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Problem 64: Odd period square roots
(see projecteuler.net/problem=64)
All square roots are periodic when written as continued fractions and can be written in the form:
sqrt{N} = a_0 + dfrac{1}{a_1 + frac{1}{a_2 + frac{1}{a_3 + ...}}}
For example, let us consider sqrt{23}:
sqrt{23} = 4 + sqrt{23} - 4 = 4 + dfrac{1}{frac{1}{sqrt{23}-4}} = 4 + dfrac{1}{1 + frac{sqrt{23}-3}{7}}
If we continue we would get the following expansion:
sqrt{23} = 4 + dfrac{1}{1 + dfrac{1}{3 + dfrac{1}{1 + frac{1}{8 + ...}}}}
The process can be summarised as follows:
a_0 = 4, dfrac{1}{sqrt{23}-4} = dfrac{sqrt{23}+4}{7} = 1 + dfrac{sqrt{23}-3}{7}
a_1 = 1, dfrac{7}{sqrt{23}-3} = dfrac{7(sqrt{23}+3)}{14} = 3 + dfrac{sqrt{23}-3}{2}
a_2 = 3, dfrac{2}{sqrt{23}-3} = dfrac{2(sqrt{23}+3)}{14} = 1 + dfrac{sqrt{23}-4}{7}
a_3 = 1, dfrac{7}{sqrt{23}-4} = dfrac{7(sqrt{23}+4)}{7} = 8 + dfrac{sqrt{23}-4}{1}
a_4 = 8, dfrac{1}{sqrt{23}-4} = dfrac{sqrt{23}+4}{7} = 1 + dfrac{sqrt{23}-3}{7}
a_5 = 1, dfrac{7}{sqrt{23}-3} = dfrac{7(sqrt{23}+3)}{14} = 3 + dfrac{sqrt{23}-3}{2}
a_6 = 3, dfrac{2}{sqrt{23}-3} = dfrac{2(sqrt{23}+3)}{14} = 1 + dfrac{sqrt{23}-4}{7}
a_7 = 1, dfrac{7}{sqrt{23}-4} = dfrac{7(sqrt{23}+4)}{7} = 8 + dfrac{sqrt{23}-4}{1}
It can be seen that the sequence is repeating. For conciseness, we use the notation sqrt{23} = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.
The first ten continued fraction representations of (irrational) square roots are:
sqrt{2} = [1;(2)], period=1
sqrt{3} = [1;(1,2)], period=2
sqrt{5} = [2;(4)], period=1
sqrt{6} = [2;(2,4)], period=2
sqrt{7} = [2;(1,1,1,4)], period=4
sqrt{8} = [2;(1,4)], period=2
sqrt{10} = [3;(6)], period=1
sqrt{11} = [3;(3,6)], period=2
sqrt{12} = [3;(2,6)], period=2
sqrt{13} = [3;(1,1,1,1,6)], period=4
Exactly four continued fractions, for N <= 13, have an odd period.
How many continued fractions for N <= 10000 have an odd period?
My Algorithm
I didn't know anything about continued fractions before I saw this problem. The Wikipedia article is pretty long: en.wikipedia.org/wiki/Continued fraction
Even more important, there is another Wikipedia article about the relationship of square roots and continued fractions: en.wikipedia.org/wiki/Methods of computing square roots
The whole algorithm can be found in my function getPeriodLength
.
My first step is to find the integer part of the square root. That's pretty easy:
root = sqrt(x)
→ if root*root == x
then x
is a perfect square and we can abort.
In the example above, x=4 and root=4. Even though a_0 was the first step, there is a step before it (kind of a_{-1}):
sqrt{23} = 0 + dfrac{sqrt{23} - 0}{1}
The variable numerator = 0
refers to the zero which is subtracted in the numerator.
The variable denominator = 1
refers to the 1 in the denominator.
The variable a = root
will be a_0.
