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## Persistent Homology and Sensor Networks

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**Persistent Homology and Sensor Networks**Persistent homology motivated by an application to sensor nets**Outline**• A word about sensor nets • Basic coverage criterion • Better coverage criterion using persistence • Introduce Persistent Homology • Correspondence Theorem • Computing the groups! • Other Applications**August 29, 2005**Hurricane Katrina hits New Orleans**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!?**Other uses of sensor networks**• Environmental monitoring • Security systems • Battlefield monitoring and communications • Large mechanical systems • Find Sarah Connor**Hole in sensor coverage area**Sarah Connor escapes!**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**Basic Coverage Criterion**Part 1.2**The problem to be solved:**rb rc Each node has sensors that can cover a circular region of radius rc Each node can detect other nodes Within its broadcast radius rb rc ≥ rb/√(3)**The problem to be solved:**Each node has sensors that can cover a circular region of radius rc Each node can detect other nodes Within 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**What we don’t have:**• Nodes don’t know their absolute or relative positions • All we get is the connectivity graph**3-simplex**It would be nice to have the Cech Complex Def: For a collection of sets U={Ua}, the Cech ComplexC(U) is the simplical complex where each non-empty intersection of (k+1) of the Ua correspond to a k-simplex.**We have just enough to build the Rips Complex**• Let X be a collection of points in a metric space • Rips complexRe(X) contains a simplex for every collection of points that are pairwise within distance e • 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**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 dD } U= Region covered by the sensors R= Rips complex of the communication graph F= Fence subcomplex R**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 [a] H2(R,F) such that ∂a ≠ 0**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**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 [a] H2(R,F) such that ∂a ≠ 0 But why require ∂a ≠ 0 ?? Why not “if and only if” ??**Pitfalls of the Rips complex**Bound was rc ≥ rb/√(3) 1/√ (3) ≈ 0.57 rb rb Therefore it’s possible to have a rectangle that is completely covered, but not triangulated in the communication graph So the conditions of the theorem are sufficient, but not necessary, to guarantee coverage.**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!**Eliminating the fence subcomplex**• The assumption of the nice fence sub-complex is unrealistic • Can we replace it with some other assumptions?**The new situation:**rw rs rc Each node has sensors that can cover a circular region of radius rc Each node can detect its neighbors via a strong signal (rs) or a weak signal (rw). rc ≥ rs/√(2) rw ≥ rs √(10) Remember: strong <---> “short” weak <---> “wlong”**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 domainD-C is connected, where C = {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**The new situation (cont…):**Domain D The fence “collar”, C The boundary ∂D rf restricted domain D-C S**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 RsRw**(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 subcomplexFs rf**What we’d like to see**Conjecture: For a set of nodes X in a domain DRd satisfying the new assumptions (rc, rs, rw, rf, fence subcomplex etc), the sensor cover U contains D-C if there exists [a] Hd(Rs,Fs) such that ∂a ≠ 0**Why it fails**• It’s possible to get “phantom” d-cycles in the relative homology that have non-zero boundary By comparing to the “weak” Rips complex, we can see which of these cycles are phantom and which are legitimate rf**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.**The “Squeezing” Theorem**For a set of points X in a domain DRd Re(X)Ce(X) Re(X) whenever e/e ≥ √(2d/(d+1)) • Note that for d=2 this means e ≥ 1.15 e • 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**Persistence**Part 2**∂**∂ ∂ ∂ ∂ ∂ Ck(X) Ck-1(X) C1(X) C0(X) 0 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