Persistent Homology and Sensor Networks
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Transcript of Persistent Homology and Sensor Networks
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Persistent Homology and Sensor Networks
Persistent homology motivated by an application to sensor nets
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Outline• A word about sensor nets
• Basic coverage criterion
• Better coverage criterion using persistence
• Introduce Persistent Homology
• Correspondence Theorem
• Computing the groups!
• Other Applications
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A Word About Sensor Nets
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August 29, 2005
Hurricane Katrina hits New Orleans
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Power and Communications Knocked Out
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Broken Levees
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City Flooded
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Inaccessible from the ground
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Law Enforcement
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Rescue Workers
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Replace live turkey with a parachute
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Result: Useful sensor network
• Measure conditions on the ground at many locations
• Relay messages to and from rescue workers
• Instant infrastructure
• Low power/auto-power
• Cheap!?
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Other uses of sensor networks
• Environmental monitoring• Security systems• Battlefield monitoring and communications• Large mechanical systems• Find Sarah Connor
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Hole in sensor coverage area
Sarah Connor escapes!
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Identifying holes in the network
• De Silva and Ghrist have developed a method for identifying gaps in sensor coverage
• Method is based on Algebraic Topology• Computing and examining Simplical
Homology groups• Theoretical underpinings allow you to do so
much more
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Basic Coverage Criterion
Part 1.2
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rc
rb
The problem to be solved:
Each node has sensors that cancover a circular region of radius rc
Each node can detect other nodesWithin its broadcast radius rb
rc ≥ rb/√(3)
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The problem to be solved:
Each node has sensors that cancover a circular region of radius rc
Each node can detect other nodesWithin its broadcast radius rb
rc ≥ rb/√(3)
Nodes lie in compact connected planar domain with piecewise linear boundary. Fence nodes at the vertices
All fence nodes know their neighbors’ identities and are no more than rb apart
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What we don’t have:• Nodes don’t know their absolute or relative
positions
• All we get is the connectivity graph
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It would be nice to have the Cech Complex
Def: For a collection of sets U={U}, the Cech Complex C(U) is the simplical complex where each non-empty intersection of (k+1) of the U correspond to a k-simplex.
3-simplex
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We have just enough to build the Rips Complex
•Let X be a collection of points in a metric space
•Rips complex R(X) contains a simplex for every collection of points that are pairwise within distance
•Even though our domain is planar, a dense graph can lead to simplices with arbitrary dimension
• In our case, we are building Rrb(X)
• Every complete k-subgraph of the communication graph becomes a simplex in the Rips Complex
• Also, it’s the maximal simplicial complex that has the connectivity graph as its 1-skeleton
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Picture of a Rips Complex
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Recap:X= { set of nodes }
rc = sensor radius
rb = broadcast radius
D = domain to be covered
∂D = boundary of D
Xf= { fence nodes that lie on D }
R= Rips complex of the communication graph
U= Region covered by the sensors
F= Fence subcomplex R
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Theorem (De Silva & Ghrist):
For a set of nodes X in a planar domain D satisfying the assumptions (rc, rb, fence nodes etc), the sensor cover Uc
contains D if there exists [] H2(R,F) such that ∂ ≠ 0
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What about a generator of H2(R,F)?
A generator will look like some linear combination of 2-simplices
i.e. Some triangulation of the domain D
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Theorem (De Silva & Ghrist):
For a set of nodes X in a planar domain D satisfying the assumptions (rc, rb, fence nodes etc), the sensor cover Uc contains D if there exists [] H2(R,F) such that ∂ ≠ 0
But why require ∂ ≠ 0 ??
Why not “if and only if” ??
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Pitfalls of the Rips complex
Bound was rc ≥ rb/√(3)
1/√ (3) ≈ 0.57
Therefore it’s possible to have a rectangle that is completely covered, but not triangulated in the communication graph
rbrb
So the conditions of the theorem are sufficient, but not necessary, to guarantee coverage.
