<< problem 209 - Circular Logic | Flea Circus - problem 213 >> |
Problem 211: Divisor Square Sum
(see projecteuler.net/problem=211)
For a positive integer n, let \sigma(n) be the sum of the squares of its divisors. For example,
\sigma^2(10) = 1 + 4 + 25 + 100 = 130.
Find the sum of all n, 0 < n < 64,000,000 such that \sigma^2(n) is a perfect square.
My Algorithm
My approach works like a sieve:
- allocate enough memory for 64000000 numbers, each 64 bit (=> 512 MByte)
- fill it with zeros
- for each number 0 < i < 64000000: add i*i to each cell that is a multiple of i
- when done, check each cell whether it is a perfect square
However, its memory consumption was much, much higher than any of my other solutions.
Therefore I decided to move the algorithm to a new function
processSlice
which works a bit smarter:instead of processing everything at once, it only looks at all numbers [
from, to
].Reducing
to - from
to about 4 million doesn't slow down the algorithm at all.But I wanted to keep it below 20 MByte (for no good reason ...) and chose a
sliceSize = 2000000
which takes about 14 seconds (40% slower).
Alternative Approaches
You can play around with prime factorization. This should be a bit faster at the cost of probably doubling the code size.
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho 100 | ./211
Output:
(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.
#include <iostream>
#include <vector>
#include <cmath>
// determine the sum of all numbers between "from" and "to" (inclusive both) which match the problem statement
unsigned long long processSlice(unsigned int from, unsigned int to)
{
std::vector<unsigned long long> sumSquares(to - from + 1, 0);
// like a prime sieve: add square of all divisors
for (unsigned long long i = 1; i <= to; i++)
{
// position of smallest multiple of i >= from
auto pos = (from / i) * i;
if (pos < from)
pos += i;
// add i^2 to all multiples of i
for (; pos <= to; pos += i)
sumSquares[pos - from] += i*i;
}
// find all sums that are perfect squares
unsigned int sum = 0;
for (size_t i = 0; i < sumSquares.size(); i++)
{
auto number = i + from;
auto current = sumSquares[i];
// compute integer square root
unsigned long long root = sqrt(current);
// iff root^2 = current then it's a perfect square
if (root * root == current)
sum += number;
}
return sum;
}
int main()
{
unsigned int limit = 64000000;
std::cin >> limit;
// how many number should be analyzed at once (=> influences memory consumption)
unsigned int sliceSize = 2000000;
// total sum
unsigned int sum = 0;
// start of current slice
unsigned int from = 1;
while (from < limit)
{
// end of current slice
auto to = from + sliceSize - 1;
if (to >= limit)
to = limit;
// process current slice
sum += processSlice(from, to);
// next slice
from = to + 1;
}
// print result
std::cout << sum << std::endl;
return 0;
}
This solution contains 10 empty lines, 14 comments and 3 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in 14.6 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 18 MByte.
(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
May 31, 2017 submitted solution
May 31, 2017 added comments
Difficulty
Project Euler ranks this problem at 50% (out of 100%).
Links
projecteuler.net/thread=211 - the best forum on the subject (note: you have to submit the correct solution first)
Java github.com/nayuki/Project-Euler-solutions/blob/master/java/p211.java (written by Nayuki)
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own. Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
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 a new problem is published | |
the flashing problem is the one I solved most recently |
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I scored 13,486 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.
Copyright
I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.
All of my solutions can be used for any purpose and I am in no way liable for any damages caused.
You can even remove my name and claim it's yours. But then you shall burn in hell.
The problems and most of the problems' images were created by Project Euler.
Thanks for all their endless effort !!!
<< problem 209 - Circular Logic | Flea Circus - problem 213 >> |