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|
C Copyright (c) 2003-2010 University of Florida
C
C This program is free software; you can redistribute it and/or modify
C it under the terms of the GNU General Public License as published by
C the Free Software Foundation; either version 2 of the License, or
C (at your option) any later version.
C This program is distributed in the hope that it will be useful,
C but WITHOUT ANY WARRANTY; without even the implied warranty of
C MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
C GNU General Public License for more details.
C The GNU General Public License is included in this distribution
C in the file COPYRIGHT.
#include "flags.h"
SUBROUTINE SYMMETRY_AUTO(SCRATCH, QTMP, NEWQ, IT, IDEGEN,
& ORIEN2, NOSILENT)
C
C SUBROUTINE DOES SYMMETRY ANALYSIS AND PUTS MOLECULE IN AN
C ORIENTATION THAT ACES2 CAN DEAL WITH
C
IMPLICIT DOUBLE PRECISION (A-H, O-Z)
#include "mxatms.par"
#include "fnamelen.par"
#include "coord.com"
DOUBLE PRECISION IT(3,3),CM(3),IV(3,3),NEWQ(NX),dtmp
DOUBLE PRECISION RM(3,3),QTMP(NX),TATB(3)
DOUBLE PRECISION SCRATCH(3*NX),MOLWT,ORIEN2(9)
LOGICAL ALLDONE,ITWOAX, TSSEARCH, NOSILENT
cSB START
logical olddone,tetrah
cSB END
LOGICAL OPTRES
CHARACTER*3 SYMSTR(6)
CHARACTER*4 some_string_func, JNKSTR, TMPGRP
CHARACTER*2 XYZP(3)
CHARACTER*1 XYZ(3),DOSTR(3),CHRTMP
CHARACTER*(fnamelen) FNAME
INTEGER ORDNEW,ORDOLD,IORGRP
LOGICAL YESNO,SAXIS,SKIP,XYZIN,NWFINDIF
DIMENSION NORD(3*MXATMS),IORBPOP(MXATMS)
DIMENSION IORBREF(MXATMS)
COMMON /FLAGS/ IFLAGS(100),IFLAGS2(500)
COMMON /TOLERS/ SYMTOL,DEGTOL
COMMON/RESTART2/OPTRES
COMMON /USINT/ NX, NXM6, IARCH, NCYCLE, NUNIQUE, NOPT
C Main OPTIM control data
C IPRNT Print level - not used yet by most routines
C INR Step-taking algorithm to use
C IVEC Eigenvector to follow (TS search)
C IDIE Ignore negative eigenvalues
C ICURVY Hessian is in curviliniear coordinates
C IMXSTP Maximum step size in millibohr
C ISTCRT Controls scaling of step
C IVIB Controls vibrational analysis
C ICONTL Negative base 10 log of convergence criterion.
C IRECAL Tells whether Hessian is recalculated on each cyc
C INTTYP Tells which integral program is to be used
C = 0 Pitzer
C = 1 VMol
C XYZTol Tolerance for comparison of cartesian coordinates
C
COMMON /OPTCTL/ IPRNT,INR,IVEC,IDIE,ICURVY,IMXSTP,ISTCRT,IVIB,
$ ICONTL,IRECAL,INTTYP,IDISFD,IGRDFD,ICNTYP,ISYM,IBASIS,
$ XYZTol
COMMON /INPTYP/ XYZIN, NWFINDIF
C
#include "cbchar.com"
C
CHARACTER*(5*MXATMS) ZSYMUNI
C
INTEGER LuArc
PARAMETER (LuArc = 77)
CHARACTER*(*) ArcFil
PARAMETER (ArcFil = 'OPTARC ')
#include "io_units.par"
C Symmetry Information
C FPGrp Full point group
C BPGrp Largest Abelian subgroup
C PGrp "Computational" point group
Character*4 FPGrp, BPGrp, PGrp
Common /PtGp_com/ FPGrp, BPGrp, PGrp
Common /Orient/ Orient(3,3)
COMMON /MACHSP/ IINTLN,IFLTLN,IINTFP,IALONE,IBITWD
DATA XYZ /'X','Y','Z'/
DATA XYZP /'YZ','XZ','XY'/
DATA ONE / 1.0D+00/
DATA ONEM /-1.0D+00/
DATA ZILCH / 0.0D+00/
DATA IONE /1/
IND(I,J) = 3*(I-1)+J
c
c
CALL GETCREC(-1, 'JOBARC', "PTGP ", 4, FPGRP)
TMPGRP=FPGRP
FPGRP=' '
DTOR=DACOS(ONEM)/180.D0
C
C INITIALIZE MILLIONS OF VARIABLES THAT CFT77 SAYS AREN'T INITIALIZED.
C
OLDDONE=.FALSE.
IHIGH=0
IROT=0
IREF=0
IINV=0
IDGRP=0
ILINEAR=0
ICOMP=0
ICOUNT=0
IDONE=0
ISINV=0
ICOMPQ=0
IIAX=0
ORDIS=0
IERR=0
ISET=0
NUMB=0
ICOMP2=0
IHIGHX=0
ISAXIS=0
JCOMPX=0
JCOMP=0
ICOMPX=0
ISKIP=0
DOSTR(1)='X'
DOSTR(2)='Y'
DOSTR(3)='Z'
TSSEARCH = (INR .EQ. 4 .OR. INR .EQ. 5 .OR. INR .EQ. 6)
c CALL IZERO(IORBPOP,MXATMS)
c CALL IZERO(IORBREF,MXATMS)
do i = 1, mxatms
iorbpop(i) = 0
iorbref(i) = 0
enddo
IF(IDEGEN.EQ.0)THEN
C IDEGEN is zero for Abelian point groups. We can transfer the
C logic to statement 9000 (only symmetry operations
C that we need to deal with is rotation, reflection, and inversion).
C It looks like that this goto 9000 can not be easily avoided.
GOTO 9000
ENDIF
C This 777 entry point is used when the degeneracy found by
C inertia matrix was accidental.
777 CONTINUE
C
C****************************************************************
C
C CODE FOR POINT GROUPS WITH DOUBLY DEGENERATE REPRESENTATIONS.
C
C****************************************************************
IF(IDEGEN.EQ.1)THEN
C
C IF DOUBLY DEGENERATE, CHECK TO MAKE SURE THAT THE PRINCIPAL AXES
C ARE PARALLEL TO X,Y AND Z. IF NOT, ROTATE ORIENTATION TO THIS
C POINT.
