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btree.f
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************************************************************************
* D. Everhart
* 22 NOV 2016
* BTREE
************************************************************************
* B I N A R Y T R E E A L G O R I T H M S
************************************************************************
* The MIT License (MIT)
*
* Copyright (c) 2016 Daniel Everhart
*
* Permission is hereby granted, free of charge, to any person obtaining
* a copy of this software and associated documentation files (the
* "Software"), to deal in the Software without restriction, including
* without limitation the rights to use, copy, modify, merge, publish,
* distribute, sublicense, and/or sell copies of the Software, and to
* permit persons to whom the Software is furnished to do so, subject
* to the following conditions:
*
* The above copyright notice and this permission notice shall be
* included in all copies or substantial portions of the Software.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
* EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
* MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.
* IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY
* CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT,
* TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE
* SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
*
*************************************************************************
* +------+ +------+------+
* IND |VALUES| |BEFORE|AFTER | 44
* +------+ +------+------+ / \
* 1 | 44 | | 4 | 2 | 42 64
* 2 | 64 | | 7 | 3 | / / \
* 3 | 87 | | 5 | 0 | 14 60 87
* 4 | 42 | | 6 | 0 | \ /
* 5 | 85 | | 8 | 0 | 18 85
* 6 | 14 | | 0 | 10 | /
* 7 | 60 | | 0 | 0 | 78
* 8 | 78 | | 9 | 0 | /
* 9 | 72 | | 0 | 0 | 72
* 10 | 18 | | 0 | 0 |
* +------+ +------+------+
*
************************************************************************
PROGRAM BTREE
IMPLICIT NONE
LOGICAL BTEXISTS
INTEGER SZ,CT,RES,BTADDHC
INTEGER HCI1,HCI2,HCI4,HCI8,HCR4,HCR8,HCSTR
INTEGER I,J
INTEGER*1 IDIV
INTEGER*1 INT1
INTEGER*2 INT2
INTEGER*4 INT4
INTEGER*8 INT8
REAL*4 REAL4
REAL*8 REAL8
CHARACTER STR12*12,STR16*16,STR20*20,STR14*14
PARAMETER(IDIV=11,SZ=64)
INTEGER BT(3,SZ),SD(SZ)
DATA REAL4,REAL8 / 4.0634579236475D8,4.0634579236475D8/
DATA BT,CT,SD /192*0, 0, SZ*0/
* 123456789012345678901234
INT1 = HUGE(INT1) / IDIV
INT2 = HUGE(INT2) / IDIV
INT4 = HUGE(INT4) / IDIV
INT8 = HUGE(INT8) / IDIV
* 123456789012345678901234
STR12 = ' Some String'
STR16 = ' Some String '
STR20 = '1 34 tmp xyz t '
STR14 = ' ABCDEFGHIJK '
*
WRITE(6,8100)
WRITE(6,8101) INT1, HCI1( INT1)
WRITE(6,8102) INT2, HCI2( INT2)
WRITE(6,8104) INT4, HCI4( INT4)
WRITE(6,8108) INT8, HCI8( INT8)
WRITE(6,8200)
WRITE(6,8204) REAL4, HCR4(REAL4)
WRITE(6,8208) REAL8, HCR8(REAL8)
WRITE(6,8300)
WRITE(6,8312) STR12, HCSTR(STR12)
WRITE(6,8316) STR16, HCSTR(STR16)
WRITE(6,8320) STR20, HCSTR(STR20)
WRITE(6,8324) STR14, HCSTR(STR14)
RES = BTADDHC(BT,SZ,CT,44)
RES = BTADDHC(BT,SZ,CT,64)
RES = BTADDHC(BT,SZ,CT,87)
RES = BTADDHC(BT,SZ,CT,42)
RES = BTADDHC(BT,SZ,CT,85)
RES = BTADDHC(BT,SZ,CT,14)
RES = BTADDHC(BT,SZ,CT,60)
RES = BTADDHC(BT,SZ,CT,78)
RES = BTADDHC(BT,SZ,CT,72)
RES = BTADDHC(BT,SZ,CT,18)
DO 1001 I = 1, CT
WRITE(6,'(3I8)') (BT(J,I),J=1,3)
1001 CONTINUE
I = 1
CT = 0
CALL HLSORT(HLSORT,BT,SZ,I,SD,CT,.TRUE.)
WRITE(6,*)
DO 1011 I = 1, CT
WRITE(6,'(I8)') SD(I)
1011 CONTINUE
I = 1
CT = 0
CALL HLSORT(HLSORT,BT,SZ,I,SD,CT,.FALSE.)
