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