1: #ifndef lint 2: static char *sccsid = "@(#)spline.c 4.3 (Berkeley) 9/21/85"; 3: #endif 4: 5: #include <stdio.h> 6: #include <math.h> 7: 8: #define NP 1000 9: #define INF HUGE 10: 11: struct proj { int lbf,ubf; float a,b,lb,ub,quant,mult,val[NP]; } x,y; 12: float *diag, *r; 13: float dx = 1.; 14: float ni = 100.; 15: int n; 16: int auta; 17: int periodic; 18: float konst = 0.0; 19: float zero = 0.; 20: 21: /* Spline fit technique 22: let x,y be vectors of abscissas and ordinates 23: h be vector of differences hi=xi-xi-1 24: y" be vector of 2nd derivs of approx function 25: If the points are numbered 0,1,2,...,n+1 then y" satisfies 26: (R W Hamming, Numerical Methods for Engineers and Scientists, 27: 2nd Ed, p349ff) 28: hiy"i-1+2(hi+hi+1)y"i+hi+1y"i+1 29: 30: = 6[(yi+1-yi)/hi+1-(yi-yi-1)/hi] i=1,2,...,n 31: 32: where y"0 = y"n+1 = 0 33: This is a symmetric tridiagonal system of the form 34: 35: | a1 h2 | |y"1| |b1| 36: | h2 a2 h3 | |y"2| |b2| 37: | h3 a3 h4 | |y"3| = |b3| 38: | . | | .| | .| 39: | . | | .| | .| 40: It can be triangularized into 41: | d1 h2 | |y"1| |r1| 42: | d2 h3 | |y"2| |r2| 43: | d3 h4 | |y"3| = |r3| 44: | . | | .| | .| 45: | . | | .| | .| 46: where 47: d1 = a1 48: 49: r0 = 0 50: 51: di = ai - hi2/di-1 1<i<_n 52: 53: ri = bi - hiri-1/di-1i 1<_i<_n 54: 55: the back solution is 56: y"n = rn/dn 57: 58: y"i = (ri-hi+1y"i+1)/di 1<_i<n 59: 60: superficially, di and ri don't have to be stored for they can be 61: recalculated backward by the formulas 62: 63: di-1 = hi2/(ai-di) 1<i<_n 64: 65: ri-1 = (bi-ri)di-1/hi 1<i<_n 66: 67: unhappily it turns out that the recursion forward for d 68: is quite strongly geometrically convergent--and is wildly 69: unstable going backward. 70: There's similar trouble with r, so the intermediate 71: results must be kept. 72: 73: Note that n-1 in the program below plays the role of n+1 in the theory 74: 75: Other boundary conditions_________________________ 76: 77: The boundary conditions are easily generalized to handle 78: 79: y0" = ky1", yn+1" = kyn" 80: 81: for some constant k. The above analysis was for k = 0; 82: k = 1 fits parabolas perfectly as well as stright lines; 83: k = 1/2 has been recommended as somehow pleasant. 84: 85: All that is necessary is to add h1 to a1 and hn+1 to an. 86: 87: 88: Periodic case_____________ 89: 90: To do this, add 1 more row and column thus 91: 92: | a1 h2 h1 | |y1"| |b1| 93: | h2 a2 h3 | |y2"| |b2| 94: | h3 a4 h4 | |y3"| |b3| 95: | | | .| = | .| 96: | . | | .| | .| 97: | h1 h0 a0 | | .| | .| 98: 99: where h0=_ hn+1 100: 101: The same diagonalization procedure works, except for 102: the effect of the 2 corner elements. Let si be the part 103: of the last element in the ith "diagonalized" row that 104: arises from the extra top corner element. 