C
FPGRP = ' '
IF(IPRNT .GE. 3 .AND. NOSILENT)WRITE(LUOUT,8002)
DO 1541 I=1,3
1541 IF(DABS(IT(I,I)).LT.SYMTOL)ILINEAR=1
IF(ILINEAR.EQ.1)THEN
IF(IPRNT .GE. 3 .AND. NOSILENT)WRITE(LUOUT,8004)
C
C Linear molecules can be handled by rotation, reflection,
C and inversion, so transfer the logic to 9000
C
GOTO 9000
ENDIF
C
C IDENTIFY HIGHEST ORDER ROTATIONAL AXIS (CHECK THROUGH 24 )
C
DO 10001 I = 2,24
ZANG = 360.D0/DFLOAT(I)
DO 10000 J = 3,3
CALL DOSYOP('C',I,1,J,NATOMS,NEWQ,SCRATCH,IERR,0)
CALL COMPARE(NEWQ,SCRATCH,ATMASS,NORD,NATOMS,ICOMP,
& SYMTOL)
IF(ICOMP.EQ.0)THEN
IF (XYZIN) CALL IPUTREC(20, "JOBARC", "SYMEQUIV",
& NATOMS, NORD)
IHIGH = I
IHIGHX = J
IF(IPRNT.GE.4 .AND. NOSILENT)WRITE(LUOUT,3150)XYZ(J),I
3150 FORMAT(T3,'@SYMMETRY-I, ',A1,' is a C[',I2,'] axis.')
ENDIF
10000 CONTINUE
10001 continue
C
IF (XYZIN) CALL IPUTREC(20, "JOBARC", "HIGHSTAX", 1, IHIGH)
C
IF(IPRNT .GE. 4 .AND. NOSILENT)WRITE(LUOUT,800)IHIGH,
& XYZ(IHIGHX)
800 FORMAT(T3,'@SYMMETRY-I, The highest order rotational ',
& 'axis is C[',i2,'] about ',A1,'.')
IF(IHIGH.EQ.0)THEN
IF (IPRNT .GE. 4 .AND. NOSILENT) WRITE(6,9301)
9301 FORMAT(T3,'@SYMMETRY-W, An accidental degeneracy is present.',
& ' Analysis might be wrong.')
C
CALL PERPOP2('C',NEWQ,SCRATCH,ATMASS,NATOMS,NORD,SYMTOL,
& IFOUND)
IF(IFOUND.NE.1)THEN
CALL PERPOP2('P',NEWQ,SCRATCH,ATMASS,NATOMS,NORD,SYMTOL,
& IFOUND)
ENDIF
IDEGEN=0
GOTO 777
ENDIF
C
C CHECK FOR S(2n) AXIS
C
CALL DOSYOP ('S',2*IHIGH,1,IHIGHX,NATOMS,NEWQ,SCRATCH,IERR,
& IMODE)
CALL COMPARE(SCRATCH,NEWQ,ATMASS,NORD,NATOMS,ICOMP,SYMTOL)
SAXIS=(ICOMP.EQ.0)
IF (SAXIS) CALL IPUTREC(20, "JOBARC", "SYMEQUIV", NATOMS, NORD)
IF(IPRNT.GT.10.AND.SAXIS .AND. NOSILENT)WRITE(6,9503)
9503 FORMAT(T3,'@SYMMETRY-I, S(2n) axis found.')
C
C NOW CHECK FOR PERPENDICULAR C2 AXES. THIS SEPARATES D AND C
C GROUPS. THE LATTER ARE HANDLED FIRST.
C
CALL PERPOP('C',NEWQ,SCRATCH,ATMASS,NATOMS,NORD,SYMTOL,IFOUND)
IF(IFOUND.EQ.1)THEN
IF(IPRNT.GT.10 .AND. NOSILENT)WRITE(6,9501)
9501 FORMAT(T3,'@SYMMETRY-I, Perpendicular C2 axis found.')
C
FPGRP='DN '
C
C CHECK FOR HORIZONTAL PLANE NOW
C
CALL DOSYOP('P',1,1,3,NATOMS,NEWQ,SCRATCH,IERR,IMODE)
CALL COMPARE(NEWQ,SCRATCH,ATMASS,NORD,NATOMS,ICOMPX,SYMTOL)
C
IF(ICOMPX.EQ.0)THEN
IF (XYZIN) CALL IPUTREC(20, "JOBARC", "SAMEPLNE", NATOMS,
& NORD)
FPGRP='DNh '
IF(IPRNT.GT.10 .AND. NOSILENT)WRITE(6,9502)
9502 FORMAT(T3,'@SYMMETRY-I, Horizontal plane found.')
ENDIF
IF(SAXIS)FPGRP='DNd '
IF (SAXIS .AND. XYZIN) THEN
CALL DOSYOP('C', 8, 1, 3, NATOMS, NEWQ, SCRATCH, IERR, 0)
CALL DCOPY(3*NATOMS, NEWQ, 1, SCRATCH(3*NATOMS+1), 1)
CALL REFLECT(SCRATCH,NEWQ,NATOMS,1)
CALL COMPARE(NEWQ, SCRATCH, ATMASS, NORD, NATOMS, ICOMP,
& SYMTOL)
CALL DCOPY(3*NATOMS, SCRATCH(3*NATOMS+1), 1, NEWQ, 1)
CALL IPUTREC(20, "JOBARC", "SAMEPLNE", NATOMS,
& NORD)
ENDIF
ENDIF
C
IF(FPGRP.EQ.' ')THEN
C
C THIS IS NOT A D GROUP, SO WE KNOW THAT IT IS EITHER CN, CNv, CNh
C OR MAYBE SN. PROCEED WITH LOGIC.
C
C FIRST CHECK FOR HORIZONTAL PLANE - THIS ESTABLISHES THE POINT GROUP
C AS CNh
C
CALL DOSYOP ('P',1,1,3,NATOMS,NEWQ,SCRATCH,IERR,IMODE)
CALL COMPARE(NEWQ,SCRATCH,ATMASS,NORD,NATOMS,ICOMP,SYMTOL)
IF(ICOMP.EQ.0)THEN
FPGRP='CNh '
IF (XYZIN) CALL IPUTREC(20, "JOBARC", "SAMEPLNE", NATOMS,
& NORD)
IF(IPRNT.GT.10 .AND. NOSILENT)WRITE(6,9502)
GOTO 9000
ENDIF
C
C CNv OR CN ? THIS CAN BE DETERMINED BY PRESENCE OR ABSENCE OF
C PLANES CONTAINING THE SYMMETRY AXIS. CHECK FOR THESE NOW.
C
CALL PERPOP('P',NEWQ,SCRATCH,ATMASS,NATOMS,NORD,SYMTOL,IFOUND)
IF(IFOUND.EQ.1)FPGRP='CNv '
ENDIF
IF(FPGRP.EQ.' ')THEN
FPGRP='CN '
IF(SAXIS)THEN
FPGRP='SN '
IHIGH=2*IHIGH
ENDIF
ENDIF
ENDIF
C****************************************************************
C
C CODE FOR CUBIC POINT GROUPS. BY FAR THE MOST COMPLICATED PART.
C
C****************************************************************
IF(IDEGEN.EQ.2)THEN
ALLDONE=.FALSE.
cSB the following is no longer necessary, it is left only so that
c the cartesian coordinates are exactly identical to the old
c version of joda
itwoax =.false.
tetrah =.false.
cSB
IF(IPRNT .GE. 3 .AND. NOSILENT)WRITE(LUOUT,8003)
C
C CHECK FOR INVERSION SYMMETRY.
C
DO 14002 I = 1,NATOMS*3
14002 SCRATCH(I) = -NEWQ(I)
CALL COMPARE(SCRATCH,NEWQ,ATMASS,NORD,NATOMS,ISINV,SYMTOL)
C
C NOW PICK A REFERENCE ATOM IN THE MOLECULE **WHICH DOES NOT LIE AT
C THE ORIGIN** AND THEN LOOP OVER ALL OTHER ATOMS HAVING THE SAME
C ATOMIC NUMBER AND COMPARABLE DISTANCE FROM THE COM. CHECK FOR
C AXES OF ORDER 3, 4 AND 5.
