8/31/2006
The code for\FromRobmin\solve.for contains changes to the bracketing procedure
that were found relevant in fitting nuclear spectra. Origfor.zip contains the original
fortran described below. The codes linked to below have been updated and thus changed
slightly. The ide is for\brack.wpj.
Fortran - for\Brack.for - and C- cpp\Brack.c - bracketing codes,
This is an annotated listing of the bracketing codes, followed by an assignment.
#include <stdlib.h> C header files always needed, in principle allow you
#include <stdio.h> to eliminate unneeded libraries.
#include <math.h>
double ftan; defines a global variable -- compare to Fortran common
double fun1(double x); function definitions
char brack(double (*fun)(double),double beg,double endc,double *x1,double *x2,
double *f1,double *f2);
double anraph(double x1,double f1,double x2, double f2,double (*fun)(double));
double bnraph(double beg,double endc,double (*fun)(double));
The main codes
main()
{double beg,endc,ftest; all variables must be defined in C
int itest;
// the test function is FUN1 The // is not always a legal way of making a comment
loop: I find that it always pays to loop through a test code.
printf(" enter the value of the tangent\n"); The \n is a carriage return line feed
scanf("%lg",&ftan); scanf is so bad that ...
endc=3.14159/2;
beg=-endc;
ftest=bnraph(beg,endc,fun1); note the passing of the external function
printf("\n tan(%lg) = %lg \n",ftest,ftan);
scanf("%d",&itest);
if(itest != -1)goto loop; The != is a not = , a logical operator
return;}
5
IMPLICIT REAL*8 (A-H,O-Z) required second line of all codes
COMMON /PASS/FTAN
supplementary method for passing variables
EXTERNAL FUN1 note the passing of the external function
PRINT*,' ENTER tan value'
READ(*,*)FTAN
ENDC=3.14159/2
BEG=-ENDC
XTEST=BNRAPH(BEG,ENDC,FUN1)
PRINT*,' TAN ',XTEST,' = ',FTAN
END
double fun1(double x) The function tan(x) - the global ftan in C
{return tan(x)-ftan;}
FUNCTION FUN1(X) The function being zeroed
IMPLICIT REAL*8 (A-H,O-Z)
COMMON /PASS/FTAN The common is needed to pass ftan
FUN1=TAN(X)-FTAN
RETURN
END
char brack(double (*fun)(double),double beg,double endc,double *x1,double *x2,
double *f1,double *f2) Note the pointers.
{double fact=1.17,alamda,abound,dup,dum,f20,fmult;
int i,itest;
/*C brack tries to find values x1 and x2 such that fun(x1)*fun(x2)<0
the return is 'y' if this has been is the case and 'n' if not */
alamda = *x1;
abound = endc-*x1;
f20 = pow(fact,20.);
dup = abound/f20;
abound = *x1 - beg;
dum = abound/f20;
*f1=fun(*x1);
