semidefinite programming and universal rigidityyyye/sdp-rigidity.pdf · sensor network localization...

Introduction to Semidefinite ProgrammingSDP Solution Rank Theorems

Sensor Network Localization and Graph RealizationSDP Relaxation and Localizability

Semidefinite Programming and Universal Rigidity

Yinyu Ye

Department of Management Science and Engineering, andInstitute of Computational and Mathematical Engineering

Stanford University

Based on joint work mostly with Abdo Alfakih, Pratik Biswas, Anthony So, NicoleTaheri, and Zhisu Zhu

Rigidity Workshop, Cornell, February 5, 2010

Yinyu Ye Rigidity Microworkshop, Cornell 2011

Outline

Introduction to Semidefinite Programming

SDP Solution Rank Theorems

Sensor Network Localization and Graph Realization

SDP Relaxation and Localizability



Semidefinite Programming Problem (SDP)

Find a real n × n symmetric matrix X ∈ Sn that satisfies

Ai • X = bi , ∀i = 1, . . . ,m, X � 0

where A1, . . . ,Am are given real symmetric matrices with realscalars (b1, . . . , bm),

A • X =∑i ,j

aijxij ,

and � represents the matrix inequality of positivesemi-definiteness.




Semidefinite Programming Problem (SDP)

Find a real n × n symmetric matrix X ∈ Sn that satisfies

Ai • X = bi , ∀i = 1, . . . ,m, X � 0

where A1, . . . ,Am are given real symmetric matrices with realscalars (b1, . . . , bm),

A • X =∑i ,j

aijxij ,

and � represents the matrix inequality of positivesemi-definiteness.

x1 + x2 + x3 = 1,(x1 x2x2 x3

)� 0.




Semidefinite Programming Optimization Problem

(SDP) inf C • Xsubject to Ai • X = bi i = 1, . . . ,m,

X � 0.

where C is a given (objective) symmetric matrix. The dualproblem to (SDP) can be written as:

(SDD) sup b · ysubject to S = C −∑m

i yiAi � 0,

where y ∈ Rm.




Duality Theorems for SDPs

Theorem( Weak duality theorem in SDP) Let Fp and Fd , the feasible setsfor the primal and dual, be non-empty. Then,

C • X ≥ b · y, ∀ X ∈ Fp , (y,S) ∈ Fd .




Duality Theorems for SDPs

Theorem( Weak duality theorem in SDP) Let Fp and Fd , the feasible setsfor the primal and dual, be non-empty. Then,

C • X ≥ b · y, ∀ X ∈ Fp , (y,S) ∈ Fd .

Theorem(Strong duality theorem in SDP) Let Fp and Fd be non-empty andat least one of them has an interior. Then, X is optimal for (SDP)and (y,S) optimal for (SDD) if and only if the following conditionshold:

i) X ∈ Fp and (y,S) ∈ Fd ;

ii) C • X = b · y or X • S = 0 or XS = 0.


Outline







SDP Solution Rank

� There are optimal solutions of X ∗ and S∗ such that the ranksof X ∗ and S∗ are maximal, respectively.




SDP Solution Rank


� But the sum of the two ranks should be no more than n. Ifthe sum is n, then we say the problem admits a strictlycomplementary solution pair.




SDP Solution Rank



� There are optimal solutions of X ∗ and S∗ such that the ranksof X ∗ and S∗ are minimal, respectively.




SDP Solution Rank




� In applications, we like to find either a max-rank or min-ranksolution or/and to prove it exist.




SDP Solution Rank




� In applications, we like to find either a max-rank or min-ranksolution or/and to prove it exist.

If there is S∗ such that rank(S∗) ≥ n− d , then the rank of any X ∗

is bounded above by d .




SDP Solution Uniqueness: Zhu 2010

TheoremLet X ∗ be a max-rank solution of (SDP) and let X ∗ = PTP whereP ∈ Rr×n. Then, X ∗ is the unique solution for (SDP) if and only ifthe null space of the linear space spanned by PAiP

T , i = 1, . . . ,m,is {0}.Corollary

If all (SDP) solutions share the same rank, then (SDP) has aunique solution.




