<< problem 65 - Convergents of e Maximum path sum II - problem 67 >>

# Problem 66: Diophantine equation

Consider quadratic Diophantine equations of the form:
x^2 - D * y^2 = 1

For example, when D=13, the minimal solution in x is 649^2 - 13 * 1802 = 1.

It can be assumed that there are no solutions in positive integers when D is square.

By finding minimal solutions in x for D = { 2, 3, 5, 6, 7 }, we obtain the following:

3^2 - 2 * 2^2 = 1
2^2 - 3 * 1^2 = 1
9^2 - 5 * 4^2 = 1
5^2 - 6 * 2^2 = 1
8^2 - 7 * 3^2 = 1

Hence, by considering minimal solutions in x for D <= 7, the largest x is obtained when D=5.

Find the value of D <= 1000 in minimal solutions of x for which the largest value of x is obtained.

# Algorithm

I didn't know anything about Pell's equation before I started solving this problem.
There is a Wikipedia article about it (en.wikipedia.org/wiki/Pell's_equation) where it becomes obvious that some numbers will be large.
In fact, too large for C++'s native data types. That means that my BigNum class (see my toolbox) has to be used.

My program computes the continuous fractions of x and y (see problem 64) and stops as soon as it finds a solution:
x^2 - D * y^2 = 1
I wasn't willing to add the code for subtraction to my BigNum class because the formula can be rewritten as:
x^2 = 1 + D * y^2

# My code

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

       #include <cmath>
#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
{
// faster multiplication possible ?
if (factor == 0)
return 0;
if (factor == 1)
return *this;
if (factor == MaxDigit)
{
auto result = *this;
result.insert(result.begin(), 0);
return result;
}
// might be slower but avoids nasty overflows
if (factor > MaxDigit)
return *this * BigNum(factor);

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;
}

// multiply two big numbers
BigNum operator*(const BigNum& other) const
{
// multiply single digits of "other" with the current object
BigNum result = 0;
for (int i = (int)other.size() - 1; i >= 0; i--)
result = result * MaxDigit + (*this * other[i]);

return result;
}

// compare two big numbers
bool operator<(const BigNum& other) const
{
if (size() < other.size())
return true;
if (size() > other.size())
return false;
for (int i = (int)size() - 1; i >= 0; i--)
{
if (operator[](i) < other[i])
return true;
if (operator[](i) > other[i])
return false;
}
return false;
}
};

int main()
{
unsigned int limit = 1000;
std::cin >> limit;

// initial solutions
unsigned int bestD = 2;
BigNum bestX = 3;

// solve for all values of D
for (unsigned int d = 3; d <= limit; d++)
{
unsigned int root = sqrt(d);
// exclude squares
if (root * root == d)
continue;

// see problem 64
unsigned int a = root;
unsigned int numerator   = 0;
unsigned int denominator = 1;

// keep only the most recent 3 numerators and denominators while diverging
BigNum x[3] = { 0, 1, root }; // numerators
BigNum y[3] = { 0, 0, 1 };    // denominators

// find better approximations until the exact solution is found
while (true)
{
numerator   = denominator * a - numerator;
denominator = (d - numerator * numerator) / denominator;
a = (root + numerator) / denominator;

// x_n = a * x_n_minus_1 + x_n_minus_2
x[0] = std::move(x[1]);
x[1] = std::move(x[2]);
x[2] = x[1] * a + x[0];

// y_n = a * y_n_minus_1 + y_n_minus_2
y[0] = std::move(y[1]);
y[1] = std::move(y[2]);
y[2] = y[1] * a + y[0];

// avoid subtraction (to keep BigNum's code short)
// x*x - d*y*y = 1
// x*y         = 1 + d*y*y
auto leftSide  = x[2] * x[2];
auto rightSide = y[2] * y[2] * d + 1;

// solved it
if (leftSide == rightSide)
break;
}

// biggest x so far ?
if (bestX < x[2])
{
bestX = x[2];
bestD = d;
}
}

// print D where x was maximized
std::cout << bestD << std::endl;
return 0;
}


This solution contains 25 empty lines, 30 comments and 3 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 7 | ./66

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 0.02 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 12, 2017 submitted solution

# Hackerrank

My code solves 6 out of 6 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 25% (out of 100%).

Hackerrank describes this problem as hard.

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=66 - 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-66-diophantine-equation/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p066.java (written by Nayuki)
Go: github.com/frrad/project-euler/blob/master/golang/Problem066.go (written by Frederick Robinson)
Scala: github.com/samskivert/euler-scala/blob/master/Euler066.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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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.
 << problem 65 - Convergents of e Maximum path sum II - problem 67 >>
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