Problem 278: Linear Combinations of Semiprimes

(see projecteuler.net/problem=278)

Given the values of integers 1 < a_1 < a_2 < ... < a_n, consider the linear combination
q_1 a_1 + q_2 a_2 + ... + q_n a_n = b, using only integer values q_k >= 0.

Note that for a given set of a_k, it may be that not all values of b are possible.
For instance, if a_1 = 5 and a_2 = 7, there are no q_1 >= 0 and q_2 >= 0 such that b could be
1, 2, 3, 4, 6, 8, 9, 11, 13, 16, 18 or 23.
In fact, 23 is the largest impossible value of b for a_1 = 5 and a_2 = 7.
We therefore call f(5, 7) = 23.
Similarly, it can be shown that f(6, 10, 15) = 29 and f(14, 22, 77) = 195.

Find sum{f(pq,pr,qr)}, where p, q and r are prime numbers and p < q < r < 5000.

My Algorithm

Assume I have two numbers x and y where gcd(x,y)=1.
The value m = xy - x - y can't be represented with some coefficients m = px + qy because:
xy - x - y = px + qy
xy = px + qy + x + y
xy = (p+1)x + (q+1)y

xy is a multiple of x and (p+1)x is a multiple of x, hence (q+1)y should be a multiple of x, too.
xy is a multiple of y and (q+1)y is a multiple of y, hence (p+1)x should be a multiple of y, too.
But gcd(x,y)=1 so y can't be a multiple of x and therefore q+1 should be a multiple of x.
And for the same reason x can't be a multiple of y and therefore p+1 should be a multiple of y.
Possible values for q+1 would be x, 2x, 3x, ... (and for p+1: y, 2y, 3y, ...)
If I assume the lowest value p+1=y and q+1=x then the equation becomes
xy = y * x + x * y
xy = 2xy → contradition !

Therefore m = xy - x - y actually can't be represented with some coefficients m = px + qy.

With three numbers x,y,z and gcd(x,y,z)=1 the idea is very similar:
if there would be some coefficients p, q and r such that m = pxy + qxz + ryz represents m = 2xyz - xy - xz - yz then
(2xyz - xy - xz - yz) mod x = -yz
pxy + qxz + ryz = 2xyz - xy - xz - yz
2xyz = pxy + qxz + ryz + xy + xz + yz
2xyz = (py+qz+y+z)x + (rz + z)y
Hence rz + z = (r+1)z must be a multiple of x. z can't be such a multiple (because of gcd(x,y,z) = 1).
The same idea for y and z gives that p+1 must be a multiple of z and q+1 a multiple of y.
As before - if I choose the smallest possible p+1=z, q+1=y and r+1=x:
2xyz = zxy + yxz + xyz + xy + xz + yz
2xyz = 3xyz + xy + xz + yz → contradiction

I didn't come up with the full solution, I just know how to use search engines :-;
I found the problem in the 24th International Mathemtical Olympiad held 1983 in Paris, France
somewhat cryptic solution: www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln833.html
I stumbled across it while reading the German Wikipedia de.wikipedia.org/wiki/M├╝nzproblem
unfortunately, the English page misses that special case en.wikipedia.org/wiki/Coin_problem
but it can be derived from their n=2 explanations (pretty much what I have done above)

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 10 | ./278

Output:

(please click 'Go !')

Note: the original problem's input 5000 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>
 
int main()
{
unsigned int limit = 5000;
std::cin >> limit;
 
// simple prime sieve from my toolbox
std::vector<unsigned long long> primes = { 2 };
for (unsigned int i = 3; i <= limit; i += 2)
{
bool isPrime = true;
 
// test against all prime numbers we have so far (in ascending order)
for (auto x : primes)
{
// prime is too large to be a divisor
if (x*x > i)
break;
 
// divisible => not prime
if (i % x == 0)
{
isPrime = false;
break;
}
}
 
// yes, we have a prime
if (isPrime)
primes.push_back(i);
}
 
// all combinations of primes
unsigned long long sum = 0;
for (size_t i = 0; i < primes.size(); i++)
for (size_t j = i + 1; j < primes.size(); j++)
for (size_t k = j + 1; k < primes.size(); k++)
{
auto p = primes[i];
auto q = primes[j];
auto r = primes[k];
sum += 2*p*q*r - p*q - p*r - q*r;
}
 
std::cout << sum << std::endl;
return 0;
}

This solution contains 7 empty lines, 6 comments and 2 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 0.11 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

July 11, 2017 submitted solution
July 11, 2017 added comments

Difficulty

50% Project Euler ranks this problem at 50% (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
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
The 306 solved problems (that's level 12) had an average difficulty of 32.5% 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 !