| 1 | SUBROUTINE SLALSA( ICOMPQ, SMLSIZ, N, NRHS, B, LDB, BX, LDBX, U,
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| 2 | $ LDU, VT, K, DIFL, DIFR, Z, POLES, GIVPTR,
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| 3 | $ GIVCOL, LDGCOL, PERM, GIVNUM, C, S, WORK,
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| 4 | $ IWORK, INFO )
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| 5 | *
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| 6 | * -- LAPACK routine (version 3.1) --
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| 7 | * Univ. of Tennessee, Univ. of California Berkeley and NAG Ltd..
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| 8 | * November 2006
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| 9 | *
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| 10 | * .. Scalar Arguments ..
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| 11 | INTEGER ICOMPQ, INFO, LDB, LDBX, LDGCOL, LDU, N, NRHS,
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| 12 | $ SMLSIZ
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| 13 | * ..
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| 14 | * .. Array Arguments ..
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| 15 | INTEGER GIVCOL( LDGCOL, * ), GIVPTR( * ), IWORK( * ),
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| 16 | $ K( * ), PERM( LDGCOL, * )
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| 17 | REAL B( LDB, * ), BX( LDBX, * ), C( * ),
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| 18 | $ DIFL( LDU, * ), DIFR( LDU, * ),
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| 19 | $ GIVNUM( LDU, * ), POLES( LDU, * ), S( * ),
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| 20 | $ U( LDU, * ), VT( LDU, * ), WORK( * ),
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| 21 | $ Z( LDU, * )
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| 22 | * ..
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| 23 | *
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| 24 | * Purpose
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| 25 | * =======
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| 26 | *
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| 27 | * SLALSA is an itermediate step in solving the least squares problem
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| 28 | * by computing the SVD of the coefficient matrix in compact form (The
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| 29 | * singular vectors are computed as products of simple orthorgonal
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| 30 | * matrices.).
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| 31 | *
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| 32 | * If ICOMPQ = 0, SLALSA applies the inverse of the left singular vector
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| 33 | * matrix of an upper bidiagonal matrix to the right hand side; and if
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| 34 | * ICOMPQ = 1, SLALSA applies the right singular vector matrix to the
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| 35 | * right hand side. The singular vector matrices were generated in
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| 36 | * compact form by SLALSA.
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| 37 | *
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| 38 | * Arguments
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| 39 | * =========
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| 40 | *
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| 41 | *
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| 42 | * ICOMPQ (input) INTEGER
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| 43 | * Specifies whether the left or the right singular vector
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| 44 | * matrix is involved.
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| 45 | * = 0: Left singular vector matrix
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| 46 | * = 1: Right singular vector matrix
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| 47 | *
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| 48 | * SMLSIZ (input) INTEGER
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| 49 | * The maximum size of the subproblems at the bottom of the
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| 50 | * computation tree.
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| 51 | *
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| 52 | * N (input) INTEGER
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| 53 | * The row and column dimensions of the upper bidiagonal matrix.
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| 54 | *
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| 55 | * NRHS (input) INTEGER
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| 56 | * The number of columns of B and BX. NRHS must be at least 1.
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| 57 | *
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| 58 | * B (input/output) REAL array, dimension ( LDB, NRHS )
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| 59 | * On input, B contains the right hand sides of the least
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| 60 | * squares problem in rows 1 through M.
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| 61 | * On output, B contains the solution X in rows 1 through N.
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| 62 | *
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| 63 | * LDB (input) INTEGER
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| 64 | * The leading dimension of B in the calling subprogram.
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| 65 | * LDB must be at least max(1,MAX( M, N ) ).
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| 66 | *
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| 67 | * BX (output) REAL array, dimension ( LDBX, NRHS )
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| 68 | * On exit, the result of applying the left or right singular
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| 69 | * vector matrix to B.
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| 70 | *
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| 71 | * LDBX (input) INTEGER
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| 72 | * The leading dimension of BX.
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| 73 | *
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| 74 | * U (input) REAL array, dimension ( LDU, SMLSIZ ).
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| 75 | * On entry, U contains the left singular vector matrices of all
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| 76 | * subproblems at the bottom level.
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| 77 | *
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| 78 | * LDU (input) INTEGER, LDU = > N.
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| 79 | * The leading dimension of arrays U, VT, DIFL, DIFR,
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| 80 | * POLES, GIVNUM, and Z.
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| 81 | *
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| 82 | * VT (input) REAL array, dimension ( LDU, SMLSIZ+1 ).
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| 83 | * On entry, VT' contains the right singular vector matrices of
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| 84 | * all subproblems at the bottom level.
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| 85 | *
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| 86 | * K (input) INTEGER array, dimension ( N ).
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| 87 | *
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| 88 | * DIFL (input) REAL array, dimension ( LDU, NLVL ).
