[Context lost to top posting.]
On Thu, 22 Jul 2010 15:56:32 -0700 (PDT)
logan wrote:
> I tried defn-memo .. it still does not appear to memoize the original
> fib call? I tried this using clojure 1.2 beta. Reading the code
> definition for defn-memo it seems like it should work, but calling
> (f
Yes, because the code I posted still makes me wonder what's happening ...
2010/7/23 Meikel Brandmeyer
> Hi,
>
> On Jul 23, 3:02 am, logan wrote:
>
> > paul what version did you use? your version was the first one I tried,
> > and on 1.2 at least it does not work.
>
> It works on 1.1. I guess th
Hi,
On Jul 23, 3:02 am, logan wrote:
> paul what version did you use? your version was the first one I tried,
> and on 1.2 at least it does not work.
It works on 1.1. I guess there is some issue with the direct linking
introduced in 1.2.
Sincerely
Meikel
--
You received this message because
paul what version did you use? your version was the first one I tried,
and on 1.2 at least it does not work.
Jules version works, except lambda should be fn in clojure.
On Jul 22, 4:51 pm, Paul Mooser wrote:
> Why not simply do:
>
> (defn fib [n]
> (println "called with " n)
> (if (> n 2)
>
I tried defn-memo .. it still does not appear to memoize the original
fib call? I tried this using clojure 1.2 beta. Reading the code
definition for defn-memo it seems like it should work, but calling
(fib 41) still takes a long time after calling (fib 40), when it
should basically be an instantane
On 23 Lip, 01:51, Paul Mooser wrote:
> Why not simply do:
>
> (defn fib [n]
> (println "called with " n)
> (if (> n 2)
> (+ (fib (- n 2)) (fib (- n 1)))
> 1))
>
> (def fib (memoize fib))
>
> I inserted the println to verify when we were actually calling the
> function, and I believe t
Here is what I get:
(and BTW, my original version with #' seems to work, I'm a little bit
puzzled ...)
user=> (defn fib [n]
(println "called with " n)
(if (> n 2)
(+ (fib (- n 2)) (fib (- n 1)))
1))
(def fib (memoize fib))
#'user/fib
user=> user=> #'user/fib
user=> (fib 6)
called with 6
Why not simply do:
(defn fib [n]
(println "called with " n)
(if (> n 2)
(+ (fib (- n 2)) (fib (- n 1)))
1))
(def fib (memoize fib))
I inserted the println to verify when we were actually calling the
function, and I believe this works - fib only seems to get invoked a
single time for
nevermind, the following code does not work.
Jules' one is the right one
2010/7/22 Laurent PETIT
> Another solution, use the var, and then use memoize on your function as
> usual:
>
> (defn fib[n]
> (if (> n 2)
> (+ (#'fib (- n 2)) (#'fib (- n 1
>
> (of course this was to answer close
(def fib (memoize (lambda ...)))
On Jul 22, 1:25 pm, Laurent PETIT wrote:
> Another solution, use the var, and then use memoize on your function as
> usual:
>
> (defn fib[n]
> (if (> n 2)
> (+ (#'fib (- n 2)) (#'fib (- n 1
>
> (of course this was to answer closely to the question. Nobod
Another solution, use the var, and then use memoize on your function as
usual:
(defn fib[n]
(if (> n 2)
(+ (#'fib (- n 2)) (#'fib (- n 1
(of course this was to answer closely to the question. Nobody would write
fib like that in practice : good general question, bad example)
HTH,
--
L
On Wed, 21 Jul 2010 14:47:12 -0700 (PDT)
logan wrote:
> Lets say I have the following function
>
> (defn fib[n]
> (if (> n 2)
> (+ (fib (- n 2)) (fib (- n 1)))
> 1))
>
> and I want to memoize it, what is the right way to do it?
Use defn-memo from clojure.contrib.def.
You should make a LazySeq to momoize intermediate result:
(defn fib[n]
(if (> n 2)
(+ (fib (- n 2)) (fib (- n 1)))
1))
(def fib (memoize fib))
(def fib-seq (map fib (iterate inc 0)))
then take the result by nth:
user=> (nth fib-seq 45)
1134903170
user=> (nth fib-seq 46)
1836311903
user
Lets say I have the following function
(defn fib[n]
(if (> n 2)
(+ (fib (- n 2)) (fib (- n 1)))
1))
and I want to memoize it, what is the right way to do it?
Using the default memoize does not work correctly. the reason is even
though the first call to fib is memoized, the recursive ca
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