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Robust Principal Components Analysis

Robust Principal Components Analysis. IT530 Lecture Notes. Basic Question. Consider a matrix M of size n1 x n2 that is the sum of two components – L (a low-rank components) and S (a component with sparse but unknown support). Can we recover L and S given only M ?

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Robust Principal Components Analysis

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  1. Robust Principal Components Analysis IT530 Lecture Notes

  2. Basic Question • Consider a matrix M of size n1 x n2 that is the sum of two components – L (a low-rank components) and S (a component with sparse but unknown support). • Can we recover L and S given only M? • Note: support of S means all those matrix entries (i,j) where S(i,j) ≠ 0.

  3. Why ask this question? Scenario 1 • Consider a video taken with a static camera (like at airports or on overhead bridges on expressways). • Such videos contain a “background” portion that is static or changes infrequently, and a moving portion called “foreground” (people, cars etc.). • In many applications (esp. surveillance for security or traffic monitoring), we need to detect and track the moving foreground.

  4. Why ask this question? Scenario 1 • Let us represent this video as a N x T matrix, where the number of columns T equals the number of frames in the video, and where each column containing a single video-frame of N = N1N2 pixels converted to a vector form. • This matrix can clearly be expressed as the sum of a low-rank background matrix and a sparse foreground matrix.

  5. Why ask this question? Scenario 2 • It is well-known that the images of a convex Lambertian object taken from the same viewpoint, but under K different lighting directions, span a low-dimensional space (ref: Basri and Jacobs, “Lambertian reflectance and linear subspaces”, IEEE TPAMI, 2003). • This means that a matrix M of size N x K (each column of which is an image captured under one of the K lighting directions and represented in vectorized form) has a low approximate rank – in fact at most rank 9.

  6. Why ask this question? Scenario 2 • But many objects of interest (such as human faces) are neither convex nor exactly Lambertian. • Effects such as shadows, specularities or saturation cause violation of the low-rank assumption. • But these errors have sparse spatial support, even though their magnitudes are large. • Here again, M can be represented as a low-rank background and a sparse foreground.

  7. Ref: Candes et al, “Robust principal components analysis?”, Journal of ACM, 2009. Theorem 1 (Informal Statement) • Consider a matrix M of size n1 by n2 which is the sum of a “sufficiently low-rank” component L and a “sufficiently sparse” component S whose support is uniformly randomly distributed in the entries of M. • Then the solution of the following optimization problem (known as principal component pursuit) yields exact estimates of L and S with “very high” probability:

  8. Be careful! • There is an implicit assumption in here that the low-rank component is not sparse (e.g.: a matrix containing zeros everywhere except the right top corner won’t do!). • And that the sparse component is not low-rank (that’s why we said that it’s support is drawn from a uniform random distribution).

  9. Be careful! • In addition, the low-rank component L must obey the assumptions A0 and A1 (required in matrix completion theory).

  10. Theorem 1 (Formal Statement)

  11. Comments on Theorem 1 • If the value m is smaller, the upper bound on the maximum rank allowed is larger. • We require the support of S to be sparse and uniformly distributed but make no assumption about the magnitude/sign of the non-zero elements in S. But we don’t know the support of S. • In the matrix completion problem, we knew which entries in the matrix were “missing”. Here we do not know the support of S. And the entries in S are not “missing” but unknown and grossly corrupted!

  12. Matrix Recovery with Corrupted as well as missing data • In some cases, we may have missing entries in the matrix (with known location) in addition to corrupted entries (whose location and magnitude we DON’T know). • In such cases, we solve:

  13. Theorem 2 (Formal Statement)

  14. Comments on Theorem 2 • If all entries were observed (i.e. m = n1n2), then this reduces to Theorem 1. • If t = 0, this becomes a pure matrix completion problem (no sparse component), but which is solved using a slightly different objective function (question: what’s that difference?). • In the t = 0case, the minimization routine is guaranteed to give us a low-rank component as usual, but a “sparse” component that contains all zeros.

  15. Results

  16. Application: foreground/background segregation in videos • The exact theory may not be applicable: the support of the sparse moving foreground may not be modelled well by an i.i.d. Bernoulli model! • Nonetheless, the principal components pursuit method yields a very good results! • Video of 176 x 144 x 200 (200 frames) processed in 40 minutes (806 iterations) on a 2.3 GHz machine with 2GB RAM.

  17. Application: removal of shadows and specularities from face images • The exact theory may not be applicable: the support of the sparse moving foreground may not be modelled well by an i.i.d. Bernoulli model! • Nonetheless, the principal components pursuit method yields a very good results! • Some results shown on a group of 58 images of a person (under the same pose but changing illumination directions) each of size 192 x 168 from the Yale Face Database.

  18. Ref: Zhou et al, “Stable principal component pursuit”, International Symposium on Information Theory, 2010 Extensions to handle noise • Consider that the matrix M is given as M = L + S + Z where Z is a noise matrix with Frobenius norm upper bounded by d. • Then the solution to the following minimization problem recovers L and S from M with upper bounded by Cn2d2:

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