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| 89 | * where NLVL = INT(log_2 (N/(SMLSIZ+1))) + 1.
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| 90 | *
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| 91 | * DIFR (input) REAL array, dimension ( LDU, 2 * NLVL ).
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| 92 | * On entry, DIFL(*, I) and DIFR(*, 2 * I -1) record
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| 93 | * distances between singular values on the I-th level and
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| 94 | * singular values on the (I -1)-th level, and DIFR(*, 2 * I)
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| 95 | * record the normalizing factors of the right singular vectors
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| 96 | * matrices of subproblems on I-th level.
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| 97 | *
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| 98 | * Z (input) REAL array, dimension ( LDU, NLVL ).
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| 99 | * On entry, Z(1, I) contains the components of the deflation-
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| 100 | * adjusted updating row vector for subproblems on the I-th
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| 101 | * level.
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| 102 | *
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| 103 | * POLES (input) REAL array, dimension ( LDU, 2 * NLVL ).
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| 104 | * On entry, POLES(*, 2 * I -1: 2 * I) contains the new and old
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| 105 | * singular values involved in the secular equations on the I-th
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| 106 | * level.
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| 107 | *
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| 108 | * GIVPTR (input) INTEGER array, dimension ( N ).
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| 109 | * On entry, GIVPTR( I ) records the number of Givens
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| 110 | * rotations performed on the I-th problem on the computation
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| 111 | * tree.
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| 112 | *
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| 113 | * GIVCOL (input) INTEGER array, dimension ( LDGCOL, 2 * NLVL ).
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| 114 | * On entry, for each I, GIVCOL(*, 2 * I - 1: 2 * I) records the
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| 115 | * locations of Givens rotations performed on the I-th level on
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| 116 | * the computation tree.
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| 117 | *
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| 118 | * LDGCOL (input) INTEGER, LDGCOL = > N.
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| 119 | * The leading dimension of arrays GIVCOL and PERM.
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| 120 | *
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| 121 | * PERM (input) INTEGER array, dimension ( LDGCOL, NLVL ).
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| 122 | * On entry, PERM(*, I) records permutations done on the I-th
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| 123 | * level of the computation tree.
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| 124 | *
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| 125 | * GIVNUM (input) REAL array, dimension ( LDU, 2 * NLVL ).
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| 126 | * On entry, GIVNUM(*, 2 *I -1 : 2 * I) records the C- and S-
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| 127 | * values of Givens rotations performed on the I-th level on the
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| 128 | * computation tree.
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| 129 | *
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| 130 | * C (input) REAL array, dimension ( N ).
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| 131 | * On entry, if the I-th subproblem is not square,
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| 132 | * C( I ) contains the C-value of a Givens rotation related to
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| 133 | * the right null space of the I-th subproblem.
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| 134 | *
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| 135 | * S (input) REAL array, dimension ( N ).
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| 136 | * On entry, if the I-th subproblem is not square,
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| 137 | * S( I ) contains the S-value of a Givens rotation related to
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| 138 | * the right null space of the I-th subproblem.
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| 139 | *
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| 140 | * WORK (workspace) REAL array.
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| 141 | * The dimension must be at least N.
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| 142 | *
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| 143 | * IWORK (workspace) INTEGER array.
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| 144 | * The dimension must be at least 3 * N
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| 145 | *
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| 146 | * INFO (output) INTEGER
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| 147 | * = 0: successful exit.
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| 148 | * < 0: if INFO = -i, the i-th argument had an illegal value.
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| 149 | *
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| 150 | * Further Details
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| 151 | * ===============
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| 152 | *
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| 153 | * Based on contributions by
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| 154 | * Ming Gu and Ren-Cang Li, Computer Science Division, University of
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| 155 | * California at Berkeley, USA
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| 156 | * Osni Marques, LBNL/NERSC, USA
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| 157 | *
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| 158 | * =====================================================================
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| 159 | *
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| 160 | * .. Parameters ..
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| 161 | REAL ZERO, ONE
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| 162 | PARAMETER ( ZERO = 0.0E0, ONE = 1.0E0 )
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| 163 | * ..
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| 164 | * .. Local Scalars ..
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| 165 | INTEGER I, I1, IC, IM1, INODE, J, LF, LL, LVL, LVL2,
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| 166 | $ ND, NDB1, NDIML, NDIMR, NL, NLF, NLP1, NLVL,
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| 167 | $ NR, NRF, NRP1, SQRE
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| 168 | * ..
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| 169 | * .. External Subroutines ..
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| 170 | EXTERNAL SCOPY, SGEMM, SLALS0, SLASDT, XERBLA
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| 171 | * ..
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| 172 | * .. Executable Statements ..
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| 173 | *
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| 174 | * Test the input parameters.