SDP Computational Complexity and Solution Rank

� Let the SDP problem have a finite solution pair. Then, theSDP interior-point algorithm finds an ε-approximate solutionwhere solution time is linear in log(1/ε) and polynomial in mand n.






� The approximate solution pair (X (ε),S(ε)) computed by theinterior-point algorithm converges to a max-rank solution pairfor the primal and dual (Guler and Y 1993, Y 1995).







� Barvinok (1995) showed that if the problem is solvable, thenthere exists a solution X whose rank r satisfies r(r +1) ≤ 2m,and it is essentially tight.







� Barvinok (1995) showed that if the problem is solvable, thenthere exists a solution X whose rank r satisfies r(r +1) ≤ 2m,and it is essentially tight.

� Given any solution, a such low-rank solution can be found inpolynomial time based on Caratheodory’s theorem (Pataki1999, Alfakih/Wolkowicz 1999).




More on SDP Solution Rank

� We are interested in finding a fixed low–rank (say d) solutionto the SDP problem.






� However, there are some issues:� Such a solution may not exist!� Even if it does, one may not be able to find it efficiently.






� However, there are some issues:� Such a solution may not exist!� Even if it does, one may not be able to find it efficiently.

� Suppose we consider an approximate low-rank solution to

Ai • X = bi , ∀i = 1, . . . ,m, X � 0,

where Ai � 0 for all i .




Approximate Low-Rank SDP Solution

More precisely, find an X � 0 with rank at most d that satisfiesthe SDP constraints approximately:

β(m, n, d) · bi ≤ Ai • X ≤ α(m, n, d) · bi ∀ i = 1, . . . ,m.







Here, α(·) ≥ 1 and β(·) ∈ (0, 1] are called the distortion factors.







Here, α(·) ≥ 1 and β(·) ∈ (0, 1] are called the distortion factors.

Clearly, the closer are both to 1, the better the approximationquality.




Approximation Result, So, Y and Zhang 2007

TheoremFor any d ≥ 1, there exists an X � 0 with rank(X ) ≤ d such that

α(m, n, d) =

⎧⎪⎪⎨⎪⎪⎩

1 +18 ln(2m)

dfor 1 ≤ d ≤ 18 ln(2m)

1 +

√18 ln(2m)

dfor d > 18 ln(2m)

,

β(m, n, d) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

1

e(2m)2/dfor 1 ≤ d ≤ 4 ln(2m)

max

{1

e(2m)2/d, 1−

√4 ln(2m)

d

}for d > 4 ln(2m)




Some Remarks

� There is always a low-rank approximate SDP solution withrespect to a bounded distortion. As the allowable rankincreases, the distortion bounds become smaller and smaller.




Some Remarks


� Moreover, there exists an efficient randomized algorithm forfinding such an X .




Some Remarks



� The distortion factors are independent of n.




Some Remarks




� The factors are sharp; but they can be improved if we onlyconsider one–sided inequalities.




Some Remarks




� The factors are sharp; but they can be improved if we onlyconsider one–sided inequalities.

� This result contains as special cases several well-known resultsin the literature.


Outline








Given a graph G = (V ,E ) and sets of non–negative weights, say{dij : (i , j) ∈ E} on edges, the goal is to compute a realization ofG in the Euclidean space Rd for a given low dimension d . That is,






� to place the vertexes of G in Rd such that







� the Euclidean distance between a pair of adjacent vertexes(i , j) equals to (or bounded by) the prescribed weight dij ∈ E .







� the Euclidean distance between a pair of adjacent vertexes(i , j) equals to (or bounded by) the prescribed weight dij ∈ E .

The positions of a few vertexes are known and they are calledanchors for SNL.




50-node 2-D Sensor Localization

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0

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Quadratic Equality and Inequality Systems

Given graph G = (V ,E ) and dij ∈ E , find xj ∈ Rd such that

‖xi − xj‖2 (≤) = (≥) d2ij , ∀ (i , j) ∈ E , i < j .




