<< problem 69 - Totient maximum | Ordered fractions - problem 71 >> |
Problem 70: Totient permutation
(see projecteuler.net/problem=70)
Euler's Totient function, phi(n) (sometimes called the phi function), is used to determine the number of positive numbers
less than or equal to n which are relatively prime to n.
For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, phi(9) = 6.
The number 1 is considered to be relatively prime to every positive number, so phi(1) = 1.
Interestingly, phi(87109) = 79180, and it can be seen that 87109 is a permutation of 79180.
Find the value of n, 1 < n < 10^7, for which phi(n) is a permutation of n and the ratio dfrac{n}{phi(n)} produces a minimum.
My Algorithm
The function phi(x)
computes phi(x):
- divide
x
by all prime numbers - count how many distinct prime numbers were divisors of
x
result = x * (1 - 1/prime1) * (1 - 1/prime2) * (1 - 1/prime3) * ...
e.g. if x == 10
result = 10 * (1 - 1/2) * (1 - 1/5) = 4
When initializing
result = x
then the multiplication can be reduced to a subtraction:whenever we find a prime factor
p
, then result -= result/p
.For phi(10) we have the prime factors 2 and 5
10 * (1 - 1/2) * (1 - 1/5)
= (10 * (1 - 1/2)) * (1 - 1/5)
= (10 - 10/2) * (1 - 1/5) →
result -= result/2
= 5 * (1 - 1/5)
= 5 - 5/5 →
result -= result/5
= 4
In this iterative algorithm,
result
is becoming smaller and smaller after each step.If it becomes obvious that the result is too small (
minQuotient < x/result = result * minQuotient < x
), then phi
aborts.This small optimization gives a 2.5x performance boost.
My
fingerprint
function was used in several Project Euler problems before, e.g. problem 49, problem 52, problem 62, ...
Alternative Approaches
I bet you can severely speed up the program by incorporating some observations:
- minimizing n/phi(n) means maximizing phi(n)
- phi(n) is maximized for prime numbers
- if i is not a prime, then phi(n) is still big if it contains just a few prime factors and those are close to sqrt(n)
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho 100 | ./70
Output:
Note: the original problem's input 10000000
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>
// precompute all primes less than sqrt(10^7)
std::vector<unsigned int> primes;
// return phi(x) if x/phi(x) <= minQuotient (else the result is undefined but > minQuotient)
unsigned int phi(unsigned int x, double minQuotient)
{
// 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;
// abort, this number can't be relevant because the quotient is already too high
if (result * minQuotient < x)
return result; // number is garbage but always >= its correct value
}
// prime number (result is still equal to x because we couldn't find any prime factors)
if (result == x)
return x - 1;
// (actually this case would be properly handled by the next if-clause, too)
// 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)
return result - result / reduced;
else
return result;
}
// count digits, two numbers have the same fingerprint if they are permutations of each other
unsigned long long fingerprint(unsigned int x)
{
unsigned long long result = 0;
while (x > 0)
{
auto digit = x % 10;
x /= 10;
unsigned long long shift = 1;
for (unsigned int i = 0; i < digit; i++)
shift *= 10;
result += shift;
}
return result;
}
int main()
{
unsigned int last;
std::cin >> last;
// step 1: the usual prime sieve
primes.push_back(2);
for (unsigned int i = 3; i*i < last; i += 2)
{
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;
}
}
// yes, we have a prime number
if (isPrime)
primes.push_back(i);
}
// step 2: analyze all phi(n)
unsigned int bestNumber = 2;
double minQuotient = 999999;
for (unsigned int n = 3; n < last; n++)
{
auto phi_n = phi(n, minQuotient);
double quotient = n / double(phi_n);
// already have a better quotient ?
if (minQuotient <= quotient)
continue;
// check if phi(n) is a permutation of n
if (fingerprint(phi_n) == fingerprint(n))
{
minQuotient = quotient;
bestNumber = n;
}
}
// print result
std::cout << bestNumber << std::endl;
return 0;
}
This solution contains 19 empty lines, 24 comments and 2 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in 1.5 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
March 11, 2017 submitted solution
May 2, 2017 added comments
Hackerrank
see https://www.hackerrank.com/contests/projecteuler/challenges/euler070
My code solves 11 out of 11 test cases (score: 100%)
Difficulty
Project Euler ranks this problem at 20% (out of 100%).
Hackerrank describes this problem as easy.
Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.
Links
projecteuler.net/thread=70 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
C# www.mathblog.dk/project-euler-70-investigate-values-of-n-for-which-φn-is-a-permutation-of-n/ (written by Kristian Edlund)
Java github.com/nayuki/Project-Euler-solutions/blob/master/java/p070.java (written by Nayuki)
Go github.com/frrad/project-euler/blob/master/golang/Problem070.go (written by Frederick Robinson)
Scala github.com/samskivert/euler-scala/blob/master/Euler070.scala (written by Michael Bayne)
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own. Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
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 |
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 |
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.
Copyright
I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.
All of my solutions can be used for any purpose and I am in no way liable for any damages caused.
You can even remove my name and claim it's yours. But then you shall burn in hell.
The problems and most of the problems' images were created by Project Euler.
Thanks for all their endless effort !!!
<< problem 69 - Totient maximum | Ordered fractions - problem 71 >> |