cbms lecture 2b sensor network coverage · cbms lecture 2b june 12, 2017 theorem (vds, ghrist,...
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CBMS Lecture 2B Sensor Network Coverage
Vin de SilvaPomona College
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Sensor networks
‣ We deploy a large number of independent robotic agents‣ dozens, hundreds, thousands, ...
‣ Each robot has limited physical and computational capabilities‣ optical/aural sensing
‣ locomotion
‣ communication with nearby robots
‣ Attempt to solve global problems using local algorithms‣ each robot has simple behaviour rules
‣ “whole is greater than sum of parts”
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The coverage problem
‣ 2D domain bounded by fence
‣ Robots populate the domain
‣ Each robot has a coverage area‣ signal transmission
‣ surveillance
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The coverage problem
‣ 2D domain bounded by fence
‣ Robots populate the domain
‣ Each robot has a coverage area‣ signal transmission
‣ surveillance
Is the entire domain covered?
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Attacking the coverage problem
‣ Exact knowledge of domain shape‣ “exploring the known”
‣ Exact knowledge of robot positions‣ e.g. using GPS systems
‣ Centralised information gathering and computation‣ “mission control”
What could we use to solve the problem?
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Attacking the coverage problem
‣ Exact knowledge of domain shape‣ “exploring the known”
‣ Exact knowledge of robot positions‣ e.g. using GPS systems
‣ Centralised information gathering and computation‣ “mission control”
What could we use to solve the problem?
using topology
unknownunknown domain shape‣ with mild constraints
crude proximity information‣ identify nearby robots and fence
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Theorem (VdS, Ghrist, Muhammad 2005)
‣ Assumptions‣ The coverage area of each robot is a
circular disk of radius rc
‣ Each robot can identify all robots which are near it (distance ≤ rs)
‣ Each robot can identify all robots which are at midrange (distance ≤ rw)
‣ Each robot knows if it is close to the fence (distance ≤ rf)
‣ rc ≥ rs√(1/3) and rw ≥ rs√(13/3)
‣ The domain is not “pinched”
‣ The fence is “not too wiggly”
‣ Conclusion‣ There is a test for global coverage which
gives no false positives, and “not too many” false negatives
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Theorem (VdS, Ghrist, Muhammad 2005)
‣ Assumptions‣ The coverage area of each robot is a
circular disk of radius rc
‣ Each robot can identify all robots which are near it (distance ≤ rs)
‣ Each robot can identify all robots which are at midrange (distance ≤ rw)
‣ Each robot knows if it is close to the fence (distance ≤ rf)
‣ rc ≥ rs√(1/3) and rw ≥ rs√(13/3)
‣ The domain is not “pinched”
‣ The fence is “not too wiggly”
‣ Conclusion‣ There is a test for global coverage which
gives no false positives, and “not too many” false negatives
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Theorem (VdS, Ghrist, Muhammad 2005)
‣ Assumptions‣ The coverage area of each robot is a
circular disk of radius rc
‣ Each robot can identify all robots which are near it (distance ≤ rs)
‣ Each robot can identify all robots which are at midrange (distance ≤ rw)
‣ Each robot knows if it is close to the fence (distance ≤ rf)
‣ rc ≥ rs√(1/3) and rw ≥ rs√(13/3)
‣ The domain is not “pinched”
‣ The fence is “not too wiggly”
‣ Conclusion‣ There is a test for global coverage which
gives no false positives, and “not too many” false negatives
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Theorem (VdS, Ghrist, Muhammad 2005)
‣ Assumptions‣ The coverage area of each robot is a
circular disk of radius rc
‣ Each robot can identify all robots which are near it (distance ≤ rs)
‣ Each robot can identify all robots which are at midrange (distance ≤ rw)
‣ Each robot knows if it is close to the fence (distance ≤ rf)
‣ rc ≥ rs√(1/3) and rw ≥ rs√(13/3)
‣ The domain is not “pinched”
‣ The fence is “not too wiggly”
‣ Conclusion‣ There is a test for global coverage which
gives no false positives, and “not too many” false negatives
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
What does the global test look like?
We construct vector spaces V and Wand a linear map T: V→W
T ≠ 0 : coverage predicted
T = 0 : coverage not predicted
Validity of the test‣ No false positives. Coverage is guaranteed when T ≠ 0.
Slightly fiddly proof using algebraic topology.
‣ Some false negatives: when T = 0 coverage may happen (but is not robust).Easy to construct. Showing that false negatives are unstable requires some technology.
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The 1-dimensional problem
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The 1-dimensional problem
‣ Represent the configuration with a graph‣ Represent each robot by a node of a graph.
‣ If two robots are close (distance ≤ R = 2r) join the nodes by an edge.
‣ If a robot is close (distance ≤ r) to the left boundary, colour its node BLUE.
‣ If a robot is close (distance ≤ r) to the right boundary, colour its node RED.
‣ Solve a combinatorial problem on the graph‣ Does there exist a path from BLUE to RED along edges of the graph?
‣ Exercise: we have coverage if, and only if, such a path exists.
Solution (assuming the robots can tell left from right)
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The 1-dimensional problem: marked boundary
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The 1-dimensional problem: marked boundary
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The 1-dimensional problem: marked boundary
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The 1-dimensional problem: marked boundary
no path
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The 1-dimensional problem
‣ Represent the configuration with a graph‣ Represent each robot by a node of a graph.
‣ If two robots are close (distance ≤ R = 2r) join the nodes by an edge.
‣ If a robot is close (distance ≤ r) to the left boundary, colour its node BLUE.
‣ If a robot is close (distance ≤ r) to the right boundary, colour its node RED.
