Problem 57: Square root convergents

(see projecteuler.net/problem=57)

It is possible to show that the square root of two can be expressed as an infinite continued fraction.
sqrt{2} = 1 + frac{1}{2 + frac{1}{2 + frac{1}{2 + ... }}} = 1.414213...

By expanding this for the first four iterations, we get:

1 + dfrac{1}{2} = dfrac{3}{2} = 1.5

1 + dfrac{1}{2 + frac{1}{2}} = dfrac{7}{5} = 1.4

1 + dfrac{1}{2 + frac{1}{2 + frac{1}{2}}} = dfrac{17}{12} = 1.41666...

1 + dfrac{1}{2 + frac{1}{2 + frac{1}{2 + frac{1}{2}}}} = dfrac{41}{29} = 1.41379...

The next three expansions are dfrac{99}{70} , dfrac{239}{169}, and dfrac{577}{408}, but the eighth expansion, dfrac{1393}{985},
is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?

Algorithm

An iteration/expansion can be described as:
f(n+1) = 1 + dfrac{1}{2 + f(n)}

Each iteration/expansion is a fraction, let's call it f(n) = dfrac{a(n)}{b(n)}
f(n+1) = 1 + dfrac{1}{1 + frac{a(n)}{b(n)}}
= 1 + dfrac{1}{frac{b(n) + a(n)}{b(n)}}

= 1 + dfrac{b(n)}{b(n) + a(n)}

= dfrac{b(n) + a(n) + b(n)}{b(n) + a(n)}

dfrac{a(n+1)}{b(n+1)} = dfrac{2b(n) + a(n)}{b(n) + a(n)}

so it's actually pretty easy to continuously compute numerator and denominator:
a(n+1) = 2b(n) + a(n)
b(n+1) = b(n) + a(n)

inital values:
x(0) = 1 + 1/2 = 1 + 1/(1+1)
a(0) = 1
b(0) = 1

The BigNum was copied from problem 56. When MaxDigit = 10 then each element of the array is one digit and I can compare a.size() > b.size().

Modifications by HackerRank

I have to print each iteration's ID where the numerator has more digits than the denominator.

My code

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

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

#include <vector>
#include <iostream>
 
// 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>
{
// must be 10 for this problem: a single "cell" store one digit 0 <= digit < 10
static const unsigned int MaxDigit = 10;
 
// 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;
}
};
 
#define ORIGINAL
 
int main()
{
unsigned int iterations = 1000;
std::cin >> iterations;
 
// both values have one digit initialized with 1
BigNum a = 1;
BigNum b = 1;
 
unsigned int count = 0;
for (unsigned int i = 0; i <= iterations; i++)
{
// different number of digits ?
if (a.size() > b.size())
{
#ifdef ORIGINAL
count++;
#else
std::cout << i << std::endl;
#endif
}
 
// a(n+1) = 2*b(n) + a(n)
// b(n+1) = b(n) + a(n)
auto twoB = b + b;
auto nextA = a + twoB;
auto nextB = b + a;
 
a = std::move(nextA);
b = std::move(nextB);
}
 
#ifdef ORIGINAL
std::cout << count << std::endl;
#endif
 
return 0;
}

This solution contains 15 empty lines, 13 comments and 8 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 14 | ./57

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)

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

February 28, 2017 submitted solution
April 24, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler057

My code solves 8 out of 8 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 5% (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.

Similar problems at Project Euler

Problem 56: Powerful digit sum

Note: I'm not even close to solving all problems at Project Euler. Chances are that similar problems do exist and I just haven't looked at them.

Links

projecteuler.net/thread=57 - 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-57-square-root-two/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p057.java (written by Nayuki)
Go: github.com/frrad/project-euler/blob/master/golang/Problem057.go (written by Frederick Robinson)
Scala: github.com/samskivert/euler-scala/blob/master/Euler057.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 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 163 solved problems had an average difficulty of 22.2% at Project Euler and I scored 11,907 points (out of 13200) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
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