105: 106: s1 = h1 107: 108: si = -si-1hi/di-1 2<_i<_n+1 109: 110: After "diagonalizing", the lower corner element remains. 111: Call ti the bottom element that appears in the ith colomn 112: as the bottom element to its left is eliminated 113: 114: t1 = h1 115: 116: ti = -ti-1hi/di-1 117: 118: Evidently ti = si. 119: Elimination along the bottom row 120: introduces further corrections to the bottom right element 121: and to the last element of the right hand side. 122: Call these corrections u and v. 123: 124: u1 = v1 = 0 125: 126: ui = ui-1-si-1*ti-1/di-1 127: 128: vi = vi-1-ri-1*ti-1/di-1 2<_i<_n+1 129: 130: The back solution is now obtained as follows 131: 132: y"n+1 = (rn+1+vn+1)/(dn+1+sn+1+tn+1+un+1) 133: 134: y"i = (ri-hi+1*yi+1-si*yn+1)/di 1<_i<_n 135: 136: Interpolation in the interval xi<_x<_xi+1 is by the formula 137: 138: y = yix+ + yi+1x- -(h2i+1/6)[y"i(x+-x+3)+y"i+1(x--x-3)] 139: where 140: x+ = xi+1-x 141: 142: x- = x-xi 143: */ 144: 145: float 146: rhs(i){ 147: int i_; 148: double zz; 149: i_ = i==n-1?0:i; 150: zz = (y.val[i]-y.val[i-1])/(x.val[i]-x.val[i-1]); 151: return(6*((y.val[i_+1]-y.val[i_])/(x.val[i+1]-x.val[i]) - zz)); 152: } 153: 154: spline(){ 155: float d,s,u,v,hi,hi1; 156: float h; 157: float D2yi,D2yi1,D2yn1,x0,x1,yy,a; 158: int end; 159: float corr; 160: int i,j,m; 161: if(n<3) return(0); 162: if(periodic) konst = 0; 163: d = 1; 164: r[0] = 0; 165: s = periodic?-1:0; 166: for(i=0;++i<n-!periodic;){ /* triangularize */ 167: hi = x.val[i]-x.val[i-1]; 168: hi1 = i==n-1?x.val[1]-x.val[0]: 169: x.val[i+1]-x.val[i]; 170: if(hi1*hi<=0) return(0); 171: u = i==1?zero:u-s*s/d; 172: v = i==1?zero:v-s*r[i-1]/d; 173: r[i] = rhs(i)-hi*r[i-1]/d; 174: s = -hi*s/d; 175: a = 2*(hi+hi1); 176: if(i==1) a += konst*hi; 177: if(i==n-2) a += konst*hi1; 178: diag[i] = d = i==1? a: 179: a - hi*hi/d; 180: } 181: D2yi = D2yn1 = 0; 182: for(i=n-!periodic;--i>=0;){ /* back substitute */ 183: end = i==n-1; 184: hi1 = end?x.val[1]-x.val[0]: 185: x.val[i+1]-x.val[i]; 186: D2yi1 = D2yi; 187: if(i>0){ 188: hi = x.val[i]-x.val[i-1]; 189: corr = end?2*s+u:zero; 190: D2yi = (end*v+r[i]-hi1*D2yi1-s*D2yn1)/ 191: (diag[i]+corr); 192: if(end) D2yn1 = D2yi; 193: if(i>1){ 194: a = 2*(hi+hi1); 195: if(i==1) a += konst*hi; 196: if(i==n-2) a += konst*hi1; 197: d = diag[i-1]; 198: s = -s*d/hi; 199: }} 200: else D2yi = D2yn1; 201: if(!periodic) { 202: if(i==0) D2yi = konst*D2yi1; 203: if(i==n-2) D2yi1 = konst*D2yi; 204: } 205: if(end) continue; 206: m = hi1>0?ni:-ni; 207: m = 1.001*m*hi1/(x.ub-x.lb); 208: if(m<=0) m = 1; 209: h = hi1/m; 210: