Problem 268: Counting numbers with at least four distinct prime factors less than 100

(see projecteuler.net/problem=268)

It can be verified that there are 23 positive integers less than 1000 that are divisible by at least four distinct primes less than 100.

Find how many positive integers less than 10^16 are divisible by at least four distinct primes less than 100.

My Algorithm

There are 25 primes less than 100.
I start a loop running from 0 to 2^25. I multiply all the n-th primes if the n-th bit of the counter (named mask) is set.
If mask has at least four bits set then the product of those primes needs to be further analyzed:

I have to be careful to avoid counting the same numbers twice:
for example 2310 is divisible by 2,3,5,7 and 11 (in fact 2*3*5*7*11 = 2310). But 2310 is also divisible by (2,3,5,7), (2,3,5,11), (2,5,7,11) and (3,5,7,11).
That means (2,3,5,7,11) "overlaps" all combinations of its primes where I take one prime away.
The tetrahedral numbers can compute this "overlap" (see en.wikipedia.org/wiki/Tetrahedral_number) together with
the inclusion-exclusion principle (see en.wikipedia.org/wiki/Inclusion–exclusion_principle).

In plain English: I add the count of numbers divisible by 4 primes, subtract the count divisible by 5 primes, add 6s, subtract 7s, etc.
→ if numBits is even, then add else subtract, the fast way is just to look at the lowest bit of the numBits counter

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 "4 25 1000" | ./268

Output:

(please click 'Go !')

Note: the original problem's input 4 25 10000000000000000 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>
 
// all 25 primes less than 100
std::vector<unsigned int> primes = { 2, 3, 5, 7, 11,
13, 17, 19, 23, 29,
31, 37, 41, 43, 47,
53, 59, 61, 67, 71,
73, 79, 83, 89, 97 };
 
// count numbers with at least "minPrimes" prime factors
// slow version - helping me while debugging the fast code below
unsigned int bruteForce(unsigned int limit, unsigned int minPrimes)
{
unsigned int result = 0;
for (unsigned int i = 1; i <= limit; i++)
{
// count distinct prime factors
unsigned int numPrimeFactors = 0;
for (auto p : primes)
if (i % p == 0)
numPrimeFactors++;
 
// at least four prime factors ?
if (numPrimeFactors >= minPrimes)
result++;
}
return result;
}
 
// return the n-th tetrahedral number
unsigned int tetrahedral(unsigned int n)
{
// see https://en.wikipedia.org/wiki/Tetrahedral_number
return n * (n + 1) * (n + 2) / 6;
}
 
int main()
{
// at least 4 primes, search up to 10^16
unsigned int minPrimes = 4;
unsigned int numPrimes = primes.size();
unsigned long long limit = 10000000000000000ULL;
 
std::cin >> minPrimes >> numPrimes >> limit;
 
// find result for small limits
//std::cout << bruteForce(limit, minPrimes) << std::endl;
 
// precompute tetrahedral numbers
std::vector<unsigned long long> count(numPrimes + 1, 0);
for (unsigned int i = minPrimes; i < count.size(); i++)
count[i] = tetrahedral(i - minPrimes + 1); // same as choose(i - 1, minPrimes - 1)
 
unsigned long long sum = 0;
unsigned int maxMask = (1 << numPrimes) - 1;
for (unsigned int mask = 0; mask < maxMask; mask++)
{
// multiply all primes
unsigned long long product = 1;
// but don't exceed 10^16
bool tooLarge = false;
// at least four bits must be set (=> pick at least four different primes)
unsigned int numBits = 0;
// look at each prime
for (unsigned int current = 0; current < numPrimes; current++)
{
// is this prime used ?
unsigned int bitpos = 1 << current;
if ((mask & bitpos) == 0)
continue;
 
// include this prime factor
numBits++;
product *= primes[current];
 
// avoid too large products (beware of 64 bit overflows ...)
if (product > limit)
{
tooLarge = true;
 
// speed optimization:
// if the next mask has the same number of bits set (or more), then it will exceed the limit as well
// because the primes are getting larger
// => get closer to the next number with the less bits set
// add the lowest set bit to the current number
// => worst case: same number of bits if the last bit is a single bit
// => best case: several bits next to the lowest bit are set to and all flip to zeros (and a zero becomes a one)
auto withLowestBit = mask & (mask - 1);
auto lowestBit = mask - withLowestBit;
mask += lowestBit;
// hope for the best case ...
while (mask & (lowestBit << 1))
{
lowestBit <<= 1;
mask += lowestBit;
}
 
mask--; // because the for-loop will add 1 in each iteration
break;
}
}
 
// skip if too few primes or their product became too large
if (tooLarge || numBits < minPrimes)
continue;
 
// count numbers divisible by the current primes
auto divisible = limit / product;
// they were part of products with less primes, too, hence compute the "overlap"
divisible *= count[numBits];
 
// add if even number of primes, else subtract (inclusion-exclusion principle)
if (numBits & 1)
sum -= divisible;
else
sum += divisible;
}
 
// that's it !
std::cout << sum << std::endl;
return 0;
}

This solution contains 17 empty lines, 31 comments and 2 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 0.9 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(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 27, 2017 submitted solution
September 27, 2017 added comments

Difficulty

70% Project Euler ranks this problem at 70% (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
701 702 703 704 705 706 707 708 709 710 711 712 713 714 715 716 717 718 719 720 721 722 723 724 725
726 727 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750
751 752 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 771 772 773 774 775
776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 795 796 797 798 799 800
801 802 803 804 805 806 807 808 809 810 811 812 813 814 815 816
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.

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 !