Quadratic Equality and Inequality Systems

Given graph G = (V ,E ) and dij ∈ E , find xj ∈ Rd such that

‖xi − xj‖2 (≤) = (≥) d2ij , ∀ (i , j) ∈ E , i < j .

Or given ak ∈ Rd , dij ∈ Nx , and dkj ∈ Na, find xi ∈ Rd such that

‖xi − xj‖2 (≤) = (≥) d2ij , ∀ (i , j) ∈ Nx , i < j ,

‖ak − xj‖2 (≤) = (≥) d2kj , ∀ (k , j) ∈ Na;

that is, edge (ij) (or (kj)) connects sensors i and j (or anchor kand sensor j) with the Euclidean length equal to dij (or dkj).




Key Questions Related to SNL

Let the “bar system” be:

‖xi − xj‖2 = d2ij , ∀ (i , j) ∈ Nx , i < j ,

‖ak − xj‖2 = d2kj , ∀ (k , j) ∈ Na,






‖xi − xj‖2 = d2ij , ∀ (i , j) ∈ Nx , i < j ,

‖ak − xj‖2 = d2kj , ∀ (k , j) ∈ Na,

� Does the system have a localization or configuration for allxj ’s?






‖xi − xj‖2 = d2ij , ∀ (i , j) ∈ Nx , i < j ,

‖ak − xj‖2 = d2kj , ∀ (k , j) ∈ Na,


� Is the localization/configuration unique, and it can be(approximately) certified?






‖xi − xj‖2 = d2ij , ∀ (i , j) ∈ Nx , i < j ,

‖ak − xj‖2 = d2kj , ∀ (k , j) ∈ Na,


� Is the localization/configuration unique, and it can be(approximately) certified?

� Is the system partially localizable with an (approximate)certification?


Outline







Matrix Representation

Find Y = XTX , where X is d × n, such that

(0; ei − ej)(0; ei − ej )T •(

I XXT Y

)= d2

ij , ∀ i , j ∈ Nx , i < j ,

(ak ;−ej )(ak ;−ej )T •

(I X

XT Y

)= d2

kj , ∀ k , j ∈ Na.

where ej is the vector of all zeros except 1 at the jth position.




SDP Relaxation of SNL, Biswas and Y 2004

Relax Y = XTX to Y � XTX ;




SDP Relaxation of SNL, Biswas and Y 2004

Relax Y = XTX to Y � XTX ;or equivalently

Z :=

(I X

XT Y

)� 0.

Then, we face a standard SDP (feasibility) problem.

(0; ei − ej)(0; ei − ej )T • Z = d2

ij , ∀ i , j ∈ Nx , i < j ,

(ak ;−ej )(ak ;−ej )T • Z = d2

kj , ∀ k , j ∈ Na,

Z � 0.




Properties of the SDP Relaxation

� If the network is connected and it has at least one anchor,then the solution set must be bounded.






� A solution matrix Z has rank at least d .







� It’s d if and only if Y = XTX and it produces onelocalization to the original SNL problem.







� It’s d if and only if Y = XTX and it produces onelocalization to the original SNL problem.

� If there exists a dual solution with rank n, then the rank ofany primal solution Z must be d .




The Dual of the SDP Relaxation

minimize I • V +∑

i<j∈Nxwijd

2ij +

∑k,j∈Na

wkj d2kj

subject to

(V 00 0

)+

∑i<j∈Nx

wij(0; ei − ej)(0; ei − ej)T

+∑

k,j∈Nawkj(ak ;−ej)(ak ;−ej)

T = S � 0,

where variable matrix V ∈ Sd , variable wij is the (stress) weighton edge between xi and xj , and wkj is the (stress) weight on edgebetween ak and xj .






i<j∈Nxwijd

2ij +

∑k,j∈Na

wkj d2kj

subject to

(V 00 0

)+

∑i<j∈Nx


+∑


T = S � 0,


� The dual is strictly feasible if the network is connected andhas at least one anchor.






i<j∈Nxwijd

2ij +

∑k,j∈Na

wkj d2kj

subject to

(V 00 0

)+

∑i<j∈Nx


+∑


T = S � 0,



� The max-rank of any optimal dual stress matrix S is n.






i<j∈Nxwijd

2ij +

∑k,j∈Na

wkj d2kj

subject to

(V 00 0

)+

∑i<j∈Nx


+∑


T = S � 0,



� The max-rank of any optimal dual stress matrix S is n.

