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c977aa998f
Replace amos with slatec
340 lines
10 KiB
Fortran
340 lines
10 KiB
Fortran
*DECK HPSORT
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SUBROUTINE HPSORT (HX, N, STRBEG, STREND, IPERM, KFLAG, WORK, IER)
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C***BEGIN PROLOGUE HPSORT
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C***PURPOSE Return the permutation vector generated by sorting a
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C substring within a character array and, optionally,
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C rearrange the elements of the array. The array may be
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C sorted in forward or reverse lexicographical order. A
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C slightly modified quicksort algorithm is used.
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C***LIBRARY SLATEC
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C***CATEGORY N6A1C, N6A2C
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C***TYPE CHARACTER (SPSORT-S, DPSORT-D, IPSORT-I, HPSORT-H)
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C***KEYWORDS PASSIVE SORTING, SINGLETON QUICKSORT, SORT, STRING SORTING
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C***AUTHOR Jones, R. E., (SNLA)
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C Rhoads, G. S., (NBS)
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C Sullivan, F. E., (NBS)
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C Wisniewski, J. A., (SNLA)
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C***DESCRIPTION
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C
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C HPSORT returns the permutation vector IPERM generated by sorting
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C the substrings beginning with the character STRBEG and ending with
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C the character STREND within the strings in array HX and, optionally,
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C rearranges the strings in HX. HX may be sorted in increasing or
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C decreasing lexicographical order. A slightly modified quicksort
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C algorithm is used.
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C
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C IPERM is such that HX(IPERM(I)) is the Ith value in the
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C rearrangement of HX. IPERM may be applied to another array by
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C calling IPPERM, SPPERM, DPPERM or HPPERM.
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C
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C An active sort of numerical data is expected to execute somewhat
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C more quickly than a passive sort because there is no need to use
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C indirect references. But for the character data in HPSORT, integers
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C in the IPERM vector are manipulated rather than the strings in HX.
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C Moving integers may be enough faster than moving character strings
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C to more than offset the penalty of indirect referencing.
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C
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C Description of Parameters
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C HX - input/output -- array of type character to be sorted.
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C For example, to sort a 80 element array of names,
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C each of length 6, declare HX as character HX(100)*6.
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C If ABS(KFLAG) = 2, then the values in HX will be
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C rearranged on output; otherwise, they are unchanged.
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C N - input -- number of values in array HX to be sorted.
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C STRBEG - input -- the index of the initial character in
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C the string HX that is to be sorted.
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C STREND - input -- the index of the final character in
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C the string HX that is to be sorted.
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C IPERM - output -- permutation array such that IPERM(I) is the
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C index of the string in the original order of the
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C HX array that is in the Ith location in the sorted
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C order.
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C KFLAG - input -- control parameter:
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C = 2 means return the permutation vector resulting from
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C sorting HX in lexicographical order and sort HX also.
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C = 1 means return the permutation vector resulting from
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C sorting HX in lexicographical order and do not sort
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C HX.
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C = -1 means return the permutation vector resulting from
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C sorting HX in reverse lexicographical order and do
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C not sort HX.
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C = -2 means return the permutation vector resulting from
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C sorting HX in reverse lexicographical order and sort
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C HX also.
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C WORK - character variable which must have a length specification
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C at least as great as that of HX.
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C IER - output -- error indicator:
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C = 0 if no error,
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C = 1 if N is zero or negative,
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C = 2 if KFLAG is not 2, 1, -1, or -2,
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C = 3 if work array is not long enough,
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C = 4 if string beginning is beyond its end,
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C = 5 if string beginning is out-of-range,
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C = 6 if string end is out-of-range.
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C
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C E X A M P L E O F U S E
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C
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C CHARACTER*2 HX, W
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C INTEGER STRBEG, STREND
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C DIMENSION HX(10), IPERM(10)
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C DATA (HX(I),I=1,10)/ '05','I ',' I',' ','Rs','9R','R9','89',
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C 1 ',*','N"'/
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C DATA STRBEG, STREND / 1, 2 /
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C CALL HPSORT (HX,10,STRBEG,STREND,IPERM,1,W)
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C PRINT 100, (HX(IPERM(I)),I=1,10)
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C 100 FORMAT (2X, A2)
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C STOP
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C END
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C
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C***REFERENCES R. C. Singleton, Algorithm 347, An efficient algorithm
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C for sorting with minimal storage, Communications of
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C the ACM, 12, 3 (1969), pp. 185-187.