‣ Solve a combinatorial problem on the graph‣ Does there exist a path from BLUE to RED along edges of the graph?
‣ Exercise: we have coverage if, and only if, such a path exists.
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The 1-dimensional problem
‣ Represent the configuration with a graph‣ Represent each robot by a node of a graph.
‣ If two robots are close (distance ≤ R = 2r) join the nodes by an edge.
‣ If a robot is close (distance ≤ r) to the left boundary, colour its node BLUE.
‣ If a robot is close (distance ≤ r) to the right boundary, colour its node RED.
‣ Solve a linear algebra problem‣ Select a red robot A and a blue robot B.
‣ Let C0 be the vector space generated by the robots.
‣ Let C1(R) be the vector space generated by edges of length ≤ R between robots.
‣ Let ∂: C1(R) → C0 denote the boundary map.
‣ Theorem: full coverage ⇔ the equation [A]-[B]=∂γ has a solution γ ∈ C1(R).
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
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Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
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solution
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
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Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
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inconsistent!
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
‣ sensors scattered in polygonal domain
‣ ‘guards’ posted on the boundary‣ cyclic sequence b0, b1, b2, ..., bn -1, bn = b0
‣ consecutive guards are close to each other:
|bi-1 - bi| ≤ R (the communication radius)
‣ radial coverage problem‣ x is covered if it lies within r = R/√3 (the
coverage radius) of a sensor
‣ Vietoris–Rips coverage problem‣ x is covered if it lies inside a triangle of sensors
with all three sides at most R.
b0=bn
b1b2b3
bn-1
Simplified 2D coverage: controlled boundary
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Global test for Vietoris–Rips coverage
‣ No false positives. Coverage is guaranteed when γ = Tσ has a solution σ ∈ V.Easily proved using what we have done in class.
‣ No false negatives. Incomplete coverage when γ = Tσ has no solution.Difficult to prove.
Validity of the test
We construct vector spaces V and W,linear map T: V→W, and vector γ∈W
γ ∈ Im(T) : coverage predicted
γ ∉ Im(T) : coverage not predicted
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
‣ W = C1(R), the vector space generated by the edges of length ≤ R.
‣ V = C2(R), the vector space generated by triangles with sides of length ≤ R.
‣ T is the boundary map ∂: C2(R) → C1(R).
Global test for Vietoris–Rips coverage
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
‣ W = C1(R), the vector space generated by the edges of length ≤ R.
‣ V = C2(R), the vector space generated by triangles with sides of length ≤ R.
‣ T is the boundary map ∂: C2(R) → C1(R).
‣ γ is the 1-cycle [b0, b1]+[b1, b2]+...+ [bn-1, b0] ∈ C1(R), which goes around the fence.
Global test for Vietoris–Rips coverage
Does there exist a 2-chain σ ∈ C2(R) such that γ=∂σ?
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
‣ W = C1(R), the vector space generated by the edges of length ≤ R.
‣ V = C2(R), the vector space generated by triangles with sides of length ≤ R.
‣ T is the boundary map ∂: C2(R) → C1(R).
‣ γ is the 1-cycle [b0, b1]+[b1, b2]+...+ [bn-1, b0] ∈ C1(R), which goes around the fence.
Global test for Vietoris–Rips coverage
Does there exist a 2-chain σ ∈ C2(R) such that γ=∂σ?
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Global test for Vietoris–Rips coverage
Does there exist a 2-chain v ∈ C2(R) such that γ=∂σ?
γ = [b0, b1]+[b1, b2]+...+ [bn-1, b0] ∈ C1(R)
If YES:
⇒ |σ| contains every x with w(γ,x) ≠ 0
⇒ |σ| contains all points in the interior of the polygonal domain
⇒ coverage!
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Global test for radial coverage
‣ No false positives. Coverage is guaranteed when γ = Tσ has a solution σ ∈ V.Easily deduced from Vietoris–Rips test, using the √3 Lemma (next slide).
‣ Some false negatives. Coverage may happen even if γ = Tσ has no solution.Easy to construct.
Validity of the test
We construct vector spaces V and W,linear map T: V→W, and vector γ∈W
γ ∈ Im(T) : coverage predicted
γ ∉ Im(T) : coverage not predicted
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The √3 Lemma
‣ Definition An R-triangle is a triangle where all three sides have length ≤ R.
‣ Lemma If three robots lie at the vertices of an R-triangle, and if r ≥ R/√3 then the three coverage disks of radius r meet and cover the entire triangle.
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The √3 Lemma
‣ Definition An R-triangle is a triangle where all three sides have length ≤ R.
‣ Lemma If three robots lie at the vertices of an R-triangle, and if r ≥ R/√3 then the three coverage disks of radius r meet and cover the entire triangle.
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The √3 Lemma
‣ Definition An R-triangle is a triangle where all three sides have length ≤ R.
‣ Lemma If three robots lie at the vertices of an R-triangle, and if r ≥ R/√3 then the three coverage disks of radius r meet and cover the entire triangle.‣ worst case: equilateral triangle
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
The √3 Lemma
‣ Definition An R-triangle is a triangle where all three sides have length ≤ R.
‣ Lemma If three robots lie at the vertices of an R-triangle, and if r ≥ R/√3 then the three coverage disks of radius r meet and cover the entire triangle.‣ worst case: equilateral triangle
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Vietoris–Rips coverage ⇒ radial coverage
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Vietoris–Rips coverage ⇒ radial coverage
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Vietoris–Rips coverage ⇒ radial coverage
Vin de Silva Pomona College
CBMS Lecture 2B June 12, 2017
Robert GhristUniversity of Pennsylvania
Philadelphia PA
Abubakr MuhammadLUMS School of Science & Engineering
Lahore, Pakistan
Collaborators