Clojure Euler: Problem 021

Evaluate the sum of all the amicable numbers under 10000.

Permalink: http://projecteuler.net/problem=21

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).

If d(a) = b and d(b) = a, where a != b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

We can do it bruteforcely, but we won't.

Let start with the function d. We call it sum-of-proper-divisors

(defn sum-of-proper-divisors [n]
  (let [base (filter #(zero? (mod n %)) (range 2 (Math/sqrt n)))]
    (reduce + 1 (concat (map #(/ n %) base) base))))

Bad approach to find all divisors: to iterate on all numbers from 1 to n and check if it is divisor or not. Instead, we iterate from 1 to sqrt(n), find one divisor and calculate another symmetric divisor.

For example, take number 10. We iterating from 2 to (sqrt 10). Number 2 is divisor, so we divide (/ 10 2), get 5 which is also divisor. And so on.

This is predicate checks where two numbers amicable or not:

(defn amicable? [a b]
  (and (not (= a b))
       (= a (sum-of-proper-divisors b))
       (= b (sum-of-proper-divisors a))))

Obvious enough, just consider that we skip case where (= a b).

The last part is to use our functions to obtain result.

(reduce +
  (let [sums (vec (map sum-of-proper-divisors (range 1 10000)))]
    (for [i (range 1 10000)]
      (if (amicable? i (nth sums (dec i))) i 0))))

Just generate all sums, and check if another number i produce the amicable pair. Done!

Code

P.S. Take a look at the last snippet. vec is really important there.

mishadoff 05 October 2013
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