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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



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