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C***ROUTINES CALLED XERMSG
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C***REVISION HISTORY (YYMMDD)
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C 761101 DATE WRITTEN
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C 761118 Modified by John A. Wisniewski to use the Singleton
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C quicksort algorithm.
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C 811001 Modified by Francis Sullivan for string data.
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C 850326 Documentation slightly modified by D. Kahaner.
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C 870423 Modified by Gregory S. Rhoads for passive sorting with the
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C option for the rearrangement of the original data.
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C 890620 Algorithm for rearranging the data vector corrected by R.
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C Boisvert.
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C 890622 Prologue upgraded to Version 4.0 style by D. Lozier.
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C 920507 Modified by M. McClain to revise prologue text.
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C 920818 Declarations section rebuilt and code restructured to use
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C IF-THEN-ELSE-ENDIF. (SMR, WRB)
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C***END PROLOGUE HPSORT
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C .. Scalar Arguments ..
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INTEGER IER, KFLAG, N, STRBEG, STREND
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CHARACTER * (*) WORK
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C .. Array Arguments ..
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INTEGER IPERM(*)
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CHARACTER * (*) HX(*)
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C .. Local Scalars ..
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REAL R
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INTEGER I, IJ, INDX, INDX0, IR, ISTRT, J, K, KK, L, LM, LMT, M,
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+ NN, NN2
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C .. Local Arrays ..
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INTEGER IL(21), IU(21)
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C .. External Subroutines ..
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EXTERNAL XERMSG
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C .. Intrinsic Functions ..
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INTRINSIC ABS, INT, LEN
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C***FIRST EXECUTABLE STATEMENT HPSORT
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IER = 0
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NN = N
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IF (NN .LT. 1) THEN
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IER = 1
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CALL XERMSG ('SLATEC', 'HPSORT',
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+ 'The number of values to be sorted, N, is not positive.',
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+ IER, 1)
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RETURN
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ENDIF
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KK = ABS(KFLAG)
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IF (KK.NE.1 .AND. KK.NE.2) THEN
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IER = 2
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CALL XERMSG ('SLATEC', 'HPSORT',
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+ 'The sort control parameter, KFLAG, is not 2, 1, -1, or -2.',
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+ IER, 1)
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RETURN
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ENDIF
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C
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IF(LEN(WORK) .LT. LEN(HX(1))) THEN
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IER = 3
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CALL XERMSG ('SLATEC',' HPSORT',
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+ 'The length of the work variable, WORK, is too short.',
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+ IER, 1)
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RETURN
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ENDIF
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IF (STRBEG .GT. STREND) THEN
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IER = 4
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CALL XERMSG ('SLATEC', 'HPSORT',
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+ 'The string beginning, STRBEG, is beyond its end, STREND.',
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+ IER, 1)
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RETURN
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ENDIF
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IF (STRBEG .LT. 1 .OR. STRBEG .GT. LEN(HX(1))) THEN
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IER = 5
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CALL XERMSG ('SLATEC', 'HPSORT',
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+ 'The string beginning, STRBEG, is out-of-range.',
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+ IER, 1)
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RETURN
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ENDIF
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IF (STREND .LT. 1 .OR. STREND .GT. LEN(HX(1))) THEN
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IER = 6
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CALL XERMSG ('SLATEC', 'HPSORT',
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+ 'The string end, STREND, is out-of-range.',
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+ IER, 1)
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RETURN
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ENDIF
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C
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C Initialize permutation vector
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C
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DO 10 I=1,NN
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IPERM(I) = I
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10 CONTINUE
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C
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C Return if only one value is to be sorted
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C
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IF (NN .EQ. 1) RETURN
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C
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C Sort HX only
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C
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M = 1
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I = 1
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J = NN
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R = .375E0
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C
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20 IF (I .EQ. J) GO TO 70
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IF (R .LE. 0.5898437E0) THEN
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R = R+3.90625E-2
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ELSE
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R = R-0.21875E0
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ENDIF
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C
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30 K = I
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C
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C Select a central element of the array and save it in location L
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C
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IJ = I + INT((J-I)*R)
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LM = IPERM(IJ)
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C
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C If first element of array is greater than LM, interchange with LM
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C
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IF (HX(IPERM(I))(STRBEG:STREND) .GT. HX(LM)(STRBEG:STREND)) THEN
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IPERM(IJ) = IPERM(I)
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IPERM(I) = LM
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LM = IPERM(IJ)
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ENDIF
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L = J
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C
