I think I badly explained myself, or maybe i completely misunderstood the code:
> next=.5&(0 3 + *)
the verb next simply applies the "multiply by five and add 3" to is right
argument a number of times given by is left argument, right?
> (i. 20) next 1 0
this instruction, first applies 0 times "next" to 1 0, then applies once "next"
to 1 0, then applies twice "next" to 1 0, ... , then applies 20 times "next" to
1 0, right?
this mean that S(0) has been calculated 20 times, S(1) has been calculated 19
times, S(n) has been calculated 20-n times since every time the function next
restarts from
the first term (which is 1 0), right?
What i mean is that with something like the following pseudo-code S(n) is
calculated once avoiding a huge waste of time and memory:
S(1)=1 0
for i=2 to 20 do
S(i)=.3+5*S(i-1)
end
My question then was if the code:
> next=.0 3+*&5
> 20 next@]^:(i.@[) 1 0
has the same problem or not and, in case of a positive answer, how can i
translate my pseudo-code in J avoiding an explicit loop.
It may be that I completely misunderstood your code and the problem that is
worring me doesn't exist at all but I'm really a beginner in J...
Alessandro
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