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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?

# 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 solved **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 never 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).

gray problems are already solved but I haven't published my solution yet.

blue problems are already solved and there wasn't a Hackerrank version of it (at the time I solved it) or I didn't care about it because it differed too much.

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