##// END OF EJS Templates
lagp.f update 7
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r1626:cb57cf1f88fc
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1 1 subroutine lagp(plag,wl,r0,dr,nl,nrange)
2 2 c
3 3 c predict lagged product matrix using state parameters stored
4 4 c as cubic b-splines
5 5 c read weights for numeric integration from a file
6 6 c
7 7 c faster version - assumes samples taken at half-integer lags, ranges
8 8 c
9 9 include 'fitter.h'
10 10 include 'spline.h'
11 11 include 'bfield.h'
12 12 integer nl,nrange
13 13 real plag(nl,nrange),taus(nl)
14 14 parameter(nw=1000)
15 15 integer ilag(nw)
16 16 real weight(nw),dtau(nw),drange(nw)
17 17 logical first
18 18 real store(2*nl,2*nrange,2)
19 19
20 20 character(1024) :: fqual_temp
21 21 character(:), allocatable :: fqual
22 22 character bfmodel_str*8
23 23
24 24 data first/.true./
25 25 c
26 26
27 27 c write(*,*) "dens_before: ",dens
28 28 c call exit
29 29 c zero out array
30 30 c write(*,*) "Starting lagp"
31 31 do i=1,nrange
32 32 do j=1,nl
33 33 plag(j,i)=0.0
34 34 end do
35 35 end do
36 36
37 37 c
38 38 c compute all required acfs
39 39 c
40 40 do i=1,2*nrange
41 41 ii=(i-1)/2+1
42 42 alt=r0+float(i-1)*dr*0.5 ! half integer ranges
43 43 c write(*,*) "dens_before: ",dens
44 44 c call exit
45 45 c write(*,*) "alt: ",alt
46 46 c write(*,*) "dens: ",dens
47 47 c write(*,*) "te: ",te
48 48 c write(*,*) "ti1: ",ti1
49 49 c write(*,*) "hf: ",hf
50 50 c write(*,*) "hef: ",hef
51 51 c call exit
52 52 call get_spline(alt,dens,te,ti1,hf,hef)
53 53 c write(*,*) "dens_after: ",dens
54 54 c call exit
55 55 do k=1,nion
56 56 ti(k)=ti1
57 57 end do
58 58 fi(2)=hf
59 59 fi(3)=hef
60 60 fi(1)=1.0-hf-hef
61 61
62 62 do j=1,2*nl
63 63 tau=float(j-1)*(dr/1.5e5)*0.5 ! half integer lags
64 64
65 65 c just consider a single azimuth angle for now
66 66
67 67 c call acf2(wl,tau,te,ti,fi,ven,vin,wi,nion,
68 68 c & alpha_prof(ii)+1.74e-2,dens,bfld_prof(ii),rho)
69 69 c store(j,i,1)=rho*dens*(100.0/alt)**2
70 70
71 71 c call acf2(wl,tau,te,ti,fi,ven,vin,wi,nion,
72 72 c & alpha_prof(ii)-1.74e-2,dens,bfld_prof(ii),rho)
73 73 c store(j,i,2)=rho*dens*(100.0/alt)**2
74 74
75 75 call acf2(wl,tau,te,ti,fi,ven,vin,wi,nion,
76 76 & alpha_prof(ii),dens,bfld_prof(ii),rho)
77 77 store(j,i,1)=rho*dens*(100.0/alt)**2
78 78 c write(*,*) "store", store(j,i,1)
79 79 c call exit
80 80 end do
81 81 end do
82 82
83 83 c construct lagged products
84 84 c
85 85 c write(*,*) "before weights"
86 86 c write(*,*) first
87 87 c write(*,*) "weight before: ",weight
88 88 c write(*,*) "BEFORE GET_PATH"
89 89 call get_path(fqual_temp)
90 90 c write(*,*) "L_BEF: ", fqual_temp, "L_BEF_end"
91 91 fqual = TRIM(fqual_temp)
92 bfmodel = 'bfmodel/'
93 fqual(1:len_trim(fqual)-len_trim(bfmodel))
92 bfmodel_str = 'bfmodel/'
93 fqual(1:len_trim(fqual)-len_trim(bfmodel_str))
94 94 fqual = fqual//'weights.dat'
95 95 c write(*,*) "Final fqual",fqual,"FINAL"
96 96
97 97 if(first) then
98 98 open(unit=25,file=fqual,
99 99 & status='old')
100 100 read (25,*) nwt
101 101 do i=1,nwt
102 102 read(25,*) weight(i),dtau(i),drange(i),ilag(i)
103 103 c write(*,*) "ilag: ",ilag(i)
104 104 end do
105 105 c flush(25)
106 106 close(25)
107 107 c first=.false.