WRITE(6,*)
DO 1021 I = 1, CT
WRITE(6,'(I8)') SD(I)
1021 CONTINUE
WRITE(6,*) '44 exists ->', BTEXISTS(44,BT,SZ)
WRITE(6,*) '12 exists ->', BTEXISTS(12,BT,SZ)
STOP
8100 FORMAT('0*** HASHTEST - INTEGER ********')
8101 FORMAT(' HASHCODE OF INT1 ',I20, ' = ',I11)
8102 FORMAT(' HASHCODE OF INT2 ',I20, ' = ',I11)
8104 FORMAT(' HASHCODE OF INT4 ',I20, ' = ',I11)
8108 FORMAT(' HASHCODE OF INT8 ',I20, ' = ',I11)
8200 FORMAT('0*** HASHTEST - REAL ********')
8204 FORMAT(' HASHCODE OF REAL4 ',G20.10, ' = ',I11)
8208 FORMAT(' HASHCODE OF REAL8 ',G20.10, ' = ',I11)
8300 FORMAT('0*** HASHTEST - STRING ********')
8312 FORMAT(' HASHCODE OF STR12 ',12X,A12,' = ',I11)
8316 FORMAT(' HASHCODE OF STR16 ', 8X,A16,' = ',I11)
8320 FORMAT(' HASHCODE OF STR20 ', 4X,A20,' = ',I11)
8324 FORMAT(' HASHCODE OF STR14 ',10X,A14,' = ',I11)
END PROGRAM BTREE
************************************************************************
LOGICAL FUNCTION BTEXISTS(VALUE,TREE,SZ)
IMPLICIT NONE
INTEGER VALUE,TREE(3,SZ),SZ
INTEGER J
BTEXISTS = .FALSE.
IF (SZ.GT.0) THEN
J =1
DO 111 WHILE (J.LE.SZ)
IF (VALUE.EQ.TREE(1,J)) THEN
BTEXISTS = .TRUE.
RETURN
ELSE IF (VALUE.LT.TREE(1,J)) THEN
J = TREE(2,J)
ELSE
J = TREE(3,J)
END IF
IF (J.EQ.0) EXIT
111 CONTINUE
END IF
999 RETURN
END FUNCTION BTEXISTS
********************************************************************************
SUBROUTINE HLSORT(SUB,BT,SZ,I,ARRAY,C,ASCENDING)
IMPLICIT NONE
EXTERNAL SUB
INTEGER SZ,BT(3,SZ),I
INTEGER ARRAY(SZ),C
LOGICAL ASCENDING
IF (ASCENDING) THEN
IF (BT(2,I).NE.0) CALL SUB(SUB,BT,SZ,BT(2,I),ARRAY,C,ASCENDING)
IF (C.GT.SZ) RETURN
C = C + 1
ARRAY(C) = BT(1,I)
IF (BT(3,I).NE.0) CALL SUB(SUB,BT,SZ,BT(3,I),ARRAY,C,ASCENDING)
ELSE
IF (BT(3,I).NE.0) CALL SUB(SUB,BT,SZ,BT(3,I),ARRAY,C,ASCENDING)
IF (C.GT.SZ) RETURN
C = C + 1
ARRAY(C) = BT(1,I)
IF (BT(2,I).NE.0) CALL SUB(SUB,BT,SZ,BT(2,I),ARRAY,C,ASCENDING)
END IF
END SUBROUTINE HLSORT
********************************************************************************
INTEGER FUNCTION BTADDHC(HL,SZ,CT,HC)
IMPLICIT NONE
INTEGER :: HL(3,SZ),SZ,CT,HC
INTEGER :: I,J
*
** Special case where this is the first value.
*
IF (CT.EQ.0) THEN
CT = 1
HL(1,CT) = HC
HL(2,1) = 0
HL(3,1) = 0
BTADDHC = CT
RETURN
ELSE IF (CT.GE.SZ) THEN
WRITE(0,9001)
BTADDHC = -1
RETURN
ENDIF
*
** Start at top of the table.
*
J = 1
DO 1001 WHILE (J.LE.CT)
IF (J.EQ.0) RETURN
IF (HC.LT.HL(1,J)) THEN
I = 2
ELSE IF (HC.GT.HL(1,J)) THEN
I = 3
ELSE
I = 0
END IF
BTADDHC = J
IF (I.EQ.0) RETURN
IF (HL(I,J).EQ.0) THEN
CT = CT + 1
HL(1,CT) = HC
HL(I,J) = CT
BTADDHC = CT
J = 0
ELSE
J = HL(I,J)