C
IREF=-1
IOFF=1
DO 5000 IATOM=1,NATOMS
X=XDNRM2(3,NEWQ(IOFF),1)*ATMASS(IATOM)
IF(X.GT.SYMTOL.AND.IREF.EQ.-1)THEN
IREF=IATOM
XREF=X
ZREF=X/ATMASS(IATOM)
IBOT=1+(IREF-1)*3
ENDIF
IOFF=IOFF+3
5000 CONTINUE
C
C HANDLE ATOMIC CALCULATION LOGIC
C
IF(IREF.EQ.-1)THEN
FPGRP='I h'
GOTO 9000
ENDIF
DO 5001 IATOM=1,NATOMS
IBOT2=1+(IATOM-1)*3
X=XDNRM2(3,NEWQ(IBOT2),1)*ATMASS(IATOM)
IF(ABS(X-XREF).LT.SYMTOL)THEN
IF(IPRNT.GT.100)WRITE(LUOUT,50000)IREF,IATOM
50000 FORMAT(T3,'@SYMMETRY-I, Checking atoms ',I3,' and ',I3,'.')
C
C FIRST ROTATE MOLECULE SO THAT THE INTERATOMIC DISTANCE VECTOR
C IS PARALLEL TO THE Z-AXIS, WITH BISECTOR ALONG X-AXIS.
C THIS MEANS THAT THE TWO ATOMS LIE IN THE XZ PLANE.
C STRUCTURE NOW IN SCRATCH(1). HERE ONE HAS TO ALLOW FOR THE
C POSSIBILITY THAT ATOMS I AND J ARE 180 DEGREES APART; IF THIS
C IS THE CASE, JUST GO TO BOTTOM OF LOOP SINCE THESE DO NOT
C DEFINE ANY AXIS UNIQUELY.
C
CALL VEC(NEWQ(IBOT),NEWQ(IBOT2),SCRATCH(1),0)
dtmp = xdot(3,SCRATCH(1),1,SCRATCH(1),1)
DIST = DSQRT(dtmp)
CALL VADD(SCRATCH(1),NEWQ(IBOT),NEWQ(IBOT2),NX,ONE)
C
C DEAL WITH THE LINEAR PROBLEM RIGHT NOW.
C
BILEN = xdot(3,SCRATCH(1),1,SCRATCH(1),1)
IF(BILEN.LT.1.D-12)GOTO 5001
CALL SIAZ(SCRATCH(1),RM,1)
CALL ZERO(SCRATCH,NX*3)
C CALL MATMULV(SCRATCH(NX+1),NEWQ,RM,NATOMS,3,3)
CALL XGEMM('T','N',3,NATOMS,3,ONE,RM,3,NEWQ,3,ZILCH,
& SCRATCH(NX+1),3)
C
C NOW YOU HAVE BISECTOR ALONG X. ROTATE ABOUT X TO BRING THE TWO
C ATOMS INTO POSITION PARALLEL TO Z!
C
dtmp = xdot(2,SCRATCH(NX+IBOT+1),1,SCRATCH(NX+IBOT+1),1)
DIP = DSQRT(dtmp)
ARGU=SCRATCH(NX+IBOT+2)/DIP
ANGLE2=-DACOSX(ARGU,1.D-10)/DTOR
IF(SCRATCH(NX+IBOT+1).GT.0.D0)ANGLE2=-ANGLE2
CALL ROTM(1,ANGLE2,1,RM)
C CALL MATMULV(SCRATCH,SCRATCH(NX+1),RM,NATOMS,3,3)
CALL XGEMM('T','N',3,NATOMS,3,ONE,RM,3,SCRATCH(NX+1),3,
& ZILCH,SCRATCH,3)
C
C NOW FIND NEW VECTOR WHICH BISECTS THE TWO ATOMS IN QUESTION -
C IT HAD BEST BE X!
C
CALL VADD(TATB,SCRATCH(IBOT),SCRATCH(IBOT2),
& 3,ONE)
C
C 1. TEST IF THE TWO ATOMS ARE CONNECTED BY SYMMETRY AXES, LOOPING
C OVER POSSIBILITIES (2,3,4 AND 5).
C
DO 5002 IAXORD=5,2,-1
ANGIAX=360.D0/DFLOAT(IAXORD)
ANGMG=ANGMAG(ZREF,DIST,IAXORD,IERR)
IF(IERR.EQ.0.AND..NOT.ALLDONE)THEN
IF(IPRNT.GE.3 .AND. NOSILENT)WRITE(LUOUT,50001)ANGMG
50001 FORMAT(T3,' ANGMAG angle is ',f8.4)
IF(IPRNT.GE.13 .AND. NOSILENT)WRITE(LUOUT,50002)IERR,IAXORD
50002 FORMAT(T3,' IERR is ',i2,' for ',i1)
IF(IPRNT.GE.200 .AND. NOSILENT)THEN
WRITE(LUOUT,*)' IN ROTATIONAL LOOP, LOOKING FOR ',
& 'ORDER ',IAXORD
WRITE(LUOUT,50003)IREF,IATOM
50003 FORMAT(' PLAYING WITH ATOMS ',I2,' AND ',I2)
WRITE(LUOUT,*)' IBOT AND IBOT2 ARE ',IBOT,IBOT2
WRITE(LUOUT,*)' ORBIT DISTANCE IS ',ORDIS
WRITE(LUOUT,*)' INTERATOMIC DISTANCE IS ',DIST
WRITE(LUOUT,*)' MAGIC ANGLE IS ',ANGMG
WRITE(LUOUT,*)' MAGIC VECTOR IS ',(TATB(JJ),JJ=1,3)
ENDIF
CALL ROTM(3,ANGMG,1,RM)
IF(IPRNT.GE.200 .AND. NOSILENT)WRITE(LUOUT,80)
& (SCRATCH(JZ),JZ=1,NX)
C CALL MATMULV(SCRATCH(NX+4),TATB,RM,1,3,3)
CALL XGEMM('T','N',3,1,3,ONE,RM,3,TATB,3,ZILCH,
& SCRATCH(NX+4),3)
CALL SIAZ(SCRATCH(NX+4),RM,3)
CALL ZERO(SCRATCH(NX+1),NATOMS*3*2)
C CALL MATMULV(SCRATCH(NX+1),SCRATCH,RM,NATOMS,3,3)
CALL XGEMM('T','N',3,NATOMS,3,ONE,RM,3,SCRATCH,3,ZILCH,
& SCRATCH(NX+1),3)
CALL ROTM(3,ANGIAX,1,RM)
C CALL MATMULV(SCRATCH(2*NX+1),SCRATCH(NX+1),
C & RM,NATOMS,3,3)
CALL XGEMM('T','N',3,NATOMS,3,ONE,RM,3,SCRATCH(NX+1),
& 3,ZILCH,SCRATCH(2*NX+1),3)
call compare (scratch(nx+1),scratch(2*nx+1),atmass,
& nord,natoms,iiax,symtol)
cOLD WRITE(6,*) "The first one"
cOLD WRITE(6,*) (NORD(I), I=1, NATOMS)
IF(XYZIN) CALL IPUTREC(20, "JOBARC", "SYMEQUIV", NATOMS,
& NORD)
IF(IIAX.EQ.0)THEN
cSB START
c The following assumes that we will run into an O or I rotation before
c running into at least 1 C2 and 1 C3 rotation. An example where this
c is not the case is an Oh molecule defined as two superimposed tetrahedrons
c (one the inverse of the other)
cSSS IF(IAXORD.EQ.2.AND.TETRAH)THEN
cSSS ALLDONE=.TRUE.
