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Problem 64: Odd period square roots

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 - numeratornumerator_1 = 1 * 4 - 0 = 4

denominator = \lfloor dfrac{x - numerator^2}{denominator} \rfloordenominator_1 = \lfloor dfrac{23 - 4^2}{1} \rfloor = 7

a = \lfloor dfrac{root + numerator}{denominator} \rfloora_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:

Input data (separated by spaces or newlines):

This is equivalent to
echo 13 | ./64

Output:

(please click 'Go !')

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 #include // 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

Hackerrank

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.

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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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
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.

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