subroutine newnot ( break, coef, l, k, brknew, lnew, coefg )
c from * a practical guide to splines * by c. de boor
c returns lnew+1 knots in brknew which are equidistributed on (a,b)
c = (break(1),break(l+1)) wrto a certain monotone fctn g related to
c the k-th root of the k-th derivative of the pp function f whose pp-
c representation is contained in break, coef, l, k .
c
c****** i n p u t ******
c break, coef, l, k.....contains the pp-representation of a certain
c function f of order k . Specifically,
c d**(k-1)f(x) = coef(k,i) for break(i).le. x .lt.break(i+1)
c lnew.....number of intervals into which the interval (a,b) is to be
c sectioned by the new breakpoint sequence brknew .
c
c****** o u t p u t ******
c brknew.....array of length lnew+1 containing the new breakpoint se-
c quence
c coefg.....the coefficient part of the pp-repr. break, coefg, l, 2
c for the monotone p.linear function g wrto which brknew will
c be equidistributed.
c
c****** optional p r i n t e d o u t p u t ******
c coefg.....the pp coeffs of g are printed out if iprint is set
c .gt. 0 in data statement below.
c
c****** m e t h o d ******
c The k-th derivative of the given pp function f does not exist
c (except perhaps as a linear combination of delta functions). Never-
c theless, we construct a p.constant function h with breakpoint se-
c quence break which is approximately proportional to abs(d**k(f)).
c Specifically, on (break(i), break(i+1)),
c
c abs(jump at break(i) of pc) abs(jump at break(i+1) of pc)
c h = -------------------------- + ----------------------------
c break(i+1) - break(i-1) break(i+2) - break(i)
c
c with pc the p.constant (k-1)st derivative of f .
c Then, the p.linear function g is constructed as
c
c g(x) = integral of h(y)**(1/k) for y from a to x
c
c and its pp coeffs. stored in coefg .
c then brknew is determined by
c
c brknew(i) = a + g**(-1)((i-1)*step) , i=1,...,lnew+1
c
c where step = g(b)/lnew and (a,b) = (break(1),break(l+1)) .
c In the event that pc = d**(k-1)(f) is constant in (a,b) and
c therefore h = 0 identically, brknew is chosen uniformly spaced.
c
integer k,l,lnew, i,iprint,j
real break(1),brknew(1),coef(k,l),coefg(2,l), dif,difprv,oneovk
* ,step,stepi
c dimension break(l+1), brknew(lnew+1)
current fortran standard makes it impossible to specify the dimension
c of break and brknew without the introduction of additional
c otherwise superfluous arguments.
data iprint /0/
c
brknew(1) = break(1)
brknew(lnew+1) = break(l+1)
c if g is constant, brknew is uniform.
if (l .le. 1) go to 90
c construct the continuous p.linear function g .
oneovk = 1./float(k)
coefg(1,1) = 0.
difprv = abs(coef(k,2) - coef(k,1))/(break(3)-break(1))
do 10 i=2,l
dif = abs(coef(k,i) - coef(k,i-1))/(break(i+1) - break(i-1))
coefg(2,i-1) = (dif + difprv)**oneovk
coefg(1,i) = coefg(1,i-1)+coefg(2,i-1)*(break(i)-break(i-1))
10 difprv = dif
coefg(2,l) = (2.*difprv)**oneovk
c step = g(b)/lnew
step = (coefg(1,l)+coefg(2,l)*(break(l+1)-break(l)))/float(lnew)
c
if (iprint .gt. 0) print 600, step,(i,coefg(1,i),coefg(2,i),i=1,l)
600 format(7h step =,e16.7/(i5,2e16.5))
c if g is constant, brknew is uniform .
if (step .le. 0.) go to 90
c
c For i=2,...,lnew, construct brknew(i) = a + g**(-1)(stepi),
c with stepi = (i-1)*step . this requires inversion of the p.lin-
c ear function g . For this, j is found so that
c g(break(j)) .le. stepi .le. g(break(j+1))
c and then
c brknew(i) = break(j) + (stepi-g(break(j)))/dg(break(j)) .
c The midpoint is chosen if dg(break(j)) = 0 .
j = 1
do 30 i=2,lnew
stepi = float(i-1)*step
21 if (j .eq. l) go to 27
if (stepi .le. coefg(1,j+1))go to 27
j = j + 1
go to 21
27 if (coefg(2,j) .eq. 0.) go to 29
brknew(i) = break(j) + (stepi - coefg(1,j))/coefg(2,j)
go to 30
29 brknew(i) = (break(j) + break(j+1))/2.
30 continue
return
c
c if g is constant, brknew is uniform .
90 step = (break(l+1) - break(1))/float(lnew)
do 93 i=2,lnew
93 brknew(i) = break(1) + float(i-1)*step
return
end