<< problem 341 - Golomb's self-describing sequence Matrix Sum - problem 345 >>

Problem 343: Fractional Sequences

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:

• start with numerator = 1 and denominator = k, then repeat until denominator == 1:
• add 1 to the numerator and subtract 1 from the denominator
• call gcd to check whether they share a factor, if yes, divide both by that factor
• the sum numerator + denominator remains constant if gcd == 1
• the gcd is always 1 if numerator + denominator is a prime number
• if gcd > 1 then gcd == numerator
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:
• my standard prime sieve finds all primes below 2 * 10^6 → the largest factor of a prime is the prime itself
• trial division using these primes
• divide the current number by each factor to reduce it as fast as possible
• if the reduced number becomes 1, then I'm finished
• if the square of the current prime is larger than the reduced number, then the reduced number must be a prime
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:

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. Or just jump to my GitHub repository.

       #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).
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 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

Difficulty

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 [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200
 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300
 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400
 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500
 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600
 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 642 643 644 645 646 647 648 649 650 651 652 653 654 655 656 657 658 659 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702 703 704 705 706 707 708
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 341 - Golomb's self-describing sequence Matrix Sum - problem 345 >>
more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
thanks to the KaTeX team for their great typesetting library !