Clojure and me has moved.

## Tuesday, June 10, 2008

### Google Treasure Hunt, Question #4

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A friend of mine asked me how I solved the fourth question of Google Treasure Hunt 2008 using Clojure. I didn't keep the original code around, so below is how I could have done it.

First, define a primes seq.

Second, define a function which returns the sequence of sums of N consecutive primes:
`(defn sum-primes [n] (map #(apply + %) (partition n 1 primes)))`

Third, define a function which, taking a list of increasing sequences, returns the first common value.
`(defn find= [seqs]  (if (apply == (map first seqs))    (ffirst seqs)    (let [[s1 & etc] (sort #(- (first %1) (first %2)) seqs)]      (recur (cons (rest s1) etc)))))`

Last, use them! Here is a sample question:
Find the smallest number that can be expressed as
the sum of 3 consecutive prime numbers,
the sum of 5 consecutive prime numbers,
the sum of 11 consecutive prime numbers,
the sum of 1493 consecutive prime numbers,
and is itself a prime number.

And here is how to compute the answer:
`(find= (cons primes (map sum-primes [3, 5, 11, 1493])))`
returns 9174901 in twenty seconds or so.

(Right now this code may throw a StackOverflow exception, please use one of those definition of `partition`.)

David said...

Very impressive. What's your implementation of primes? My Haskell implementation, with all compiler optimizations turned on, took about 16 minutes. (I'm a Haskell novice, but still, 20 seconds on JVM vs. 16 minutes for native code says something).

import Primes
import Data.List hiding (find)

sums n = map (\m -> sum . take n \$ drop m primes) [0..]

find ss@((x:_):_) =
if all (== x) (map head ss) then x
else let ss' = map snd \$ sort \$ zip (map head ss) ss in
find \$ tail (head ss') : tail ss'

main = print \$ find (primes : map sums [3, 5, 11, 1493])

David said...
Christophe Grand said...

@david: my implementation of primes can be found here: http://clj-me.blogspot.com/2008/06/primes.html

To be fair, I retimed:
user=> (time (find= (cons primes (map sum-primes [3, 5, 11, 1493]))))
"Elapsed time: 88650.283232 msecs"
9174901
user=> (time (find= (cons primes (map sum-primes [3, 5, 11, 1493]))))
"Elapsed time: 5894.601122 msecs"
9174901

I guess that I got 20s by having some primes computed from a prior invocation. So nearly 90s with no primes precomputed and 6s when all required primes are already computed. Anonymous said...

Well, here is result of mine ruby implementation:
Solving...
Found: 9174901. Testing...solved.
Time elapsed: 3.343
:)

David said...

Ok, saw a canonical solution in Haskell by a more proficient author :) -- in defense of my language du jour, the canonical solution terminated in a couple seconds with no compiler optimizations as far as I know. I'm fairly certain that my bottleneck was my sums function, which potentially recalculated the first n primes to create a sum containing the n+1th prime.

Christophe Grand said...

@david: ah, it fits better in the natural order of all things :-)