printf(" x1 = %lg f1 = %lg \n",*x1,*f1);
if(*f1 == 0 )
{*x2=*x1;
*f2=*f1;
return 'f';}
fmult=fact;
for(i=0;i<20;++i)
{*x2=alamda-dum*fmult;
*f2=fun(*x2);
printf(" x2 = %lg f2 = %lg \n",*x2,*f2);
if(*f2 == 0) Note that == is the same as .eq. in fortran not the same as =
{*x1=*x2;
*f1=*f2;
return 'f';}
if (*f1 * *f2 < 0)return 'y'; Note the abiguity in the *’s C actually needs a bit more
*x2 = alamda + dup * fmult; space than Fortran for this reason
*f2=fun(*x2);
printf(" x2 = %lg f2 = %lg \n",*x2,*f2);
if(*f2 == 0)
{*x1=*x2;
*f1=*f2;
return 'f';}
if (*f1 * *f2 < 0)return 'y'; must have a space for readability only
fmult *= fact;} No need to say fmult = fmult * fact.
return 'y';}
The routine below dates from 2001. It finds tries a successively finer grid of
points until finally the range between beg and end has been broken into 46 intervals with
no sign change.
SUBROUTINE BRACK(FUNC,BEG,ENDC,X1,X2,F1,F2,SUCCES)
C BRACK TRIES TO FIND VALUES X1 AND X2 SUCH THAT FUNC(X1)*FUNC(X2)<0
C SUCCES IS RETURNED AS 'Y' IF THIS IS THE CASE AND 'N' IF NOT
C SUCCES IS RETURNED AS '0' IF FUNC(X1)=0
5
10
IMPLICIT REAL*8 (A-H,O-Z)
CHARACTER*1 SUCCES
DIMENSION MFRACT(10)
DATA MFRACT/2,3,5,7,9,11,13,17,19,23/
JM=1
IM=1
SUCCES='N'
X1=BEG
F1=FUNC(BEG)
IF(F1.EQ.0)THEN
SUCCES='0'
RETURN
ENDIF
X2=ENDC
CONTINUE
F2=FUNC(X2)
IF(F2.EQ.0)THEN
SUCCES='0'
X1=ENDC
RETURN
ELSEIF(F1*F2.LT.0)THEN
SUCCES='Y'
RETURN
ELSE
X2=BEG+(ENDC-BEG)*IM/MFRACT(JM)
IM=IM+1
IF(IM.EQ.MFRACT(JM))THEN
IF(JM.GE.10)GOTO 10
JM=JM+1
IM=1
ENDIF
GOTO 5
ENDIF
CONTINUE
RETURN
END
double anraph(double x1,double f1,double x2, double f2,double (*fun)(double))
{double xp[4],xm[4],x,xn,xd,ft,xt,xlower,xupper;
int np=0,nm=0,itest;
loop5:
x = x1-f1*(x2-x1)/(f2-f1);
xt=max(fabs(x),1e-12); watch the fabs, abs will not work nor will it give an error
if(fabs(x-x1)/xt < 1e-37)return x;
loop7:
ft = fun(x); note the definition of fun above
printf("x = %lg ft = %lg \n",x,ft);
printf("x1 = %lg x2=%lg",x1,x2);
if(ft*ft < 1e-37)return x;
if(f1*ft > 0)
{np += 1;
nm=0;
if(np > 3)
// Aitkin's extrapolation positive
{xn = xp[3]*xp[1]-xp[2]*xp[2];
xd = xp[3]+xp[1]-2*xp[2];
if(xd != 0)x=xn/xd;
else x=xp[3];
xlower = min (x1,x2);
xupper = max (x1,x2);
if(x < xlower || x > xupper)x = .5 *(x1+x2);
np=0;
goto loop7;}
xp[np]=x;
f1=ft;
if(x == x1)return x;
x1=x;
goto loop5;}
else
{nm += 1;
np=0;
if(nm > 3)
// Aitkin's extrapolation negative
{xn = xm[3]*xm[1]-xm[2]*xm[2];
xd = xm[3]+xm[1]-2*xm[2];
if(xd != 0)
{x=xn/xd;
if(x < x1 || x > x2 ) x = .5*(x1+x2);}
else x=xm[3];
xlower = min (x1,x2);
xupper = max (x1,x2);
