<< problem 539 - Odd elimination | Robot Welders - problem 563 >> |
Problem 549: Divisibility of factorials
(see projecteuler.net/problem=549)
The smallest number m such that 10 divides m! is m=5.
The smallest number m such that 25 divides m! is m=10.
Let s(n) be the smallest number m such that n divides m!.
So s(10)=5 and s(25)=10`.
Let S(n) be sum{s(i)} for 2 <= i <= n.
S(100)=2012.
Find S(10^8).
My Algorithm
As always I wrote a simple function to solve the problem for small values.
The function naive(n)
returns s(n) and easily verifies that s(2) + s(3) + s(4) + ... + s(100) = 2012.
Unfortunately it way too slow to find s(10^8) in a reasonable amount of time.
(If you had enough spare time it would produce the correct result eventually).
naive(n)
computes 1! mod n, then 2! mod n, 3! mod n, ... until it finds a value result
such that result! mod n == 0.
It took me a while to realize that
s(n) = s(p^{e_p} * other) = max(s(p^{e_p}), s(other) )
In plain English: if I factorize n into its prime factors p_1, p_2, ... then I only need to find a way to compute s(p^{e_p})
where e_p is the exponent of the prime factors.
An example:
24 = 2^3 * 3^1
s(24) = s(2^3 * 3) = max(s(2^3), s(3)) = max(4, 3) = 4
Resolving the recursive structure of the formula shown above:
s(n) = max(s(p_1^{e_1}), s(p_2^{e_2}), s(p_3^{e_3}), ...)
where p_i are the prime factors of n and e_i the exponents of those prime factors.
Obviously for each prime p we have s(p^1) = p because p! is the smallest factorial which contains p and thus can be divided by p.
A modified version of the naive
algorithm is pretty fast when it comes to finding s(p^{e_p}):
instead of looking at each consecutive factorial 1!, 2!, 3!, ... I only look at each factorial that is a multiple of p:
p! mod p^{e_p}, (2p)! mod p^{e_p}, (3p)! mod p^{e_p}, ... until I find some (x * p)! mod p^{e_p} == 0.
My cache
contains only 2633 such values prime^{2 ... x} < 10^8.
Whenever the main()
function finds a prime number, then it adds its powers to the cache
.
For all composite numbers it calls getSmallestFactorial
which performs a prime factorization and returns the maximum value of any prime power encountered.
Alternative Approaches
It's possible to write faster solutions using some special properties of the Kempner function (see en.wikipedia.org/wiki/Kempner_function).
However, I wasn't aware of it and therefore didn't look it up.
Note
I'm surprised that this problem has a rating of only 10%. There are many easier problems with a higher percentage.
In my personal opinion its rating should be something like 40%.
Copying all prime numbers to a dense std::vector
gives a little speed boost (almost 3x) in getSmallestFactorial
at the cost of about 40 MByte RAM consumption.
The high execution time (about 30 seconds) combined with an increased memory usage puts my solution in the top spot of the "most expensive solutions" (as of August 2017) -
I didn't expect that when I saw the 10% rating ...
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 | ./549
Output:
Note: the original problem's input 100000000
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)
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 <unordered_map>
#include <algorithm>
// ---------- standard prime sieve from my toolbox ----------
// odd prime numbers are marked as "true" in a bitvector
std::vector<bool> sieve;
// return true, if x is a prime number
bool isPrime(unsigned int x)
{
// handle even numbers
if ((x & 1) == 0)
return x == 2;
// lookup for odd numbers
return sieve[x >> 1];
}
// find all prime numbers from 2 to size
void fillSieve(unsigned int size)
{
// store only odd numbers
const unsigned int half = (size >> 1) + 1;
// allocate memory
sieve.resize(half, true);
// 1 is not a prime number
sieve[0] = false;
// process all relevant prime factors
for (unsigned int i = 1; 2*i*i < half; i++)
// do we have a prime factor ?
