Problem 343: Fractional Sequences

(see projecteuler.net/problem=343)

For any positive integer k, a finite sequence a_i of fractions dfrac{x_i}{y_i} is defined by:
a_1 = dfrac{1}{k} and

a_i = dfrac{x_{i-1}+1}{y_{i-1} - 1} reduced to lowest terms for i>1.

When a_i reaches some integer n, the sequence stops. (That is, when y_i=1.)
Define f(k) = n.
For example, for k = 20:

1/20 → 2/19 → 3/18 → 1/6 → 2/5 → 3/4 → 4/3 → 5/2 → 6/1 = 6

So f(20) = 6.

Also f(1) = 1, f(2) = 2, f(3) = 1 and sum{f(k^3)} = 118937 for 1 <= k <= 100.

Find sum{f(k^3)} for 1 <= k <= 2 * 10^6.

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 bruteForce() function was written in a few minutes - and as expected is too slow for 2 * 10^6 but nevertheless proved to be a good help while debugging my faster algorithm:

I made a few observations:
That means the numerator will slowly increase from 1 to the smallest prime factor of numerator + denominator, then reset to 1,
then slowly increase to the next prime factor, etc.
Since the initial value of numerator + denominator is 1 + k, we are basically factorizing 1 + k.
In the very last cycle, numerator + denominator will be the largest prime factor of 1 + k.
Since the algorithm stops if denominator == 1 I know that the last numerator is the largest prime factor of 1 + k, minus 1.

My function getMaxFactor() returns the largest prime factor of its argument.
Therefore getMaxFactor(k + 1) - 1 is equivalent to f(k).

One problem remains: factorization of large numbers is hard.
Unfortunately, the problem statement asks for k^3 where k can be 2 * 10^6 such that I have to factorize 8 * 10^18 + 1
(by the way: the largest prime factor of that number is 666667).

Wolfram Alpha told me that x^3 + 1 can be simplified:
x^3 + 1 = (x + 1) (x^2 - x + 1)
So instead of factorizing a single huge number x^3 + 1 I can factorize two much smaller numbers.
The largest prime factor found in x + 1 or x^2 - x + 1 is the largest prime factor of x^3 + 1.

I tried several tricks to further speed up the factorization:
However, the program still needed about 2.5 minutes (at least the result was correct ...).
Desperately I search the web for factorization methods that are fast AND easy to implement.

Fermat's factorization method was a nice candidate - but I can't measure any speedup (see en.wikipedia.org/wiki/Fermat's factorization method).
Maybe the getFermatFactors() turns out to be more useful in another Project Euler problem ... I keep it here for the sake of completeness.

My experiments with Pollard's rho algorithm were a desaster: the program became much slower (maybe I had a bug somewhere, see en.wikipedia.org/wiki/Pollard's rho algorithm).

Each computation of getMaxFactor is independent of each other. That's a perfect use case for multi-core CPUs !
If you uncomment //#define PARALLEL then most modern C++ compilers (G++, Visual C++, Intel C++, etc.) will distribute the workload to all available CPUs.
In my case I achieved an almost perfect performance boost: the solution is found 5.1 times faster on my 8-core computer (in about 28 seconds).
I consider this speedup to be "almost linear" because my computer isn't a true 8-core system, it has 4 cores plus hyperthreading.

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 200 | ./343

Output:

(please click 'Go !')

Note: the original problem's input 10000 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.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
 
// ---------- a few routines from my toolbox ----------
 
// greatest common divisor
template <typename T>
T gcd(T a, T b)
{
while (a != 0)
{
T c = a;
a = b % a;
b = c;
}
return b;
}
 
// odd prime numbers are marked as "true" in a bitvector
std::vector<bool> sieve;
unsigned int primeLimit = 0;
 
// return true, if x is a prime number
bool isSmallPrime(unsigned int x)
{
// handle even numbers
if ((x & 1) == 0)
return x == 2;
 
// lookup for odd numbers
return sieve[x >> 1];
}
 
// find all prime numbers from 2 to size
void fillSieve(unsigned int size)
{
// store only odd numbers
const unsigned int half = (size >> 1) + 1;
 
// allocate memory
sieve.resize(half, true);
// 1 is not a prime number
sieve[0] = false;
 
// process all relevant prime factors
for (unsigned int i = 1; 2*i*i < half; i++)
// do we have a prime factor ?
if (sieve[i])
{
// mark all its multiples as false
unsigned int current = 3*i+1;
while (current < half)
{
sieve[current] = false;
current += 2*i+1;
}
}
}
 
// 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)
{
// modified for problem 343: invoke isSmallPrime where appropriate
if (p < primeLimit)
return isSmallPrime(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 ----------
 
// brute-force: just follow the chain and reduce fractions as soon as possible
unsigned long long bruteForce(unsigned long long k)
{
auto numerator = 1ULL;
auto denominator = k;
 