108 108 end if
109 109
110 110 c write(*,*) "after weights"
111 111
112 112 c write(*,*) "weight(3): ",weight(3)
113 113 c call exit
114 114 c write(*,*) "nl: ",nl,"nrange: ",nrange,"nwt: ",nwt
115 115
116 116 do i=1+nl,nrange ! don't worry about redundancy here
117 117
118 118 do j=1,nwt
119 119 c write(*,*) "j: ",j
120 120 alt=r0+(float(i-1)+drange(j))*dr
121 121 tau=abs(dtau(j)*dr/1.5e5)
122 122 lag=ilag(j)+1
123 123 c write(*,*) tau
124 124
125 125 k=1+2.0*(tau/(dr/1.5e5))
126 126 l=1+2.0*(alt-r0)/dr
127 127
128 128 do m=1,1 ! 2
129 129
130 130 c call exit
131 131 c write(*,*) i
132 132 c call exit
133 133 c if (i .eq. nrange) then
134 134 c if (j .eq. 227) then
135 135
136 136 c write(*,*) "lag: ",lag
137 137 c write(*,*) "ilag: ",ilag(j)
138 138 c call exit
139 139 c write(*,*) "plag: ",plag(lag,i)
140 140 c write(*,*) "weight: ",weight
141 141 c
142 142 c write(*,*) "weight: ",weight(j)
143 143 c write(*,*) "store: ",store(k,l,m)
144 144 c call exit
145 145 c end if
146 146 c end if
147 147 plag(lag,i)=plag(lag,i)+weight(j)*store(k,l,m)
148 148 c if (i .eq. nrange) then
149 149 c write(*,*) "lag: ",lag
150 150 c write(*,*) "plag: ",plag(lag,i)
151 151 c write(*,*) "plag: ",plag(lag,i)
152 152 c call exit
153 153 c end if
154 154 end do
155 155 c write(*,*) "plag: ",plag(lag,i)
156 156 c write(*,*) "plag: ",plag(12,72)
157 157 c call exit
158 158 end do
159 159 c call exit
160 160 c write(*,*) "plag: ",plag(12,72)
161 161 end do
162 162 c write(*,*) "plag: ",plag(12,72)
163 163 c call exit
164 164 c write(*,*) "End LAGP"
165 165 return
166 166 end
167 167
168 168 subroutine lagp_old(plag,wl,r0,dr,nl,nrange)
169 169 c
170 170 c predict lagged product matrix using state parameters stored
171 171 c as cubic b-splines
172 172 c read weights for numeric integration from a file
173 173 c
174 174 c general version - samples can be anywhere
175 175 c
176 176 include 'fitter.h'
177 177 include 'spline.h'
178 178 include 'bfield.h'
179 179 integer nl,nrange
180 180 real plag(nl,nrange),taus(nl)
181 181 parameter(nw=1000)
182 182 integer ilag(nw)
183 183 real weight(nw),dtau(nw),drange(nw)
184 184 logical first
185 185 data first/.true./
186 186 c
187 187
188 188 c zero out array
189 189
190 190 do i=1,nrange
191 191 do j=1,nl
192 192 plag(j,i)=0.0
193 193 end do
194 194 end do
195 195
196 196 c
197 197 c construct lagged products
198 198 c
199 199
200 200 if(first) then
201 201 open(unit=25,file='weights.dat',status='old')
202 202 read (25,*) nwt
203 203 do i=1,nwt
204 204 read(25,*) weight(i),dtau(i),drange(i),ilag(i)
205 205 end do
206 206 close(25)
207 207 first=.false.
