It is wonderful that people are so willing to help with a specific
problem, and I encourage you to continue doing so, but I don't think
anyone has answered the real question. Is there material out there
that describes some of the mechanisms, tools, ideas, etc. that will
allow the average clojure
On Nov 4, 2010, at 1:32 PM, cej38 wrote:
It is wonderful that people are so willing to help with a specific
problem, and I encourage you to continue doing so, but I don't think
anyone has answered the real question. Is there material out there
that describes some of the mechanisms, tools,
Sorry for the delayed response.
OK... here's an example; my solution to problem 187:
(defn divides? [n d]
(zero? (rem n d))
)
(declare prime? prime-seq)
(defn prime? [n]
(if ( n 2)
false
(every? #(not (divides? n %)) (take-while #(= (* % %) n) prime-
seq)))
)
(def prime? (memoize
Your Clojure implementation of your particular approach is reasonable,
but you need a cleverer approach. I'll send you a hint offline.
--
You received this message because you are subscribed to the Google
Groups Clojure group.
To post to this group, send email to clojure@googlegroups.com
Note
One idea that I had was to inline the factoring logic into twice-
composite. Who cares about the factors? We just want to know if there
are two or not:
(defn twice-composite? [n]
(loop [ps prime-seq
tmp n
p-count 0]
(if ( 2 p-count)
false
(if (= 1 tmp)
All the time spent factoring is wasted. You can *construct* and/or
count the semiprimes far more efficiently than your method of
factoring each number one a time in order to filter the semiprimes
from the entire range of numbers up to 10^8. Your approach is
fundamentally slow; this has nothing
If you want to do this, you can do it simpler, without the loop and temp var:
(defn twice-composite? [n]
(- (take-while #( % n) prime-seq)
(every #(or (divides? (* 2 %) n)
(not (divides? % n
but this is not what you want. See the hints I sent you off the list.
On Wed,
If you're looking to generate all the numbers with exactly two prime
factors (counting multiplicity), I have clojure code that generates
the ones up to 10^8 in under two minutes:
com.example.sbox= (time (count (semiprimes-to 1)))
Elapsed time: 106157.33292 msecs
41803068
It's
Hello fellow clojurians...
I've been using Clojure now fairly regularly for the past two months
solving problems on Project Euler. I've been fairly successful solving
them but there are times where the performance of my code either
stinks (the answer may come back in 5-10 minutes) or not at all
Usually it's more about the algorithm than the language. Java can
generally do things faster than clojure, simply because it has fewer
layers, but the speedup is a linear factor. If the java solution for
1000 elements takes 5ms and your clojure code takes even a second,
it's likely that clojure
Yes, on the handful of Project Euler exercises I've attempted, my
algorithm choice was frequently the source of poor performance.
Are you familiar with the clojure-euler wiki at
http://clojure-euler.wikispaces.com/
? After I do an Euler exercise I compare my code and runtime with
other Clojure
11 matches
Mail list logo