cSSS FPGRP='T '
cSSS IF(ISINV.EQ.0)FPGRP='T h'
cSSS CALL XDCOPY (3*NATOMS,SCRATCH(NX+1),1,NEWQ,1)
cSSS CALL PERPOP('C',NEWQ,SCRATCH,ATMASS,NATOMS,NORD,
cSSS & SYMTOL,IFOUND)
cSSS IF(IFOUND.EQ.0)THEN
cSSS WRITE(6,50004)FPGRP
cSSS CALL ERREX
cSSS ENDIF
cSSS CALL DOSYOP('S',4,1,3,NATOMS,NEWQ,SCRATCH,IERR,0)
cSSS CALL COMPARE(NEWQ,SCRATCH,ATMASS,NORD,NATOMS,ICOMP,
cSSS & SYMTOL)
cSSS IF(ICOMP.EQ.0.AND.ISINV.NE.0)THEN
cSSS FPGRP='T d'
cSSSC
cSSSC COMMENT OUT TWO LINES BELOW TO GET Td TO RUN AS D2.
cSSSC
cSSS CALL DOSYOP('C',8,1,3,NATOMS,NEWQ,SCRATCH,IERR,0)
cSSS CALL XDCOPY (NX,SCRATCH,1,NEWQ,1)
cSSS ENDIF
cSSS ELSEIF(IAXORD.EQ.2.AND..NOT.TETRAH)THEN
cSSS ITWOAX=.TRUE.
cSSS CALL XDCOPY(3*NATOMS,SCRATCH(NX+1),1,QTMP,1)
cSSS ELSEIF(IAXORD.EQ.3)THEN
cSSS TETRAH=.TRUE.
cSSS IF(ITWOAX)THEN
cSSS ALLDONE=.TRUE.
cSSS FPGRP='T '
cSSS IF(ISINV.EQ.0)FPGRP='T h'
cSSS CALL XDCOPY (NX,QTMP,1,NEWQ,1)
cSSS CALL PERPOP('C',NEWQ,SCRATCH,ATMASS,NATOMS,NORD,
cSSS & SYMTOL,IFOUND)
cSSS IF(IFOUND.EQ.0)THEN
cSSS WRITE(6,50004)FPGRP
cSSS CALL ERREX
cSSS ENDIF
cSSS CALL DOSYOP('S',4,1,3,NATOMS,NEWQ,SCRATCH,IERR,0)
cSSS CALL COMPARE(NEWQ,SCRATCH,ATMASS,NORD,NATOMS,
cSSS & ICOMP,SYMTOL)
cSSS IF(ICOMP.EQ.0.AND.ISINV.NE.0)FPGRP='T d'
cSSS ENDIF
c We replace the above with the following which essentiall ignores all
c C2 and C3 rotations. If, when we are done, we have not encountered a
c O or I rotation, we know it must be a T type molecule.
c
c All of the logic involving olddone, tetrah, and iaxord is optional.
c It is included only to make sure that the cartesian coordinates are
c exactly identical (as opposed to being off by a symmetry rotation).
if (iaxord.eq.2 .and. .not.olddone) then
call xdcopy(3*natoms,scratch(nx+1),1,qtmp,1)
if (tetrah) then
olddone=.true.
else
itwoax=.true.
endif
elseif (iaxord.eq.3 .and. .not.olddone) then
tetrah=.true.
if (itwoax) olddone=.true.
cSB END
ELSEIF(IAXORD.EQ.5)THEN
FPGRP='I '
IF(ISINV.EQ.0)FPGRP='I h'
CALL XDCOPY (3*NATOMS,SCRATCH(NX+1),1,NEWQ,1)
C
C IF WE HAVE FOUND A FIVE-FOLD AXIS, FIRST REORIENT MOLECULE SO
C THAT C2 AXES LIE ALONG X, Y AND Z
C
CALL PERPOP('C',NEWQ,SCRATCH,ATMASS,NATOMS,NORD,
& SYMTOL,IFOUND)
IF(IFOUND.EQ.0)THEN
WRITE(6,50004)FPGRP
CALL ERREX
ENDIF
CALL ROTM(2,90.D0,1,RM)
C CALL MATMULV(SCRATCH,NEWQ,RM,NATOMS,3,3)
CALL XGEMM('T','N',3,NATOMS,3,ONE,RM,3,NEWQ,
& 3,ZILCH,SCRATCH,3)
CALL XDCOPY (3*NATOMS,SCRATCH,1,NEWQ,1)
CALL PERPOP('C',NEWQ,SCRATCH,ATMASS,NATOMS,NORD,
& SYMTOL,IFOUND)
IF(IFOUND.EQ.0)THEN
WRITE(6,50004)FPGRP
CALL ERREX
ENDIF
CALL ROTM(2,90.D0,1,RM)
C CALL MATMULV(SCRATCH,NEWQ,RM,NATOMS,3,3)
CALL XGEMM('T','N',3,NATOMS,3,ONE,RM,3,NEWQ,
& 3,ZILCH,SCRATCH,3)
CALL XDCOPY (3*NATOMS,SCRATCH,1,NEWQ,1)
C
C NOW MAKE SURE THAT C5 ORIENTATION IS CORRECT
C
SQ5=ONE/DSQRT(5.D0)
ARGX=DSQRT(0.5D0*(ONE-SQ5))
ANGL=DACOS(ARGX)
CALL ROTM(1,ANGL,0,RM)
C CALL MATMULV(SCRATCH,NEWQ,RM,NATOMS,3,3)
CALL XGEMM('T','N',3,NATOMS,3,ONE,RM,3,NEWQ,
& 3,ZILCH,SCRATCH,3)
CALL ROTM(3,72.D0,1,RM)
C CALL MATMULV(SCRATCH(NX+1),SCRATCH,RM,NATOMS,3,3)
CALL XGEMM('T','N',3,NATOMS,3,ONE,RM,3,NEWQ,
& 3,ZILCH,SCRATCH,3)
call compare(scratch(nx+1),scratch,atmass,
& nord,natoms,icomp,symtol)
IF(ICOMP.NE.0)THEN
CALL ROTM(3,90.D0,1,RM)
C CALL MATMULV(SCRATCH,NEWQ,RM,NATOMS,3,3)
CALL XGEMM('T','N',3,NATOMS,3,ONE,RM,3,NEWQ,
& 3,ZILCH,SCRATCH,3)
CALL XDCOPY(NX,SCRATCH,1,NEWQ,1)
ENDIF
ALLDONE=.TRUE.
ELSEIF(IAXORD.EQ.4)THEN
C
C IF WE HAVE FOUND A FOUR-FOLD AXIS, THEN WE ARE DONE. THE MOLECULE
C BELONGS TO EITHER O OR Oh AND REORIENTATION IS COMPLETELY
C STRAIGHTFORWARD.