208 208 end if
209 209
210 210 c can avoid recalculating splines and/or ACFs when parameters repeat
211 211
212 212 altp=-1.0
213 213 taup=-1.0
214 214
215 215 do i=1+nl,nrange
216 216 do j=1,nwt
217 217 alt=r0+(float(i-1)+drange(j))*dr
218 218 ii=(alt-r0)/dr+1
219 219
220 220 if(alt.ne.altp) then
221 221 call get_spline(alt,dens,te,ti1,hf,hef)
222 222
223 223 c write(*,*) alt,dens,te,ti1,hf,hef
224 224
225 225 do k=1,nion
226 226 ti(k)=ti1
227 227 end do
228 228 fi(2)=hf
229 229 fi(3)=hef
230 230 fi(1)=1.0-hf-hef
231 231 end if
232 232
233 233 tau=abs(dtau(j)*dr/1.5e5)
234 234 lag=ilag(j)+1
235 235
236 236 c write(*,*) i,alpha_prof(i),bfld_prof(i)
237 237
238 238 if(tau.ne.taup.or.alt.ne.altp) then
239 239
240 240 c two- or three- point Gauss Hermite quadrature rule
241 241
242 242 c call acf2(wl,tau,te,ti,fi,ven,vin,wi,nion,
243 243 c & alpha_prof(ii)+1.74e-2,dens,bfld_prof(ii),rho1)
244 244
245 245 c call acf2(wl,tau,te,ti,fi,ven,vin,wi,nion,
246 246 c & alpha_prof(ii)-1.74e-2,dens,bfld_prof(ii),rho2)
247 247
248 248 c call acf2(wl,tau,te,ti,fi,ven,vin,wi,nion,
249 249 c & alpha_prof(ii),dens,bfld_prof(ii),rho1)
250 250
251 251 c call acf2(wl,tau,te,ti,fi,ven,vin,wi,nion,
252 252 c & alpha_prof(ii)+1.9e-2,dens,bfld_prof(ii),rho2)
253 253
254 254 c call acf2(wl,tau,te,ti,fi,ven,vin,wi,nion,
255 255 c & alpha_prof(ii)-1.9e-2,dens,bfld_prof(ii),rho3)
256 256
257 257 call acf2(wl,tau,te,ti,fi,ven,vin,wi,nion,
258 258 & alpha_prof(ii),dens,bfld_prof(ii),rho)
259 259
260 260 c rho=(rho1+rho2)/2.0
261 261 c rho=(rho2+rho3)*0.29541+rho1*1.18164)/1.77246
262 262
263 263 end if
264 264
265 265 plag(lag,i)=plag(lag,i)+weight(j)*rho*dens*(100.0/alt)**2
266 266
267 267 altp=alt
268 268 taup=tau
269 269
270 270 end do
271 271 end do
272 272
273 273 return
274 274 end
275 275
276 276 function atanh(x)
277 277 c
278 278 real atanh,x
279 279 c
280 280 atanh=log(sqrt((1.+x)/(1.-x)))
281 281 return
282 282 end
283 283
284 284 subroutine get_spline(alt,dens,te,ti,hf,hef)
285 285 c
286 286 c routines for handling cubic b-spline interpolation
287 287 c gets values consistent with stored coefficients
288 288 c
289 289
290 290 include 'spline.h'
291 291
292 292 c
293 293 c bspline values from 200-1500 km in 15-km intervals
294 294 c must specify five knots above ceiling
295 295 c note offset ... start accessing splines above bottom two knots
296 296 c
297 297 c five banks of splines, for Ne, Te, Ti, H+, and He+ versus altitude
298 298 c need to initialize ta somewhere
299 299 c
300 300 c write(*,*) "ta: ",ta
301 301 c write(*,*) "bcoef(1,1): ",bcoef(1,1)
302 302 c write(*,*) "nspline: ",nspline
303 303 c write(*,*) "norder: ",norder
304 304 c write(*,*) "alt: ",alt
305 305 dens=bvalue(ta,bcoef(1,1),nspline,norder,alt,0)
306 306 c write(*,*) "dens_inside_get_spline: ",dens
307 307 c write(*,*) dens
308 308 c call exit
309 309 dens=10.0**MAX(dens,2.0)
310 310 c write(*,*) dens
311 311 c call exit
312 312 te=bvalue(ta,bcoef(1,2),nspline,norder,alt,0)
313 313 te=t0+t1*(1.0+tanh(te))/2.0 ! DLH 10/14 was 3500
314 314 c tr=bvalue(ta,bcoef(1,3),nspline,norder,alt,0)
315 315 c tr=exp(tr)
316 316 c ti=te/tr
317 317 ti=bvalue(ta,bcoef(1,3),nspline,norder,alt,0)
318 318 ti=t0+t1*(1.0+tanh(ti))/2.0 ! DLH 10/14 was 3500
319 319 hf=bvalue(ta,bcoef(1,4),nspline,norder,alt,0)
320 320 hf=(1.0+tanh(hf))/2.0
321 321 hef=bvalue(ta,bcoef(1,5),nspline,norder,alt,0)
322 322 hef=(1.0+tanh(hef))/2.0
323 323 return
324 324 end
325 325
326 326
327 327 real function bvalue ( t, bcoef, n, k, x, jderiv )
328 328 c from * a practical guide to splines * by c. de boor
329 329 calls interv
330 330 c
331 331 calculates value at x of jderiv-th derivative of spline from b-repr.