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| 175 | *
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| 176 | INFO = 0
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| 177 | *
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| 178 | IF( ( ICOMPQ.LT.0 ) .OR. ( ICOMPQ.GT.1 ) ) THEN
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| 179 | INFO = -1
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| 180 | ELSE IF( SMLSIZ.LT.3 ) THEN
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| 181 | INFO = -2
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| 182 | ELSE IF( N.LT.SMLSIZ ) THEN
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| 183 | INFO = -3
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| 184 | ELSE IF( NRHS.LT.1 ) THEN
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| 185 | INFO = -4
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| 186 | ELSE IF( LDB.LT.N ) THEN
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| 187 | INFO = -6
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| 188 | ELSE IF( LDBX.LT.N ) THEN
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| 189 | INFO = -8
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| 190 | ELSE IF( LDU.LT.N ) THEN
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| 191 | INFO = -10
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| 192 | ELSE IF( LDGCOL.LT.N ) THEN
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| 193 | INFO = -19
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| 194 | END IF
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| 195 | IF( INFO.NE.0 ) THEN
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| 196 | CALL XERBLA( 'SLALSA', -INFO )
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| 197 | RETURN
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| 198 | END IF
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| 199 | *
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| 200 | * Book-keeping and setting up the computation tree.
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| 201 | *
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| 202 | INODE = 1
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| 203 | NDIML = INODE + N
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| 204 | NDIMR = NDIML + N
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| 205 | *
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| 206 | CALL SLASDT( N, NLVL, ND, IWORK( INODE ), IWORK( NDIML ),
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| 207 | $ IWORK( NDIMR ), SMLSIZ )
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| 208 | *
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| 209 | * The following code applies back the left singular vector factors.
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| 210 | * For applying back the right singular vector factors, go to 50.
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| 211 | *
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| 212 | IF( ICOMPQ.EQ.1 ) THEN
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| 213 | GO TO 50
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| 214 | END IF
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| 215 | *
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| 216 | * The nodes on the bottom level of the tree were solved
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| 217 | * by SLASDQ. The corresponding left and right singular vector
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| 218 | * matrices are in explicit form. First apply back the left
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| 219 | * singular vector matrices.
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| 220 | *
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| 221 | NDB1 = ( ND+1 ) / 2
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| 222 | DO 10 I = NDB1, ND
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| 223 | *
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| 224 | * IC : center row of each node
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| 225 | * NL : number of rows of left subproblem
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| 226 | * NR : number of rows of right subproblem
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| 227 | * NLF: starting row of the left subproblem
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| 228 | * NRF: starting row of the right subproblem
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| 229 | *
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| 230 | I1 = I - 1
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| 231 | IC = IWORK( INODE+I1 )
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| 232 | NL = IWORK( NDIML+I1 )
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| 233 | NR = IWORK( NDIMR+I1 )
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| 234 | NLF = IC - NL
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| 235 | NRF = IC + 1
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| 236 | CALL SGEMM( 'T', 'N', NL, NRHS, NL, ONE, U( NLF, 1 ), LDU,
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| 237 | $ B( NLF, 1 ), LDB, ZERO, BX( NLF, 1 ), LDBX )
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| 238 | CALL SGEMM( 'T', 'N', NR, NRHS, NR, ONE, U( NRF, 1 ), LDU,
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| 239 | $ B( NRF, 1 ), LDB, ZERO, BX( NRF, 1 ), LDBX )
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| 240 | 10 CONTINUE
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| 241 | *
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| 242 | * Next copy the rows of B that correspond to unchanged rows
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| 243 | * in the bidiagonal matrix to BX.
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| 244 | *
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| 245 | DO 20 I = 1, ND
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| 246 | IC = IWORK( INODE+I-1 )
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| 247 | CALL SCOPY( NRHS, B( IC, 1 ), LDB, BX( IC, 1 ), LDBX )
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| 248 | 20 CONTINUE
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| 249 | *
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| 250 | * Finally go through the left singular vector matrices of all
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| 251 | * the other subproblems bottom-up on the tree.
