<< problem 215 - Crack-free Walls Perfect right-angled triangles - problem 218 >>

# Problem 216: Investigating the primality of numbers of the form 2n^2-1

Consider numbers t(n) of the form t(n) = 2n^2-1 with n > 1.
The first such numbers are 7, 17, 31, 49, 71, 97, 127 and 161.
It turns out that only 49 = 7 * 7 and 161 = 7 * 23 are not prime.
For n <= 10000 there are 2202 numbers t(n) that are prime.

How many numbers t(n) are prime for n <= 50,000,000 ?

# Very inefficient solution

My code needs more than 60 seconds to find the correct result. (scroll down to the benchmark section)
Apparantly a much smarter algorithm exists - or my implementation is just inefficient.

# My Algorithm

The Miller-Rabin primality test from my toolbox can easily solve this problem.
The only drawback: it takes xyz minutes.

I still can't solve this problem in under a minute but at least reduced the runtime considerably by observing that
if t(n) = 2n^2 - 1 is a multiple of number k then t(n+k) = 2(n + k)^2 - 1 is a multiple of k too:
t(n + k) - t(n)
= (2(n + k)^2 - 1) - (2n^2 - 1)
= 2(n + k)^2 - 2n^2
= 2(n^2 + 2nk + k^2) - 2n^2
= 2n^2 + 4nk + 2k^2 - 2n^2
= 4nk + 2k^2
= k * (4n + 2k)

In fact, if t(n) is a multiple of such k then not only t(n + k) but also t(n + 2k) and t(n + 3k) and so on are multiples of k.
And that means they can't be prime - reducing the number of Miller-Rabin tests to about one fifth.

I don't have a fast factorization algorithm in my toolbox yet, so I perform simple trial division to find prime factors.
After trying several values I found that it doesn't pay off to check for prime factors larger than one million (see MaxSievePrime).
Even more, I hardly found any candidates above two millions that are divisible by a prime which wasn't already a factor of a smaller candidates (see FilterThreshold).
Both constants were heuristically determined by lots of trial'n'error.

## Alternative Approaches

The Tonelli-Shanks algorithm (which I wasn't aware of) is much faster.
I probably should translate Wikipedia's pseudo-code to C++ and add it to my toolbox.

## Note

OpenMP gives a nice speed-up but I still need about 6 minutes to find the correct result (see #define PARALLEL).
By the way: if I would use only my Miller-Rabin test (without the optimizations mentioned above), the program finishes after 55 minutes.

Reading the forums, the vast majority of solvers seem to have a simple loop invoking the prime test available in Java, Mathematica, etc.
They neither wrote the prime test nor looked for optimizations.
In my opinion, this is a quite hard problem if you really want to stick to the "one-minute rule".

Looking at the high number of solvers and the low difficulty rating I expected that I missed something big - but actually only a small number of people
discovered/knew the most appropriate way to solve this problem, the Tonelli-Shanks algorithm.

# Interactive test

This feature is not available for the current problem.

# My code

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

       #include <iostream>
#include <vector>
#include <algorithm>

// ---------- copied the Miller-Rabin primality test from my toolbox ----------

// return (a*b) % modulo
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
// (a * b) % modulo = (a % modulo) * (b % modulo) % modulo
a %= modulo;
b %= modulo;

// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;

// we might encounter overflows (slow path)
// the number of loops depends on b, therefore try to minimize b
if (b > a)
std::swap(a, b);

// bitwise multiplication
unsigned long long result = 0;
while (a > 0 && b > 0)
{
// b is odd ? a*b = a + a*(b-1)
if (b & 1)
{
result += a;
result %= modulo;
// skip b-- because the bit-shift at the end will remove the lowest bit anyway
}

// b is even ? a*b = (2*a)*(b/2)
a <<= 1;
a  %= modulo;

// next bit
b >>= 1;
}

return result;
}

// return (base^exponent) % modulo
unsigned long long powmod(unsigned long long base, unsigned long long exponent, unsigned long long modulo)
{
unsigned long long result = 1;
while (exponent > 0)
{
// fast exponentation:
// odd exponent ? a^b = a*a^(b-1)
if (exponent & 1)
result = mulmod(result, base, modulo);

// even exponent ? a^b = (a*a)^(b/2)
base = mulmod(base, base, modulo);
exponent >>= 1;
}
return result;
}

// Miller-Rabin-test
bool isPrime(unsigned long long p)
{
// IMPORTANT: requires mulmod(a, b, modulo) and powmod(base, exponent, modulo)

// some code from             https://ronzii.wordpress.com/2012/03/04/miller-rabin-primality-test/
// with optimizations from    http://ceur-ws.org/Vol-1326/020-Forisek.pdf
// good bases can be found at http://miller-rabin.appspot.com/

