Problem 66: Diophantine equation

(see projecteuler.net/problem=66)

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);
}
 
// 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
{
// 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:

(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 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
May 3, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler066

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.

Links

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).
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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+.
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