<< problem 64 - Odd period square roots Diophantine equation - problem 66 >>

# Problem 65: Convergents of e

The square root of 2 can be written as an infinite continued fraction.
sqrt{2} = 1 + dfrac{1}{2 + dfrac{1}{2 + dfrac{1}{2 + frac{1}{2 + ...}}}}

The infinite continued fraction can be written, sqrt{2} = [1;(2)], (2) indicates that 2 repeats ad infinitum.
In a similar way, sqrt{23} = [4;(1,3,1,8)].

It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations.
Let us consider the convergents for sqrt{2}.

1 + dfrac{1}{2} = dfrac{3}{2}

1 + dfrac{1}{2 + dfrac{1}{2}} = dfrac{7}{5}

1 + dfrac{1}{2 + dfrac{1}{2 + dfrac{1}{2}}} = dfrac{17}{12}

1 + dfrac{1}{2 + dfrac{1}{2 + dfrac{1}{2 + dfrac{1}{2}}}} = dfrac{41}{29}

Hence the sequence of the first ten convergents for sqrt{2} are:

1, dfrac{3}{2}, dfrac{7}{5}, dfrac{17}{12}, dfrac{41}{29}, dfrac{99}{70}, dfrac{239}{169}, dfrac{577}{408}, dfrac{1393}{985}, dfrac{3363}{2378}, ...

What is most surprising is that the important mathematical constant,
e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].

The first ten terms in the sequence of convergents for e are:
2, 3, dfrac{8}{3}, dfrac{11}{4}, dfrac{19}{7}, dfrac{87}{32}, dfrac{106}{39}, dfrac{193}{71}, dfrac{1264}{465}, dfrac{1457}{536}, ...

The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17.

Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.

# Algorithm

Let's compare the first 10 numerators and the continuous fractions (values taken from problem statement):

knumeratorcontinuous fraction
121
232
381
4111
5194
6871
71061
81936
912641
1014571

After staring at the numbers for 5 minutes I saw that:
nominator_i = nominator_{i-2} + nominator_{i-1} * fraction_{i-1}

All I have to do is writing a simple for-loop that keeps tracks of the two most recent numerators.
The continuous fraction is 1 if index \mod 3 != 2 and 2k if index \mod 3 == 2.

The numerators can grow pretty fast and exceed 2^64: the numerator for k=100 has 58 decimal digits.
That's why I have to use my BigNum class. The code was used previously in several solutions, too, e.g. problem 56.
There is a minor change: the constant MaxDigit=10 was replaced by the highest power-of-10 that is below 2^32.
The only reason is an improved performance (about 10x faster than MaxDigit = 10).

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

       #include <iostream>
#include <vector>

// store single digits with lowest digits first
// e.g. 1024 is stored as { 4,2,0,1 }
// only non-negative numbers supported
struct BigNum : public std::vector<unsigned int>
{
// string conversion works only properly when MaxDigit is a power of 10
static const unsigned int MaxDigit = 1000000000;

// store a non-negative number
BigNum(unsigned long long x = 0)
{
do
{
push_back(x % MaxDigit);
x /= MaxDigit;
} while (x > 0);
}

BigNum operator+(const BigNum& other) const
{
auto result = *this;
// add in-place, make sure it's big enough
if (result.size() < other.size())
result.resize(other.size(), 0);

unsigned int carry = 0;
for (size_t i = 0; i < result.size(); i++)
{
carry += result[i];
if (i < other.size())
carry += other[i];
else
if (carry == 0)
return result;

if (carry < MaxDigit)
{
// no overflow
result[i] = carry;
carry     = 0;
}
else
{
// yes, we have an overflow
result[i] = carry - MaxDigit;
carry = 1;
}
}

if (carry > 0)
result.push_back(carry);

return result;
}

// multiply a big number by an integer
BigNum operator*(unsigned int factor) const
{
unsigned long long carry = 0;
auto result = *this;
for (auto& i : result)
{
carry += i * (unsigned long long)factor;
i      = carry % MaxDigit;
carry /= MaxDigit;
}
// store remaining carry in new digits
while (carry > 0)
{
result.push_back(carry % MaxDigit);
carry /= MaxDigit;
}

return result;
}
};

int main()
{
unsigned int lastIndex;
std::cin >> lastIndex;

// to save memory we dont keep all numerators, only the latest three
BigNum numerators[3] = { 0,   // dummy, will be overwritten immediately
1,   // always 1
2 }; // the first number of the continuous fraction ("before the semicolon")

for (unsigned int index = 2; index <= lastIndex; index++)
{
// e = [2; 1,2,1, 1,4,1, ... 1,2k,1, ...]
unsigned int fractionNumber = 1;
if (index % 3 == 0)
fractionNumber = (index / 3) * 2;

// keep only the latest two numerators
numerators[0] = std::move(numerators[1]);
numerators[1] = std::move(numerators[2]);
// and generate the next one
if (fractionNumber == 1)
numerators[2] = numerators[0] + numerators[1];
else
numerators[2] = numerators[0] + numerators[1] * fractionNumber;
}

// add all digits of the last numerator
unsigned int sum = 0;
for (auto x : numerators[2])
// when MaxDigit != 10 then we have to split into single digits
while (x > 0)
{
sum += x % 10;
x   /= 10;
}

std::cout << sum << std::endl;
return 0;
}


This solution contains 16 empty lines, 17 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:

Input data (separated by spaces or newlines):

This is equivalent to
echo 10 | ./65

Output:

Note: the original problem's input 100 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

# Hackerrank

My code solved 9 out of 9 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 15% (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.

projecteuler.net/thread=65 - 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-65-numerator-continued-fraction-e/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p065.java (written by Nayuki)
Go: github.com/frrad/project-euler/blob/master/golang/Problem065.go (written by Frederick Robinson)
Scala: github.com/samskivert/euler-scala/blob/master/Euler065.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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The 133 solved problems had an average difficulty of 16.9% at Project Euler and I scored 11,174 points (out of 12300) at Hackerrank's Project Euler+.
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