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Pitfalls of the Rips complex
It’s possible to have an arrangement of nodes whose Rips complex is the surface of an octahedron.
This has non-zero H2, but its boundary is zero!
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Better coverage criterion using persistence
Part 1.3
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Eliminating the fence subcomplex• The assumption of the nice fence sub-complex is
unrealistic• Can we replace it with some other assumptions?
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The new situation:
Each node has sensors that cancover a circular region of radius rc
Each node can detect its neighborsvia a strong signal (rs) or a weaksignal (rw).
rc ≥ rs/√(2)rw ≥ rs √(10)
rc
rs
rw
Remember:strong <---> “short”weak <---> “wlong”
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The new situation (cont…):rc ≥ rs/√(2) rw ≥ rs √(10)
Nodes lie in a compact connected domain D in Rd
Nodes can detect the presence of ∂D within distance rf
The restricted domain D-C is connected, whereC = {x D ||x-∂D|| ≤ rf + rs/√(2)
The fence-detection hypersurface = {x D ||x-∂D|| = rf}Has internal injectivity radius ≥ rs/√(2) external injectivity radius ≥ rs
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The new situation (cont…):
rf
The fence “collar”, C
restricted domainD-C
The boundary ∂D
Domain D
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New complexes
• We get two communication graphs now, corresponding to rs and rw
• One gives us the “strong” Rips Complex, Rs
• The other gives the “weak” Rips complex Rw
• Note that Rs Rw
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(more) New complexes
• We also get a subcomplex based on the nodes that lie within rf of ∂D
• Build this as a subcomplex of Rs
• Call it the (strong) fence subcomplex Fs
rf
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What we’d like to see
Conjecture:For a set of nodes X in a domain D Rd satisfying the new assumptions (rc, rs, rw, rf, fence subcomplex etc), the sensor cover
U contains D-C if there exists [] Hd(Rs,Fs) such that ∂ ≠ 0
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Why it fails
rf
By comparing to the “weak” Rips complex, we can see which of these cycles are phantom and which are legitimate
• It’s possible to get “phantom” d-cycles in the relative homology that have non-zero boundary
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Theorem (De Silva & Ghrist):
For a set of nodes X in a domain D in Rd satisfying the new assumptions (rc, rs, rw, rf, fence subcomplex etc), the sensor cover U contains D-C if the homomorphism
i*: Hd(Rs,Fs) ----> Hd(Rw,Fw) induced by the inclusion i: (Rs,Fs) ----> (Rw,Fw) is nonzero.
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The “Squeezing” Theorem
For a set of points X in a domain D Rd
R(X) C(X) R(X)
whenever / ≥ √(2d/(d+1))
•Note that for d=2 this means ≥ 1.15
• This means that if you can enlarge (or shrink) the radius of your Rips complex a little, and the complex doesn’t change, then you actually have a Cech complex
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Persistence
Part 2
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The Usual Homology
• Have a single topological space, X, and a PID, R
• Get a chain complex
• For k=0, 1, 2, …
compute Hk(X)
• Hk=Zk/Bk
Ck(X) C1(X) C0(X)Ck-1(X) 0∂ ∂∂ ∂ ∂ ∂
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How about a filtration of spaces?
