Problem 23: Non-abundant sums

(see projecteuler.net/problem=23)

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number.
For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24.
By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers.
However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

My Algorithm

The function getSum was taken from problem 21.
It can be used to find all abundant numbers below 28124 (see problem statement) which I store in abundant.

The function isAbundantSum returns true if it finds at least one combination of two abundant numbers.
For each element i of abundant it computes other = x - i. If other is an element of abundant, then it return true.

Modifications by HackerRank

Again, the original problem was heavily modified by Hackerrank:
instead of asking for the sum it just asks to figure out for a single number whether it can be written as the sum
of two abundant numbers or not and print YES or NO accordingly.

Note

The "live test" is based on the Hackerrank problem.

Interactive test

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

This live test is based on the Hackerrank problem.

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 24" | ./23

Output:

(please click 'Go !')

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

My code

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

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

#include <iostream>
#include <set>
 
// constant according to problem statement
const unsigned int EverythingsASumFromHere = 28124;
 
// will contain all abundant numbers
std::set<unsigned int> abundant;
 
// generate sum of all divisor's of x
unsigned int getSum(unsigned int x)
{
// note: code very similar to problem 21
 
// 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"
unsigned int otherDivisor = x / divisor;
// except square numbers
if (otherDivisor != divisor)
divisorSum += otherDivisor;
}
 
return divisorSum;
}
 
// return true if parameter can be written as the sum of two abundant numbers
bool isAbundantSum(unsigned int x)
{
// big numbers are always an abundant sum
if (x >= EverythingsASumFromHere)
return true;
 
// look at all small abundant numbers in ascending order
for (auto i : abundant)
{
// abort if there is no sum possible
if (i >= x) // or even faster: if (i > x/2)
return false;
 
// is its partner abundant, too ?
unsigned int other = x - i;
if (abundant.count(other) == 0)
continue;
 
// yes, we found a valid combination
return true;
}
 
// nope
return false;
}
 
int main()
{
// precomputation step:
// find all abundant numbers <= 28123
for (unsigned int i = 1; i < EverythingsASumFromHere; i++) // actually, we could start at 12
// divisor's sum bigger than the number itself ?
if (getSum(i) > i)
abundant.insert(i);
 
//#define ORIGINAL
#ifdef ORIGINAL
unsigned long long sum = 0;
for (unsigned int i = 0; i < EverythingsASumFromHere; i++)
{
// sum of all numbers which cannot be written as an abundant sum
if (!isAbundantSum(i))
sum += i;
}
std::cout << sum << std::endl;
 
#else
 
unsigned int tests;
std::cin >> tests;
while (tests--)
{
// find out whether a certain number can be written as an abundant sum
unsigned int x;
std::cin >> x;
std::cout << (isAbundantSum(x) ? "YES" : "NO") << std::endl;
}
#endif
return 0;
}

This solution contains 16 empty lines, 24 comments and 5 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 0.27 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++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/euler023

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

Difficulty

5% 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 rarely an option.

Similar problems at Project Euler

Problem 21: Amicable numbers

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.

Heatmap

Please click on a problem's number to open my solution to that problem:

green   solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too
yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily)
gray problems are already solved but I haven't published my solution yet
blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much
orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte
red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too
black problems are solved but access to the solution is blocked for a few days until the next problem is published
[new] the flashing problem is the one I solved most recently
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The 306 solved problems (that's level 12) had an average difficulty of 32.5% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

more about me can be found on my homepage, especially in my coding blog.
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