       Re: Finding length in recursive definition?

• To: mathgroup at smc.vnet.net
• Subject: [mg60884] Re: Finding length in recursive definition?
• From: albert <awnl at arcor.de>
• Date: Sun, 2 Oct 2005 01:54:36 -0400 (EDT)
• References: <dhlcj2\$d16\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Jose  Reckoner wrote:

> I have something like:
> f = 1
> f = 3
> f[n_] := f[n] = f[n - 1] + f[n - 2]
>
> and in the course of work, f[n] gets evaluated an unknown number of
> times resulting in
>
>>> ?f
> f = 1
> f = 3
> f = 4
> f[n_] := f[n] = f[n - 1] + f[n - 2]
>
> I want to figure out the greatest integer n such that f[n] has already
> been computed and is stored. In this case, it is 3.
>
> How can I do this?

I think this will do what you want:

Max[Cases[DownValues[f], HoldPattern[_[_[f[nn_Integer]], _]] :> nn]]

all patterns for f are stored in DownValues[f], the Cases extracts all
patterns where the argument of f is an integer and replaces them by this
integer...

Albert

```

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