X1 X2 X3 … Xn
a b a b
cd
a b
cd
a b
cd
a b
cd
a b
cd
a, b c, d, ab, bc cd, ad ac abc acd
t=0 t=1 t=2 t=3 t=4 t=5
• We restrict to simplical complexes (so we can compute)
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Leads to a Persistence Complex
X1 X2 X3 … Xn
C0k C0
1 C00C0
k-10
∂∂∂ ∂ ∂ ∂
C1k C1
1 C10C1
k-10
∂∂∂ ∂ ∂ ∂
Cnk Cn
1 Cn0Cn
k-10
∂∂∂ ∂ ∂ ∂
• Columns are inclusion maps
• Inclusion is a chain map, and so induces a map on homology
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Induces a map on homology
a b a b
cd
a b
cd
a b
cd
a b
cd
a b
cda, b c, d, ab, bc cd, ad ac abc acd
t=0 t=1 t=2 t=3 t=4 t=5
• For each dimension k=0,1,2,…
• Consider a generator []Hik
• We may want to consider where in the filtration that generator first appears (created), and when it first becomes bounding (destroyed)
H0k H1
k
i*H2
k
i*Hn
k
i*
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Concept: P-interval
a b a b
cd
a b
cd
a b
cd
a b
cd
a b
cda, b c, d, ab, bc cd, ad ac abc acd
t=0 t=1 t=2 t=3 t=4 t=5
• A P-interval is an ordered pair (i, j) with 0≤i<j ≤∞
• Consider a generator []Hik
• We can encode information about the creation and destruction time of [] as a P-interval
•For example [ab+bc-ac] H1 has P-interval (3, 4)
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Definition: Persistent Homology
a b a b
cd
a b
cd
a b
cd
a b
cd
a b
cda, b c, d, ab, bc cd, ad ac abc acd
t=0 t=1 t=2 t=3 t=4 t=5
Hki,p =
• Start with the k-cycles at t=i
Zki
• “fast-forward” the boundaries to some future time, i+p
Bki+p
• Intersect the denominator with Zki so it’s well-defined
Zki
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Too much work?
a b a b
cd
a b
cd
a b
cd
a b
cd
a b
cda, b c, d, ab, bc cd, ad ac abc acd
t=0 t=1 t=2 t=3 t=4 t=5
• This is interesting, but for an N-step filtration of dimension D, this means we have to compute O(N2D) homology groups!
• And how can we tell what a generator at one time step becomes at the next timestep?
• We need compatible bases for the whole filtration!
Hki,p = Zk
i
Bki+p Zk
i
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Definition: Persistence ModuleLet R be a commutative PID
A persistence module is a collection of R-modules, Mi, together with R-module homomorphisms i such that i:Mi ---> Mi+1
M = {Mi, i}
A persistence module M is said to be of finite type if the individual Mi are finitely generated, and N such that n ≥ N i:Mi Mi+1
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Correspondence TheoremLet R be a commutative PID, and M = {Mi, i} a persistence module of finite type over R
Define a functor
Where the R-module structure on the Mi is the sum of the individual components, and the action of t is given by
t·(m0, m1, m2, …) = (0, 0(m0), 1(m1), 2(m2), …)
R-persistence modules of finite type
Finitely generated non-negatively graded R[t] modules
i=0
∞(M) = Mi
Proof: “the Artin-Rees theory in commutative algebra”?
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Correspondence TheoremLet R be a commutative PID, and M = {Mi, i} a persistence module over R
Define a functor
If R=F is a field, then F[t] is a graded PID and we have a structure theorem for its finitely-generated graded modules
R-persistence modules of finite type
Finitely generated non-negatively graded R[t] modules
i=0
∞
(M) = Mi
i=1
n
= _i F[t] _j F[t]/(tn_j) m
j=1
free part torsion part
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Example: Homology of a filtration
The homology groups Hkl (for a fixed k) of a finite filtration {Xl}, along with the maps
induced by inclusions are a persistence module of finite type.
Hk = {Hkl, i*
l}
In the corresponding graded R[t] module M=(Hk), each stage in the filtration corresponds to a particular degree.
a b a b
cd
a b
cd
a b
cd
a b
cd
a b
cda, b c, d, ab, bc cd, ad ac abc acd
t0 t1 t2 t3 t4 t5
The element [ab+bc-ac] H13 has degree 3
But t·[ab+bc-ac] 0 in H14
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Visualization: “Barcodes”a b a b
cd
a b
cd
a b
cd
a b
cd
a b
cda, b c, d, ab, bc cd, ad ac abc acd
t0 t1 t2 t3 t4 t5
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Computing simplicial homology
The boundary operators of the chain complex are linear operators operating on chain groups which are free R-modules
Therefore they can be represented as matrices relative to some bases.