if (sieve[i])
{
// mark all its multiples as false
unsigned int current = 3*i+1;
while (current < half)
{
sieve[current] = false;
current += 2*i+1;
}
}
}
// ---------- problem specific code ----------
// compute all factorials until factorial % n == 0
unsigned int naive(unsigned int n)
{
unsigned long long factorial = 1;
unsigned int result = 0;
while (factorial % n != 0)
{
result++;
factorial *= result;
factorial %= n;
}
return result;
}
// all prime numbers < 10^8
std::vector<unsigned int> primes;
// cache for i^2, i^3, i^4, ... where i is prime
std::unordered_map<unsigned int, unsigned int> cache;
// compute s(n)
unsigned int getSmallestFactorial(unsigned int n)
{
// will be the result
unsigned int best = 0;
// split off all prime factors
for (auto p : primes)
{
// p is not a prime factor of the current number ?
if (n % p != 0)
continue;
// extract the current prime factor as often as possible
// e.g. => 24 => 2^3 * 3 => primePower will be 8 and reduced = 3
unsigned int primePower = 1;
do
{
n /= p;
primePower *= p;
} while (n % p == 0);
// higher result ?
best = std::max(best, cache[primePower]);
// no further factorization possible ?
if (n == 1)
return best;
if (isPrime(n))
// s(prime) = prime
return std::max(best, n);
}
return best;
}
int main()
{
unsigned int limit = 100000000;
std::cin >> limit;
unsigned long long sum = 0;
// simple algorithm, too slow
//for (unsigned int i = 2; i <= 100; i++)
// sum += naive(i);
// and now the more sophisticated approach
// find all primes below 10^8
fillSieve(limit);
// copy those 5761455 primes to a dense array for faster access
for (unsigned int i = 2; i < limit; i++)
if (isPrime(i))
primes.push_back(i);
// find result for numbers with are powers of a single prime
for (unsigned int i = 2; i <= limit; i++)
{
if (isPrime(i))
{
// pre-compute all values of i^2, i^3, ... where i is prime and store in cache[]
unsigned long long power = i * (unsigned long long) i;
for (unsigned int exponent = 2; power <= limit; exponent++)
{
// optimized version of naive(), skip i numbers in each iteration
unsigned long long factorial = i;
unsigned int result = i;
do
{
result += i;
factorial *= result;
factorial %= power;
} while (factorial % power != 0);
cache[power] = result;
// next exponent
power *= i;
}
// s(prime) = prime
sum += i;
}
else
{
// compute s(non prime)
sum += getSmallestFactorial(i);
}
}
// and display the result
std::cout << sum << std::endl;
return 0;
}
This solution contains 26 empty lines, 38 comments and 4 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in 26.8 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 41 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
August 10, 2017 submitted solution
August 10, 2017 added comments
Difficulty
Project Euler ranks this problem at 10% (out of 100%).
Links
projecteuler.net/thread=549 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
Python github.com/HaochenLiu/My-Project-Euler/blob/master/549.py (written by Haochen Liu)
Python github.com/Meng-Gen/ProjectEuler/blob/master/549.py (written by Meng-Gen Tsai)
Python github.com/nayuki/Project-Euler-solutions/blob/master/python/p549.py (written by Nayuki)
Python github.com/smacke/project-euler/blob/master/python/549.py (written by Stephen Macke)
C++ github.com/evilmucedin/project-euler/blob/master/euler549/549.cpp (written by Den Raskovalov)
C++ github.com/roosephu/project-euler/blob/master/549.cpp (written by Yuping Luo)
Java github.com/nayuki/Project-Euler-solutions/blob/master/java/p549.java (written by Nayuki)
Java github.com/thrap/project-euler/blob/master/src/Java/Problem549.java (written by Magnus Solheim Thrap)
Perl github.com/shlomif/project-euler/blob/master/project-euler/549/euler-549-v2.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
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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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 !!!
<< problem 539 - Odd elimination | Robot Welders - problem 563 >> |