332 332 c the spline is taken to be continuous from the right, EXCEPT at the
333 333 c rightmost knot, where it is taken to be continuous from the left.
334 334 c
335 335 c****** i n p u t ******
336 336 c t, bcoef, n, k......forms the b-representation of the spline f to
337 337 c be evaluated. specifically,
338 338 c t.....knot sequence, of length n+k, assumed nondecreasing.
339 339 c bcoef.....b-coefficient sequence, of length n .
340 340 c n.....length of bcoef and dimension of spline(k,t),
341 341 c a s s u m e d positive .
342 342 c k.....order of the spline .
343 343 c
344 344 c w a r n i n g . . . the restriction k .le. kmax (=20) is imposed
345 345 c arbitrarily by the dimension statement for aj, dl, dr below,
346 346 c but is n o w h e r e c h e c k e d for.
347 347 c
348 348 c x.....the point at which to evaluate .
349 349 c jderiv.....integer giving the order of the derivative to be evaluated
350 350 c a s s u m e d to be zero or positive.
351 351 c
352 352 c****** o u t p u t ******
353 353 c bvalue.....the value of the (jderiv)-th derivative of f at x .
354 354 c
355 355 c****** m e t h o d ******
356 356 c The nontrivial knot interval (t(i),t(i+1)) containing x is lo-
357 357 c cated with the aid of interv . The k b-coeffs of f relevant for
358 358 c this interval are then obtained from bcoef (or taken to be zero if
359 359 c not explicitly available) and are then differenced jderiv times to
360 360 c obtain the b-coeffs of (d**jderiv)f relevant for that interval.
361 361 c Precisely, with j = jderiv, we have from x.(12) of the text that
362 362 c
363 363 c (d**j)f = sum ( bcoef(.,j)*b(.,k-j,t) )
364 364 c
365 365 c where
366 366 c / bcoef(.), , j .eq. 0
367 367 c /
368 368 c bcoef(.,j) = / bcoef(.,j-1) - bcoef(.-1,j-1)
369 369 c / ----------------------------- , j .gt. 0
370 370 c / (t(.+k-j) - t(.))/(k-j)
371 371 c
372 372 c Then, we use repeatedly the fact that
373 373 c
374 374 c sum ( a(.)*b(.,m,t)(x) ) = sum ( a(.,x)*b(.,m-1,t)(x) )
375 375 c with
376 376 c (x - t(.))*a(.) + (t(.+m-1) - x)*a(.-1)
377 377 c a(.,x) = ---------------------------------------
378 378 c (x - t(.)) + (t(.+m-1) - x)
379 379 c
380 380 c to write (d**j)f(x) eventually as a linear combination of b-splines
381 381 c of order 1 , and the coefficient for b(i,1,t)(x) must then be the
382 382 c desired number (d**j)f(x). (see x.(17)-(19) of text).
383 383 c
384 384 parameter (kmax = 40)
385 385 integer jderiv,k,n, i,ilo,imk,j,jc,jcmin,jcmax,jj,kmj,km1
386 386 * ,mflag,nmi,jdrvp1
387 387 c integer kmax
388 388 C real bcoef(n),t(1),x, aj(20),dl(20),dr(20),fkmj
389 389 real bcoef(n),x, aj(kmax),dl(kmax),dr(kmax),fkmj
390 390 dimension t(n+k)
391 391 c former fortran standard made it impossible to specify the length of t
392 392 c precisely without the introduction of otherwise superfluous addition-
393 393 c al arguments.
394 394 bvalue = 0.