if(x < xlower || x > xupper)x = .5 *(x1+x2);
nm=0;
goto loop7;}
else
{xm[nm]=x;
f2=ft;
if(x == x2)return x;
x2=x;
goto loop5;}}} The multiple parens can be a problem
5
SUBROUTINE NRAF(X1,F1,X2,F2,FUNC)
IMPLICIT REAL*8 (A-H,O-Z)
DIMENSION XP(3),XM(3)
NP=0
NM=0
NCALL=0
X=X1-F1*(X2-X1)/(F2-F1) Newtons method
c in robmin, I found that there is a problem in that x2 can
be almost x1 while f1 and f2 are different.
XT=MAX(ABS(X1),ABS(X2),1D-5)
IF(ABS((X1-X2)/XT).LT.1D-12)RETURN
7
FT=FUNC(X)
NCALL=NCALL+1
IF(NCALL.GT.500)RETURN
PRINT*,' X,FT',X,FT
IF(F1*FT.GT.0)THEN
NP=NP+1
NM=0
IF(NP.GT.3)THEN
XN=XP(3)*XP(1)-XP(2)*XP(2) AITKINS'S EXTRAPOLATION
XD=XP(3)+XP(1)-2*XP(2)
X=XP(3)
IF(XD.NE.0)X=XN/XD
XLOWER=MIN(X1,X2)
XUPPER=MAX(X1,X2)
IF(X.LT.XLOWER.OR.X.GT.XUPPER)X=(X1+X2)/2
NP=0
GOTO 7
ENDIF
XP(NP)=X
F1=FT
IF(X.EQ.X1)THEN
X1=X
X2=X
F2=FT
RETURN
ENDIF
X1=X
GOTO 5
ELSE
NM=NM+1
NP=0
IF(NM.GT.3)THEN
XN=XM(3)*XM(1)-XM(2)*XM(2) AITKINS'S EXTRAPOLATION
XD=XM(3)+XM(1)-2*XM(2)
X=XM(3)
IF(XD.NE.0)X=XN/XD
XLOWER=MIN(X1,X2)
XUPPER=MAX(X1,X2)
IF(X.LT.XLOWER.OR.X.GT.XUPPER)X=(X1+X2)/2
NM=0
GOTO 7
ENDIF
XM(NM)=X
F2=FT
IF(X.EQ.X2)THEN
X1=X
X2=X
RETURN
ENDIF
X2=X
GOTO 5
ENDIF
PRINT*,' ANOM RETURN'
RETURN
END
double bnraph(double beg,double endc,double (*fun)(double))
{double x1,x2,f1,f2,ftest;
int itest;
char succes;
// the test function is FUN1
x1=.5*(beg+endc);
succes=brack(*fun,beg,endc,&x1,&x2,&f1,&f2); Note the addresses being passed
printf(" succes = %c \n",succes);
if(succes == 'n')
{fprintf(stderr,"brack could find no evidence of a solution \n"); always leave an out
fprintf(stderr,"in the range %lg to %lg \n",beg,endc); try to describe the error
exit(2);}
if(succes == 'f')return x1;
ftest=anraph(x1,f1,x2,f2,fun1);
return ftest;}
FUNCTION BNRAPH(BEG,ENDC,FUN1)
IMPLICIT REAL*8 (A-H,O-Z)
CHARACTER*1 SUCCES
EXTERNAL FUN1
X1=.5*(BEG+ENDC)
CALL BRACK(FUN1,BEG,ENDC,X1,X2,F1,F2,SUCCES)
IF(SUCCES.NE.'Y')THEN
PRINT*,' CANNOT FIND A ZERO CROSSING BETWEEN ',BEG,'
AND',ENDC
STOP 'STOPPED IN BNRAPH'
ENDIF
PRINT*,X1,F1
PRINT*,X2,F2
PRINT*,SUCCES
IF(SUCCES.EQ.'Y')CALL NRAF(X1,F1,X2,F2,FUN1)
PRINT*,' AFTER NRAF X1,F1=',X1,F1
PRINT*,' AFTER NRAF X2,F2=',X2,F2
PRINT*,' FUN1(X1)=',FUN1(X1)
BNRAPH=.5D0*(X1+X2)
RETURN
END
Assignment


1
FA T ; T0 , d , p   

 T0  T
1

exp

 d








p
Use the brack routine to find T such that FA T ; T0 , d , p   0.975 for T0 = 1.17, d=0.5,
p=1.
Write the code such that the uniform grid in T(FA) can be found for FA = (j-1d0/2)*.01 for
j between 1 and 100. Send me the code. Plot T(FA) – not the line.