Re: iteration question

*To*: mathgroup at smc.vnet.net*Subject*: [mg122608] Re: iteration question*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Thu, 3 Nov 2011 03:45:45 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201111021121.GAA03532@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

What could be simpler or more efficient than Length, if you want to know how long a list is? Bobby On Wed, 02 Nov 2011 06:21:44 -0500, Francisco Gutierrez <fgutiers2002 at yahoo.com> wrote: > Dear Group: > > I have a function, and I iterate it using fixedpoint. Something of this > sort: > > FixedPointList[ > function[arg1,arg2,arg3, #] &, arg4, > SameTest -> (Max[Abs[Flatten[#1] - Flatten[#2]]] < 0.01 &)] > > > I would like to know how many steps it takes this function to converge. > I have tried with EvaluationMonitor to no avail: > > Block[{veamos, c = 0}, > veamos = FixedPoint[ > function[arg1,arg2,arg3, #] &, arg4, > SameTest -> (Max[Abs[Flatten[#1] - Flatten[#2]]]< 0.01 &)]; > EvaluationMonitor :> c++; {veamos, c}] > > > The iterator c simply does not move, and the previous function works ok > but returns the value of c as 0. > > I tried if this was true with Length[FixedPointList[...]] and the answer > was 20 (which should be the value of c in the immediately previous > function). In principle, this would solve my problem, but it seems > rather inefficient, especially when the convergence criterion is severe > (not 0.01, but say 10^-4). > > Is there an efficient and nice way to solve this? > > Thanks, > > Francisco -- DrMajorBob at yahoo.com

**References**:**iteration question***From:*Francisco Gutierrez <fgutiers2002@yahoo.com>