395 395 if (jderiv .ge. k) go to 99
396 396 c
397 397 c *** Find i s.t. 1 .le. i .lt. n+k and t(i) .lt. t(i+1) and
398 398 c t(i) .le. x .lt. t(i+1) . If no such i can be found, x lies
399 399 c outside the support of the spline f , hence bvalue = 0.
400 400 c (The asymmetry in this choice of i makes f rightcontinuous, except
401 401 c at t(n+k) where it is leftcontinuous.)
402 402 call interv ( t, n+k, x, i, mflag )
403 403 if (mflag .ne. 0) go to 99
404 404 c *** if k = 1 (and jderiv = 0), bvalue = bcoef(i).
405 405 km1 = k - 1
406 406 if (km1 .gt. 0) go to 1
407 407 bvalue = bcoef(i)
408 408 go to 99
409 409 c
410 410 c *** store the k b-spline coefficients relevant for the knot interval
411 411 c (t(i),t(i+1)) in aj(1),...,aj(k) and compute dl(j) = x - t(i+1-j),
412 412 c dr(j) = t(i+j) - x, j=1,...,k-1 . set any of the aj not obtainable
413 413 c from input to zero. set any t.s not obtainable equal to t(1) or
414 414 c to t(n+k) appropriately.
415 415 1 jcmin = 1
416 416 imk = i - k
417 417 if (imk .ge. 0) go to 8
418 418 jcmin = 1 - imk
419 419 do 5 j=1,i
420 420 5 dl(j) = x - t(i+1-j)
421 421 do 6 j=i,km1
422 422 aj(k-j) = 0.
423 423 6 dl(j) = dl(i)
424 424 go to 10
425 425 8 do 9 j=1,km1
426 426 9 dl(j) = x - t(i+1-j)
427 427 c
428 428 10 jcmax = k
429 429 nmi = n - i
430 430 if (nmi .ge. 0) go to 18
431 431 jcmax = k + nmi
432 432 do 15 j=1,jcmax
433 433 15 dr(j) = t(i+j) - x
434 434 do 16 j=jcmax,km1
435 435 aj(j+1) = 0.
436 436 16 dr(j) = dr(jcmax)
437 437 go to 20
438 438 18 do 19 j=1,km1
439 439 19 dr(j) = t(i+j) - x
440 440 c
441 441 20 do 21 jc=jcmin,jcmax
442 442 21 aj(jc) = bcoef(imk + jc)
443 443 c
444 444 c *** difference the coefficients jderiv times.
445 445 if (jderiv .eq. 0) go to 30
446 446 do 23 j=1,jderiv
447 447 kmj = k-j
448 448 fkmj = float(kmj)
449 449 ilo = kmj
450 450 do 23 jj=1,kmj
451 451 aj(jj) = ((aj(jj+1) - aj(jj))/(dl(ilo) + dr(jj)))*fkmj
452 452 23 ilo = ilo - 1
453 453 c
454 454 c *** compute value at x in (t(i),t(i+1)) of jderiv-th derivative,
455 455 c given its relevant b-spline coeffs in aj(1),...,aj(k-jderiv).
456 456 30 if (jderiv .eq. km1) go to 39
457 457 jdrvp1 = jderiv + 1
458 458 do 33 j=jdrvp1,km1
459 459 kmj = k-j
460 460 ilo = kmj
461 461 do 33 jj=1,kmj
462 462 aj(jj) = (aj(jj+1)*dl(ilo) + aj(jj)*dr(jj))/(dl(ilo)+dr(jj))
463 463 33 ilo = ilo - 1
464 464 39 bvalue = aj(1)
465 465 c
466 466 99 return
467 467 end
468 468 subroutine interv ( xt, lxt, x, left, mflag )
469 469 c from * a practical guide to splines * by C. de Boor
470 470 computes left = max( i : xt(i) .lt. xt(lxt) .and. xt(i) .le. x ) .
471 471 c
472 472 c****** i n p u t ******
473 473 c xt.....a real sequence, of length lxt , assumed to be nondecreasing
474 474 c lxt.....number of terms in the sequence xt .
475 475 c x.....the point whose location with respect to the sequence xt is
476 476 c to be determined.