C
cSSS FPGRP='O '
cSSS IF(ISINV.EQ.0)FPGRP='O h'
CALL XDCOPY (3*NATOMS,SCRATCH(NX+1),1,NEWQ,1)
CALL PERPOP('C',NEWQ,SCRATCH,ATMASS,NATOMS,NORD,
& SYMTOL,IFOUND)
IF(IFOUND.EQ.0)THEN
WRITE(6,50004)FPGRP
50004 FORMAT(T3,'@SYMMETRY-F, Reorientation failure ',
& 'for point group ',A,'.')
CALL ERREX
ENDIF
cSB START
c At this point, we have either an O, Oh, or Th molecule. In either
c case, the molecule is oriented in one of two ways (visualized with a
c cube).
c A. 4 vertices in the XZ plane, 4 in the YZ plane (X and Y axis
c point to edges of a cube, Z axis points to a face)
c B. 45 degree rotation from this (all axis point to faces)
c In the B orientation, there is no way (using an X, Y, Z rotation or
c an XY, XZ, YZ reflection) to distinguish between Oh and Th. In The
c A orientation, an XY, XZ, YZ reflection fails for Th. Also, in the
c B orientation (but NOT the A orientation), there is a C4 rotation
c along the X axis.
c
c If the molecule is O or Oh, we want to rotate it to the B orientation
c in the end. If it is a Th molecule, we need to exit without having
c modified the orientation. In this case, the above XDCOPY and PERPOP
c may cause a problem (but it may not).
c Look for the C4(x) rotation (icomp=0 if B orientation))
CALL DOSYOP('C',4,1,1,NATOMS,NEWQ,SCRATCH,IERR,0)
CALL COMPARE(NEWQ,SCRATCH,ATMASS,NORD,NATOMS,ICOMP,
& SYMTOL)
c The B orientation.
if (icomp.eq.0) then
c In the B orientation, an inversion center means either Oh or Th.
c Rotate to the A orientation and check for a any reflection plane.
if (isinv.eq.0) then
call dosyop('C',8,1,3,natoms,newq,scratch,
& ierr,0)
call dosyop('P',1,1,3,natoms,scratch,
& scratch(nx+1),ierr,0)
call compare(scratch,scratch(nx+1),atmass,
& nord,natoms,icomp,symtol)
cSSS WRITE(6,*) "The First one"
cSSS WRITE(6,*) (NORD(II), II =1, NATOMS)
cSSS CALL IPUTREC(20, "JOBARC", "SYMEQUIV", NATOMS,
cSSS & NORD)
if (icomp.eq.0) then
fpgrp='O h'
alldone=.true.
endif
c In the B orientation, no inversion center means O.
else
fpgrp='O '
alldone=.true.
endif
c The A orientation.
else
c In the A orientation, an inversion center means either Oh or Th.
c We need to check for a reflection plane. If found, rotate to
c B orientation.
if (isinv.eq.0) then
call dosyop('P',1,1,3,natoms,newq,scratch(nx+1),
& ierr,0)
call compare(newq,scratch(nx+1),atmass,nord,
& natoms,icomp,symtol)
cSSS WRITE(6,*) "The Second One"
cSSS WRITE(6,*) (NORD(II), II =1, NATOMS)
cSSS CALL IPUTREC(20, "JOBARC", "SYMEQUIV", NATOMS,
cSSS & NORD)
if (icomp.eq.0) then
call dosyop('C',8,1,3,natoms,newq,scratch,
& ierr,0)
call xdcopy (nx,scratch,1,newq,1)
fpgrp='O h'
alldone=.true.
endif
c In the A orientation, no inversion center means O. Rotate it
c to the B orientation.
else
call dosyop('C',8,1,3,natoms,newq,scratch,
& ierr,0)
call xdcopy (nx,scratch,1,newq,1)
fpgrp='O '
alldone=.true.
endif
endif
c The old logic was wrong
cSSS CALL DOSYOP('C',4,1,1,NATOMS,NEWQ,SCRATCH,IERR,0)
cSSS CALL COMPARE(NEWQ,SCRATCH,ATMASS,NORD,NATOMS,ICOMP,
cSSS & SYMTOL)
cSSS IF(ICOMP.NE.0)THEN
cSSS CALL DOSYOP('C',8,1,3,NATOMS,NEWQ,SCRATCH,IERR,0)
cSSS CALL XDCOPY (NX,SCRATCH,1,NEWQ,1)
cSSS ALLDONE=.TRUE.
cSSS ENDIF
cSSSCJDW 12/22/94
cSSSC This lets the Cr(CO)6 job run. CrC6 ends up with ICOMP=6 and so
cSSSC ALLDONE is set to true. The C coordinates are different in Cr(CO)6
cSSSC and CrC6. In the former all C atoms lie along Cartesian axes, while
cSSSC in the latter four of the C atoms bisect axes. This is a somewhat
cSSSC empirical fix, and a rigorous explanation would be appreciated.
cSSS ALLDONE = .TRUE.
cSB END
ENDIF
ENDIF
ENDIF
5002 CONTINUE
ENDIF
5001 CONTINUE
IF(.NOT.ALLDONE)THEN
cSB START
c It used to be that if we did not get the ALLDONE flag, there was
c an error. Now it just means that we are in a T-type group.
cSSS WRITE(6,50005)
cSSS50005 FORMAT(T3,'@SYMMETRY-F, Cubic group not determined.')
cSSS CALL ERREX
FPGRP='T '
IF(ISINV.EQ.0)FPGRP='T h'
CALL XDCOPY (NX,QTMP,1,NEWQ,1)
CALL PERPOP('C',NEWQ,SCRATCH,ATMASS,NATOMS,NORD,
& SYMTOL,IFOUND)
IF(IFOUND.EQ.0)THEN
WRITE(6,50004)FPGRP
CALL ERREX
ENDIF
CALL DOSYOP('S',4,1,3,NATOMS,NEWQ,SCRATCH,IERR,0)
CALL COMPARE(NEWQ,SCRATCH,ATMASS,NORD,NATOMS,ICOMP,
& SYMTOL)
IF(ICOMP.EQ.0.AND.ISINV.NE.0)THEN
FPGRP='T d'
IF (XYZIN) CALL IPUTREC(20, "JOBARC", "SYMEQUIV", NATOMS,
& NORD)
C
C COMMENT OUT TWO LINES BELOW TO GET Td TO RUN AS D2.
C
CALL DOSYOP('C',8,1,3,NATOMS,NEWQ,SCRATCH,IERR,0)
CALL XDCOPY (NX,SCRATCH,1,NEWQ,1)
ENDIF
cSB END
ENDIF
ENDIF
C
C CUBIC POINT GROUP NOW DETERMINED. ON WITH THE SHOW.
C
80 FORMAT((4X,3(2X,F16.12)))
8001 FORMAT(T3,'@SYMMETRY-I, The molecule belongs to an ',
& 'Abelian group.')
8002 FORMAT(T3,'@SYMMETRY-I, The molecule belongs to a point ',
& 'group with doubly',
& ' degenerate representations.')
8004 FORMAT(T3,'@SYMMETRY-I, The molecule is linear.')
8003 FORMAT(T3,'@SYMMETRY-I, The molecule belongs to a cubic ',
& 'point group.')
