, adding π time

p1 p2 · · · pk p′

1

p′

2

p′

k

· · · · · · · · · Sk . . . . . . . . . . . . mirror(Sk) . . . . . . . . . . . . pj p′

j

· · · · · · · · · · · · p1 p2 pk pj p′

1

p′

2

p′

k

p′

j

pj p′

j

final rounds: scan bipartite graph between left and right

Example: S2 → S4 → S3 → S6

p1 p2 S2

Example: S2 → S4 → S3 → S6

p1 p2 p3 p4 S4

Example: S2 → S4 → S3 → S6

p1 p2 p3 p4 S4 p1 p2 p3 p4 p1 p2 p3 p4

Example: S2 → S4 → S3 → S6

p1 p2 p3 p4 S4 p1 p2 p3 p4 p1 p2 p3 p4

Example: S2 → S4 → S3 → S6

p1 p2 p3 p1 p2 p3 p1 p2 p3 S3

Example: S2 → S4 → S3 → S6

p1 p2 p3 p1 p2 p3 p1 p2 p3 S6 p6 p5 p4 p6 p5 p4 p6 p5 p4

Example: S2 → S4 → S3 → S6

p1 p2 p3 p1 p2 p3 p1 p2 p3 S6 p6 p5 p4 p6 p5 p4 p6 p5 p4

Example: S2 → S4 → S3 → S6

p1 p2 p3 p1 p2 p3 p1 p2 p3 S6 p6 p5 p4 p6 p5 p4 p6 p5 p4 p1 p2 p3 p6 p5 p4 p1 p2 p3 p6 p5 p4

Example: S2 → S4 → S3 → S6

p1 p2 p3 p1 p2 p3 p1 p2 p3 S6 p6 p5 p4 p6 p5 p4 p6 p5 p4 p1 p2 p3 p6 p5 p4 p1 p2 p3 p6 p5 p4

Conclusion

Results: Work in progress:

• synchronous solutions for more general cases
• complexity of finding optimal solutions in 2D

line/1D uniform circle general 2D async. π(⌈log n⌉ − 1) ∼ π log n π( 3

2⌈log n⌉ − 1 2)

sync. π(⌈log n⌉ − 1) 2π log n a – LB π(⌈log n⌉ − 1) ∼ π log n –

awhen n is a power of 2

line/1D uniform circle general 2D async. π(⌈log n⌉ − 1) ∼ π log n π( 3

2⌈log n⌉ − 1 2)

sync. π(⌈log n⌉ − 1) 2π log n a – LB π(⌈log n⌉ − 1) ∼ π log n –

awhen n is a power of 2

line/1D uniform circle general 2D async. π(⌈log n⌉ − 1) ∼ π log n π( 3

2⌈log n⌉ − 1 2)

sync. π(⌈log n⌉ − 1) 2π log n a – LB π(⌈log n⌉ − 1) ∼ π log n –

awhen n is a power of 2