Rank Revealing QR factorization F. Guyomarc’h, D. Mezher and B. Philippe 1
Outline • Introduction • Classical Algorithms ⋆ Full matrices ⋆ Sparse matrices • Rank-Revealing QR • Conclusion CSDA 2005, Cyprus 2
Situation of the problem See Bjorck SIAM96 : Numerical Methods for Least Squares Problems. Data : A ∈ R m × n , b ∈ R m . Problem P : Find x ∈ S such that � x � minimum where S = { x ∈ R n | � b − Ax � minimum } . ( � . � ≡ � . � 2 ) x = A + b . The solution is unique : ˆ Property : A T ( b − Ax ) = 0 , x ∈ S iff A T Ax = A T b. iff The last equation is consistent since R ( A T A ) = R ( A T ). CSDA 2005, Cyprus 3
Rank-Revealing QR factorization Theorem : A ∈ R m × n with rank( A ) = r ( r ≤ min( m, n )). There exist Q , E , R 11 and R 12 such that • Q ∈ R m × m is orthogonal, • E ∈ R n × n is a permutation, • R 11 ∈ R r × r is upper-triangular with positive diagonal entries, • R 12 ∈ R r × ( n − r ) , � � R 11 R 12 and AE = Q . 0 0 RRQR The factorization is not unique. Let RRQR be any of them. CSDA 2005, Cyprus 4
Actually, if we consider a complete orthogonal decomposition � � T 0 V T A = Q 0 0 where Q and V are orthogonal and T triangular with positive diagonal � T − 1 � 0 entries, we have A + = V Q T . 0 0 Such a factorization can be obtained from the previous one by per- � R T � 11 forming a QR factorization of R T 12 CSDA 2005, Cyprus 5
Column pivoting strategy Householder QR factorization using Householder reflections : H 1 A = A = v = A 2 .. 12 , 2 + || A 2 .. 12 , 2 || e 1 v = A 1 .. 12 , 1 + || A 1 .. 12 , 1 || e 1 H 2 = I 11 − 2 vv t H 1 = I 12 − 2 vv t v t v v t v CSDA 2005, Cyprus 6
Sparse QR factorization Factorizing a sparse matrix implies fill-in in the factors. The situation is worse with QR than with LU since when updating the tailing matrix : → y = ( I − αve T • LU : the elementary transformation x − k ) x keeps x invariant when x k = 0. → y = ( I − αvv T ) x keeps x • QR : the elementary transformation x − invariant when x ⊥ v . Since A T A = R T R , the QR factorization A is related to the Cholesky factorization of A T A . It is known that a symmetric permutation on a sparse s.p.d. matrix changes the level of fill-in. CSDA 2005, Cyprus 7
Therefore, a permutation of the columns of A changes the fill-in of R . ⇒ there is conflict between pivoting to minimize fill-ins and pivoting associated with numerical properties. We choose to decouple the sparse factorization phase and the rank- revealing phase For a standard QR factorization of a sparse matrix : Multi-frontal QR method [Amestoy-Duff-Puglisi ’94] Routine MA49AD in Library HSL. CSDA 2005, Cyprus 8
Column pivoting strategy Householder QR factorization using Householder reflections : H 1 A = A = v = A 2 .. 12 , 2 + || A 2 .. 12 , 2 || e 1 v = A 1 .. 12 , 1 + || A 1 .. 12 , 1 || e 1 H 2 = I 11 − 2 vv t H 1 = I 12 − 2 vv t v t v v t v Householder QR with Column Pivoting (Businger and Golub) : At step k , the column j k ≥ k defining the Householder transformation is chosen so that � A ( k : m, j k ) � ≥ � A ( k : m, j ) � for j ≥ k . CSDA 2005, Cyprus 9
Some properties on R 11 At step k : R ( k ) R ( k ) A ( k ) ≡ H k · · · H 1 AE ( k ) = . 11 12 A ( k ) 0 22 R ( k ) 12 ( k, :) satisfies R ( k ) Any column a of 11 ( k, k ) ≥ � a � . A ( k ) 22 � A � ≥ � R ( k ) 11 � ≥ R ( k ) Moreover 11 (1 , 1) R ( k ) 11 ( k, k ) ≥ σ min ( R ( k ) 11 ) ≥ σ min ( A ), 11 ) ≥ R ( k ) 11 (1 , 1) This implies that : cond ( A ) ≥ cond ( R ( k ) 11 ( k,k ) . R ( k ) CSDA 2005, Cyprus 10