I do the following to compute a_1:
numerator = denominator * a - numerator → numerator_1 = 1 * 4 - 0 = 4
denominator = \lfloor dfrac{x - numerator^2}{denominator} \rfloor → denominator_1 = \lfloor dfrac{23 - 4^2}{1} \rfloor = 7
a = \lfloor dfrac{root + numerator}{denominator} \rfloor → a_1 = \lfloor dfrac{4 + 4}{7} \rfloor = 1
As you can see in the problem statement, the first line contained the fraction dfrac{sqrt{23}+4}{7}
And for a_2:
numerator_2 = 7 * 1 - 4 = 3
denominator_2 = \lfloor dfrac{23 - 3^2}{7} \rfloor = 2
a_2 = \lfloor dfrac{4 + 3}{2} \rfloor = 3
The mathematical reasoning is based on the general concept that (a - b) * (a + b) = a^2 - b^2.
If a = sqrt{x} and b = y then (sqrt{x} - y) * (sqrt{x} + y) = x - y^2
For x = 23 and y = \lfloor sqrt{x} \rfloor = \lfloor sqrt{23} \rfloor = 4:
dfrac{1}{(sqrt{23} - 4)}
= dfrac{sqrt{23} + 4}{(sqrt{23} - 4) * (sqrt{23} + 4)}
= dfrac{sqrt{23} + 4}{23 - 4^2}
= dfrac{sqrt{23} + 4}{7}
Then the largest possible integer such that 0 < numerator < denominator:
= dfrac{sqrt{23} + 4 - 7 + 7}{7}
= 1 + dfrac{sqrt{23} - 3}{7}
A loop can be identified by keeping track of the tupel (a, numerator, denominator)
. If it appears a second time, we have entered a loop.
We could create a data structure - something like std::set<std::tupel>
- but Wikipedia mentions a neat trick that I don't understand:
The equation a == 2 * root
becomes true as soon as we enter a loop.
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho 13 | ./64
Output:
Note: the original problem's input 1000
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++ and can be compiled with G++, Clang++, Visual C++. You can download it, too.
#include <cmath>
#include <iostream>
// return length of period or 0 for perfect squares
unsigned int getPeriodLength(unsigned int x)
{
// without any fractional part yet ...
unsigned int root = sqrt(x);
// exclude perfect squares (no period)
if (root * root == x)
return 0;
// the integer part of sqrt(x)
unsigned int a = root;
// let's use a variable numerator to store what we subtract
unsigned int numerator = 0; // initially zero, e.g. 4 will appear in second iteration of sqrt(23)
unsigned int denominator = 1; // initially one, e.g. 7 will appear in second iteration of sqrt(23)
// count how long it takes until the next period starts
unsigned int period = 0;
// terminate when we see the same triplet (a, numerator, denominator) a second time
// to me it wasn't obvious that this happens exactly when a == 2 * root
// but thanks to Wikipedia for that trick ...
while (a != 2 * root)
{
numerator = denominator * a - numerator;
// must be integer divisions !
denominator = (x - numerator * numerator) / denominator;
a = (root + numerator) / denominator;
period++;
}
return period;
}
int main()
{
unsigned int last;
std::cin >> last;
// count all odd periods
unsigned int numOdd = 0;
for (unsigned int i = 2; i <= last; i++) // 0 and 1 are perfect squares
{
unsigned int period = getPeriodLength(i);
// count number of odd lengths (if not a perfect square)
if (period % 2 == 1)
numOdd++;
// branchless: numOdd += period & 1;
}
// print result
std::cout << numOdd << std::endl;
return 0;
}
This solution contains 11 empty lines, 14 comments and 2 preprocessor commands.
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
March 8, 2017 submitted solution
April 27, 2017 added comments
Hackerrank
see https://www.hackerrank.com/contests/projecteuler/challenges/euler064
My code solves 8 out of 8 test cases (score: 100%)
Difficulty
Project Euler ranks this problem at 20% (out of 100%).
Hackerrank describes this problem as easy.
Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.
Links
projecteuler.net/thread=64 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
C# www.mathblog.dk/project-euler-continued-fractions-odd-period/ (written by Kristian Edlund)
Java github.com/nayuki/Project-Euler-solutions/blob/master/java/p064.java (written by Nayuki)
Go github.com/frrad/project-euler/blob/master/golang/Problem064.go (written by Frederick Robinson)
Scala github.com/samskivert/euler-scala/blob/master/Euler064.scala (written by Michael Bayne)
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 !!!
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