dlamch.f

InsightToolkit-1.4.0(有大量的优化算法程序)

         LEPS = A
*
*        Try some tricks to see whether or not this is the correct  EPS.
*
         B = TWO / 3
         HALF = ONE / 2
         SIXTH = DLAMC3( B, -HALF )
         THIRD = DLAMC3( SIXTH, SIXTH )
         B = DLAMC3( THIRD, -HALF )
         B = DLAMC3( B, SIXTH )
         B = ABS( B )
         IF( B.LT.LEPS )
     $      B = LEPS
*
         LEPS = 1
*
*+       WHILE( ( LEPS.GT.B ).AND.( B.GT.ZERO ) )LOOP
   10    CONTINUE
         IF( ( LEPS.GT.B ) .AND. ( B.GT.ZERO ) ) THEN
            LEPS = B
            C = DLAMC3( HALF*LEPS, ( TWO**5 )*( LEPS**2 ) )
            C = DLAMC3( HALF, -C )
            B = DLAMC3( HALF, C )
            C = DLAMC3( HALF, -B )
            B = DLAMC3( HALF, C )
            GO TO 10
         END IF
*+       END WHILE
*
         IF( A.LT.LEPS )
     $      LEPS = A
*
*        Computation of EPS complete.
*
*        Now find  EMIN.  Let A = + or - 1, and + or - (1 + BASE**(-3)).
*        Keep dividing  A by BETA until (gradual) underflow occurs. This
*        is detected when we cannot recover the previous A.
*
         RBASE = ONE / LBETA
         SMALL = ONE
         DO 20 I = 1, 3
            SMALL = DLAMC3( SMALL*RBASE, ZERO )
   20    CONTINUE
         A = DLAMC3( ONE, SMALL )
         CALL DLAMC4( NGPMIN, ONE, LBETA )
         CALL DLAMC4( NGNMIN, -ONE, LBETA )
         CALL DLAMC4( GPMIN, A, LBETA )
         CALL DLAMC4( GNMIN, -A, LBETA )
         IEEE = .FALSE.
*
         IF( ( NGPMIN.EQ.NGNMIN ) .AND. ( GPMIN.EQ.GNMIN ) ) THEN
            IF( NGPMIN.EQ.GPMIN ) THEN
               LEMIN = NGPMIN
*            ( Non twos-complement machines, no gradual underflow;
*              e.g.,  VAX )
            ELSE IF( ( GPMIN-NGPMIN ).EQ.3 ) THEN
               LEMIN = NGPMIN - 1 + LT
               IEEE = .TRUE.
*            ( Non twos-complement machines, with gradual underflow;
*              e.g., IEEE standard followers )
            ELSE
               LEMIN = MIN( NGPMIN, GPMIN )
*            ( A guess; no known machine )
               IWARN = .TRUE.
            END IF
*
         ELSE IF( ( NGPMIN.EQ.GPMIN ) .AND. ( NGNMIN.EQ.GNMIN ) ) THEN
            IF( ABS( NGPMIN-NGNMIN ).EQ.1 ) THEN
               LEMIN = MAX( NGPMIN, NGNMIN )
*            ( Twos-complement machines, no gradual underflow;
*              e.g., CYBER 205 )
            ELSE
               LEMIN = MIN( NGPMIN, NGNMIN )
*            ( A guess; no known machine )
               IWARN = .TRUE.
            END IF
*
         ELSE IF( ( ABS( NGPMIN-NGNMIN ).EQ.1 ) .AND.
     $            ( GPMIN.EQ.GNMIN ) ) THEN
            IF( ( GPMIN-MIN( NGPMIN, NGNMIN ) ).EQ.3 ) THEN
               LEMIN = MAX( NGPMIN, NGNMIN ) - 1 + LT
*            ( Twos-complement machines with gradual underflow;
*              no known machine )
            ELSE
               LEMIN = MIN( NGPMIN, NGNMIN )
*            ( A guess; no known machine )
               IWARN = .TRUE.
            END IF
*
         ELSE
            LEMIN = MIN( NGPMIN, NGNMIN, GPMIN, GNMIN )
*         ( A guess; no known machine )
            IWARN = .TRUE.
         END IF
***
* Comment out this if block if EMIN is ok
         IF( IWARN ) THEN
            FIRST = .TRUE.
            WRITE( 6, FMT = 9999 )LEMIN
         END IF
***
*
*        Assume IEEE arithmetic if we found denormalised  numbers above,
*        or if arithmetic seems to round in the  IEEE style,  determined
*        in routine DLAMC1. A true IEEE machine should have both  things
*        true; however, faulty machines may have one or the other.
*
         IEEE = IEEE .OR. LIEEE1
*
*        Compute  RMIN by successive division by  BETA. We could compute
*        RMIN as BASE**( EMIN - 1 ),  but some machines underflow during
*        this computation.
*
         LRMIN = 1
         DO 30 I = 1, 1 - LEMIN
            LRMIN = DLAMC3( LRMIN*RBASE, ZERO )
   30    CONTINUE
*
*        Finally, call DLAMC5 to compute EMAX and RMAX.
