<< problem 277 - A Modified Collatz sequence | Triangles with integral sides and an integral ... - problem 279 >> |
Problem 278: Linear Combinations of Semiprimes
(see projecteuler.net/problem=278)
Given the values of integers 1 < a_1 < a_2 < ... < a_n, consider the linear combination
q_1 a_1 + q_2 a_2 + ... + q_n a_n = b, using only integer values q_k >= 0.
Note that for a given set of a_k, it may be that not all values of b are possible.
For instance, if a_1 = 5 and a_2 = 7, there are no q_1 >= 0 and q_2 >= 0 such that b could be
1, 2, 3, 4, 6, 8, 9, 11, 13, 16, 18 or 23.
In fact, 23 is the largest impossible value of b for a_1 = 5 and a_2 = 7.
We therefore call f(5, 7) = 23.
Similarly, it can be shown that f(6, 10, 15) = 29 and f(14, 22, 77) = 195.
Find sum{f(pq,pr,qr)}, where p, q and r are prime numbers and p < q < r < 5000.
My Algorithm
Assume I have two numbers x and y where gcd(x,y)=1.
The value m = xy - x - y can't be represented with some coefficients m = px + qy because:
xy - x - y = px + qy
xy = px + qy + x + y
xy = (p+1)x + (q+1)y
xy is a multiple of x and (p+1)x is a multiple of x, hence (q+1)y should be a multiple of x, too.
xy is a multiple of y and (q+1)y is a multiple of y, hence (p+1)x should be a multiple of y, too.
But gcd(x,y)=1 so y can't be a multiple of x and therefore q+1 should be a multiple of x.
And for the same reason x can't be a multiple of y and therefore p+1 should be a multiple of y.
Possible values for q+1 would be x, 2x, 3x, ... (and for p+1: y, 2y, 3y, ...)
If I assume the lowest value p+1=y and q+1=x then the equation becomes
xy = y * x + x * y
xy = 2xy → contradition !
Therefore m = xy - x - y actually can't be represented with some coefficients m = px + qy.
With three numbers x,y,z and gcd(x,y,z)=1 the idea is very similar:
if there would be some coefficients p, q and r such that m = pxy + qxz + ryz represents m = 2xyz - xy - xz - yz then
(2xyz - xy - xz - yz) mod x = -yz
pxy + qxz + ryz = 2xyz - xy - xz - yz
2xyz = pxy + qxz + ryz + xy + xz + yz
2xyz = (py+qz+y+z)x + (rz + z)y
Hence rz + z = (r+1)z must be a multiple of x. z can't be such a multiple (because of gcd(x,y,z) = 1).
The same idea for y and z gives that p+1 must be a multiple of z and q+1 a multiple of y.
As before - if I choose the smallest possible p+1=z, q+1=y and r+1=x:
2xyz = zxy + yxz + xyz + xy + xz + yz
2xyz = 3xyz + xy + xz + yz → contradiction
I didn't come up with the full solution, I just know how to use search engines :-;
I found the problem in the 24th International Mathemtical Olympiad held 1983 in Paris, France
somewhat cryptic solution: www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln833.html
I stumbled across it while reading the German Wikipedia de.wikipedia.org/wiki/Münzproblem
unfortunately, the English page misses that special case en.wikipedia.org/wiki/Coin_problem
but it can be derived from their n=2 explanations (pretty much what I have done above)
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho 10 | ./278
Output:
Note: the original problem's input 5000
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. Or just jump to my GitHub repository.
#include <iostream>
#include <vector>
int main()
{
unsigned int limit = 5000;
std::cin >> limit;
// simple prime sieve from my toolbox
std::vector<unsigned long long> primes = { 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 x : primes)
{
// prime is too large to be a divisor
if (x*x > i)
break;
// divisible => not prime
if (i % x == 0)
{
isPrime = false;
break;
}
}
// yes, we have a prime
if (isPrime)
primes.push_back(i);
}
// all combinations of primes
unsigned long long sum = 0;
for (size_t i = 0; i < primes.size(); i++)
for (size_t j = i + 1; j < primes.size(); j++)
for (size_t k = j + 1; k < primes.size(); k++)
{
auto p = primes[i];
auto q = primes[j];
auto r = primes[k];
sum += 2*p*q*r - p*q - p*r - q*r;
}
std::cout << sum << std::endl;
return 0;
}
This solution contains 7 empty lines, 6 comments and 2 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in 0.11 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
July 11, 2017 submitted solution
July 11, 2017 added comments
Difficulty
Project Euler ranks this problem at 50% (out of 100%).
Links
projecteuler.net/thread=278 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
C++ github.com/Meng-Gen/ProjectEuler/blob/master/278.cc (written by Meng-Gen Tsai)
C++ github.com/roosephu/project-euler/blob/master/278.cpp (written by Yuping Luo)
C github.com/LaurentMazare/ProjectEuler/blob/master/e278.c (written by Laurent Mazare)
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 |
I stopped working on Project Euler problems around the time they released 617.
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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 277 - A Modified Collatz sequence | Triangles with integral sides and an integral ... - problem 279 >> |