<< problem 240 - Top Dice Squares under a hyperbola - problem 247 >>

# Problem 243: Resilience

A positive fraction whose numerator is less than its denominator is called a proper fraction.
For any denominator, d, there will be d-1 proper fractions; for example, with d=12:
1/12 , 2/12 , 3/12 , 4/12 , 5/12 , 6/12 , 7/12 , 8/12 , 9/12 , 10/12 , 11/12 .

We shall call a fraction that cannot be cancelled down a resilient fraction.
Furthermore we shall define the resilience of a denominator, R(d), to be the ratio of its proper fractions that are resilient;
for example, R(12) = 4/11 .
In fact, d = 12 is the smallest denominator having a resilience R(d) < 4/10 .

Find the smallest denominator d, having a resilience R(d) < 15499/94744 .

# My Algorithm

If a fraction is not a proper fraction then gcd(numerator, denominator) > 1, that means both can be divided by the same integer.
The Euler totient function phi(x) returns how many numbers 0 < y < x are relatively prime to x.
phi(12) = 4 because 12 is relatively prime to 1, 5, 7 and 11.

And indeed, 1/12, 5/12, 7/12 and 11/12 are the only proper fraction where the denominator is 12.
Therefore the program should find a denominator d such that
dfrac{phi(d)}{d-1} < dfrac{15499}{94744}

The code of my phi() function is shared with problem 214 (and was originally written for problem 70).
I had to change it's parameter type from unsigned int to unsigned long long but everything else is identical to problem 214.

Unfortunately it's much too slow to check every single number, beginning with 12 (hint: the correct solution is a nine-digit number).
But I did exactly that "to get a feeling" - and obviously stopped after a few hundred numbers.
It turns out that phi(x)/(x-1) is low when x is the product of the first n prime numbers: x_n = 2 * 3 * 5 * 7 * ....

My current implementation multiplies all prime numbers until phi(x_n)/(x_n-1) is smaller than 15499/94744 (x_n is my variable current).
Then it reverts the last step (now phi(x_{n-1})/(x_{n-1}-1) is bigger than 15499/94744). current is updated to hold x_{n-1}.
And finally all multiples of x_{n-1} (→ current) are compared against the ratio until it's below 15499/94744.

I consider the last step to be a "lucky guess". Yes, it produces the correct result but my mathematical knowledge isn't sufficient to know why.

# 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>

std::vector<unsigned int> primes;

// return phi(x) if x/phi(x) <= minQuotient (else the result is undefined but > minQuotient)
unsigned long long phi(unsigned long long x)
{
// totient function can be computed by finding all prime factors p
// and subtracting them from x
auto result  = x;
auto reduced = x;
for (auto p : primes)
{
// prime factors have to be p <= sqrt
if (p*p > reduced)
break;

// not a prime factor ...
if (reduced % p != 0)
continue;

// prime factors may occur multiple times, remove them all
do
{
reduced /= p;
} while (reduced % p == 0);

// but subtract from result only once
result -= result / p;
}

// we only checked prime factors <= sqrt(x)
// there might exist one (!) prime factor > sqrt(x)
// e.g. 3 is a prime factor of 6, and 3 > sqrt(6)
if (reduced > 1)
result -= result / reduced;

return result;
}

// return true if a1/b1 < a2/b2
bool isLess(unsigned long long a1, unsigned long long b1, unsigned long long a2, unsigned long long b2)
{
// a1/b1 < a2/b2 is the same as
// a1*b2 < a2*b1 if all numbers are positive
return a1*b2 < a2*b1;
}

int main()
{
unsigned int numerator   = 15499;
unsigned int denominator = 94744;

unsigned long long current = 1;
for (unsigned int i = 2; ; i++)
{
bool isPrime = true;

// test against all prime numbers we have so far (in ascending order)
for (auto p : primes)
{
// next prime is too large to be a divisor ?
if (p*p > i)
break;

// divisible ? => not prime
if (i % p == 0)
{
isPrime = false;
break;
}
}

if (!isPrime)
continue;

// yes, we have a prime number
primes.push_back(i);

// multiply prime numbers until the ratio becomes too small
current *= i;
if (isLess(phi(current), current - 1, numerator, denominator))
break;
}

// undo last prime
current /= primes.back();

// simple iterative scan
for (unsigned int i = 1; ; i++)
{
auto next = current * i;

// below threshold ?
if (isLess(phi(next), next - 1, numerator, denominator))
{
std::cout << next << std::endl;
break;
}
}

return 0;
}


This solution contains 19 empty lines, 21 comments and 2 preprocessor commands.

# Interactive test

This feature is not available for the current problem.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 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

June 19, 2017 submitted solution

# Difficulty

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

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.
red problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte.

Please click on a problem's number to open my solution to that problem:

 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
The 235 solved problems (level 9) had an average difficulty of 29.1% at Project Euler and
I scored 13,183 points (out of 15300 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.
Look at my progress and performance pages to get more details.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

 << problem 240 - Top Dice Squares under a hyperbola - problem 247 >>
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 !