subroutine interv ( xt, lxt, x, left, mflag ) c from * a practical guide to splines * by C. de Boor computes left = max( i : xt(i) .lt. xt(lxt) .and. xt(i) .le. x ) . c c****** i n p u t ****** c xt.....a real sequence, of length lxt , assumed to be nondecreasing c lxt.....number of terms in the sequence xt . c x.....the point whose location with respect to the sequence xt is c to be determined. c c****** o u t p u t ****** c left, mflag.....both integers, whose value is c c 1 -1 if x .lt. xt(1) c i 0 if xt(i) .le. x .lt. xt(i+1) c i 0 if xt(i) .lt. x .eq. xt(i+1) .eq. xt(lxt) c i 1 if xt(i) .lt. xt(i+1) .eq. xt(lxt) .lt. x c c In particular, mflag = 0 is the 'usual' case. mflag .ne. 0 c indicates that x lies outside the CLOSED interval c xt(1) .le. y .le. xt(lxt) . The asymmetric treatment of the c intervals is due to the decision to make all pp functions cont- c inuous from the right, but, by returning mflag = 0 even if C x = xt(lxt), there is the option of having the computed pp function c continuous from the left at xt(lxt) . c c****** m e t h o d ****** c The program is designed to be efficient in the common situation that c it is called repeatedly, with x taken from an increasing or decrea- c sing sequence. This will happen, e.g., when a pp function is to be c graphed. The first guess for left is therefore taken to be the val- c ue returned at the previous call and stored in the l o c a l varia- c ble ilo . A first check ascertains that ilo .lt. lxt (this is nec- c essary since the present call may have nothing to do with the previ- c ous call). Then, if xt(ilo) .le. x .lt. xt(ilo+1), we set left = c ilo and are done after just three comparisons. c Otherwise, we repeatedly double the difference istep = ihi - ilo c while also moving ilo and ihi in the direction of x , until c xt(ilo) .le. x .lt. xt(ihi) , c after which we use bisection to get, in addition, ilo+1 = ihi . c left = ilo is then returned. c integer left,lxt,mflag, ihi,ilo,istep,middle real x,xt(lxt) data ilo /1/ save ilo ihi = ilo + 1 if (ihi .lt. lxt) go to 20 if (x .ge. xt(lxt)) go to 110 if (lxt .le. 1) go to 90 ilo = lxt - 1 ihi = lxt c 20 if (x .ge. xt(ihi)) go to 40 if (x .ge. xt(ilo)) go to 100 c c **** now x .lt. xt(ilo) . decrease ilo to capture x . istep = 1 31 ihi = ilo ilo = ihi - istep if (ilo .le. 1) go to 35 if (x .ge. xt(ilo)) go to 50 istep = istep*2 go to 31 35 ilo = 1 if (x .lt. xt(1)) go to 90 go to 50 c **** now x .ge. xt(ihi) . increase ihi to capture x . 40 istep = 1 41 ilo = ihi ihi = ilo + istep if (ihi .ge. lxt) go to 45 if (x .lt. xt(ihi)) go to 50 istep = istep*2 go to 41 45 if (x .ge. xt(lxt)) go to 110 ihi = lxt c c **** now xt(ilo) .le. x .lt. xt(ihi) . narrow the interval. 50 middle = (ilo + ihi)/2 if (middle .eq. ilo) go to 100 c note. it is assumed that middle = ilo in case ihi = ilo+1 . if (x .lt. xt(middle)) go to 53 ilo = middle go to 50 53 ihi = middle go to 50 c**** set output and return. 90 mflag = -1 left = 1 return 100 mflag = 0 left = ilo return 110 mflag = 1 if (x .eq. xt(lxt)) mflag = 0 left = lxt 111 if (left .eq. 1) return left = left - 1 if (xt(left) .lt. xt(lxt)) return go to 111 end