Problem 65: Convergents of e

(see projecteuler.net/problem=65)

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);
}
 
// add two big numbers
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:

(please click 'Go !')

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
April 29, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler065

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.

Links

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)

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