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>
- iteration question