Problem 21: Amicable numbers

(see 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.

Evaluate the sum of all the amicable numbers under 10000.

Algorithm

My function getSum returns the sum of all proper divisors of x.
The brute-force approach works but turns out to be too slow: a for-loop from 2 to x-1.

But we can easily reduce the number of iterations:
For each proper divisor i of x there is another divisor j=x/i, except when i^2=x.
More interesting, if we assume i<j then i<=sqrt{x}.

A simple loop running from 2 to sqrt{x} check whether x mod i == 0 (proper divisor)
and then adds i as well as j (if i!=j) to the result.

A precomputation step in main finds the sums of all divisors of all numbers i below 100000 and calls it sibling = getSum(i).
If getSum(sibling) = i then i,sibling is an amicable pair.

Those two numbers are stored in an std::set named amicables.
For each test case, the sum of all relevant numbers is printed.

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include <iostream>
#include <set>
 
// generate sum of all divisor's of x (where x > 1)
unsigned int getSum(unsigned int x)
{
// find all factors:
// look only for the "smaller" divisors <= sqrt(x)
// and they have a "bigger" brother x / divisor
 
// 1 is always a divisor, but not the number x itself
unsigned int divisorSum = 1;
// therefore start at 2
for (unsigned int divisor = 2; divisor * divisor <= x; divisor++)
if (x % divisor == 0)
{
divisorSum += divisor;
 
// add the "bigger brother"
auto otherDivisor = x / divisor;
// except square numbers
if (otherDivisor != divisor)
divisorSum += otherDivisor;
}
 
return divisorSum;
}
 
int main()
{
// contain all numbers which are part of an amicable pair
std::set<unsigned int> amicables;
 
// precomputation step:
// find all amicable numbers <= 100000
const unsigned int MaxAmicable = 100000;
for (unsigned int i = 2; i <= MaxAmicable; i++)
{
auto sibling = getSum(i);
 
// found a pair ?
if (i == getSum(sibling) && i != sibling)
{
amicables.insert(i);
amicables.insert(sibling);
}
}
 
// and now start processing input
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int x;
std::cin >> x;
 
// just look up all suitables numbers
unsigned int sum = 0;
for (auto i : amicables)
{
// discard those that are too big
if (i > x)
break;
// note: an set::set is sorted ascendingly by default
 
// yes, accept that amicable number
sum += i;
}
 
std::cout << sum << std::endl;
}
return 0;
}

This solution contains 11 empty lines, 17 comments and 2 preprocessor commands.

Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 300" | ./21

Output:

(please click 'Go !')

Note: the original problem's input 10000 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

Benchmark

The correct solution to the original Project Euler problem was found in 0.2 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

Changelog

February 24, 2017 submitted solution
April 4, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler021

My code solved 4 out of 4 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is never an option.

Similar problems at Project Euler

Problem 23: Non-abundant sums

Note: I'm not even close to solving all problems at Project Euler. Chances are that similar problems do exist and I just haven't looked at them.

Links

projecteuler.net/thread=21 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-21-sum-of-amicable-pairs/ (written by Kristian Edlund)
Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p021.hs (written by Nayuki)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p021.java (written by Nayuki)
Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p021.mathematica (written by Nayuki)
C: github.com/eagletmt/project-euler-c/blob/master/20-29/problem21.c (written by eagletmt)
Go: github.com/frrad/project-euler/blob/master/golang/Problem021.go (written by Frederick Robinson)
Javascript: github.com/dsernst/ProjectEuler/blob/master/21 Amicable numbers.js (written by David Ernst)
Scala: github.com/samskivert/euler-scala/blob/master/Euler021.scala (written by Michael Bayne)

Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are already solved and there wasn't a Hackerrank version of it (at the time I solved it) or I didn't care about it because it differed too much.

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The 133 solved problems had an average difficulty of 16.9% at Project Euler and I scored 11,174 points (out of 12300) at Hackerrank's Project Euler+.
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