Bad new The quantity R ( k ) 11 (1 , 1) 11 ( k,k ) cannot be considered as an estimator of cond ( R ( k ) 11 ) R ( k ) since it can be of different order of magnitude : Example (Kahan’s matrix of order n) : for θ ∈ (0 , π 2 ), c = cos θ and s = arcsin θ : d=c .ˆ [0:n-1] ; M=diag(d)*(eye(n)-s*triu(ones(n),1)); MATLAB : n=100 ; θ = arcsin(0 . 2) : σ min ( R 11 ) = 3 . 7 e − 9 ≪ R 11 (100 , 100) = 1 . 3 e − 1 Therefore a better estimate of cond ( R ( k ) 11 ) is needed. CSDA 2005, Cyprus 11
Incremental Condition Estimator (ICE) [Bischof 90] which is implemented in LAPACK : it estimates cond ( R ( k ) 11 ) from an estimation of cond ( R ( k − 1) ). 11 However, the strategy which consists in stopping the factorization when 11 ) < 1 cond ( R ( k ) ǫ ≤ cond ( R ( k +1) ) 11 may fail in very special situations : Counterexample : M = Kahan’s matrix of order 30 with θ = arccos(0 . 2) cond ( R (16) ǫ < cond ( R (17) ) < 1 ) 11 11 indicates a numerical rank equal to 16 although the numerical rank of M computed by SVD is 22. CSDA 2005, Cyprus 12
Pivoting strategies Convert R to reveal the rank by pushing the singularities towards the right end • Apply an Incremental Condition Estimator to evaluate σ max and σ min of R 1 → i, 1 → i , • if σ max /σ min > τ , move column i towards the right end and re- orthogonalise R using Givens rotations CSDA 2005, Cyprus 13
Pivoting strategies • RRQR uses a set of thresholds to overcome numerical singularities, • A predifined set of thresholds might fail 2 0 0 0 10 − 6 A = 0 1 1 10 − 3 − 10 − 3 0 0 rank=2 if τ = { 10 7 } , rank=3 if τ = { 10 4 , 10 7 } • RRQR adapts the thresholds to avoid failure � � � � � < σ max ⋆ Upon completion, check if � R 22 ≃ σ min � � � � τ � � ⋆ On failure, insert new thresholds and restart algorithm CSDA 2005, Cyprus 14
Numerical results 50 10 smax/smin 45 cond([q,r,p]=qr(A) 10 RRQR, thres=1e3,1e7,1e12,1e25 RRQR, thres=1e3,1e9,1e25 40 10 35 10 30 condition number 10 25 10 20 10 15 10 10 10 5 10 0 10 0 5 10 15 20 25 30 35 40 45 50 kahan(50,acos(0.2)), cond=2.4721e+49, rank=20, size=50x50, req=0 CSDA 2005, Cyprus 15
Numerical results 50 10 smax/smin 45 10 cond([q,r,p]=qr(A) RRQR, thres=1e30 40 10 RRQR, thres=1e20 RRQR, thres=1e25, req=2 35 10 30 condition number 10 25 10 20 10 15 10 10 10 5 10 0 10 0 5 10 15 20 25 30 35 40 45 50 kahan(50,acos(0.2)), cond=2.4721e+49, rank=20, size=50x50 CSDA 2005, Cyprus 16
Second strategy The k + 1 column is not the worst usually for the condition number. Can we select the worst one for the conditionement of R 11 ? Then we reject this worst column to the end. k+1 j k+1 We need to recompute the singular values and vectors estimates. Reverse ICE CSDA 2005, Cyprus 17
Second strategy Random matrix, n=500 20 10 Exact Estimated tol=1e−12 Estimated tol=1e−7 15 10 10 10 5 10 0 10 0 100 200 300 400 500 CSDA 2005, Cyprus 18
Conclusion Work is still on progress : Case where R is sparse • ICE might fail, so does the Reverse ICE. • Using the elimination tree, we can try to keep R as sparse as possible. CSDA 2005, Cyprus 19
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