END IF
1001 CONTINUE
RETURN
9001 FORMAT('0*** HASH IS FULL.')
END FUNCTION BTADDHC
********************************************************************************
INTEGER*4 FUNCTION HCI1(VALUE)
INTEGER*1 VALUE
INTEGER*1 VALUES(4)
INTEGER*4 RVALUE
EQUIVALENCE(VALUES,RVALUE)
RVALUE = 0
VALUES(1) = VALUE
HCI1 = RVALUE
END FUNCTION HCI1
********************************************************************************
INTEGER*4 FUNCTION HCI2(VALUE)
INTEGER*2 VALUE
INTEGER*2 VALUES(2)
INTEGER*4 RVALUE
EQUIVALENCE(VALUES,RVALUE)
RVALUE = 0
VALUES(1) = VALUE
HCI2 = RVALUE
END FUNCTION HCI2
********************************************************************************
INTEGER*4 FUNCTION HCI4(VALUE)
INTEGER*4 VALUE
HCI4 = VALUE
END FUNCTION HCI4
********************************************************************************
INTEGER*4 FUNCTION HCI8(VALUE)
INTEGER*8 VALUE
INTEGER*4 VALUES(2)
INTEGER*8 SETVALUE
EQUIVALENCE(VALUES,SETVALUE)
SETVALUE = VALUE
HCI8 = IEOR(VALUES(1),VALUES(2))
END FUNCTION HCI8
********************************************************************************
INTEGER*4 FUNCTION HCR4(VALUE)
REAL*4 VALUE,RVALUE
INTEGER*4 IVALUE,HCI4
EQUIVALENCE(RVALUE,IVALUE)
RVALUE = VALUE
HCR4 = HCI4(IVALUE)
END FUNCTION HCR4
********************************************************************************
INTEGER*4 FUNCTION HCR8(VALUE)
REAL*8 VALUE,RVALUE
INTEGER*8 IVALUE
INTEGER*4 HCI8
EQUIVALENCE(RVALUE,IVALUE)
RVALUE = VALUE
HCR8 = HCI8(IVALUE)
END FUNCTION HCR8
********************************************************************************
* HASHCODE_STRING
*
* This subroutine will loop through in 4-byte words calculating the hashcode
* and IEOR (XOR) it with a running hashcode. This is done ignoring leading and
* trailing space characters.
*
* To illustrate, consider the following 14 character string:
*
* ' ABCDEFGHI '
*
* +---+---+---+---+---+---+---+---+---+---+---+---+---+---+
* | | | A | B | C | D | E | F | G | H | I | J | K | |
* +---+---+---+---+---+---+---+---+---+---+---+---+---+---+
*
********************************************************************************
INTEGER*4 FUNCTION HCSTR(VALUE)
CHARACTER*(*) VALUE
INTEGER*4 IVALUE,JVALUE
CHARACTER*4 CVALUE
INTEGER*4 I,ISTART,IEND,IREM
EQUIVALENCE(CVALUE,JVALUE)
********************************************************************************
* First, the start and end are found excluding spaces on the ends:
* ISTART IEND
* 3 13
* +---+---+---+---+---+---+---+---+---+---+---+---+---+---+
* | | | A | B | C | D | E | F | G | H | I | J | K | |
* +---+---+---+---+---+---+---+---+---+---+---+---+---+---+
* |<--- WORD ---->|<--- WORD ---->|<-- IREM ->|
* IREM = MOD(13-3+1,4) = 3
********************************************************************************
ISTART = VERIFY(VALUE,' ',.FALSE.)
IEND = VERIFY(VALUE,' ', .TRUE.)
IREM = MOD(IEND-ISTART+1,4)
IVALUE = 0
********************************************************************************
* Loop through first part of string excluding the remainder.
* In our example, the indicies will will be: 3:6, 7:10
* The 4-byte char words to be XORed are: 'ABCD', 'EFGH'
* The corresponding integers will be: 1145258561, 201589764
********************************************************************************
DO 1001 I = ISTART,IEND-IREM,4
CVALUE = VALUE(I:I+3)
IVALUE = IEOR(IVALUE,JVALUE)
* WRITE(0,'(I4,X,A4,X,I16)') I,CVALUE,IVALUE
1001 CONTINUE
********************************************************************************
* If there is a remainder put it into CVALUE LEFT JUSTIFIED.
* Note that I = 11 from the termination of the previous loop. 11 + 3 - 1 = 13
* In our example, the indicies will will be: 11:13
* The 4-byte character word to be XORed is: 'IJK'
* 743394893
********************************************************************************
IF (IREM.GT.0) THEN
CVALUE = VALUE(I:I+IREM-1)
IVALUE = IEOR(IVALUE,JVALUE)
* WRITE(0,'(I4,X,A4,X,I16)') I,CVALUE,IVALUE
END IF
HCSTR = IVALUE
END FUNCTION HCSTR
********************************************************************************