c
c _____________________________
c
c
	subroutine disquat(qq,index1,index2,qresult,index1a,index2a)
	real qq(4),qqn(4),qresult(4),quint(4),quintn(4),qmax
	integer index1,index2,index1a,index2a,index3,iq4neg,i,j,k,ip,jp
c
c       PI=3.14159265
c
c  algorithm for finding disorientation corresponding to 
c  input quaternion, QQ (from COMQUAT):  resultant quat. is in QRESULT
c  indices for symmetry operators are in INDEX1, INDEX2
c
	index1=0
	index2=0
	index1a=0
	index2a=0
	qmax=0.
	ip=1
	jp=1
c
	do 200, i=1,24
		 call presymm(qq,i,quint)
c
		do 190, j=1,24
c  loop over the two sides of the grain
			call postsymm(quint,j,quintn)
c
		do 185, kjl=1,2
		    switch=1.0
			if(kjl.eq.2) then
			switch=-1.0
			endif
				do 180, k=1,4
					qqn(k)=quintn(k)*switch
180					continue
c  can take negative of Q because identity = +/-(0,0,0,1)
c
	do 210, ijk=1,2
		qresult(1)=qqn(1)
		qresult(2)=qqn(2)
		qresult(3)=qqn(3)
		qresult(4)=qqn(4)
			iq4neg=ijk
	if(ijk.eq.2) qresult(4)=-1.*qqn(4)
c  take the inverse rotation, i.e. apply the switching symmetry
c
			if(qresult(1).ge.0.0) then
c			  if((qresult(1)-qresult(2)).lt.1e-5) then
c			    if((qresult(2)-qresult(3)).lt.1e-5) then
c			      if((qresult(3)-qresult(4)).lt.1e-5) then
			  if(qresult(1).le.qresult(2)) then
			    if(qresult(2).le.qresult(3)) then
			      if(qresult(3).le.qresult(4)) then
c   these IFs ensure that 0<=q1<q2<q3<q4 to place it in the fundamental zone
			        if(qresult(4).gt.qmax) then
c  so now we test to see if we have a smaller angle
			        qmax=qresult(4)
			        index1=i
			        index2=j
c			        index3=0
				index1a=0
			        if(ijk.eq.2) index1a=1
			        index2a=0
			        if(kjl.eq.2) index2a=1
c  record the result if we have found a quat in the fund. zone, and a smaller angle
				  endif
				endif
			       endif
			      endif
			     endif
210		continue
185				continue
190				continue
200			continue
c
	if(index1.eq.0.or.index2.eq.0) stop 'DISQUAT failed!'
c
		 call presymm(qq,index1,quint)
c
		 call postsymm(quint,index2,qqn)
c		 write(*,"('DISQUAT: qqn = ',4f8.3)") qqn
		do 280, k=1,4
			if(index2a.eq.1) then
c  if we had to take the negative of the Q, then do so
c  to the answer
c
				qresult(k)=qqn(k)*-1.0
c  can take negative of Q because identity = +/-(0,0,0,1)
			else
				qresult(k)=qqn(k)
			endif
280		continue
c
	if(index1a.eq.1) then
		qresult(4)=-1.*qresult(4)
c  restore the 4th component if disorient found for ijk=1
	endif
c
c	write(*,*) 'DISQUAT: qmax= ',qmax
	return
	end
c
c
c  ----------------
c