<< problem 20 - Factorial digit sum | Names scores - problem 22 >> |

# 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:

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

Output:

*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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