*
         CALL DLAMC5( LBETA, LT, LEMIN, IEEE, LEMAX, LRMAX )
      END IF
*
      BETA = LBETA
      T = LT
      RND = LRND
      EPS = LEPS
      EMIN = LEMIN
      RMIN = LRMIN
      EMAX = LEMAX
      RMAX = LRMAX
*
      RETURN
*
 9999 FORMAT( / / ' WARNING. The value EMIN may be incorrect:-',
     $      '  EMIN = ', I8, /
     $      ' If, after inspection, the value EMIN looks',
     $      ' acceptable please comment out ',
     $      / ' the IF block as marked within the code of routine',
     $      ' DLAMC2,', / ' otherwise supply EMIN explicitly.', / )
*
*     End of DLAMC2
*
      END
*
************************************************************************
*
      DOUBLE PRECISION FUNCTION DLAMC3( A, B )
*
*  -- LAPACK auxiliary routine (version 2.0) --
*     Univ. of Tennessee, Univ. of California Berkeley, NAG Ltd.,
*     Courant Institute, Argonne National Lab, and Rice University
*     October 31, 1992
*
*     .. Scalar Arguments ..
      DOUBLE PRECISION   A, B
*     ..
*
*  Purpose
*  =======
*
*  DLAMC3  is intended to force  A  and  B  to be stored prior to doing
*  the addition of  A  and  B ,  for use in situations where optimizers
*  might hold one of these in a register.
*
*  Arguments
*  =========
*
*  A, B    (input) DOUBLE PRECISION
*          The values A and B.
*
* =====================================================================
*
*     .. Executable Statements ..
*
      DLAMC3 = A + B
*
      RETURN
*
*     End of DLAMC3
*
      END
*
************************************************************************
*
      SUBROUTINE DLAMC4( EMIN, START, BASE )
*
*  -- LAPACK auxiliary routine (version 2.0) --
*     Univ. of Tennessee, Univ. of California Berkeley, NAG Ltd.,
*     Courant Institute, Argonne National Lab, and Rice University
*     October 31, 1992
*
*     .. Scalar Arguments ..
      INTEGER            BASE, EMIN
      DOUBLE PRECISION   START
*     ..
*
*  Purpose
*  =======
*
*  DLAMC4 is a service routine for DLAMC2.
*
*  Arguments
*  =========
*
*  EMIN    (output) EMIN
*          The minimum exponent before (gradual) underflow, computed by
*          setting A = START and dividing by BASE until the previous A
*          can not be recovered.
*
*  START   (input) DOUBLE PRECISION
*          The starting point for determining EMIN.
*
*  BASE    (input) INTEGER
*          The base of the machine.
*
* =====================================================================
*
*     .. Local Scalars ..
      INTEGER            I
      DOUBLE PRECISION   A, B1, B2, C1, C2, D1, D2, ONE, RBASE, ZERO
*     ..
*     .. External Functions ..
      DOUBLE PRECISION   DLAMC3
      EXTERNAL           DLAMC3
*     ..
*     .. Executable Statements ..
*
      A = START
      ONE = 1
      RBASE = ONE / BASE
      ZERO = 0
      EMIN = 1
      B1 = DLAMC3( A*RBASE, ZERO )
      C1 = A
      C2 = A
      D1 = A
      D2 = A
*+    WHILE( ( C1.EQ.A ).AND.( C2.EQ.A ).AND.
*    $       ( D1.EQ.A ).AND.( D2.EQ.A )      )LOOP
   10 CONTINUE
      IF( ( C1.EQ.A ) .AND. ( C2.EQ.A ) .AND. ( D1.EQ.A ) .AND.
     $    ( D2.EQ.A ) ) THEN
         EMIN = EMIN - 1
         A = B1
         B1 = DLAMC3( A / BASE, ZERO )
         C1 = DLAMC3( B1*BASE, ZERO )
         D1 = ZERO
         DO 20 I = 1, BASE
            D1 = D1 + B1
   20    CONTINUE
         B2 = DLAMC3( A*RBASE, ZERO )
         C2 = DLAMC3( B2 / RBASE, ZERO )
         D2 = ZERO
         DO 30 I = 1, BASE
            D2 = D2 + B2
   30    CONTINUE
         GO TO 10
      END IF
*+    END WHILE
*
      RETURN
*
*     End of DLAMC4
*
      END
*
************************************************************************
*
      SUBROUTINE DLAMC5( BETA, P, EMIN, IEEE, EMAX, RMAX )
*
*  -- LAPACK auxiliary routine (version 2.0) --
*     Univ. of Tennessee, Univ. of California Berkeley, NAG Ltd.,
*     Courant Institute, Argonne National Lab, and Rice University
*     October 31, 1992
*
*     .. Scalar Arguments ..
      LOGICAL            IEEE
      INTEGER            BETA, EMAX, EMIN, P
      DOUBLE PRECISION   RMAX
*     ..
