168 lines
4.7 KiB
FortranFixed
168 lines
4.7 KiB
FortranFixed
SUBROUTINE CPBEQU( UPLO, N, KD, AB, LDAB, S, SCOND, AMAX, INFO )
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*
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* -- LAPACK routine (version 3.1) --
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* Univ. of Tennessee, Univ. of California Berkeley and NAG Ltd..
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* November 2006
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*
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* .. Scalar Arguments ..
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CHARACTER UPLO
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INTEGER INFO, KD, LDAB, N
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REAL AMAX, SCOND
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* ..
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* .. Array Arguments ..
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REAL S( * )
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COMPLEX AB( LDAB, * )
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* ..
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*
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* Purpose
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* =======
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*
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* CPBEQU computes row and column scalings intended to equilibrate a
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* Hermitian positive definite band matrix A and reduce its condition
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* number (with respect to the two-norm). S contains the scale factors,
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* S(i) = 1/sqrt(A(i,i)), chosen so that the scaled matrix B with
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* elements B(i,j) = S(i)*A(i,j)*S(j) has ones on the diagonal. This
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* choice of S puts the condition number of B within a factor N of the
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* smallest possible condition number over all possible diagonal
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* scalings.
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*
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* Arguments
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* =========
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*
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* UPLO (input) CHARACTER*1
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* = 'U': Upper triangular of A is stored;
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* = 'L': Lower triangular of A is stored.
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*
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* N (input) INTEGER
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* The order of the matrix A. N >= 0.
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*
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* KD (input) INTEGER
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* The number of superdiagonals of the matrix A if UPLO = 'U',
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* or the number of subdiagonals if UPLO = 'L'. KD >= 0.
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*
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* AB (input) COMPLEX array, dimension (LDAB,N)
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* The upper or lower triangle of the Hermitian band matrix A,
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* stored in the first KD+1 rows of the array. The j-th column
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* of A is stored in the j-th column of the array AB as follows:
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* if UPLO = 'U', AB(kd+1+i-j,j) = A(i,j) for max(1,j-kd)<=i<=j;
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* if UPLO = 'L', AB(1+i-j,j) = A(i,j) for j<=i<=min(n,j+kd).
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*
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* LDAB (input) INTEGER
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* The leading dimension of the array A. LDAB >= KD+1.
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*
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* S (output) REAL array, dimension (N)
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* If INFO = 0, S contains the scale factors for A.
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*
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* SCOND (output) REAL
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* If INFO = 0, S contains the ratio of the smallest S(i) to
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* the largest S(i). If SCOND >= 0.1 and AMAX is neither too
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* large nor too small, it is not worth scaling by S.
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*
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* AMAX (output) REAL
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* Absolute value of largest matrix element. If AMAX is very
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* close to overflow or very close to underflow, the matrix
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* should be scaled.
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*
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* INFO (output) INTEGER
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* = 0: successful exit
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* < 0: if INFO = -i, the i-th argument had an illegal value.
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* > 0: if INFO = i, the i-th diagonal element is nonpositive.
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*
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* =====================================================================
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*
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* .. Parameters ..
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REAL ZERO, ONE
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PARAMETER ( ZERO = 0.0E+0, ONE = 1.0E+0 )
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* ..
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* .. Local Scalars ..
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LOGICAL UPPER
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INTEGER I, J
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REAL SMIN
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* ..
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* .. External Functions ..
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LOGICAL LSAME
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EXTERNAL LSAME
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* ..
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* .. External Subroutines ..
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EXTERNAL XERBLA
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* ..
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* .. Intrinsic Functions ..
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INTRINSIC MAX, MIN, REAL, SQRT
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* ..
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* .. Executable Statements ..
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*
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* Test the input parameters.
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*
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INFO = 0
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UPPER = LSAME( UPLO, 'U' )
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IF( .NOT.UPPER .AND. .NOT.LSAME( UPLO, 'L' ) ) THEN
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INFO = -1
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ELSE IF( N.LT.0 ) THEN
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INFO = -2
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ELSE IF( KD.LT.0 ) THEN
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INFO = -3
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ELSE IF( LDAB.LT.KD+1 ) THEN
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INFO = -5
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END IF
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IF( INFO.NE.0 ) THEN
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CALL XERBLA( 'CPBEQU', -INFO )
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RETURN
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END IF
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*
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* Quick return if possible
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*
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IF( N.EQ.0 ) THEN
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SCOND = ONE
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AMAX = ZERO
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RETURN
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END IF
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*
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IF( UPPER ) THEN
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J = KD + 1
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ELSE
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J = 1
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END IF
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*
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* Initialize SMIN and AMAX.
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*
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S( 1 ) = REAL( AB( J, 1 ) )
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SMIN = S( 1 )
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AMAX = S( 1 )
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*
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* Find the minimum and maximum diagonal elements.
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*
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DO 10 I = 2, N
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S( I ) = REAL( AB( J, I ) )
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SMIN = MIN( SMIN, S( I ) )
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AMAX = MAX( AMAX, S( I ) )
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10 CONTINUE
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*
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IF( SMIN.LE.ZERO ) THEN
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*
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* Find the first non-positive diagonal element and return.
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*
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DO 20 I = 1, N
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IF( S( I ).LE.ZERO ) THEN
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INFO = I
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RETURN
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END IF
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20 CONTINUE
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ELSE
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*
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* Set the scale factors to the reciprocals
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* of the diagonal elements.
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*
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DO 30 I = 1, N
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S( I ) = ONE / SQRT( S( I ) )
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30 CONTINUE
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*
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* Compute SCOND = min(S(I)) / max(S(I))
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*
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SCOND = SQRT( SMIN ) / SQRT( AMAX )
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END IF
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RETURN
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*
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* End of CPBEQU
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*
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END
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