// trivial cases
const unsigned int bitmaskPrimes2to31 = (1 <<  2) | (1 <<  3) | (1 <<  5) | (1 <<  7) |
(1 << 11) | (1 << 13) | (1 << 17) | (1 << 19) |
(1 << 23) | (1 << 29); // = 0x208A28Ac
if (p < 31)
return (bitmaskPrimes2to31 & (1 << p)) != 0;

if (p %  2 == 0 || p %  3 == 0 || p %  5 == 0 || p % 7 == 0 || // divisible by a small prime
p % 11 == 0 || p % 13 == 0 || p % 17 == 0)
return false;

if (p < 17*19) // we filtered all composite numbers < 17*19, all others below 17*19 must be prime
return true;

// test p against those numbers ("witnesses")
// good bases can be found at http://miller-rabin.appspot.com/
const unsigned int STOP = 0;
const unsigned int TestAgainst1[] = { 377687, STOP };
const unsigned int TestAgainst2[] = { 31, 73, STOP };
const unsigned int TestAgainst3[] = { 2, 7, 61, STOP };
// first three sequences are good up to 2^32
const unsigned int TestAgainst4[] = { 2, 13, 23, 1662803, STOP };
const unsigned int TestAgainst7[] = { 2, 325, 9375, 28178, 450775, 9780504, 1795265022, STOP };

// good up to 2^64
const unsigned int* testAgainst = TestAgainst7;
// use less tests if feasible
if (p < 5329)
testAgainst = TestAgainst1;
else if (p < 9080191)
testAgainst = TestAgainst2;
else if (p < 4759123141ULL)
testAgainst = TestAgainst3;
else if (p < 1122004669633ULL)
testAgainst = TestAgainst4;

// find p - 1 = d * 2^j
auto d = p - 1;
d >>= 1;
unsigned int shift = 0;
while ((d & 1) == 0)
{
shift++;
d >>= 1;
}

// test p against all bases
do
{
auto x = powmod(*testAgainst++, d, p);
// is test^d % p == 1 or -1 ?
if (x == 1 || x == p - 1)
continue;

// now either prime or a strong pseudo-prime
// check test^(d*2^r) for 0 <= r < shift
bool maybePrime = false;
for (unsigned int r = 0; r < shift; r++)
{
// x = x^2 % p
// (initial x was test^d)
x = mulmod(x, x, p);
// x % p == 1 => not prime
if (x == 1)
return false;

// x % p == -1 => prime or an even stronger pseudo-prime
if (x == p - 1)
{
// next iteration
maybePrime = true;
break;
}
}

// not prime
if (!maybePrime)
return false;
} while (*testAgainst != STOP);

// prime
return true;
}

// ---------- problem-specific code ----------

int main()
{
unsigned int limit = 50000000;
std::cin >> limit;

// count all prime 2n^2-1
unsigned int count = 0;

// if candidate[n] is false, then n can't be a prime, else there is a possibility
std::vector<bool> candidate(limit + 1, true);

// generate a few primes, use slower Miller-Rabin algorithm for convenience
const auto MaxSievePrime = std::max<unsigned int>(limit / 50, 10000);
std::vector<unsigned int> smallPrimes;
for (unsigned int p = 3; p <= MaxSievePrime; p++)
if (isPrime(p))
smallPrimes.push_back(p);
// almost all numbers above this threshold have bigger factors (or are primes)
const auto FilterThreshold = 2 * MaxSievePrime;

// accelerate with OpenMP
#define PARALLEL
#ifdef  PARALLEL
unsigned int numCores = 0; // 0 => run on all cores, 1,2,3,... => restrict to 1,2,3,... cores
#pragma omp parallel for reduction(+:count) num_threads(numCores) schedule(dynamic, 10000)
#endif
for (unsigned long long n = 2; n <= limit; n++)
{
if (!candidate[n])
continue;

// calculate 2n^2-1
auto p = 2 * n * n - 1;
// run primality test
if (isPrime(p))
{
count++;
continue;
}

// p failed the primality test, figure out which (small) prime is a factor of p
// I didn't observe many "new" small prime factors above a certain threshold
if (n < FilterThreshold)
for (auto s : smallPrimes)
if (p > s && p % s == 0) // trial division
for (auto i = n; i <= limit; i += s) // remove all multiples
candidate[i] = false;
}

// display result
std::cout << count << std::endl;
return 0;
}


This solution contains 33 empty lines, 50 comments and 7 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in  1960 seconds  (exceeding the limit of 60 seconds).
The code can be accelerated with OpenMP but the timings refer to the single-threaded version on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 9 MByte.

(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

November 1, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 45% (out of 100%).

# 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.

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