2.11BSD/src/usr.bin/spline.c
#ifndef lint
static char *sccsid = "@(#)spline.c 4.3 (Berkeley) 9/21/85";
#endif
#include <stdio.h>
#include <math.h>
#define NP 1000
#define INF HUGE
struct proj { int lbf,ubf; float a,b,lb,ub,quant,mult,val[NP]; } x,y;
float *diag, *r;
float dx = 1.;
float ni = 100.;
int n;
int auta;
int periodic;
float konst = 0.0;
float zero = 0.;
/* Spline fit technique
let x,y be vectors of abscissas and ordinates
h be vector of differences hi=xi-xi-1
y" be vector of 2nd derivs of approx function
If the points are numbered 0,1,2,...,n+1 then y" satisfies
(R W Hamming, Numerical Methods for Engineers and Scientists,
2nd Ed, p349ff)
hiy"i-1+2(hi+hi+1)y"i+hi+1y"i+1
= 6[(yi+1-yi)/hi+1-(yi-yi-1)/hi] i=1,2,...,n
where y"0 = y"n+1 = 0
This is a symmetric tridiagonal system of the form
| a1 h2 | |y"1| |b1|
| h2 a2 h3 | |y"2| |b2|
| h3 a3 h4 | |y"3| = |b3|
| . | | .| | .|
| . | | .| | .|
It can be triangularized into
| d1 h2 | |y"1| |r1|
| d2 h3 | |y"2| |r2|
| d3 h4 | |y"3| = |r3|
| . | | .| | .|
| . | | .| | .|
where
d1 = a1
r0 = 0
di = ai - hi2/di-1 1<i<_n
ri = bi - hiri-1/di-1i 1<_i<_n
the back solution is
y"n = rn/dn
y"i = (ri-hi+1y"i+1)/di 1<_i<n
superficially, di and ri don't have to be stored for they can be
recalculated backward by the formulas
di-1 = hi2/(ai-di) 1<i<_n
ri-1 = (bi-ri)di-1/hi 1<i<_n
unhappily it turns out that the recursion forward for d
is quite strongly geometrically convergent--and is wildly
unstable going backward.
There's similar trouble with r, so the intermediate
results must be kept.
Note that n-1 in the program below plays the role of n+1 in the theory
Other boundary conditions_________________________
The boundary conditions are easily generalized to handle
y0" = ky1", yn+1" = kyn"
for some constant k. The above analysis was for k = 0;
k = 1 fits parabolas perfectly as well as stright lines;
k = 1/2 has been recommended as somehow pleasant.
All that is necessary is to add h1 to a1 and hn+1 to an.
Periodic case_____________
To do this, add 1 more row and column thus
| a1 h2 h1 | |y1"| |b1|
| h2 a2 h3 | |y2"| |b2|
| h3 a4 h4 | |y3"| |b3|
| | | .| = | .|
| . | | .| | .|
| h1 h0 a0 | | .| | .|
where h0=_ hn+1
The same diagonalization procedure works, except for
the effect of the 2 corner elements. Let si be the part
of the last element in the ith "diagonalized" row that
arises from the extra top corner element.
s1 = h1
si = -si-1hi/di-1 2<_i<_n+1
After "diagonalizing", the lower corner element remains.
Call ti the bottom element that appears in the ith colomn
as the bottom element to its left is eliminated
t1 = h1
ti = -ti-1hi/di-1
Evidently ti = si.
Elimination along the bottom row
introduces further corrections to the bottom right element
and to the last element of the right hand side.
Call these corrections u and v.