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| 252 | *
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| 253 | J = 2**NLVL
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| 254 | SQRE = 0
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| 255 | *
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| 256 | DO 40 LVL = NLVL, 1, -1
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| 257 | LVL2 = 2*LVL - 1
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| 258 | *
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| 259 | * find the first node LF and last node LL on
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| 260 | * the current level LVL
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| 261 | *
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| 262 | IF( LVL.EQ.1 ) THEN
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| 263 | LF = 1
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| 264 | LL = 1
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| 265 | ELSE
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| 266 | LF = 2**( LVL-1 )
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| 267 | LL = 2*LF - 1
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| 268 | END IF
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| 269 | DO 30 I = LF, LL
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| 270 | IM1 = I - 1
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| 271 | IC = IWORK( INODE+IM1 )
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| 272 | NL = IWORK( NDIML+IM1 )
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| 273 | NR = IWORK( NDIMR+IM1 )
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| 274 | NLF = IC - NL
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| 275 | NRF = IC + 1
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| 276 | J = J - 1
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| 277 | CALL SLALS0( ICOMPQ, NL, NR, SQRE, NRHS, BX( NLF, 1 ), LDBX,
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| 278 | $ B( NLF, 1 ), LDB, PERM( NLF, LVL ),
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| 279 | $ GIVPTR( J ), GIVCOL( NLF, LVL2 ), LDGCOL,
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| 280 | $ GIVNUM( NLF, LVL2 ), LDU, POLES( NLF, LVL2 ),
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| 281 | $ DIFL( NLF, LVL ), DIFR( NLF, LVL2 ),
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| 282 | $ Z( NLF, LVL ), K( J ), C( J ), S( J ), WORK,
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| 283 | $ INFO )
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| 284 | 30 CONTINUE
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| 285 | 40 CONTINUE
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| 286 | GO TO 90
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| 287 | *
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| 288 | * ICOMPQ = 1: applying back the right singular vector factors.
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| 289 | *
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| 290 | 50 CONTINUE
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| 291 | *
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| 292 | * First now go through the right singular vector matrices of all
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| 293 | * the tree nodes top-down.
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| 294 | *
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| 295 | J = 0
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| 296 | DO 70 LVL = 1, NLVL
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| 297 | LVL2 = 2*LVL - 1
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| 298 | *
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| 299 | * Find the first node LF and last node LL on
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| 300 | * the current level LVL.
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| 301 | *
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| 302 | IF( LVL.EQ.1 ) THEN
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| 303 | LF = 1
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| 304 | LL = 1
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| 305 | ELSE
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| 306 | LF = 2**( LVL-1 )
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| 307 | LL = 2*LF - 1
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| 308 | END IF
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| 309 | DO 60 I = LL, LF, -1
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| 310 | IM1 = I - 1
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| 311 | IC = IWORK( INODE+IM1 )
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| 312 | NL = IWORK( NDIML+IM1 )
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| 313 | NR = IWORK( NDIMR+IM1 )
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| 314 | NLF = IC - NL
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| 315 | NRF = IC + 1
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| 316 | IF( I.EQ.LL ) THEN
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| 317 | SQRE = 0
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| 318 | ELSE
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| 319 | SQRE = 1
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| 320 | END IF
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| 321 | J = J + 1
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| 322 | CALL SLALS0( ICOMPQ, NL, NR, SQRE, NRHS, B( NLF, 1 ), LDB,
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| 323 | $ BX( NLF, 1 ), LDBX, PERM( NLF, LVL ),
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| 324 | $ GIVPTR( J ), GIVCOL( NLF, LVL2 ), LDGCOL,
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| 325 | $ GIVNUM( NLF, LVL2 ), LDU, POLES( NLF, LVL2 ),
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| 326 | $ DIFL( NLF, LVL ), DIFR( NLF, LVL2 ),
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| 327 | $ Z( NLF, LVL ), K( J ), C( J ), S( J ), WORK,
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| 328 | $ INFO )
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| 329 | 60 CONTINUE
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| 330 | 70 CONTINUE
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| 331 | *
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| 332 | * The nodes on the bottom level of the tree were solved
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| 333 | * by SLASDQ. The corresponding right singular vector
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| 334 | * matrices are in explicit form. Apply them back.
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| 335 | *
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| 336 | NDB1 = ( ND+1 ) / 2
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| 337 | DO 80 I = NDB1, ND
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| 338 | I1 = I - 1
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| 339 | IC = IWORK( INODE+I1 )
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| 340 | NL = IWORK( NDIML+I1 )
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| 341 | NR = IWORK( NDIMR+I1 )
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| 342 | NLP1 = NL + 1
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| 343 | IF( I.EQ.ND ) THEN
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| 344 | NRP1 = NR
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| 345 | ELSE
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| 346 | NRP1 = NR + 1
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| 347 | END IF
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| 348 | NLF = IC - NL
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| 349 | NRF = IC + 1
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| 350 | CALL SGEMM( 'T', 'N', NLP1, NRHS, NLP1, ONE, VT( NLF, 1 ), LDU,
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| 351 | $ B( NLF, 1 ), LDB, ZERO, BX( NLF, 1 ), LDBX )
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| 352 | CALL SGEMM( 'T', 'N', NRP1, NRHS, NRP1, ONE, VT( NRF, 1 ), LDU,
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| 353 | $ B( NRF, 1 ), LDB, ZERO, BX( NRF, 1 ), LDBX )
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| 354 | 80 CONTINUE
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| 355 | *
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| 356 | 90 CONTINUE
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| 357 | *
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| 358 | RETURN
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| 359 | *
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| 360 | * End of SLALSA
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| 361 | *
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| 362 | END
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