Problem 95: Amicable chains

(see projecteuler.net/problem=95)

The proper divisors of a number are all the divisors excluding the number itself.
For example, the proper divisors of 28 are 1, 2, 4, 7, and 14. As the sum of these divisors is equal to 28, we call it a perfect number.

Interestingly the sum of the proper divisors of 220 is 284 and the sum of the proper divisors of 284 is 220, forming a chain of two numbers.
For this reason, 220 and 284 are called an amicable pair.

Perhaps less well known are longer chains. For example, starting with 12496, we form a chain of five numbers:

12496 → 14288 → 15472 → 14536 → 14264 (→ 12496 → ...)

Since this chain returns to its starting point, it is called an amicable chain.

Find the smallest member of the longest amicable chain with no element exceeding one million.

Algorithm

A pre-computation step creates a container divsum such that divsum[x] is the sum of all proper divisors of x.
My function getSum() from problem 21 produces the correct sum of all proper divisors but turns out to be too slow,
therefore I searched the web for faster algorithms and found this one:
1. split a number into its prime factors x = p_1^{e_1} * p_2^{e_2} * ...
where p_1, p_2, ... are the primes and e_1, e_2, ... how often they appear in x (=exponents), for example 96 = 2^5 * 3^1
2. then the sum of all divisors is (2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5) * (3^0 + 3^1) = 252
3. but since we need only the proper divisors instead of all divisors (which include the number itself): 252 - 96 = 156
divsum[96] = 156

The chain algorithm appends chain[i] = divsum[chain[i - 1]] until:
1. it's the initial number, then an amicable chain was found or
2. it's a number lower than the initial number → abort because we should have seen this loop before, when we processed that lower number or
3. it's a number above the limit → abort according to problem description ("no element exceeding one million") or
4. it's a number that is already part of the chain → a loop

If the chain is longer than anything before than its first element is stored in smallestMember (it must be the first element of the chain).

My code

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

#include <iostream>
#include <vector>
 
int main()
{
unsigned int limit;
std::cin >> limit;
 
// the usual prime sieve
std::vector<unsigned int> primes;
primes.push_back(2);
for (unsigned int i = 3; i <= limit; i += 2)
{
bool isPrime = true;
 
// test against all prime numbers we have so far (in ascending order)
for (auto p : primes)
{
// next prime is too large to be a divisor ?
if (p*p > i)
break;
 
// divisible ? => not prime
if (i % p == 0)
{
isPrime = false;
break;
}
}
 
// yes, we have a prime number
if (isPrime)
primes.push_back(i);
}
 
// initial mapping of each number to its proper divisor's sum
std::vector<unsigned int> divsum(limit + 1, 0);
for (unsigned int i = 2; i <= limit; i++)
{
unsigned int sum = 1;
unsigned int reduce = i;
for (auto p : primes)
{
// note: reduce itself might be prime in the end
if (p*p > reduce)
break;
 
// divide by all primes
unsigned int factor = 1;
while (reduce % p == 0)
{
reduce /= p;
 
// add 1 to each exponent, e.g. p^0 + p^1 becomes p^1 + p^2
factor *= p;
// add a new term p^0
factor++;
}
sum *= factor;
}
 
// if a large prime was left over
if (reduce > 1 && reduce < i)
sum *= reduce + 1;
 
// subtract number itself if it isn't a prime
if (sum > 1)
sum -= i;
 
divsum[i] = sum;
}
 
// loop until numbers are mapped to themselves (or get stuck in a loop)
unsigned int longestChain = 0;
unsigned int smallestMember = limit;
for (unsigned int i = 1; i <= limit; i++)
{
// re-use the same vector over and over again to avoid memory re-allocations
static std::vector<unsigned int> chain;
chain.clear();
chain.push_back(i);
 
// until we:
// 1. return to i or
// 2. exceed the limit or
// 3. get stuck in a loop
while (true)
{
unsigned int add = divsum[chain.back()];
chain.push_back(add);
 
// yes, we found an amicable chain
if (add == i)
break;
 
// can't be a new shorter loop:
// we have already seen this number in an earlier iteration
if (add < i)
break;
 
// abort if limit exceeded
if (add > limit)
break;
 
// stuck in a loop ?
bool isLoop = false;
for (size_t j = 1; j < chain.size() - 1; j++) // except last element
if (add == chain[j])
{
isLoop = true;
break;
}
if (isLoop)
break;
}
 
// did we return to i ?
if (chain.back() != i)
continue;
 
// too short ?
if (chain.size() < longestChain)
continue;
 
// shorter chain ?
if (longestChain < chain.size())
{
longestChain = chain.size();
smallestMember = chain.front();
}
}
 
std::cout << smallestMember << std::endl;
return 0;
}

This solution contains 21 empty lines, 26 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:

Input data (separated by spaces or newlines):

This is equivalent to
echo 10 | ./95

Output:

(please click 'Go !')

(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 0.32 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 6 MByte.

(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

March 16, 2017 submitted solution
May 8, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler095

My code solves 8 out of 8 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 30% (out of 100%).

Hackerrank describes this problem as medium.

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.

Links

projecteuler.net/thread=95 - 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-95-longest-amicable-chain/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p095.java (written by Nayuki)
Scala: github.com/samskivert/euler-scala/blob/master/Euler095.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 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 163 solved problems had an average difficulty of 22.2% at Project Euler and I scored 11,907 points (out of 13200) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
more about me can be found on my homepage, especially in my coding blog.
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