Problem 304: Primonacci

(see projecteuler.net/problem=304)

For any positive integer n the function next_prime(n) returns the smallest prime p such that p>n.

The sequence a(n) is defined by:
a(1)=next_prime(10^14) and a(n)=next_prime(a(n-1)) for n>1.

The fibonacci sequence f(n) is defined by: f(0)=0, f(1)=1 and f(n)=f(n-1)+f(n-2) for n>1.

The sequence b(n) is defined as f(a(n)).

Find sum{b(n)} for 1<=n<=100 000. Give your answer mod 1234567891011.

My Algorithm

I need two thing for the problem: a fast Fibonacci generator and a prime test for large numbers.
My toolbox already has a Miller-Rabin test, that means I only have to find a fast Fibonacci generator.
Last week I solved problem 137 with the matrix form found on en.wikipedia.org/wiki/Fibonacci_number

Using those two tools I can easily compute F(10^14) and test each consecutive Fibonacci number until I have 100000 primes.

Alternative Approaches

When I read the Wikipedia a second time I noted a simplification of the matrix form called "fast doubling" on other websites:
F_{2n-1} = F^2_n + F^2_{n-1}
F_{2n} = F_n (2 F_{n-1} + F_n)

My function fibonacci implements that algorithm. It's about 2 to 3x faster than fibonacciMatrix (from problem 137).
Unfortunately, looking for the initial Fibonacci numbers poses a tiny fraction of the overall computation time.
Therefore it doesn't make a difference - but hopefully my new fibonacci function can be used in other problems, too.

I played around with my trusted powmod function, too. It's pretty fast when compiled with GCC but noticeable slower
with Visual C++ because of its lack of 128-bit integer arithmetic.
Based on the discussion apps.topcoder.com/forums/?module=Thread&threadID=670443&start=0&mc=10 I wrote a second
powmod which is about 30% on Visual C++ and doesn't need a mulmod function.
The old powmod still remains my fastest implementation of powmod for GCC.

All these little optimizations caused a little bit of code bloat - that's one of my longest solutions to a Project Euler problem ...
(actually it became the new number 1 spot on my list).

My code

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

#include <iostream>
 
// ---------- mulmod, powmod and Miller-Rabin 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);
 
#ifndef __GNUC__
// 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;
if (result >= modulo)
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;
if (a >= modulo)
a -= modulo;
 
// next bit
b >>= 1;
}
 
return result;
 
#else
 
// based on GCC's 128 bit implementation
return ((unsigned __int128)a * b) % modulo;
 
#endif
}
 
#ifdef __GNUC__
// return (base^exponent) % modulo => simple implementation
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;
}
 
#else
 
// return (base^exponent) % modulo => faster implementation
unsigned long long powmod(unsigned long long base, unsigned long long exponent, unsigned long long modulo)
{
// based on https://apps.topcoder.com/forums/?module=Thread&threadID=670443&start=0&mc=10
// fastest generic code (but slower than G++ optimized code)
unsigned long long result = 1;
base %= modulo;
 
while (exponent > 0)
{
// odd exponent ? a^b = a*a^(b-1)
if (exponent & 1)
{
// "unrolled" code of:
// result = mulmod(result, base, modulo);
// base = mulmod(base, base, modulo);
auto x = result;
auto y = base;
auto b = base;
 
result = 0;
base = 0;
while (b > 0)
{
if (b & 1)
{
result += x; if (result >= modulo) result -= modulo;
base += y; if (base >= modulo) base -= modulo;
}
 
x += x; if (x >= modulo) x -= modulo;
y += y; if (y >= modulo) y -= modulo;
 
b >>= 1;
}
}
else
{
// even exponent ? a^b = (a*a)^(b/2)
auto y = base;
auto b = base;
 
base = 0;
while (b > 0)
{
if (b & 1)
{
base += y; if (base >= modulo) base -= modulo;
}
 
y += y; if (y >= modulo) y -= modulo;
b >>=1;
}
}
 
exponent >>=1;
}
 
return result;
}
#endif
 
// 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;
 
// fine-tuning for the problem:
// fast check of more small primes (up to 100)
if (p % 19 == 0 || p % 23 == 0 || p % 29 == 0 || p % 31 == 0 || p % 37 == 0 ||
p % 41 == 0 || p % 43 == 0 || p % 47 == 0 || p % 53 == 0 || p % 59 == 0 ||
p % 61 == 0 || p % 67 == 0 || p % 71 == 0 || p % 73 == 0 || p % 79 == 0 ||
p % 83 == 0 || p % 89 == 0 || p % 97 == 0)
return false;
 
// 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;
}
 
// ---------- compute Fibonacci % modulo ----------
 
// version 1: matrix algorithm
unsigned long long fibonacciMatrix(unsigned long long n, unsigned long long modulo)
{
// fast exponentiation: same idea as powmod from my toolbox
 
