Problem 284: Steady Squares

(see projecteuler.net/problem=284)

The 3-digit number 376 in the decimal numbering system is an example of numbers with the special property that its square ends with the same digits: 376^2 = 141376.
Let's call a number with this property a steady square.

Steady squares can also be observed in other numbering systems. In the base 14 numbering system, the 3-digit number c37 is also a steady square: c37^2 = aa0c37,
and the sum of its digits is c+3+7=18 in the same numbering system.
The letters a, b, c and d are used for the 10, 11, 12 and 13 digits respectively, in a manner similar to the hexadecimal numbering system.

For 1 <= n <= 9, the sum of the digits of all the n-digit steady squares in the base 14 numbering system is 2d8 (582 decimal).
Steady squares with leading 0's are not allowed.

Find the sum of the digits of all the n-digit steady squares in the base 14 numbering system for
1 <= n <= 10000 (decimal) and give your answer in the base 14 system using lower case letters where necessary.

My Algorithm

This problem bugged me for a while ... the amount of code written for my solution was astonishing (and only a fraction of it survived ...).

Nevertheless, here's how I did it:
Within a few seconds I found exactly three single steady digits:
1^2 mod 14 = 1
7^2 mod 14 = 49 mod 14 == 7
8^2 mod 14 = 64 mod 14 == 8
(zero was disallowed by the problem statement)

Continuing with two digits caused a problem:
I couldn't find a digit that can be prepended to 1 - but there are such digits for 7 and 8.
As it turns out, I can "extend" 7 and 8 basically to infinitely long steady squares.
And there is always exactly one digit I can prepend to a steady square such that it becomes a larger steady square.
The BigNum class from my toolbox already supported different bases.
Adding a toString() method for base 14 was straightforward.
Then I wrote isSteady(): it multiplies the number with itself and in each step checks whether the last digits remain unchanged.

The obvious brute force algorithm is as follows:

It works pretty well for up to 100 digits.

The Wikipedia page en.wikipedia.org/wiki/Automorphic_number focusses on an algorithm based on the Chinese Remainder Theorem.
And I don't really like it ... so it looked at the output and saw that there is a certain relationship between the first steady squares of 7:
s_1 = 7_14 = 7_10
s_2 = 37_14 = 49_10 = 7^2
s_3 = c37_14 = 2401_10 = 49^2 = 7^4
s_4 = 0c37_14 = 2401_10 = 49^2 = 7^4
s_5 = a0c37_14 = 386561_10 = 7^5 * 23
Unfortunately it doesn't work for s_4 and s_5 anymore, so I stopped working and solved other Project Euler problems.
Problem 455 isn't related to the current problem but it reminded my that modulo often works in strange ways.
And indeed:
s_4 = s^2_3 mod 14^4 = 2401^2 mod 14^4 = 5764801 mod 38416 == 2401
s_5 = s^2_4 mod 14^5 = 2401^2 mod 14^5 = 2401 mod 537824 == 386561
So I have a simple way to generate all digits step-by-step.

My multiplication code was extremely slow: the square of a number with m digits has 2m digits.
But I only need m + 1 of those - so I wrote multiplyLow() which stops after enough digits were computed and discards the rest.

The Wikipedia mentions a smart way to generate the second automorphic number: "the sum of two automorphic numbers is 10^k + 1".
If I have one automorphic number s then the other one is S = 10^k + 1 - s. My findOther() function does exactly that.
The solution for 1000 digits takes half a second and the solution for 10000 digits is found after about 8 minutes.

But another formula on the Wikipedia page caught my attention (I didn't realize its importance when I read it the first time):
n' = (3n^2 - 2n^3) mod 10^{2k}
Starting with an automorphic number n I get a new automorphic number n' that has twice as many digits.
Even though this formula is intended for base 10 numbers it still works for base 14 (I have no idea about other bases).

