On Sunday, April 14, 2013 5:30:13 PM UTC+1, Jonathan Fischer Friberg wrote:
Calling fib just creates a new function, no values
> are calculated. So you're measuring the time
> it takes to create a function, and not the calculation
> of fibonacci numbers.
>
>
Oops ;)
Of course you are right. The amazing thing is that the times I observed
fitted somehow the situation (the first (fib 30) call taking much more time
than the others, the third call more than the second and fourth) that I was
tricked into believing the calculations were being done and wasn't careful
enough ... btw, why does such a thing happen if in all four cases only a
function is being created as you say?
And given that a simple let as follows cannot be used:
(let [fib (memoize
#(if (or (zero? %) (= % 1))
1
(+ (fib (- % 1)) (fib (- % 2)))))]
...)
Is this an example of something that cannot be elegantly expressed without
a letrec, and for which letfn is not enough as a substitute?
Regards,
Paulo
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