C
C This is the principal axis or computational orientation of the molecule
C
IF(OPTRES) THEN
CALL DGETREC(20,'JOBARC','COORD ',NX,NEWQ)
ENDIF
C
C This is the entry point for Abelain subgrop. All these 'GO TO 9000"
C could have been easily avoided.
9000 CONTINUE
IF(IPRNT.GE.0 .AND. NOSILENT) THEN
WRITE(LUOUT,8202)
8202 FORMAT(T3,' Principal axis orientation for molecule:')
WRITE(LUOUT,80)(NEWQ(I),I=1,NX)
ENDIF
C
C CHECK SYMMETRY OPERATIONS BELONGING TO ABELIAN GROUPS
C
C THE SYMSTR INFORMATION IS USED ONLY TO MAKE VMOL DECKS. COMMENTED
C OUT FOR NOW SINCE SYMEQV AND GRPOPS ASSUME "FUDGED" ORIENTATIONS.
C
ANG = 180.D0
IREF=0
IOPS=1
DO 873 I=1,6
SYMSTR(I)=' '
873 CONTINUE
DO 31 I = 3,1,-1
CALL REFLECT(NEWQ,SCRATCH,NATOMS,I)
call compare(newq,scratch,atmass,
& nord,natoms,icomp,symtol)
IF(ICOMP.EQ.0)THEN
C SYMSTR(IOPS)(2:2)=XYZ(I)
IF (XYZIN) THEN
CALl IGETREC(0, "JOBARC", "SAMEPLNE", LENGTH, IJUNK)
IF (LENGTH .LT. 0) CALL IPUTREC(20, "JOBARC", "SAMEPLNE",
& NATOMS, NORD)
ENDIF
IOPS=IOPS+1
IF(IPRNT .GE. 3 .AND. NOSILENT)WRITE(LUOUT,77)I
77 FORMAT(T3,' Reflection in plane ',i2,' is a valid ',
$ 'symmetry operation.')
IREF = IREF + 2**I/2
ENDIF
CALL ROTM(I,ANG,1,RM)
IF(IPRNT .GE. 5 .AND. NOSILENT)WRITE(LUOUT,80)
$ ((RM(IX,JX),JX = 1,3),IX = 1,3)
C CALL MATMULV(SCRATCH,NEWQ,RM,NATOMS,3,3)
CALL XGEMM('T','N',3,NATOMS,3,ONE,RM,3,NEWQ,3,ZILCH,SCRATCH,3)
call compare(newq,scratch,atmass,
& nord,natoms,icomp,symtol)
IF(ICOMP.EQ.0)THEN
C SYMSTR(IOPS)(2:3)=XYZP(I)
IOPS=IOPS+1
IF(IPRNT .GE. 3 .AND. NOSILENT)WRITE(LUOUT,78)I
78 FORMAT(T3,' Rotation about ',i2,' is a valid symmetry ',
$ 'operation ')
IROT = IROT + 2**I/2
endif
31 continue
DO 102 I = 1,NATOMS*3
102 SCRATCH(I) = -NEWQ(I)
call compare(newq,scratch,atmass,
& nord,natoms,icomp,symtol)
IF(ICOMP.EQ.0)THEN
C SYMSTR(IOPS)='XYZ'
IOPS=IOPS+1
IF(IPRNT .GE. 3 .AND. NOSILENT)WRITE(LUOUT,81)
81 FORMAT(T3,' The molecule possesses an inversion center. ')
IF (XYZIN) CALL IPUTREC(20, "JOBARC", "SYMEQUIV",
& NATOMS, NORD)
IF(ILINEAR.EQ.1)FPGRP = 'DXh '
IINV=1
ELSE
IF(ILINEAR.EQ.1)FPGRP = 'CXv '
cOLD IF (XYZIN) CALL IPUTREC(20, "JOBARC", "SYMEQUIV",
cOLD & NATOMS, NORD)
CALL IGETREC(0, "JOBARC", "SYMEQUIV", LENGTH, IJUNK)
IF (XYZIN .AND. LENGTH .LT. 0) CALL IPUTREC(20, "JOBARC",
& "SYMEQUIV", NATOMS, NORD)
ENDIF
C
C DETERMINE LARGEST ABELIAN SUBGROUP OF MOLECULE
C
IF(IROT.EQ.0.AND.IREF.EQ.0.AND.IINV.EQ.0)PGRP = 'C1 '
IF(IROT.EQ.0.AND.IREF.EQ.0.AND.IINV.EQ.1)PGRP = 'C i '
IF(IROT.EQ.0.AND.IREF.NE.0.AND.IINV.EQ.0)PGRP = 'C s '
IF(IROT.NE.0.AND.IREF.EQ.0.AND.IINV.EQ.0)PGRP = 'C2 '
IF(IROT.NE.0.AND.IREF.NE.0.AND.IINV.EQ.0)PGRP = 'C2v '
IF(IROT.NE.0.AND.IREF.NE.0.AND.IINV.EQ.1)PGRP = 'C2h '
IF(IROT.EQ.7.AND.IREF.EQ.0.AND.IINV.EQ.0)PGRP = 'D2 '
c IF(IROT.EQ.7.AND.IREF.EQ.0.AND.IINV.EQ.0)PGRP = 'C2 '
IF(IROT.EQ.7.AND.IREF.EQ.7.AND.IINV.EQ.1)PGRP = 'D2h '
IF(IPRNT .GE. 4 .AND. NOSILENT)WRITE(LUOUT,27)IREF,IROT,IINV
27 FORMAT(T3,'Symmetry bits: ',3(1x,i3))
CJDW 9/16/96. Comment out 3 lines below, as they will not be needed if
C JFS modifications work (routine SETD2XYZ). Above C2 line
C is how JFS used to handle D2, i.e. switch to C2 always
C was done.
CJDW 4/ 1/97. Putting these three lines back in. D2 is causing too much
C trouble in finite difference calculations at the moment.
c
CNO 5/11/94 Remove D2 from findif calculations and also from
C Resonance Raman Calculations.
C This restriction should no longer be needed
C (see my comments in vmlgen in vmol2ja and vib1 and geopt in joda)
C Ajith Perera, 01/2006.
CSSS
CSSS if(pgrp.eq.'D2 ') then
CSSS if(iflags(54).eq.3.or.iflags(54).eq.2.or.
CSSS & iflags2(3) .ne. 0) pgrp = 'C2 '
CSSS endif
c
C IFLAGS(85) is non-zero when the subgroup key-word is turned on
C
IF(IFLAGS(85).NE.0)THEN
IF(IPRNT .GE. 4 .AND. NOSILENT) WRITE(6,1051)
1051 FORMAT(T3,'@SYMMETRY-I, Reference coordinates used for ',
& 'subgroup specification.')