My BigNum class supports only positive numbers and 3n^2 - 2n^3 is always negative for n > 1.
So I had to rewrite the formula using the relationship -a mod b = (b-a) mod b:
n' = (3n^2 - 2n^3) mod 10^{2k}
n' = -(2n^3 - 3n^2) mod 10^{2k}
n' = (10^{2k} - (2n^3 - 3n^2)) mod 10^{2k}

See my fastDoubling() function → it finishes after 1.1 seconds.

The final step is to compute the digit sum of my steady squares:

Alternative Approaches

Dedicated large integer library such as GMP are much faster than my simple BigNum class.
With a clever use of residues you can get away with simple integer arithmetic as well and solve the problem in less than 0.1 seconds.

Note

This solution currently shares the "first place" of most code I needed to solve a problem, see my ranking.
Despite all the code (and time) needed to crack the problem, I felt that this was a really nice one.

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 9 | ./284

Output:

(please click 'Go !')

Note: the original problem's input 10000 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. Or just jump to my GitHub repository.

#include <iostream>
#include <vector>
#include <string>
 
// ---------- code taken from my toolbox, with a few modifications / extensions, see explanations above ----------
 
// store single digits with lowest digits first
// e.g. 1024 (in base 14) is stored as { 4,2,0,1 }
// only non-negative numbers supported
struct BigNum : public std::vector<unsigned char>
{
static const unsigned int MaxDigit = 14;
 
// store a non-negative number
BigNum(unsigned int x = 0)
{
do
{
push_back(x % MaxDigit);
x /= MaxDigit;
} while (x > 0);
}
 
// convert from base 14 to string
std::string toString() const
{
const char digits[] = "0123456789abcd";
 
std::string result;
for (auto x : *this)
{
// process a bucket
for (unsigned int shift = 1; shift < MaxDigit; shift *= MaxDigit)
{
auto digit = (x / shift) % MaxDigit;
result.insert(0, 1, digits[digit]);
}
}
 
// remove leading zeros
while (result.size() > 1 && result.front() == '0')
result.erase(0, 1);
 
return result;
}
 
// add two big numbers
BigNum operator+(const BigNum& other) const
{
auto result = *this;
// add in-place, make sure it's big enough
if (result.size() < other.size())
result.resize(other.size(), 0);
 
unsigned int carry = 0;
for (size_t i = 0; i < result.size(); i++)
{
carry += result[i];
if (i < other.size())
carry += other[i];
else
if (carry == 0)
return result;
 
if (carry < MaxDigit)
{
// no overflow
result[i] = carry;
carry = 0;
}
else
{
// yes, we have an overflow
result[i] = carry - MaxDigit;
carry = 1;
}
}
 
if (carry > 0)
result.push_back(carry);
 
return result;
}
 
// multiply a big number by an integer
BigNum operator*(unsigned int factor) const
{
// faster multiplication possible ?
if (factor == 0)
return 0;
if (factor == 1)
return *this;
if (factor == MaxDigit)
{
if (size() == 1 && operator[](0) == 0)
return 0;
 
auto result = *this;
result.insert(result.begin(), 0);
return result;
}
 
unsigned int carry = 0;
auto result = *this;
for (auto& i : result)
{
carry += i * factor;
i = carry % MaxDigit;
carry /= MaxDigit;
}
// store remaining carry in new digits
while (carry > 0)
{
result.push_back(carry % MaxDigit);
carry /= MaxDigit;
}
 
return std::move(result);
}
 
// subtract a smaller-or-equal number
BigNum operator-(const BigNum& other) const
{
BigNum result = *this;
int borrow = 0;
for (size_t i = 0; i < result.size(); i++)
{
int diff = (int)result[i] - borrow;
if (i < other.size())
diff -= other[i];
else
if (borrow == 0)
break;
 
if (diff < 0)
{
borrow = 1;
diff += MaxDigit;
}
else
borrow = 0;
 
result[i] = diff;
}
 
// no high zeros
while (result.size() > 1 && result.back() == 0)
result.pop_back();
 
return result;
}
 
// multiply two big numbers
BigNum operator*(const BigNum& other) const
{
if (size() < other.size())
return other * *this;
 