// matrix values from https://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form
unsigned long long fibo [2][2]= { { 1, 1 },
{ 1, 0 } };
// initially identity matrix
unsigned long long result[2][2]= { { 1, 0 }, // { { F(n+1), F(n) },
{ 0, 1 } }; // { F(n), F(n-1) } }
 
while (n > 0)
{
// fast exponentation:
// odd exponent ? a^n = a*a^(n-1)
if (n & 1)
{
// compute new values, store them in temporaries
auto t00 = mulmod(result[0][0], fibo[0][0], modulo) + mulmod(result[0][1], fibo[1][0], modulo);
auto t01 = mulmod(result[0][0], fibo[0][1], modulo) + mulmod(result[0][1], fibo[1][1], modulo);
auto t10 = mulmod(result[1][0], fibo[0][0], modulo) + mulmod(result[1][1], fibo[1][0], modulo);
auto t11 = mulmod(result[1][0], fibo[0][1], modulo) + mulmod(result[1][1], fibo[1][1], modulo);
 
if (t00 >= modulo) t00 -= modulo;
if (t01 >= modulo) t01 -= modulo;
if (t10 >= modulo) t10 -= modulo;
if (t11 >= modulo) t11 -= modulo;
 
// copy back to matrix
result[0][0] = t00; result[0][1] = t01;
result[1][0] = t10; result[1][1] = t11;
}
 
// even exponent ? a^n = (a*a)^(n/2)
 
// compute new values, store them in temporaries
auto t00 = mulmod(fibo[0][0], fibo[0][0], modulo) + mulmod(fibo[0][1], fibo[1][0], modulo);
auto t01 = mulmod(fibo[0][0], fibo[0][1], modulo) + mulmod(fibo[0][1], fibo[1][1], modulo);
auto t10 = mulmod(fibo[1][0], fibo[0][0], modulo) + mulmod(fibo[1][1], fibo[1][0], modulo);
auto t11 = mulmod(fibo[1][0], fibo[0][1], modulo) + mulmod(fibo[1][1], fibo[1][1], modulo);
 
if (t00 >= modulo) t00 -= modulo;
if (t01 >= modulo) t01 -= modulo;
if (t10 >= modulo) t10 -= modulo;
if (t11 >= modulo) t11 -= modulo;
 
// copy back to matrix
fibo[0][0] = t00; fibo[0][1] = t01;
fibo[1][0] = t10; fibo[1][1] = t11;
 
n >>= 1;
}
 
return result[0][1]; // same as result[1][0]
}
 
// version 2: fast doubling algorithm, modulo < 2^62
unsigned long long fibonacci(unsigned long long n, unsigned long long modulo)
{
// extract highest bit
auto bit = n;
while (bit & (bit - 1))
bit &= bit - 1;
 
// F(0) and F(1)
auto a = 0ULL;
auto b = 1ULL;
 
// from highest to lowest bit
while (bit != 0)
{
// F(2n) = F(n) * (2 * F(n+1) - F(n))
auto nextA = mulmod(a, 2*b + modulo - a, modulo); // plus modulo to avoid negative results
// F(2n+1) = F(n)^2 + F(n+1)^2
b = mulmod(a, a, modulo) + mulmod(b, b, modulo);
if (b >= modulo)
b -= modulo;
 
a = nextA;
 
// odd ?
if (n & bit)
{
// one step further, need F(2n+1) and F(2n+2)
auto next = a + b;
if (next >= modulo)
next -= modulo;
 
a = b;
b = next;
}
 
bit >>= 1;
}
 
return a;
}
 
int main()
{
auto n = 100000000000000ULL;
auto numPrimes = 100000;
auto Modulo = 1234567891011ULL;
std::cin >> n >> numPrimes >> Modulo;
 
// F(n) = F(n-1) + F(n-2)
// my "seed" values to find F(10^14 + x)
auto last = fibonacci(n - 1, Modulo); // replace by fibonacciMatrix for the other algorithm
auto current = fibonacci(n , Modulo);
// actually it's already in result[0][0] of matrix algorithm
// and the second result of the fast doubling algorithm
 
auto sum = 0ULL;
for (auto i = 1; i <= numPrimes; i++)
{
do
{
n++;
 
// next Fibonacci number
auto next = (last + current) % Modulo;
last = current;
current = next;
}
while (!isPrime(n));
 
// if n is prime then add F(n)
sum += current;
// don't overflow
sum %= Modulo;
}
 
// display result
std::cout << sum << std::endl;
return 0;
}

This solution contains 63 empty lines, 77 comments and 7 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):
Note: Enter the the lower limit for the first prime, the number of primes to be searched and the modulo.

This is equivalent to
echo "10000 10000 10000000000" | ./304

Output:

(please click 'Go !')

Note: the original problem's input 100000000000000 100000 1234567891011 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 2.4 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

August 7, 2017 submitted solution
August 7, 2017 added comments

Difficulty

35% Project Euler ranks this problem at 35% (out of 100%).

Heatmap

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