*DECK INTRV
      SUBROUTINE INTRV (XT, LXT, X, ILO, ILEFT, MFLAG)
C***BEGIN PROLOGUE  INTRV
C***PURPOSE  Compute the largest integer ILEFT in 1 .LE. ILEFT .LE. LXT
C            such that XT(ILEFT) .LE. X where XT(*) is a subdivision
C            of the X interval.
C***LIBRARY   SLATEC
C***CATEGORY  E3, K6
C***TYPE      SINGLE PRECISION (INTRV-S, DINTRV-D)
C***KEYWORDS  B-SPLINE, DATA FITTING, INTERPOLATION, SPLINES
C***AUTHOR  Amos, D. E., (SNLA)
C***DESCRIPTION
C
C     Written by Carl de Boor and modified by D. E. Amos
C
C     Abstract
C         INTRV is the INTERV routine of the reference.
C
C         INTRV computes the largest integer ILEFT in 1 .LE. ILEFT .LE.
C         LXT such that XT(ILEFT) .LE. X where XT(*) is a subdivision of
C         the X interval.  Precisely,
C
C                      X .LT. XT(1)                1         -1
C         if  XT(I) .LE. X .LT. XT(I+1)  then  ILEFT=I  , MFLAG=0
C           XT(LXT) .LE. X                         LXT        1,
C
C         That is, when multiplicities are present in the break point
C         to the left of X, the largest index is taken for ILEFT.
C
C     Description of Arguments
C         Input
C          XT      - XT is a knot or break point vector of length LXT
C          LXT     - length of the XT vector
C          X       - argument
C          ILO     - an initialization parameter which must be set
C                    to 1 the first time the spline array XT is
C                    processed by INTRV.
C
C         Output
C          ILO     - ILO contains information for efficient process-
C                    ing after the initial call, and ILO must not be
C                    changed by the user.  Distinct splines require
C                    distinct ILO parameters.
C          ILEFT   - largest integer satisfying XT(ILEFT) .LE. X
C          MFLAG   - signals when X lies out of bounds
C
C     Error Conditions
C         None
C
C***REFERENCES  Carl de Boor, Package for calculating with B-splines,
C                 SIAM Journal on Numerical Analysis 14, 3 (June 1977),
C                 pp. 441-472.
C***ROUTINES CALLED  (NONE)
C***REVISION HISTORY  (YYMMDD)
C   800901  DATE WRITTEN
C   890831  Modified array declarations.  (WRB)
C   890831  REVISION DATE from Version 3.2
C   891214  Prologue converted to Version 4.0 format.  (BAB)
C   920501  Reformatted the REFERENCES section.  (WRB)
C***END PROLOGUE  INTRV
C
      INTEGER IHI, ILEFT, ILO, ISTEP, LXT, MFLAG, MIDDLE
      REAL X, XT
      DIMENSION XT(*)
C***FIRST EXECUTABLE STATEMENT  INTRV
      IHI = ILO + 1
      IF (IHI.LT.LXT) GO TO 10
      IF (X.GE.XT(LXT)) GO TO 110
      IF (LXT.LE.1) GO TO 90
      ILO = LXT - 1
      IHI = LXT
C
   10 IF (X.GE.XT(IHI)) GO TO 40
      IF (X.GE.XT(ILO)) GO TO 100
C
C *** NOW X .LT. XT(IHI) . FIND LOWER BOUND
      ISTEP = 1
   20 IHI = ILO
      ILO = IHI - ISTEP
      IF (ILO.LE.1) GO TO 30
      IF (X.GE.XT(ILO)) GO TO 70
      ISTEP = ISTEP*2
      GO TO 20
   30 ILO = 1
      IF (X.LT.XT(1)) GO TO 90
      GO TO 70
C *** NOW X .GE. XT(ILO) . FIND UPPER BOUND
   40 ISTEP = 1
   50 ILO = IHI
      IHI = ILO + ISTEP
      IF (IHI.GE.LXT) GO TO 60
      IF (X.LT.XT(IHI)) GO TO 70
      ISTEP = ISTEP*2
      GO TO 50
   60 IF (X.GE.XT(LXT)) GO TO 110
      IHI = LXT
C
C *** NOW XT(ILO) .LE. X .LT. XT(IHI) . NARROW THE INTERVAL
   70 MIDDLE = (ILO+IHI)/2
      IF (MIDDLE.EQ.ILO) GO TO 100
C     NOTE. IT IS ASSUMED THAT MIDDLE = ILO IN CASE IHI = ILO+1
      IF (X.LT.XT(MIDDLE)) GO TO 80
      ILO = MIDDLE
      GO TO 70
   80 IHI = MIDDLE
      GO TO 70
C *** SET OUTPUT AND RETURN
   90 MFLAG = -1
      ILEFT = 1
      RETURN
  100 MFLAG = 0
      ILEFT = ILO
      RETURN
  110 MFLAG = 1
      ILEFT = LXT
      RETURN
      END
