<< problem 336 - Maximix Arrangements Fractional Sequences - problem 343 >>

# Problem 341: Golomb's self-describing sequence

The Golomb's self-describing sequence G(n) is the only nondecreasing sequence of natural numbers such that n appears exactly G(n) times in the sequence.
The values of G(n) for the first few n are

n123456789101112131415...
G(n)122334445556666...

You are given that G(10^3) = 86, G(10^6) = 6137.
You are also given that sum{G(n^3)} = 153506976 for 1 <= n < 10^3.

Find sum{G(n^3)} for 1 <= n < 10^6.

# My Algorithm

I brute forced the first values up to 100 (far more can be found here: oeis.org/A001462/b001462.txt) and observed a certain pattern:

• G(1) = 1 and there is exactly one G(x) = 1
• G(2) = 2 and there are two x such that G(x) = 2 (it's 2 and 3)
• G(4) = 3 and there are two x such that G(x) = 3 (it's 4 and 5)
• G(6) = 4 and there are three x such that G(x) = 4 (it's 6, 7 and 8)
• G(9) = 5 and there are three x such that G(x) = 5 (it's 9, 10 and 11)
• G(12) = 6 and there are four x such that G(x) = 6 (it's 12, 13, 14 and 15)
• ...
→ the number of identical G(x) increments by one after sum{x * G(x)} steps:
1 * G(1) = 1 * 1 = 1 → all G(x > 1) appear at least twice
1 * G(1) + 2 * G(2) = 1 + 4 = 5 → all G(x > 5) appear at least three times
1 * G(1) + 2 * G(2) + 3 * G(3) = 1 + 4 + 6 = 11 → all G(x > 11) appear at least four times
I store these products in products[] (or compute them on-the-fly in the LOW_MEMORY code path).

It turns out that sum{x * G(x)} > {10^6}^3 for x approx 10 million.
Brute-forcing the first few million values of the Golomb sequence is no problem using the "recursive" Wikipedia formula:
a(1) = 1
a(n+1) = 1 + a (n + 1 - a( a(n) ) )
which can be simplified for a(n)
a(n) = 1 + a (n - a( a(n - 1) ) )

Once I know the upper limit of the number of elements I make use of a second observation:
I already demonstrated that G(1..3) is sufficient to know that any number 6 <= x <= 11 maps to a value to appears three times in the Golomb sequence.
But I still don't know which number ! (it can be 4 or 5)
The sums G(1) + G(2) = 3 and G(1) + G(2) + G(3) = 5 have a nice property:
I know there 6 values between those sums (it's 6 to 11). If I find the relative position ratio between the relevant products, then the sum at the same relative position contains the correct value.

Let's compute G(x) for x = 10:
from = products[2] = 5
to = products[3] = 11

ratio = dfrac{10 - from}{to - from} = dfrac{10 - 5}{11 - 5} = dfrac{5}{6} approx 0.83333

low = sums[2] = G(1) + G(2) = 3
high = sums[3] = G(1) + G(2) + G(3) = 5

G(10) = low + roundUp((high - low) * ratio) = 3 + roundUp((5 - 3) * 0.83333) = 5

(that's good old en.wikipedia.org/wiki/Linear_interpolation)

My code became a mess when I replaced the temporary containers sums and produces by variables.
However, it cut the memory usage by 50%: from about 270 MByte (slightly above the inofficial limit of 256 MByte) down to 135 MByte.
#define LOW_MEMORY is also about twice as fast.

## Note

The Wikipedia page describes an O(1) formula which is unfortunately only asymptotic correct:
the result is more or less pretty close to the actual value but can't be used to solve this problem.
Nevertheless, the fast function helped me a lot in my debugging session when some intermediate values were "strange".

Let's be honest: I discovered the connections between G(x) the sums and the products more or less by chance when I spent time on the commuter train home.
It was trial'n'error all the way - and therefore it's definitely not amongst my favorite problems.

# 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 1000 | ./341

Output:

Note: the original problem's input 1000000 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>
#include <cmath>

// find the asymptotic value, quite often off by a few numbers (first four digits tend to be okay)
// note: unused code, but helped me a lot during debugging
double fast(unsigned long long n)
{
// golden ratio
const auto phi = (1 + sqrt(5.0)) / 2;
const auto constant = pow(phi, 2 - phi);
// see https://en.wikipedia.org/wiki/Golomb_sequence
return constant * pow(n, phi - 1);
}

int main()
{
unsigned int limit = 1000000;
std::cin >> limit;

// (10^6)^3 = 10^18
auto cubicLimit = (unsigned long long) limit * limit * limit;

// index 0 is not used, golomb(1) = 1
std::vector<unsigned long long> golomb = { 0, 1 };

#define LOW_MEMORY
#ifdef LOW_MEMORY
// precompute golomb[i] = G(i)
// stop when 1*G(1) + 2*G(2) + 3*G(3) + ... + i*G(i) >= 10^18
unsigned long long products = 1;
for (unsigned long long i = 2; products < cubicLimit; i++)
{
// https://en.wikipedia.org/wiki/Golomb_sequence
auto current = 1 + golomb[i - golomb[golomb[i - 1]]];
golomb.push_back(current);
products += current * i;
}

#else

// precompute golomb[i] = G(i)
// and sums[i]     =   G(1) +   G(2) +   G(3) + ... +   G(i)
// and products[i] = 1*G(1) + 2*G(2) + 3*G(3) + ... + i*G(i)
std::vector<unsigned long long> sums     = { 0, 1 };
std::vector<unsigned long long> products = { 0, 1 };
// stop when products[i] >= 10^18
for (unsigned long long i = 2; products.back() < cubicLimit; i++)
{
auto current = 1 + golomb[i - golomb[golomb[i - 1]]];

golomb  .push_back(current);
sums    .push_back(current     + sums.back());
products.push_back(current * i + products.back());
}
#endif

// will contain the result
unsigned long long sum = 0;

#ifdef LOW_MEMORY
unsigned long long lastSums     = 0;
unsigned long long sums         = 1;

unsigned long long lastProducts = 0;
products     = 1;
#endif

// find products[index - 1] < i <= products[index]
auto index = 1;
for (unsigned long long i = 1; i < limit; i++)
{
// n = i^3
auto n = i * i * i;

// find products[index - 1] < i <= products[index]
#ifdef LOW_MEMORY
while (products < n)
{
index++;
lastSums     = sums;
sums        += golomb[index];
lastProducts = products;
products    += golomb[index] * index;
}
#else
while (products[index] < n)
index++;
#endif
// note: n will be in ascending order, therefore I re-use index from previous iterations
//       in most cases it will be already the correct value

// find linear interpolation between products[index - 1] and products[index]
#ifdef LOW_MEMORY
auto from = lastProducts;
auto to   = products;
#else
auto from = products[index - 1];
auto to   = products[index];
#endif
auto ratio = (n - from) / double(to - from);

// and apply it to sums[index - 1] and sums[index]
#ifdef LOW_MEMORY
auto low  = lastSums;
auto high = sums;
#else
auto low  = sums[index - 1];
auto high = sums[index];
#endif

// round up
auto offset = ceil((high - low) * ratio);
// note: convert to integer as soon as possible to avoid losing digits due to double's limited precision
auto result = (unsigned long long)offset + low;

// finished another number ...
sum += result;
}

// solved another problem !
std::cout << sum << std::endl;
return 0;
}


This solution contains 18 empty lines, 25 comments and 18 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.
Peak memory usage was about 133 MByte.

(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

August 15, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 45% (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
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.

 << problem 336 - Maximix Arrangements Fractional Sequences - problem 343 >>
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 !