Ck(X) C1(X) C0(X)Ck-1(X) 0∂ ∂∂ ∂ ∂ ∂
By the standard basis we mean the basis where individual simplices are represented as the unit vectors in Rk
a b
cd
C0 = < a, b, c, d >
= < , , , >1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
C1 = < ab, bc, cd, ad, ac >
= < , , , , >
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
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Computing simplicial homology 2
The boundary map
k:Ck ---> Ck-1
is represented by the R-matrix Mk
a b
cd
C0 = < a, b, c, d >
= < , , , >1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
C1 = < ab, bc, cd, ad, ac >
= < , , , , >
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
-1 0 0 -1 -1
1 -1 0 0 0
0 1 -1 0 1
0 0 1 1 0
M1 =
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Computing simplicial homology 3
Then Mk can be reduced by elementary operations to a matrix, Mk in Smith Normal Form
a b
cd
-1 0 0 -1 -1
1 -1 0 0 0
0 1 -1 0 1
0 0 1 1 0
M1 =
~
a
b
c
d
ab bc cd ad ac
1 0 0
0 ... 0 0
0 0 r
0 0
Mk = ~b1
...
br
br+1
...
bm
a1 ... ar z1 .... zn-r The i’s that are >1 are the torsion coefficients of Hk-1
z1, ..., zr are a basis for kerMk = Zk
1b1, ..., rbr are a basis for imMk = Bk-1
So between Mk and Mk+1 we have enough information to compute Hk , betti numbers
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 0 0
M1 = ~a
a+b
b+c
c+d
-ab -bc -cd z1 z2
z1 = ab+bc-ac
z2 = ac+cd-ad
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Computing persistent homologyTo compute persistent homology over a field, F, do the same thing except work over the ring F[t]
a b a b
cd
a b
cd
a b
cd
a b
cd
a b
cda, b c, d, ab, bc cd, ad ac abc acd
t0 t1 t2 t3 t4 t5
Each simplex is assigned a degree according to when it got added to the complex
For example, deg(a)=0 deg(abc)=4
The boundary operator can’t map across the grading
So for a simplex Ck deg() = deg(∂k)
For example, ∂kac) = t2·c - t3·a
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Computing persistent homology 2
a b a b
cd
a b
cd
a b
cd
a b
cd
a b
cda, b c, d, ab, bc cd, ad ac abc acd
t0 t1 t2 t3 t4 t5
For a given dimension, k, there is a single boundary operator, ∂k encoding information for the entire filtration.
Note that basis elements are homogenous.
-t 0 0 -t2 -t3
t -t 0 0 0
0 1 -t 0 t2
0 0 t t 0
M1 =
a
b
c
d
ab bc cd ad ac t 0 0 0 0
-t 1 0 0 0
0 -t t 0 0
0 0 -t 0 0
M1 = ~d
c
b
a
cd bc ab z1 z2
z1 = ad - cd - t·bc - t·ab
z2 = ac - t2·bc - t2·ab
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Computing persistent homology 3a b a b
cd
a b
cd
a b
cd
a b
cd
a b
cda, b c, d, ab, bc cd, ad ac abc acd
t0 t1 t2 t3 t4 t5
t 0 0 0 0
-t 1 0 0 0
0 -t t 0 0
0 0 -t 0 0
M1 = ~d
c
b
a
cd bc ab z1 z2
z1 = ad - cd - t·bc - t·ab
z2 = ac - t2·bc - t2·ab
Torsion terms in persistent homology!
A torsion coefficient ti corresponding to a basis element of degree j gives a term in the
persistent homology group: j F[t]/(ti)
Or in other words, a P-interval (j, i+j)
An extra basis element of degree j at the bottom gives a free term:
j F[t]
IOW a P-interval (j,∞)
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Applications• When your only tool is persistent homology, every problems
starts to look like a filtered simplicial complex
1. That sensor nets thing
2. Point cloud data
3. Dimension estimation
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