<< problem 473 - Phigital number base | Double pandigital number divisible by 11 - problem 491 >> |
Problem 485: Maximum number of divisors
(see projecteuler.net/problem=485)
Let d(n) be the number of divisors of n.
Let M(n,k) be the maximum value of d(j) for n <= j <= n+k-1.
Let S(u,k) be the sum of M(n,k) for 1 <= n <= u-k+1.
You are given that S(1000, 10) = 17176.
Find S(100 000 000, 100 000).
My Algorithm
I needed to translate the short problem description into plain English (well, I'm German so actually it wasn't English ...):
to find S(1000, 10) I have to create 990 blocks with 10 elements each:
\{ d(1), d(2), ..., d(10) \},
\{ d(2), d(3), ..., d(11) \},
\{ d(3), d(4), ..., d(12) \},
...
\{ d(991), d(992), ..., d(1000) \}
and add the maximum values of all those 990 blocks.
My first step is to fill the container numDivisors
with d(1), d(2), ..., d(10^8).
The maximum value is 768 so an unsigned short
is sufficient - but it still requires 200 MByte RAM.
I implemented two algorithms to compute the number of divisors of n:
countDivisorsSlow
is based on trial division: it divides n by every number between 1 and n and counts how often the remainder is zero.
That's pretty fast if n is small (well, even 10^6 is almost okay) but much too slow for 10^8.
countDivisors
needs only one second to perform the same task:
if the number n is factorized into its prime factors n = {p_1}^e_1 * {p_2}^e_2 * {p_3}^e_3 * ...
then the number of divisors is d(n) = (e_1 + 1) * (e_2 + 1) * (e_3 + 1) * ...
My first step is to find all prime numbers below 10^8 (see below for a significant optimization / chapter "Note").
Then all d(n) are initialized with 1 and:
- iterate over all multiples m of 2^1 = 2, multiply d(m) by 1+1=2
- iterate over all multiples m of 2^2 = 4, multiply d(m) by 2+1=3 and divide by 1+1=2 (that's an undo of the previous step)
- iterate over all multiples m of 2^3 = 8, multiply d(m) by 3+1=4 and divide by 2+1=3 (that's an undo of the previous step)
- iterate over all multiples m of 2^4 = 16, multiply d(m) by 4+1=5 and divide by 3+1=4 (that's an undo of the previous step)
- ... and so on, until 2^x > 10^8
- repeat the same procedure for all other primes 3, 5, 7, 11, ...
- the obvious
bruteForce()
algorithm with two nested loops has 10^8 * 10^5 = 10^13 iterations and is too slow. - a smarter algorithm can do the same job in just 10^8 iterations !
search()
, each iteration updates mostRecent[d(n)]
with the current position n
(and enlarges mostRecent
if required).Zero is more or less a dummy element and the whole algorithm still works if I omit it but it simplifies index calculations a lot.
Only once it causes minor problems: the maximum number is the size of
mostRecent
minus 1.For example:
nd(n)mostRecent
00{ 0 }
11{ 0,1 }
22{ 0,1,2 }
32{ 0,1,3 }
43{ 0,1,3,4 }
52{ 0,1,5,4 }
64{ 0,1,5,4,6 }
72{ 0,1,7,4,6 }
84{ 0,1,7,4,8 }
93{ 0,1,7,9,8 }
104{ 0,1,7,9,10 }
If an element of
mostRecent
is too far away, that means if its value is more than blockSize
away from the current position, then it becomes invalid.That's only important for the largest indices of
mostRecent
: they will be removed, thus shrinking mostRecent
.
Alternative Approaches
The whole algorthm could be rewritten to process the data in chunks instead of all 100 million at once.
Memory consumption will drop considerably and I expect performance to remain about the same (maybe a tiny bit slower).
But the code size will grow considerably and become less reabable.
Note
d(n) is a large number if:
- n has many different prime factors or
- those prime factors have large exponents
- or both
such that d(73513440) = (5+1) * (3+1) * (1+1) * (1+1) * (1+1) * (1+1) * (1+1) = 6 * 4 * 2 * 2 * 2 * 2 * 2 = 768
In conclusion, pretty much every maximum d(n) of a block has typically very small prime factors.
Assuming that all prime factors of a maximum d(n) are less than sqrt{10^8} = 10^4, my code will still find the correct result for S(10^8, 10^5).
However, this "square-root" assumption isn't always true: it fails for S(1000, 10) (the problem is the small
blockSize
).The less primes I use as prime factors in
countDivisors
the faster the code becomes (at the risk of being off, depending on blockSize
).I manually looked for the smallest "valid" value for
primeLimit
for S(10^8, 10^5) and it turned out to be 107.Using this parameter cuts down the execution time from 1.8 to 1.1 seconds.
Interactive test
This feature is not available for the current problem.