477 477 c
478 478 c****** o u t p u t ******
479 479 c left, mflag.....both integers, whose value is
480 480 c
481 481 c 1 -1 if x .lt. xt(1)
482 482 c i 0 if xt(i) .le. x .lt. xt(i+1)
483 483 c i 0 if xt(i) .lt. x .eq. xt(i+1) .eq. xt(lxt)
484 484 c i 1 if xt(i) .lt. xt(i+1) .eq. xt(lxt) .lt. x
485 485 c
486 486 c In particular, mflag = 0 is the 'usual' case. mflag .ne. 0
487 487 c indicates that x lies outside the CLOSED interval
488 488 c xt(1) .le. y .le. xt(lxt) . The asymmetric treatment of the
489 489 c intervals is due to the decision to make all pp functions cont-
490 490 c inuous from the right, but, by returning mflag = 0 even if
491 491 C x = xt(lxt), there is the option of having the computed pp function
492 492 c continuous from the left at xt(lxt) .
493 493 c
494 494 c****** m e t h o d ******
495 495 c The program is designed to be efficient in the common situation that
496 496 c it is called repeatedly, with x taken from an increasing or decrea-
497 497 c sing sequence. This will happen, e.g., when a pp function is to be
498 498 c graphed. The first guess for left is therefore taken to be the val-
499 499 c ue returned at the previous call and stored in the l o c a l varia-
500 500 c ble ilo . A first check ascertains that ilo .lt. lxt (this is nec-
501 501 c essary since the present call may have nothing to do with the previ-
502 502 c ous call). Then, if xt(ilo) .le. x .lt. xt(ilo+1), we set left =
503 503 c ilo and are done after just three comparisons.
504 504 c Otherwise, we repeatedly double the difference istep = ihi - ilo
505 505 c while also moving ilo and ihi in the direction of x , until
506 506 c xt(ilo) .le. x .lt. xt(ihi) ,
507 507 c after which we use bisection to get, in addition, ilo+1 = ihi .
508 508 c left = ilo is then returned.
509 509 c
510 510 integer left,lxt,mflag, ihi,ilo,istep,middle
511 511 real x,xt(lxt)
512 512 data ilo /1/
513 513 save ilo
514 514 ihi = ilo + 1
515 515 if (ihi .lt. lxt) go to 20
516 516 if (x .ge. xt(lxt)) go to 110
517 517 if (lxt .le. 1) go to 90
518 518 ilo = lxt - 1
519 519 ihi = lxt
520 520 c
521 521 20 if (x .ge. xt(ihi)) go to 40
522 522 if (x .ge. xt(ilo)) go to 100
523 523 c
524 524 c **** now x .lt. xt(ilo) . decrease ilo to capture x .
525 525 istep = 1
526 526 31 ihi = ilo
527 527 ilo = ihi - istep
528 528 if (ilo .le. 1) go to 35
529 529 if (x .ge. xt(ilo)) go to 50
530 530 istep = istep*2
531 531 go to 31
532 532 35 ilo = 1
533 533 if (x .lt. xt(1)) go to 90
534 534 go to 50
535 535 c **** now x .ge. xt(ihi) . increase ihi to capture x .
536 536 40 istep = 1
537 537 41 ilo = ihi
538 538 ihi = ilo + istep
539 539 if (ihi .ge. lxt) go to 45
540 540 if (x .lt. xt(ihi)) go to 50
541 541 istep = istep*2
542 542 go to 41
543 543 45 if (x .ge. xt(lxt)) go to 110
544 544 ihi = lxt
545 545 c
546 546 c **** now xt(ilo) .le. x .lt. xt(ihi) . narrow the interval.
547 547 50 middle = (ilo + ihi)/2
548 548 if (middle .eq. ilo) go to 100
549 549 c note. it is assumed that middle = ilo in case ihi = ilo+1 .
550 550 if (x .lt. xt(middle)) go to 53
551 551 ilo = middle
552 552 go to 50
553 553 53 ihi = middle
554 554 go to 50
555 555 c**** set output and return.
556 556 90 mflag = -1
557 557 left = 1
558 558 return
559 559 100 mflag = 0
560 560 left = ilo
561 561 return
562 562 110 mflag = 1
563 563 if (x .eq. xt(lxt)) mflag = 0
564 564 left = lxt
565 565 111 if (left .eq. 1) return
566 566 left = left - 1
567 567 if (xt(left) .lt. xt(lxt)) return
568 568 go to 111
569 569 end
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