for(j=m;j>0||i==0&&j==0;j--){ /* interpolate */ 211: x0 = (m-j)*h/hi1; 212: x1 = j*h/hi1; 213: yy = D2yi*(x0-x0*x0*x0)+D2yi1*(x1-x1*x1*x1); 214: yy = y.val[i]*x0+y.val[i+1]*x1 -hi1*hi1*yy/6; 215: printf("%f ",x.val[i]+j*h); 216: printf("%f\n",yy); 217: } 218: } 219: return(1); 220: } 221: readin() { 222: for(n=0;n<NP;n++){ 223: if(auta) x.val[n] = n*dx+x.lb; 224: else if(!getfloat(&x.val[n])) break; 225: if(!getfloat(&y.val[n])) break; } } 226: 227: getfloat(p) 228: float *p;{ 229: char buf[30]; 230: register c; 231: int i; 232: extern double atof(); 233: for(;;){ 234: c = getchar(); 235: if (c==EOF) { 236: *buf = '\0'; 237: return(0); 238: } 239: *buf = c; 240: switch(*buf){ 241: case ' ': 242: case '\t': 243: case '\n': 244: continue;} 245: break;} 246: for(i=1;i<30;i++){ 247: c = getchar(); 248: if (c==EOF) { 249: buf[i] = '\0'; 250: break; 251: } 252: buf[i] = c; 253: if('0'<=c && c<='9') continue; 254: switch(c) { 255: case '.': 256: case '+': 257: case '-': 258: case 'E': 259: case 'e': 260: continue;} 261: break; } 262: buf[i] = ' '; 263: *p = atof(buf); 264: return(1); } 265: 266: getlim(p) 267: struct proj *p; { 268: int i; 269: for(i=0;i<n;i++) { 270: if(!p->lbf && p->lb>(p->val[i])) p->lb = p->val[i]; 271: if(!p->ubf && p->ub<(p->val[i])) p->ub = p->val[i]; } 272: } 273: 274: 275: main(argc,argv) 276: char *argv[];{ 277: extern char *malloc(); 278: int i; 279: x.lbf = x.ubf = y.lbf = y.ubf = 0; 280: x.lb = INF; 281: x.ub = -INF; 282: y.lb = INF; 283: y.ub = -INF; 284: while(--argc > 0) { 285: argv++; 286: again: switch(argv[0][0]) { 287: case '-': 288: argv[0]++; 289: goto again; 290: case 'a': 291: auta = 1; 292: numb(&dx,&argc,&argv); 293: break; 294: case 'k': 295: numb(&konst,&argc,&argv); 296: break; 297: case 'n': 298: numb(&ni,&argc,&argv); 299: break; 300: case 'p': 301: periodic = 1; 302: break; 303: case 'x': 304: if(!numb(&x.lb,&argc,&argv)) break; 305: x.lbf = 1; 306: if(!numb(&x.ub,&argc,&argv)) break; 307: x.ubf = 1; 308: break; 309: default: 310: fprintf(stderr, "Bad agrument\n"); 311: exit(1); 312: } 313: } 314: if(auta&&!x.lbf) x.lb = 0; 315: readin(); 316: getlim(&x); 317: getlim(&y); 318: i = (n+1)*sizeof(dx); 319: diag = (float *)malloc((unsigned)i); 320: r = (float *)malloc((unsigned)i); 321: if(r==NULL||!spline()) for(i=0;i<n;i++){ 322: printf("%f ",x.val[i]); 323: printf("%f\n",y.val[i]); } 324: } 325: numb(np,argcp,argvp) 326: int *argcp; 327: float *np; 328: char ***argvp;{ 329: double atof(); 330: char c; 331: if(*argcp<=1) return(0); 332: c = (*argvp)[1][0]; 333: if(!('0'<=c&&c<='9' || c=='-' || c== '.' )) return(0); 334: *np = atof((*argvp)[1]); 335: (*argcp)--; 336: (*argvp)++; 337: return(1); }