Problem 200: Find the 200th prime-proof sqube containing the contiguous sub-string "200"

(see projecteuler.net/problem=200)

We shall define a sqube to be a number of the form, p^2 q^3, where p and q are distinct primes.
For example, 200 = 5^2 2^3 or 120072949 = 23^2 61^3.

The first five squbes are 72, 108, 200, 392, and 500.

Interestingly, 200 is also the first number for which you cannot change any single digit to make a prime; we shall call such numbers, prime-proof.
The next prime-proof sqube which contains the contiguous sub-string "200" is 1992008.

Find the 200th prime-proof sqube containing the contiguous sub-string "200".

My Algorithm

I need two things:
- a fast primality test, suitable for moderately large numbers → Miller-Rabin test from my toolbox.
- a sorted container storing candidates in ascending order

My struct Sqube represents a number p^2 q^3. It automatically computes value == p*p * q*q*q.
Due to its member function operator<, I can insert it into a standard std::set where the left-most element will be the smallest sqube (at squbes.begin()).
The main() function starts with the two squbes { 2, 3 } and { 3, 2 }. Whenever a sqube has been processed, its "successors" { p+1, q } and { p, q+1 } will be added to the set.
(but avoid adding a sqube where p == q).

primeProof() evaluates a number whether it is prime-proof:
- convert it to a string and modify every digit separately, be careful with the first digit because it must not be zero
- run the Miller-Rabin primality test: if it returns true, then the number is not prime-proof

I added a few optimizations but they don't really improve performance:
- don't run a prime a test on even numbers (last digit is even)
- don't run a prime a test on the original number (every sqube is not prime)

Alternative Approaches

My std::set is more or less a priority queue. It contains a few thousands elements (to be precise: 15888 when the 200th sqube is found).
You can replace it by two nested loops iterating over p and q. However, you need to choose reasonable limits for p and q and sort the result afterwards.

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):
Note: Enter a value x and the program will look for the x-th sqube containing the string x

This is equivalent to
echo 10 | ./200

Output:

(please click 'Go !')

Note: the original problem's input 200 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 <string>
#include <vector>
#include <set>
#include <algorithm>
 
// ---------- Miller-Rabin prime test from my toolbox ----------
 
// 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;
if (result >= modulo)
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;
if (a >= modulo)
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)
{
// 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;
}
 
// ---------- and now my solution ----------
 
// a sqube has value = p^2 * q^3
struct Sqube
{
// note: this struct doesn't check whether p and q are different primes
const unsigned int p;
const unsigned int q;
const unsigned long long value;
 
// create a new sqube
Sqube(unsigned int p_, unsigned int q_)
: p(p_), q(q_), value((unsigned long long)p_*p_ * q_*q_*q_)
{}
 
// sort two squbes by their value, needed by std::set
bool operator<(const Sqube& other) const
{
return value < other.value;
}
};
 
// return true if changing a digit converts the number to a prime number
bool isPrimeProof(unsigned long long value)
{
auto strValue = std::to_string(value);
for (unsigned int pos = 0; pos < strValue.size(); pos++)
{
// an even number can only become prime when modifying the last digit
if (value % 2 == 0)
pos = strValue.size() - 1;
 
// change digit by digit
auto strModified = strValue;
for (auto digit = '0'; digit <= '9'; digit++)
{
// no leading zero
if (digit == '0' && pos == 0)
continue;
// last digit can't be even
if (digit % 2 == 0 && pos == strValue.size() - 1) // ASCII codes of even digits are even, too
digit++; // strictly speaking this doesn't test 2 (which is a prime)
// but the next number 3 is prime and produced the correct result
 
// no need to check the original value (a sqube is never prime)
if (digit == strValue[pos])
continue;
 
// convert from string to binary
strModified[pos] = digit;
auto modified = std::stoull(strModified);
 
// is it prime ?
if (isPrime(modified))
return false;
}
}
 
return true;
}
 
int main()
{
// count how many squbes contain "200"
unsigned int sequence = 200;
std::cin >> sequence;
 
std::string strSequence = std::to_string(sequence); // = "200"
unsigned int count = 0; // stop when count = 200
 
// the two smallest squbes, my "seed values"
std::set<Sqube> squbes = { Sqube(3, 2), Sqube(2, 3) };
 
while (true) // abort/exit condition can be found inside the loop
{
// pick smallest sqube and remove it
auto current = *(squbes.begin());
squbes.erase(squbes.begin());
 
// does it contain "200" ?
auto strCurrent = std::to_string(current.value);
if (strCurrent.find(strSequence) != std::string::npos &&
isPrimeProof(current.value))
{
// yes, another match
count++;
 
// done ?
if (count == sequence)
{
std::cout << strCurrent << std::endl;
break;
}
}
 
// add next squbes
// find a sqube with the same q but p is the next prime (not equal to q)
auto nextP = current.p + 1;
while (nextP == current.q || !isPrime(nextP))
nextP++;
squbes.insert(Sqube(nextP, current.q));
 
// find a sqube with the same p but q is the next prime (not equal to p)
auto nextQ = current.q + 1;
while (nextQ == current.p || !isPrime(nextQ))
nextQ++;
squbes.insert(Sqube(current.p, nextQ));
}
 
return 0;
}

This solution contains 42 empty lines, 61 comments and 5 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 0.11 seconds on a 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=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

August 26, 2017 submitted solution
August 26, 2017 added comments

Difficulty

65% Project Euler ranks this problem at 65% (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

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
The 270 solved problems (level 10) had an average difficulty of 31.3% at Project Euler and
I scored 13,386 points (out of 15600 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 !