// until the fraction is a natural number ...
while (denominator != 1)
{
// figure out whether the fraction can be reduced
auto sharedFactor = gcd(numerator, denominator);
if (sharedFactor == 1)
{
// already a proper fraction, increase numerator and decrease denominator
numerator++;
denominator--;
}
else
{
// reduce to a proper fraction
numerator /= sharedFactor;
denominator /= sharedFactor;
}
}
 
return numerator;
}
 
// Fermat's factorization method
std::pair<unsigned long long, unsigned long long> getFermatFactors(unsigned long long n, unsigned int maxIterations)
{
// https://en.wikipedia.org/wiki/Fermat%27s_factorization_method
 
// trivial case: even numbers
if (n % 2 == 0)
return std::make_pair(2, n / 2);
 
// already a perfect square ?
auto x = (unsigned long long)sqrt(n);
if (x * x == n)
return std::make_pair(x, x);
 
while (maxIterations-- > 0)
{
x++;
auto y2 = x*x - n;
 
// optimization: all perfect squares mod 16 are always 0,1,4,9
// based on https://www.johndcook.com/blog/2008/11/17/fast-way-to-test-whether-a-number-is-a-square/
auto mod16 = y2 % 16;
if (mod16 == 0 || mod16 == 1 || mod16 == 4 || mod16 == 9)
{
// is y2 a perfect square ?
auto y = (unsigned long long)sqrt(y2);
if (y*y == y2)
// yes, found factorization
return std::make_pair(x - y, x + y);
}
}
// aborted, no factors found
return std::make_pair(1ULL, n);
}
 
std::vector<unsigned int> smallPrimes;
 
// return the largest prime factor, if x is prime then return x
unsigned long long getMaxFactor(unsigned long long x, unsigned long long minResult = 0)
{
// no use in factorizing a prime :-)
if (isPrime(x))
return x;
 
unsigned long long result = 1;
auto reduce = x;
for (auto factor : smallPrimes)
{
// at most one prime factor left ?
if (factor * factor > reduce)
break;
 
// remove current prime factor
bool foundFactor = false;
while (reduce % factor == 0)
{
result = std::max<unsigned long long>(result, factor);
reduce /= factor;
 
// abort search, prime factors will be too small
if (reduce < minResult)
return result;
 
foundFactor = true;
}
 
if (foundFactor)
{
// only a prime left ?
if (isPrime(reduce))
break;
 
// try Fermat's factorization
#define FERMAT
#ifdef FERMAT
// limit number of iterations
auto fermat = getFermatFactors(reduce, 10);
// found a factorization ?
if (fermat.first > 1)
return std::max(getMaxFactor(fermat.first), getMaxFactor(fermat.second));
#endif
}
}
 
// x can be a prime, too
return std::max(reduce, result);
}
 
int main()
{
unsigned int limit = 2000000;
std::cin >> limit;
 
// generate enough primes
primeLimit = limit + 100; // a few extra primes
fillSieve(primeLimit);
 
// copy to a dense array
smallPrimes = { 2 };
for (unsigned int i = 3; i <= primeLimit; i += 2)
if (isPrime(i))
smallPrimes.push_back(i);
 
unsigned long long sum = 0;
 
#define PARALLEL
#ifdef PARALLEL
#pragma omp parallel for reduction(+:sum) schedule(dynamic)
#endif
for (long long i = 1; i <= limit; i++) // note: Visual C++ needs a signed data type
{
//auto slow = bruteForce(i * i * i);
 
//#define FACTORIZE_LARGE
#ifdef FACTORIZE_LARGE
// takes about 3x as long as the #else version
auto factor = getMaxFactor(i * i * i + 1) - 1;
#else
// i^3+1 = (i+1)(i^2-i+1)
// getMaxFactor(i^3+1) = max(getMaxFactor(i+1), getMaxFactor(i^2-i+1))
unsigned long long a = i + 1;
unsigned long long b = i*i - i + 1;
 
auto factor2 = getMaxFactor(b);
// abort search early if no big factor possible
auto factor1 = 1ULL;
if (factor2 < a)
factor1 = getMaxFactor(a, factor2);
 
// choose the bigger factor
auto factor = std::max(factor1, factor2);
#endif
 
// remember: fast algorithm is getMaxFactor(k + 1) - 1 => I still have to subtract 1
sum += factor - 1;
}
 
std::cout << sum << std::endl;
return 0;
}

This solution contains 59 empty lines, 86 comments and 14 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in  147 seconds  (exceeding the limit of 60 seconds) on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 3 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

September 29, 2017 submitted solution
September 29, 2017 added comments

Difficulty

35% Project Euler ranks this problem at 35% (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
  the flashing problem is the one I solved most recently

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The 289 solved problems (level 11) had an average difficulty of 32.1% at Project Euler and
I scored 13,486 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.

more about me can be found on my homepage, especially in my coding blog.
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