CALL DUMPCORD(NATOMS,NEWQ,IATNUM)
IF(FPGRP.EQ.' ')FPGRP=PGRP
IF(ISYM .EQ. 1) PGRP='C1 '
#ifdef _DEBUG_LVL0
Print*, "PGRP, FPGRP:", FPGRP, PGRP
#endif
CALL SUBGROUP(IFLAGS(85),PGRP,FPGRP,IROTATE)
C
IF(IROTATE.EQ.1)THEN
CALL ROTM(3,45.0D0,1,RM)
C CALL MATMULV(SCRATCH,NEWQ,RM,NATOMS,3,3)
CALL XGEMM('T','N',3,NATOMS,3,ONE,RM,3,NEWQ,3,ZILCH,SCRATCH,3)
CALL XDCOPY(3*NATOMS,SCRATCH,1,NEWQ,1)
ENDIF
IF(IFLAGS(86)+1.NE.1)THEN
CHRTMP=DOSTR(1)
DOSTR(1)=DOSTR(IFLAGS(86)+1)
DOSTR(IFLAGS(86)+1)=CHRTMP
ENDIF
C
C In order to do geo. optimization in a subgroup, Andrew Taube, 06/20
C
ORDNEW=IORGRP(PGRP)
ORDOLD=IORGRP(FPGRP)
IF (ORDNEW.LE.ORDOLD) FPGRP=PGRP
ENDIF
CALL ZERO(ORIENT,9)
C
C ROTATE TO DEFAULT SYMMETRY FRAME IF NEEDED - COMPLETELY DONE BY
C BRUTE FORCE. NO CUTE ALGORITHM USED HERE.
C
C Reorientation of the molecule depending on the point group.
BPGRP = PGRP
CALL ZERO(SCRATCH,NATOMS*3)
IF(PGRP.EQ.'C1 '.OR.PGRP.EQ.'C i '.OR.
& PGRP.EQ.'D2 '.OR.PGRP.EQ.'D2h ')THEN
c IF(PGRP(1:2).NE.'D2')PGRP = 'C1 '
CALL VADD(Q,NEWQ,SCRATCH(1),NATOMS*3,ONE)
GOTO 75
ENDIF
C
C C2V
C
IF(PGRP.EQ.'C2v'.OR.PGRP.EQ.'C2h'.OR.PGRP.EQ.'C2 ')THEN
C
C IF X IS ROTATION AXIS - SWITCH TO Z - AND PUT LARGEST MOMENT OF
C INERTIA AROUND X - EIGENVECTORS RETURNED FROM EIG ARE SORTED
C HIGHEST TO LOWEST FOR D2H, WE TAKE STANDARD ORIENTATION SUCH
C THAT IX>IY>IZ
C
CALL SETXYZ(Q,NEWQ,ORIENT,DOSTR(1),IROT,IOK,NATOMS)
IF(IOK.EQ.0.AND.IFLAGS(86).EQ.0)THEN
CALL SETXYZ(Q,NEWQ,ORIENT,DOSTR(2),IROT,IOK,NATOMS)
IF(IOK.EQ.0.AND.IFLAGS(86).EQ.0)THEN
CALL SETXYZ(Q,NEWQ,ORIENT,DOSTR(3),IROT,IOK,NATOMS)
ENDIF
ENDIF
C
ELSE
C
CALL SETXYZ(Q,NEWQ,ORIENT,DOSTR(1),IREF,IOK,NATOMS)
IF(IOK.EQ.0.AND.IFLAGS(86).EQ.0)THEN
CALL SETXYZ(Q,NEWQ,ORIENT,DOSTR(2),IREF,IOK,NATOMS)
IF(IOK.EQ.0.AND.IFLAGS(86).EQ.0)THEN
CALL SETXYZ(Q,NEWQ,ORIENT,DOSTR(3),IREF,IOK,NATOMS)
ENDIF
ENDIF
C
ENDIF
c IF(Mod(IRot,2).EQ.1)THEN
c DO 1001 J = 1,NATOMS
c Q(3*J) = NEWQ(3*J-2)
c Q(3*J-2) = NEWQ(3*J-1)
c 1001 Q(3*J-1) = NEWQ(3*J)
c ORIENT(1,2)=ONE
c ORIENT(2,3)=ONE
c ORIENT(3,1)=ONE
c ELSEIF(Mod(IRot/2,2).EQ.1)THEN
c DO 1002 J = 1,NATOMS
c Q(3*J) = NEWQ(3*J-1)
c Q(3*J-1) = NEWQ(3*J-2)
c 1002 Q(3*J-2) = NEWQ(3*J)
c ORIENT(3,2)=ONE
c ORIENT(2,1)=ONE
c ORIENT(1,3)=ONE
c ELSEIF(Mod(IRot/4,2).EQ.1)THEN
c CALL VADD(Q,NEWQ,SCRATCH(1),NATOMS*3,ONE)
c ORIENT(1,1)=ONE
c ORIENT(2,2)=ONE
c ORIENT(3,3)=ONE
c ENDIF
c ELSE
c IF(Mod(IRef,2).EQ.1)THEN
c DO 2001 J = 1,NATOMS
c Q(3*J) = NEWQ(3*J-2)
c Q(3*J-2) = NEWQ(3*J-1)
c 2001 Q(3*J-1) = NEWQ(3*J)
c ORIENT(3,1)=ONE
c ORIENT(1,2)=ONE
c ORIENT(2,3)=ONE
c ELSEIF(Mod(IRef/2,2).EQ.1)THEN
c DO 2002 J = 1,NATOMS
c Q(3*J) = NEWQ(3*J-1)
c Q(3*J-1) = NEWQ(3*J-2)
c 2002 Q(3*J-2) = NEWQ(3*J)
c ORIENT(3,2)=ONE
c ORIENT(2,1)=ONE
c ORIENT(1,3)=ONE
c ELSEIF(MOD(IRef/4,2).EQ.1)THEN
c CALL VADD(Q,NEWQ,SCRATCH(1),NX,ONE)
c ORIENT(1,1)=ONE
c ORIENT(2,2)=ONE
c ORIENT(3,3)=ONE
c ENDIF
c ENDIF
C
C NOW RESET ORIENT TO THE IDENTITY IF THE GROUP IS ABELIAN.
C
C This is important, for abelian groups, the ORIENT is the unit matrix
IF(IDEGEN.EQ.0)THEN
CALL ZERO(ORIENT,9)
ORIENT(1,1)=ONE
ORIENT(2,2)=ONE
ORIENT(3,3)=ONE
ENDIF
C
C FILL SYMSTR STRING AND WRITE OUT IF THIS IS A VMOL-BASED CALCULATION.
C
C
C IF THIS IS A VMOL-DRIVEN CALCULATION, WRITE SYMMETRY INFORMATION
C TO VMLSYM. ALSO FOR ARGOS INPUT
C
CJDW 9/20/96. Also for SEWARD.
CADY 5/06/04. Also for GAMESS.
C
75 IF(INTTYP.EQ.1.OR.INTTYP.EQ.2.OR.INTTYP.EQ.0.OR.INTTYP.EQ.5
& .OR.INTTYP.EQ.4)THEN
IF(PGRP.EQ.'C2v')THEN
SYMSTR(1)(2:2)='X'
SYMSTR(2)(2:2)='Y'
IOPS=3
ELSEIF(PGRP.EQ.'D2h')THEN
SYMSTR(1)(2:2)='X'
SYMSTR(2)(2:2)='Y'
SYMSTR(3)(3:3)='Z'
IOPS=4
ELSEIF(PGRP.EQ.'C2h')THEN
SYMSTR(1)(2:2)='Z'
SYMSTR(2)(2:3)='XY'
IOPS=3
ELSEIF(PGRP.EQ.'C s')THEN
SYMSTR(1)(2:2)='Z'
IOPS=2
ELSEIF(PGRP.EQ.'C i')THEN
SYMSTR(1) ='XYZ'
IOPS=2
ELSEIF(PGRP.EQ.'D2 ')THEN
CJDW 9/16/96. Modification from JFS.
