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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#Continued_fraction_expansion

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.

# 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.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This is equivalent to`echo 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)*

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

(compiled for x86_64 / Linux, GCC flags: `-O3 -march=native -fno-exceptions -fno-rtti -std=c++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:

Python: 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)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.

yellow problems score less than 100% at Hackerrank (but still solve the original problem).

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red problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte.

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I scored 12,983 points (out of 15100 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

Look at my progress and performance pages to get more details.

My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

# 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 63 - Powerful digit counts | Convergents of e - problem 65 >> |