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C If last element of array is less than LM, interchange with LM
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C
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IF (HX(IPERM(J))(STRBEG:STREND) .LT. HX(LM)(STRBEG:STREND)) THEN
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IPERM(IJ) = IPERM(J)
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IPERM(J) = LM
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LM = IPERM(IJ)
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C
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C If first element of array is greater than LM, interchange
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C with LM
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C
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IF (HX(IPERM(I))(STRBEG:STREND) .GT. HX(LM)(STRBEG:STREND))
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+ THEN
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IPERM(IJ) = IPERM(I)
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IPERM(I) = LM
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LM = IPERM(IJ)
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ENDIF
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ENDIF
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GO TO 50
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40 LMT = IPERM(L)
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IPERM(L) = IPERM(K)
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IPERM(K) = LMT
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C
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C Find an element in the second half of the array which is smaller
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C than LM
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C
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50 L = L-1
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IF (HX(IPERM(L))(STRBEG:STREND) .GT. HX(LM)(STRBEG:STREND))
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+ GO TO 50
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C
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C Find an element in the first half of the array which is greater
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C than LM
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C
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60 K = K+1
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IF (HX(IPERM(K))(STRBEG:STREND) .LT. HX(LM)(STRBEG:STREND))
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+ GO TO 60
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C
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C Interchange these elements
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C
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IF (K .LE. L) GO TO 40
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C
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C Save upper and lower subscripts of the array yet to be sorted
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C
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IF (L-I .GT. J-K) THEN
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IL(M) = I
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IU(M) = L
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I = K
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M = M+1
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ELSE
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IL(M) = K
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IU(M) = J
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J = L
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M = M+1
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ENDIF
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GO TO 80
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C
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C Begin again on another portion of the unsorted array
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C
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70 M = M-1
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IF (M .EQ. 0) GO TO 110
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I = IL(M)
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J = IU(M)
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C
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80 IF (J-I .GE. 1) GO TO 30
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IF (I .EQ. 1) GO TO 20
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I = I-1
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C
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90 I = I+1
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IF (I .EQ. J) GO TO 70
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LM = IPERM(I+1)
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IF (HX(IPERM(I))(STRBEG:STREND) .LE. HX(LM)(STRBEG:STREND))
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+ GO TO 90
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K = I
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C
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100 IPERM(K+1) = IPERM(K)
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K = K-1
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C
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IF (HX(LM)(STRBEG:STREND) .LT. HX(IPERM(K))(STRBEG:STREND))
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+ GO TO 100
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IPERM(K+1) = LM
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GO TO 90
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C
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C Clean up
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C
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110 IF (KFLAG .LE. -1) THEN
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C
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C Alter array to get reverse order, if necessary
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C
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NN2 = NN/2
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DO 120 I=1,NN2
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IR = NN-I+1
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LM = IPERM(I)
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IPERM(I) = IPERM(IR)
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IPERM(IR) = LM
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120 CONTINUE
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ENDIF
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C
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C Rearrange the values of HX if desired
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C
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IF (KK .EQ. 2) THEN
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C
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C Use the IPERM vector as a flag.
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C If IPERM(I) < 0, then the I-th value is in correct location
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C
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DO 140 ISTRT=1,NN
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IF (IPERM(ISTRT) .GE. 0) THEN
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INDX = ISTRT
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INDX0 = INDX
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WORK = HX(ISTRT)
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130 IF (IPERM(INDX) .GT. 0) THEN
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HX(INDX) = HX(IPERM(INDX))
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INDX0 = INDX
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IPERM(INDX) = -IPERM(INDX)
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INDX = ABS(IPERM(INDX))
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GO TO 130
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ENDIF
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HX(INDX0) = WORK
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ENDIF
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140 CONTINUE
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C
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C Revert the signs of the IPERM values
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C
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DO 150 I=1,NN
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IPERM(I) = -IPERM(I)
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150 CONTINUE
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C
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ENDIF
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C
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RETURN
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END
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