CALL SETD2XYZ(NATOMS,Q,ATMASS,SYMSTR(1),SYMSTR(2),IERROR)
IOPS=3
IF(IERROR.EQ.1)THEN
PGRP='C2 '
SYMSTR(1)(2:3)='XY'
IOPS=2
ENDIF
c SYMSTR(1)(2:3)='XY'
c SYMSTR(2)(2:3)='XZ'
c IOPS=3
ELSEIF(PGRP.EQ.'C2 ')THEN
SYMSTR(1)(2:3)='XY'
IOPS=2
ELSEIF(PGRP.EQ.'C1 ')THEN
IOPS=1
ENDIF
INQUIRE(FILE='VMLSYM',EXIST=YESNO,NUMBER=JUNK)
IF(YESNO)THEN
OPEN(UNIT=78,FILE='VMLSYM',STATUS='OLD',FORM='UNFORMATTED')
CLOSE(UNIT=78,STATUS='DELETE')
ENDIF
OPEN(UNIT=78,FILE='VMLSYM',STATUS='NEW',FORM='UNFORMATTED')
WRITE(78)MIN(3,IOPS-1),(SYMSTR(I),I=1,3)
CLOSE(UNIT=78,STATUS='KEEP')
ENDIF
C
C DO NUMERIC-ASCII CONVERSION OF IHIGH.
C
JnkStr=some_string_func(FPGRP,IHIGH)
FPGrp = JnkStr
IF(FPGRP.EQ.' ')FPGRP=BPGRP
C
C FIGURE OUT WHAT THE COMPUTATIONAL POINT GROUP IS FROM OPTCTL PARAMETER
C ISYM. IF IT IS SET TO 0, THEN JUST USE PGRP AS IS. IF =1, THEN USE
C C1. IF =3, USE FULL POINT GROUP{SYMM=NONE(0),OFF(1),ON(2),FULL(3)}.
C
IF(ISYM.EQ.1)PGRP='C1 '
IF(ISYM.EQ.3)PGRP=FPGRP
C The ncycle=0 refers to the first joda run during geo. optimizations
IF ( ncycle .EQ. 0 ) THEN
IF(NOSILENT)WRITE(LUOUT,788)
IF(IDEGEN.GT.0)THEN
IF(NOSILENT)WRITE(LUOUT,1881)FPGRP
ENDIF
IF(IDEGEN.EQ.0)THEN
IF(NOSILENT)
& WRITE(LUOUT,1881)some_string_func(BPGRP,IHigh)
FPGRP=BPGRP
ENDIF
1881 FORMAT(T3,' The full molecular point group is ',a,'.')
IF(NOSILENT) WRITE(LUOUT,177)BPGRP
IF(NOSILENT) WRITE(LUOUT,712)PGRP
IF(NOSILENT) WRITE(LUOUT,788)
788 FORMAT(80('*'))
712 FORMAT(T3,' The computational point group is ',A,'.')
177 FORMAT(T3,' The largest Abelian subgroup of the full ',
$ 'molecular point group is ',a,'.')
ELSE
C
C MAKE SURE THAT SYMMETRY HAS NOT BEEN REDUCED FROM PREVIOUS STEP. IF
C IT HAS, AND THIS IS NOT A TS SEARCH, THEN ABORT.
C
IF(FPGRP.NE.TMPGRP)THEN
ORDNEW=IORGRP(FPGRP)
ORDOLD=IORGRP(TMPGRP)
WRITE(LUOUT,1883)FPGRP
1883 FORMAT(T3,'@SYMMETRY-I, Point group has changed to ',A,'.')
C
C Originally test for TS searches were done by INR .NE. 2. That
C was dangerous. Now the TSSEARCH is a logical variable that
C is set to TRUE for TS searches (see above, INR=4, 5, and 6).
C Ajith Perera 08/2001.
C
C A bug fix, we must allow only the TS searches to proceed when
C there is a symmetry lowering. It is amazing that it took us
C 15 years to get this right! Ajith Perera 05/2005
C
IF (ORDNEW.LT.ORDOLD) THEN
IF (.NOT.TSSEARCH) THEN
WRITE(LUOUT,1884)
1884 FORMAT(T3,'@SYMMETRY-F, Descent in symmetry detected.')
CALL ERREX
ENDIF
ELSE
WRITE(LUOUT,1885)
1885 FORMAT(T3,'@SYMMETRY-F, Ascent in symmetry detected.')
CALL ERREX
ENDIF
ENDIF
ENDIF
C
C CHECK TO SEE IF ORIENTATION MATRIX IS EMPTY. IF IT IS, THEN SET IT EQ
C TO THE IDENTITY MATRIX.
C
ZTOTAL=0.D0
DO 1032 I=1,3
DO 1033 J=1,3
ZTOTAL=ZTOTAL+DABS(ORIENT(I,J))
1033 CONTINUE
1032 CONTINUE
IF(ZTOTAL.LT.1.D-8)THEN
DO 1034 I=1,3
ORIENT(I,I)=ONE
1034 CONTINUE
ENDIF
C
C DEFINE ORIENT TO PUT I AND Ih IN THEIR CANONICAL ORIENTATIONS (C5s
C ALONG (0,0,1) AND (0,2/SQRT(5),1/SQRT(5)).
C
IF(FPGRP.EQ.'I h '.OR.FPGRP.EQ.'I ')THEN
CALL ZERO(SCRATCH,3*NX)
SQ5=ONE/DSQRT(5.D0)
ARGX=DSQRT(0.5D0*(ONE-SQ5))
ANGL=DACOS(ARGX)
ANGL2=DACOS(SQ5)
CALL ROTM(1,ANGL,0,ORIENT)
ENDIF
C
C DUMP SOME NECESSARIES TO JOBARC
C
IF(FPGRP(2:2).EQ.'X')ILINEAR=1
#ifdef _DEBUG_LVL0
Print*, "The Orientation matrices: ORIEN2 and ORIENT"
Write(6,*)
CALL OUTPUT(ORIEN2, 1, 3, 1, 3, 3, 3, 0)
Write(6,*)
CALL OUTPUT(ORIENT, 1, 3, 1, 3, 3, 3, 0)
Write(6,*) "The coords in /COORD/ common block"
Print*, (Q(I), I=1, 3*NATOMS)
Write(6,*)
Write(6,*) "The coords NEWQ array "
Print*, (NEWQ(I), I=1, 3*NATOMS)
Write(6,*)
#endif
CALL XGEMM('N','N',3,3,3,ONE,ORIEN2,3,ORIENT,3,ZILCH,
& SCRATCH,3)
C
#ifdef _DEBUG_LVL0
Write(6,*)
CALL OUTPUT(SCRATCH, 1, 3, 1, 3, 3, 3, 0)
#endif
CALL IPUTREC(20,'JOBARC','LINEAR ',IONE,ILINEAR)
RETURN
END
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