<< 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.

# My 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

# 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)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       #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 = { 0, 1, root }; // numerators
BigNum y = { 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 = std::move(x);
x = std::move(x);
x = x * a + x;

// y_n = a * y_n_minus_1 + y_n_minus_2
y = std::move(y);
y = std::move(y);
y = y * a + y;

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

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

// biggest x so far ?
if (bestX < x)
{
bestX = x;
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.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.02 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++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 rarely an option.

# Heatmap

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

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

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