<< problem 56 - Powerful digit sum | Spiral primes - problem 58 >> |

# 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 `#ifdef`

s 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:

This is equivalent to`echo 14 | ./57`

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

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.

*Please click on a problem's number to open my solution to that problem:*

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My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

<< problem 56 - Powerful digit sum | Spiral primes - problem 58 >> |