<< problem 206 - Concealed Square | Circular Logic - problem 209 >> |
Problem 207: Integer partition equations
(see projecteuler.net/problem=207)
For some positive integers k, there exists an integer partition of the form 4^t = 2^t + k,
where 4^t, 2^t, and k are all positive integers and t is a real number.
The first two such partitions are 4^1 = 2^1 + 2 and 4^{1.5849625}... = 2^{1.5849625}... + 6.
Partitions where t is also an integer are called perfect.
For any m >= 1 let P(m) be the proportion of such partitions that are perfect with k <= m.
Thus P(6) = 1/2.
In the following table are listed some values of P(m)
P(5) = 1/1
P(10) = 1/2
P(15) = 2/3
P(20) = 1/2
P(25) = 1/2
P(30) = 2/5
...
P(180) = 1/4
P(185) = 3/13
Find the smallest m for which P(m) < 1/12345
My Algorithm
The equation can be written as:
4^t = 2^t + k
(2^t)^2 = 2^t + k
And if I replace x = 2^t:
x^2 = x + k
My program iterates over all integer values of x. Then x^2 must be an integer and k must be an integer, too.
The problem statement already gave away that x=2 is the smallest valid partition and is perfect, too (which means t=1 because 2^1=2).
My variables total
, perfect
and x
are initialized accordingly.
x = 2^t is a perfect partition if t is an integer. That means that only one bit of the binary representation of x is set
(left side is the power of two, right side how it is written in binary notation):
2^1 = 10_2
2^2 = 100_2
2^3 = 1000_2
2^4 = 10000_2
2^5 = 100000_2
2^6 = 1000000_2
2^7 = 10000000_2
...
My bit twiddling code from bits.stephan-brumme.com/isPowerOfTwo.html can figure out whether x is a power-of-two in just a few CPU cycles.
As soon as the smallest x is known such that dfrac{perfect}{total} < dfrac{1}{12345}, the formula x^2 = x + k is translated to k = x^2 - x which produces the correct result.
Note
I added a small live test. You can the numerator and denominator for P(m) = dfrac{numerator}{denominator}, e.g. dfrac{1}{1234} → 1 1234
.
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho "3 13" | ./207
Output:
Note: the original problem's input 1 12345
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>
// return true, if x is a power of two
bool isPowerOfTwo(unsigned int x)
{
// see http://bits.stephan-brumme.com/isPowerOfTwo.html
return (x & (x - 1)) == 0;
}
int main()
{
// stop when proportion perfect/total is less than numerator/denominator
unsigned int numerator = 1;
unsigned int denominator = 12345;
std::cin >> numerator >> denominator;
if (numerator > denominator || numerator == 0) // check input: for live test only
return 1;
// count all valid partitions
unsigned long long total = 1;
// count perfect partitions
unsigned long long perfect = 1;
// x=2^t
unsigned long long x = 3;
// perfect/total > numerator/denominator is the same as
// perfect*denominator > total*numerator
while (perfect * denominator > total * numerator)
{
// found another perfect partition ?
if (isPowerOfTwo(x))
perfect++;
// keep going ...
total++;
x++;
}
// find k
unsigned long long k = x * (x - 1);
std::cout << k << std::endl;
return 0;
}
This solution contains 7 empty lines, 11 comments and 1 preprocessor command.
Benchmark
The correct solution to the original Project Euler problem was found in less than 0.01 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
May 31, 2017 submitted solution
May 31, 2017 added comments
Difficulty
Project Euler ranks this problem at 40% (out of 100%).
Links
projecteuler.net/thread=207 - the best forum on the subject (note: you have to submit the correct solution first)
Go github.com/frrad/project-euler/blob/master/golang/Problem207.go (written by Frederick Robinson)
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 | |
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 206 - Concealed Square | Circular Logic - problem 209 >> |