On 09/06/11 18:11, Steven Schveighoffer wrote:
On Thu, 09 Jun 2011 13:06:54 -0400, Ben Grabham <evil.nebs...@gmail.com>
wrote:
On 09/06/11 17:57, bearophile wrote:
Ben Grabham:
import std.range;
import std.stdio;
int main() {
auto a = recurrence!("a[n-1] + a[n-2]")(0,1);
int i = 0;
foreach(int n; a) {
if(i++> 20) break;
writefln("%d", n);
}
return 0;
}
This program does something similar to yours (but it doesn't print
newlines):
import std.stdio, std.range;
void main() {
auto fib = recurrence!q{ a[n-1] + a[n-2] }(0, 1);
writeln(take(fib, 21));
}
Bye,
bearophile
Yeah, thanks
I just wanted to post a bit of code which went wrong :P
Didn't look for optimisations.
Also, how come recurrence isn't properly lazy?
If I define a recurrence and iterate over it twice with foreach, it
takes the same amount of time due to the stack size being set. Is
there a way of defining a lazy list that stores the results when
calculated?
That's not lazy, that's caching. lazy is 'calculate this when asked'.
You can cache with array:
auto cached = array(take(fib, 21));
// cached now contains the first 21 elements of fib.
-Steve
I tried that, but how can I calculate the values only when I want them?
Would I have to store them in a linked list? Since if I know that I will
need 100000 prime numbers, it takes my old pc about 4.7 seconds to
calculate them. But can I only calculate them when I need them?
I am running through the recurrence twice so I don't want to calculate
it twice.
This is theoretical so please no answers like put them both in one loop,
etc.
Thanks for all the info so far! I'm learning quite a lot!
Thanks,
Nebster