*
*  Purpose
*  =======
*
*  DLAMC5 attempts to compute RMAX, the largest machine floating-point
*  number, without overflow.  It assumes that EMAX + abs(EMIN) sum
*  approximately to a power of 2.  It will fail on machines where this
*  assumption does not hold, for example, the Cyber 205 (EMIN = -28625,
*  EMAX = 28718).  It will also fail if the value supplied for EMIN is
*  too large (i.e. too close to zero), probably with overflow.
*
*  Arguments
*  =========
*
*  BETA    (input) INTEGER
*          The base of floating-point arithmetic.
*
*  P       (input) INTEGER
*          The number of base BETA digits in the mantissa of a
*          floating-point value.
*
*  EMIN    (input) INTEGER
*          The minimum exponent before (gradual) underflow.
*
*  IEEE    (input) LOGICAL
*          A logical flag specifying whether or not the arithmetic
*          system is thought to comply with the IEEE standard.
*
*  EMAX    (output) INTEGER
*          The largest exponent before overflow
*
*  RMAX    (output) DOUBLE PRECISION
*          The largest machine floating-point number.
*
* =====================================================================
*
*     .. Parameters ..
      DOUBLE PRECISION   ZERO, ONE
      PARAMETER          ( ZERO = 0.0D0, ONE = 1.0D0 )
*     ..
*     .. Local Scalars ..
      INTEGER            EXBITS, EXPSUM, I, LEXP, NBITS, TRY, UEXP
      DOUBLE PRECISION   OLDY, RECBAS, Y, Z
*     ..
*     .. External Functions ..
      DOUBLE PRECISION   DLAMC3
      EXTERNAL           DLAMC3
*     ..
*     .. Intrinsic Functions ..
      INTRINSIC          MOD
*     ..
*     .. Executable Statements ..
*
*     First compute LEXP and UEXP, two powers of 2 that bound
*     abs(EMIN). We then assume that EMAX + abs(EMIN) will sum
*     approximately to the bound that is closest to abs(EMIN).
*     (EMAX is the exponent of the required number RMAX).
*
      LEXP = 1
      EXBITS = 1
   10 CONTINUE
      TRY = LEXP*2
      IF( TRY.LE.( -EMIN ) ) THEN
         LEXP = TRY
         EXBITS = EXBITS + 1
         GO TO 10
      END IF
      IF( LEXP.EQ.-EMIN ) THEN
         UEXP = LEXP
      ELSE
         UEXP = TRY
         EXBITS = EXBITS + 1
      END IF
*
*     Now -LEXP is less than or equal to EMIN, and -UEXP is greater
*     than or equal to EMIN. EXBITS is the number of bits needed to
*     store the exponent.
*
      IF( ( UEXP+EMIN ).GT.( -LEXP-EMIN ) ) THEN
         EXPSUM = 2*LEXP
      ELSE
         EXPSUM = 2*UEXP
      END IF
*
*     EXPSUM is the exponent range, approximately equal to
*     EMAX - EMIN + 1 .
*
      EMAX = EXPSUM + EMIN - 1
      NBITS = 1 + EXBITS + P
*
*     NBITS is the total number of bits needed to store a
*     floating-point number.
*
      IF( ( MOD( NBITS, 2 ).EQ.1 ) .AND. ( BETA.EQ.2 ) ) THEN
*
*        Either there are an odd number of bits used to store a
*        floating-point number, which is unlikely, or some bits are
*        not used in the representation of numbers, which is possible,
*        (e.g. Cray machines) or the mantissa has an implicit bit,
*        (e.g. IEEE machines, Dec Vax machines), which is perhaps the
*        most likely. We have to assume the last alternative.
*        If this is true, then we need to reduce EMAX by one because
*        there must be some way of representing zero in an implicit-bit
*        system. On machines like Cray, we are reducing EMAX by one
*        unnecessarily.
*
         EMAX = EMAX - 1
      END IF
*
      IF( IEEE ) THEN
*
*        Assume we are on an IEEE machine which reserves one exponent
*        for infinity and NaN.
*
         EMAX = EMAX - 1
      END IF
*
*     Now create RMAX, the largest machine number, which should
*     be equal to (1.0 - BETA**(-P)) * BETA**EMAX .
*
*     First compute 1.0 - BETA**(-P), being careful that the
*     result is less than 1.0 .
*
      RECBAS = ONE / BETA
      Z = BETA - ONE
      Y = ZERO
      DO 20 I = 1, P
         Z = Z*RECBAS
         IF( Y.LT.ONE )
     $      OLDY = Y
         Y = DLAMC3( Y, Z )
   20 CONTINUE
      IF( Y.GE.ONE )
     $   Y = OLDY
*
*     Now multiply by BETA**EMAX to get RMAX.
*
      DO 30 I = 1, EMAX
         Y = DLAMC3( Y*BETA, ZERO )
   30 CONTINUE
*
      RMAX = Y
      RETURN
*
*     End of DLAMC5
*
      END