� It’s n if the problem admits strict complementarity.




Unique Localizability and Uniquely Localizable Problem I

A sensor network is uniquely-localizable (UL) if there exists aunique localization in R2 and there is no nontrivial localizations,xj ∈ Rh, j = 1, ..., n, where h > 2, such that

‖xi − xj‖2 = d2ij , ∀ i , j ∈ Nx , i < j ,

‖(ak ;0)− xj‖2 = d2kj , ∀ k , j ∈ Na.




Unique Localizability and Uniquely Localizable Problem I

A sensor network is uniquely-localizable (UL) if there exists aunique localization in R2 and there is no nontrivial localizations,xj ∈ Rh, j = 1, ..., n, where h > 2, such that

‖xi − xj‖2 = d2ij , ∀ i , j ∈ Nx , i < j ,

‖(ak ;0)− xj‖2 = d2kj , ∀ k , j ∈ Na.

It basically says that the problem cannot be localized in a higherdimension space where anchor points are simply augmented to(ak ;0) ∈ Rh, k = 1, ...,m.




One sensor-Two anchors: Not localizable

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

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Unique Localizability and Uniquely Localizable Problem II

� It’s necessary to have d + 1 anchors in general positions forany network to be UL.






� It’s also necessary for every sensor to have at least d + 1edges.







� For a UL instance, if there exists a dual stress matrix withrank n, then it is called strongly-localizable (SL).







� For a UL instance, if there exists a dual stress matrix withrank n, then it is called strongly-localizable (SL).

� UL is stronger than the notion of Global Rigidity, and it ismost similar to Universal Rigidity.




Two sensor-Three anchors: Strongly Localizable

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Two sensor-Three anchors: Localizable but not Strongly

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Two sensor-Three anchors: Not localizable

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Two sensor-Three anchors: Strongly Localizable

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An Equivalence Theorem, So and Y 2005

TheoremThe following statements are equivalent:

1. The sensor network is uniquely-localizable;

2. The max-rank solution of the SDP relaxation has rank d;

3. The solution matrix has Y = XTX or Trace(Y − XTX ) = 0 .









Moreover, the localization of a uniquely localizable instance can becomputed approximately in a time polynomial in n, d, and theaccuracy log(1/ε).









Moreover, the localization of a uniquely localizable instance can becomputed approximately in a time polynomial in n, d, and theaccuracy log(1/ε).

The existence of a rank n stress matrix (or strong localizability) issufficient but not necessary for UL.




Localize All Localizable Points, So and Y 2005

TheoremIf an SNL problem (graph) contains a subproblem (subgraph) thatis strongly localizable, then the solution submatrix correspondingto the subproblem in the SDP has rank d. Thus, the SDPrelaxation computes a solution that localize all possibly localizablesensor points.




Localize All Localizable Points, So and Y 2005

TheoremIf an SNL problem (graph) contains a subproblem (subgraph) thatis strongly localizable, then the solution submatrix correspondingto the subproblem in the SDP has rank d. Thus, the SDPrelaxation computes a solution that localize all possibly localizablesensor points.

Diagonals of the “co-variance” matrix

Y − XT X ,

can be a certification for uniquely localizable sensor points. Thatis, Yjj − ‖xj‖2 = 0 if any only if the jth sensor point has a uniquelocalization.




Uniquely-Localizable Graphs

TheoremLet the SNL problem have at least d + 1 anchors and they,together with all sensors, are in general positions.

� If G is complete or every edge length is specified, then thesensor network is uniquely-localizable (Schoenberg 1942).




Uniquely-Localizable Graphs

TheoremLet the SNL problem have at least d + 1 anchors and they,together with all sensors, are in general positions.

� If G is complete or every edge length is specified, then thesensor network is uniquely-localizable (Schoenberg 1942).