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 <algorithm>
#include <cmath>
// store the number of divisors for the first 100 million numbers
typedef unsigned short Number; // => 200 MByte
std::vector<Number> numDivisors;
// slow trial division
void countDivisorsSlow(unsigned int limit)
{
// zero has no divisors
numDivisors = { 0 };
// process all numbers 1..10^8
for (unsigned int current = 1; current <= limit; current++)
{
Number count = 0;
// trial division of all numbers <= sqrt(current)
for (unsigned int divisor = 1; divisor*divisor <= current; divisor++)
{
// divisible ?
if (current % divisor != 0)
continue;
count++; // one divisor if i^2 = x
if (divisor*divisor != current) // or two if not (it's i and x/i)
count++;
}
numDivisors.push_back(count);
}
}
// similar to a prime sieve, much faster
void countDivisors(unsigned int limit, unsigned int primeLimit = 0)
{
numDivisors.resize(limit + 1, 1);
// zero has no divisors
numDivisors[0] = 0;
// accurate algorithm by default
if (primeLimit == 0)
primeLimit = limit;
// simple prime sieve (trial division because it has the shortest/most simple implementation)
std::vector<unsigned int> primes = { 2 };
for (unsigned int smallPrime = 3; smallPrime <= primeLimit; smallPrime += 2)
{
// find primes by trial division
bool isPrime = true;
for (auto p : primes)
if (smallPrime % p == 0)
{
isPrime = false;
break;
}
// yes, found another prime
if (isPrime)
primes.push_back(smallPrime);
}
// multiply all "simple" multiples of each prime by 2
// ... and multiply all multiples p^2 by 3
// ... and multiply all multiples p^3 by 4
// ... and so on
for (auto p : primes)
{
// "simple" multiples
for (auto i = p; i <= limit; i += p)
numDivisors[i] *= 2;
// multiples of p^2, p^3, ...
auto power = p * p;
auto exponent = 2;
while (power <= limit)
{
// undo the previous factor and multiply with "better" factor
for (auto i = power; i <= limit; i += power)
numDivisors[i] = (numDivisors[i] / exponent) * (exponent + 1);
// from to p^2 to p^3 to p^4 ...
power *= p;
exponent++;
}
}
}
// enough to verify the example ... (you need to fill numDivisors first)
unsigned long long bruteForce(unsigned int limit, unsigned int blockSize)
{
// for each range ...
unsigned long long result = 0;
for (unsigned int from = 1; from <= limit - blockSize + 1; from++)
{
// ... find its maximum
Number maximum = numDivisors[from];
for (unsigned int i = 1; i < blockSize; i++)
maximum = std::max(maximum, numDivisors[from + i]);
// and add it to the result
//std::cout << maximum << " ";
result += maximum;
}
return result;
}
// fast O(n) algorithm: a single pass iterating over all 10^8 elements of numDivisors (you need to fill numDivisors first)
unsigned long long search(unsigned int limit, unsigned int blockSize)
{
// store the most recent position a certain number of divisors was encountered
// actually the maximum is 768 divisors (first seen at 73513440)
std::vector<unsigned int> mostRecent;
// process number of divisors of the first block (10^5-1 elements)
for (unsigned int i = 0; i < blockSize; i++)
{
// new/updated maximum
auto current = numDivisors[i];
if (current >= mostRecent.size())
mostRecent.resize(current + 1, 0);
mostRecent[current] = i;
}
unsigned long long result = 0;
for (unsigned int i = blockSize; i <= limit; i++)
{
// remove "old" maximums
auto tooFar = i - blockSize;
while (!mostRecent.empty() && mostRecent.back() <= tooFar)
mostRecent.pop_back();
// new/updated maximum
auto current = numDivisors[i];
if (current >= mostRecent.size())
mostRecent.resize(current + 1, 0);
mostRecent[current] = i;
// highest index is equal to size-1
result += mostRecent.size() - 1;
}
return result;
}
int main()
{
unsigned int limit = 100000000; // 10^8
unsigned int blockSize = 100000; // 10^5
std::cin >> limit >> blockSize;
// compute number of divisors for all 100 million numbers
unsigned int primeLimit = limit;
// faster heuristic if possible, fails with small blockSize
if (blockSize >= 100)
primeLimit = sqrt(limit);
// lowest limit for default input: I found it by trial'n'error
if (limit == 100000000 && blockSize == 100000)
primeLimit = 107;
// compute number of divisors (might be off a little bit when primeLimit != limit)
countDivisors(limit, primeLimit);
// and now find S(10^8, 10^5)
//std::cout << bruteForce(limit, blockSize) << std::endl;
std::cout << search(limit, blockSize) << std::endl;
return 0;
}
This solution contains 25 empty lines, 39 comments and 4 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in 1.1 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 198 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
December 8, 2017 submitted solution
December 8, 2017 added comments
Difficulty
Project Euler ranks this problem at 30% (out of 100%).
Links
projecteuler.net/thread=485 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
C++ github.com/evilmucedin/project-euler/blob/master/euler485/485.cpp (written by Den Raskovalov)
Perl github.com/shlomif/project-euler/blob/master/project-euler/485/euler-485-v1.pl (written by Shlomi Fish)
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
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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 | |
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[new] | the flashing problem is the one I solved most recently |
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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.
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 !!!
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