John,
In my function f I have to access a global array (Gh and T) and I do
need the iteration count for that. The context is stochastic image
processing and I use a Simulated Annealing scheme.
My f looks like:
f=:3 : 0
'state index'=.y.
nstate=.index{Gh
TEMP=.index{T
e=. (lhood state) + state prior state
ne=. (lhood nstate) + nstate prior state
de=. ne - e
accept=. de <: - TEMP * ^. ALPHA
state=.(nstate*-.accept)+state*accept
state;>:index
)
And I call it e.g. by:
f^:n (g0;0)
Can this process be done more elegant?
Thanks in advance,
Jan.
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of John Randall
Sent: donderdag 14 december 2006 0:27
To: General forum
Subject: Re: [Jgeneral] Recursive function
Jacobs, Jan wrote:
> Ls,
> I need to evaluate gn =: f^:n (g0) where g0 is an gray-level image.
> f is a recurrent relation g(n+1)=f(g(n)) but needs the iteration
> count, one out of 0..n-1, for indexing in a global array.
> I managed to do this by f^:n (g0,0) and splitting the argument in an
> image and index by 'state index'=.y. and construct at the end (f
> state);>:index but this is not elegant.
> How do I this the right way?
> Jan.
>
I don't completely understand the question, but I think you need to look
at the Vocabulary page for ^: . You can assemble a specified number of
iterates into an array, and get all the iterates up to a fixed point:
s=:_2 0 1&p.
newton=:1 : ']- u%u d. 1'
f=:s newton
f^:(i.10) 2 NB. first 10 iterates
2 1.5 1.41667 1.41422 1.41421 1.41421 1.41421 1.41421 1.41421 1.41421
f^:(<10) 2 NB. first 10 iterates
2 1.5 1.41667 1.41422 1.41421 1.41421 1.41421 1.41421 1.41421 1.41421
f^:(<_) 2 NB. iterates to fixed point
2 1.5 1.41667 1.41422 1.41421 1.41421
John
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