� The (d + 1)-lateration graph is uniquely-localizable (So 2006and Zhu, So and Y 2009). Moreover, it is near the sparsestgraph (with only (d + 1)n edges) that is uniquely-localizable.




(d + 1)-Lateration Graph

A (d + 1)-lateration ordering for a graph G is an ordering of thevertexes 1, · · · , d + 1, d + 2, · · · , n such that the first d + 1vertexes form the complete graph, and every vertex j > d + 1 hasd + 1 edges connected to its preceding vertexes on the sequence.




(d + 1)-Lateration Graph

A (d + 1)-lateration ordering for a graph G is an ordering of thevertexes 1, · · · , d + 1, d + 2, · · · , n such that the first d + 1vertexes form the complete graph, and every vertex j > d + 1 hasd + 1 edges connected to its preceding vertexes on the sequence.

2

34

5

6

.....

1




Universal Rigidity Theorems in Generic Position

Recall that an SNL problem is strongly localizable if there exists amax-rank n PSD stress matrix.






Let’s consider a bar framework (G ,P ∈ Rd×(n+d+1)) of n + d + 1vertexes in generic positions of Rd . Then







� The existence of a max-rank PSD stress matrix implies thatthe framework is universally rigid (Connelly 1999, Alfakih2010).







� The existence of a max-rank PSD stress matrix implies thatthe framework is universally rigid (Connelly 1999, Alfakih2010).

� The framework is universally rigid if and only if there exists amax-rank PSD stress matrix (Gortler and Thurston, 2009).




Universal Rigidity Theorems in General Position

Let’s consider a bar framework (G ,P ∈ Rd×(n+d+1)) of n + d + 1vertexes in general positions of Rd . Then






� The existence of a max-rank PSD stress matrix implies thatthe framework is universally rigid (Alfakih and Y 2010).







� A framework problem that contains a spanning(d + 1)-lateration graph is universally rigid if and only if thereexists a max-rank PSD stress matrix (Alfakih, Taheri and Y2010).







� A framework problem that contains a spanning(d + 1)-lateration graph is universally rigid if and only if thereexists a max-rank PSD stress matrix (Alfakih, Taheri and Y2010).

� Given configuration/position matrix P and the laterationorder, such a PSD stress matrix can be computed exactly instrongly polynomial time.




Proof Sketch I

Let the extended position matrix

A =

(PeT

)∈ R(d+1)×(n+d+1) .




Proof Sketch I

Let the extended position matrix

A =

(PeT

)∈ R(d+1)×(n+d+1) .

Recall that symmetric matrix S is a stress matrix if and only if

orthogonality: AS = 0, (1)

andpurity: Sij = 0, ∀(i , j) �∈ E . (2)




Proof Sketch II

� We start a PSD matrix satisfies orthogonality condition (1),say

S0 = I − AT (AAT )−1A,

where the columns of A are ordered according to thelateration order, and we call it a “prestress” matrix.




Proof Sketch II


S0 = I − AT (AAT )−1A,


� We modify S0 column (row) by column (row), starting fromthe last column (row) backward, by zeroing entries not in Efrom solving a linear equation system of d + 1 variables ineach step.




Proof Sketch II


S0 = I − AT (AAT )−1A,



� In the modification process, we maintain the PSD, rank-n,and the orthogonality conditions of the “prestress” matrix.




Proof Sketch II


S0 = I − AT (AAT )−1A,



� In the modification process, we maintain the PSD, rank-n,and the orthogonality conditions of the “prestress” matrix.

� In at most n steps, we reach a “prestress” matrix that finallysatisfies purity condition (2).




Conclusions and Open Research Questions

� SDP seems to provide an efficient computation and analysismodel for certain applications of the Rigidity Theory.






� Does any UL graph contain at least one d + 1 clique?







� Sufficient and necessary conditions based on the max-rankcondition of primal SDP matrix solutions?







� Sufficient and necessary conditions based on the max-rankcondition of primal SDP matrix solutions?

� Sufficient and necessary conditions of UL for points in generalpositions?


semidefinite programming and universal rigidityyyye/sdp-rigidity.pdf · sensor network localization...

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