program llsl
C THIS program FITS Y=A+BX
C X IS THE INDEPENDENT VARIABLE ARRAY WITH N POINTS:INPUT DATA
C Y IS THE DEPENDENT VARIABLE ARRAY WITH N POINTS:INPUT DATA
C N IS THE NUMBER OF POINTS: INPUT DATA
C R IS A two LINES ARRAY WHERE:R(1)=A,R(2)=B:OUTPUT DATA
C SIG2 IS THE VARIANCE OF THE FIT:SIG2=SUM((Y OBS-Y CALC)**2)/(N-2)
C CM IS THE 2X2 COVARIANCE MATRIX:OUTPUT DATA
C USES SUBROUTINE SYMIN
C************************************************************************
C Adapted by S. O. Kepler, this version written 16 DEC 85
C Last modified ** 16 DEC 1985 **
C************************************************************************
implicit double precision(a-h,p-z)
common /x/ X(8000)
common /y/ Y(8000)
dimension R(2),A(2,2),B(2),CM(2,2),sigma(2)
DIMENSION P(20),Q(20),RR(20)
CHARACTER*20 infile
pi=3.1415926536
321 write(*,322)
322 format(1x,'input file is: ', $)
read(*,323) infile
323 format(a20)
open(2,file=infile,status='old')
c ***********************************
c read in the number of data points
read(2,*)N
c ***********************************
iii=1
write(*,521) iii
521 format(1x,'program completed phase: ', I3)
write(*,522) N
522 format(1x,'N= ',i5)
DO 1 I=1,2
DO 1 J=1,2
A(I,J)=0
B(I)=0
R(I)=0
CM(I,J)=0
1 CONTINUE
c read header***********************************
c *****************************************
c read in data - time (x) and intensity (y)
DO 325 I=1,N
325 READ(2,*) X(I),Y(I)
c *****************************************
iii=2
write(*,521) iii
DO 2 I=1,N
XI=X(I)
YI=Y(I)
PI=1
A(1,1)=A(1,1)+PI
A(1,2)=A(1,2)+PI*XI
A(2,2)=A(2,2)+PI*XI**2
B(1)=B(1)+PI*YI
B(2)=B(2)+PI*YI*XI
2 CONTINUE
a(2,1)=a(1,2)
NN=2
C SUBROUTINE SYMIN (A,NN)
C
C THIS SUBROUTINE INVERTS A SYMMETRIC MATRIX.
C ***** INPUT DATA *****
C A = A SYMMETRIC MATRIX
C NN = THE RANK OF THE MATRIX. N MUST BE LESS THAN 20.
C ***** OUTPUT DATA *****
C A = THE INVERSE OF THE INPUT MATRIX
C
C DIMENSION A(NN,NN)
C DIMENSION P(20),Q(20),R(20)
1300 FORMAT(13H0SYMIN FAILED)
ZERO = 0.0
ONE = 1.0
DO 1000 M=1,NN
1000 RR(M) = ONE
DO 200 M=1,NN
BIG = ZERO
DO 3000 L=1,NN
AB = DABS(A(L,L))
IF(AB-BIG)3000,3000,4000
4000 IF(RR(L))1400,3000,1400
1400 BIG = AB
K = L
3000 CONTINUE
IF(BIG)6000,5000,6000
5000 WRITE (*,1300)
GO TO 101
C RETURN
6000 RR(K) = ZERO
Q(K) = ONE/A(K,K)
P(K) = ONE
A(K,K) = ZERO
KM1 = K-1
IF(KM1.EQ.0) GO TO 1600
DO 7000 L=1,KM1
P(L) = A(L,K)
IF(RR(L))9000,8000,9000
8000 Q(L) = A(L,K)*Q(K)
GO TO 7000
9000 Q(L) = -A(L,K)*Q(K)
7000 A(L,K) = ZERO
1600 CONTINUE
KP1 = K+1
IF(KP1.GT.NN) GO TO 1700
DO 1500 L=KP1,NN
IF(RR(L))1200,1100,1200
1200 P(L) = A(K,L)
GO TO 10000
1100 P(L) = -A(K,L)
10000 Q(L) = (-A(K,L))*Q(K)
1500 A(K,L) = ZERO
1700 CONTINUE
DO 200 L=1,NN
DO 200 K=L,NN
200 A(L,K) = A(L,K) + P(L)*Q(K)
M = NN+1
L = NN
DO 27 K=2,NN
M = M-1
L = L-1
DO 27 J=1,L
27 A(M,J) = A(J,M)
C RETURN
C NOW THE A MATRIX IS THE INVERSE OF THE INITIAL ONE
C FINDING R(I),WHERE R(1)=A,R(2)=B
DO 4 I=1,2
DO 4 J=1,2
R(I)=R(I)+A(I,J)*B(J)
4 CONTINUE
C FINDING SIG2
RSIG=0
DO 5 I=1,N
PI=1
RSIG=RSIG+((Y(I)-R(1)-R(2)*X(I))**2)*PI
5 CONTINUE
SIG2=RSIG/(N-2)
C FINDIND THE CORRELATION MATRIX CM
DO 6 I=1,2
DO 6 J=1,2
CM(I,J)=SIG2*A(I,J)
6 CONTINUE
do 3915 i=1,2
sigma(i)=dsqrt(cm(i,i))
3915 continue
100 format('Fitting a line y=a+bx'/,'a = ',1pe12.5,/'b = ',1pe12.5,
&/'sig2= ',1pe12.3)
c write header*******************************
c *******************************************
432 write(*,433)
433 format(1x,'output file name is: ', $)
read(*,323) infile
open(3,file=infile,status='new')
write(3,99)N
99 format('npoints= ',i5)
c output results
write(3,100)r,sig2
write(3,110)(sigma(i),i=1,2)
110 format('siga ='1pe12.3,/'sigb ='1pe12.3)
write(3,107)((cm(i,j),j=1,2),i=1,2)
107 format(31x,'Correlation Matrix'/,
&29x,'a, b'/,
&1x,2(2(0pg18.6,2x)/,1x))
101 continue
end