u1 = v1 = 0
ui = ui-1-si-1*ti-1/di-1
vi = vi-1-ri-1*ti-1/di-1 2<_i<_n+1
The back solution is now obtained as follows
y"n+1 = (rn+1+vn+1)/(dn+1+sn+1+tn+1+un+1)
y"i = (ri-hi+1*yi+1-si*yn+1)/di 1<_i<_n
Interpolation in the interval xi<_x<_xi+1 is by the formula
y = yix+ + yi+1x- -(h2i+1/6)[y"i(x+-x+3)+y"i+1(x--x-3)]
where
x+ = xi+1-x
x- = x-xi
*/
float
rhs(i){
int i_;
double zz;
i_ = i==n-1?0:i;
zz = (y.val[i]-y.val[i-1])/(x.val[i]-x.val[i-1]);
return(6*((y.val[i_+1]-y.val[i_])/(x.val[i+1]-x.val[i]) - zz));
}
spline(){
float d,s,u,v,hi,hi1;
float h;
float D2yi,D2yi1,D2yn1,x0,x1,yy,a;
int end;
float corr;
int i,j,m;
if(n<3) return(0);
if(periodic) konst = 0;
d = 1;
r[0] = 0;
s = periodic?-1:0;
for(i=0;++i<n-!periodic;){ /* triangularize */
hi = x.val[i]-x.val[i-1];
hi1 = i==n-1?x.val[1]-x.val[0]:
x.val[i+1]-x.val[i];
if(hi1*hi<=0) return(0);
u = i==1?zero:u-s*s/d;
v = i==1?zero:v-s*r[i-1]/d;
r[i] = rhs(i)-hi*r[i-1]/d;
s = -hi*s/d;
a = 2*(hi+hi1);
if(i==1) a += konst*hi;
if(i==n-2) a += konst*hi1;
diag[i] = d = i==1? a:
a - hi*hi/d;
}
D2yi = D2yn1 = 0;
for(i=n-!periodic;--i>=0;){ /* back substitute */
end = i==n-1;
hi1 = end?x.val[1]-x.val[0]:
x.val[i+1]-x.val[i];
D2yi1 = D2yi;
if(i>0){
hi = x.val[i]-x.val[i-1];
corr = end?2*s+u:zero;
D2yi = (end*v+r[i]-hi1*D2yi1-s*D2yn1)/
(diag[i]+corr);
if(end) D2yn1 = D2yi;
if(i>1){
a = 2*(hi+hi1);
if(i==1) a += konst*hi;
if(i==n-2) a += konst*hi1;
d = diag[i-1];
s = -s*d/hi;
}}
else D2yi = D2yn1;
if(!periodic) {
if(i==0) D2yi = konst*D2yi1;
if(i==n-2) D2yi1 = konst*D2yi;
}
if(end) continue;
m = hi1>0?ni:-ni;
m = 1.001*m*hi1/(x.ub-x.lb);
if(m<=0) m = 1;
h = hi1/m;
for(j=m;j>0||i==0&&j==0;j--){ /* interpolate */
x0 = (m-j)*h/hi1;
x1 = j*h/hi1;
yy = D2yi*(x0-x0*x0*x0)+D2yi1*(x1-x1*x1*x1);
yy = y.val[i]*x0+y.val[i+1]*x1 -hi1*hi1*yy/6;
printf("%f ",x.val[i]+j*h);
printf("%f\n",yy);
}
}
return(1);
}
readin() {
for(n=0;n<NP;n++){
if(auta) x.val[n] = n*dx+x.lb;
else if(!getfloat(&x.val[n])) break;
if(!getfloat(&y.val[n])) break; } }
getfloat(p)
float *p;{
char buf[30];
register c;
int i;
extern double atof();
for(;;){
c = getchar();
if (c==EOF) {
*buf = '\0';
return(0);
}
*buf = c;
switch(*buf){
case ' ':
case '\t':
case '\n':
continue;}
break;}
for(i=1;i<30;i++){
c = getchar();
if (c==EOF) {
buf[i] = '\0';
break;
}
buf[i] = c;
if('0'<=c && c<='9') continue;
switch(c) {
case '.':
case '+':
case '-':
case 'E':
case 'e':
continue;}
break; }
buf[i] = ' ';
*p = atof(buf);
return(1); }
getlim(p)
struct proj *p; {
int i;
for(i=0;i<n;i++) {
if(!p->lbf && p->lb>(p->val[i])) p->lb = p->val[i];
if(!p->ubf && p->ub<(p->val[i])) p->ub = p->val[i]; }
}
main(argc,argv)
char *argv[];{
extern char *malloc();
int i;
x.lbf = x.ubf = y.lbf = y.ubf = 0;
x.lb = INF;
x.ub = -INF;
y.lb = INF;
y.ub = -INF;
while(--argc > 0) {
argv++;
again: switch(argv[0][0]) {
case '-':
argv[0]++;
goto again;
case 'a':
auta = 1;
numb(&dx,&argc,&argv);
break;
case 'k':
numb(&konst,&argc,&argv);
break;
case 'n':
numb(&ni,&argc,&argv);
break;
case 'p':
periodic = 1;
break;
case 'x':
if(!numb(&x.lb,&argc,&argv)) break;
x.lbf = 1;
if(!numb(&x.ub,&argc,&argv)) break;
x.ubf = 1;
break;
default:
fprintf(stderr, "Bad agrument\n");
exit(1);
}
}
if(auta&&!x.lbf) x.lb = 0;
readin();
getlim(&x);
getlim(&y);
i = (n+1)*sizeof(dx);
diag = (float *)malloc((unsigned)i);
r = (float *)malloc((unsigned)i);
if(r==NULL||!spline()) for(i=0;i<n;i++){
printf("%f ",x.val[i]);
printf("%f\n",y.val[i]); }
}
numb(np,argcp,argvp)
int *argcp;
float *np;
char ***argvp;{
double atof();
char c;
if(*argcp<=1) return(0);
c = (*argvp)[1][0];
if(!('0'<=c&&c<='9' || c=='-' || c== '.' )) return(0);
*np = atof((*argvp)[1]);
(*argcp)--;
(*argvp)++;
return(1); }