// multiply single digits of "other" with the current object
BigNum result = 0;
for (int i = (int)other.size() - 1; i >= 0; i--)
result = result * MaxDigit + (*this * other[i]);
 
return std::move(result);
}
 
// ---------- problem-specific code (still part of BigNum class) ----------
 
// multiply by a big number and keep only the lowest digits
void multiplyLow(const BigNum& other, size_t numDigits)
{
BigNum result;
result.resize(numDigits, 0);
 
// multiply single digits of "other" with the current object
unsigned int carry = 0;
for (size_t i = 0; i < other.size() && i < numDigits; i++)
{
carry = 0;
for (size_t j = 0; i + j < numDigits; j++)
{
// multiply two digits (if possible) and add carry
if (j >= size())
carry += result[i + j];
else
carry += result[i + j] + other[i] * operator[](j);
 
// store lowest digit of product and keep "carrying on" :-)
result[i + j] = carry % MaxDigit;
carry /= MaxDigit;
}
}
 
*this = std::move(result);
}
 
// try to find the square and check whether its lower digits remain unchanged
bool isSteady() const
{
// multiply the number with itself and abort as soon as possible
BigNum square = 0;
 
for (size_t pos = 0; pos < size(); pos++)
{
// multiply with next digit
auto digit = operator[](pos);
if (digit > 0)
{
auto product = *this * digit;
 
for (size_t appendZeros = 0; appendZeros < pos; appendZeros++)
product.insert(product.begin(), 0);
square = square + product;
}
 
// mismatch ?
if (digit != square[pos])
return false;
}
 
// yes, a steady square
return true;
}
};
 
// ---------- problem-specific code (not part of BigNum class) ----------
 
// prepend every possible digit until a steady square is fouund
BigNum bruteForce(const BigNum& number)
{
auto next = number;
next.push_back(0);
 
// prepend a digit (remember: digits are stored in reverse order)
for (unsigned int digit = 0; digit < BigNum::MaxDigit; digit++)
{
next.back() = digit;
// steady ?
if (next.isSteady())
break;
}
return next;
}
 
// compute steady square where the right-most digits are "number", result has exactly "numDigits" digits
BigNum fastDoubling(const BigNum& number, unsigned int numDigits)
{
// https://en.wikipedia.org/wiki/Automorphic_number
// n' = (3n^2 - 2n^3) mod 14^2k => a negative number, which my simple BigNum class doesn't support yet
// n' = 14^2k - (2n^3 - 3n^2) mod 14^2k
 
// after each iteration, the number of steady digits will double
auto current = number;
while (current.size() < numDigits)
{
// new number will have twice as many digits
auto twiceDigits = 2 * current.size();
 
auto square = current * current; // n^2
auto cube = square * current; // n^3
auto diff = cube * 2 - square * 3; // 3n^2 - 2n^3
 
diff.resize(twiceDigits); // (3n^2 - 2n^3) mod 14^twiceDigits
 
BigNum largeOne = 0;
largeOne.resize(twiceDigits, 0); // 14*twiceDigits zeros
largeOne.push_back(1); // and prepend a "1" => 14^twiceDigits
 
current = largeOne - diff; // 14^twiceDigits - (3n^2 - 2n^3) mod 14^twiceDigits
}
 
// remove superfluous leading digits (current has always 2^iterations digits, e.g. 16384 which will be reduced to 10000)
current.resize(numDigits);
return current;
}
 
// given an automorphic number, find its sibling
BigNum findOther(const BigNum& number)
{
// eight = (14^maxDigits + 1) - seven
BigNum one0one = 1;
one0one.resize(number.size(), 0);
one0one.push_back(1);
return one0one - number;
}
 
int main()
{
unsigned int maxDigits = 10000;
std::cin >> maxDigits;
 
// manually found those two number to be the only steady digits mod 14 (except for 1, see below)
BigNum seven = 7;
BigNum eight = 8;
 
//#define BRUTEFORCE
#ifdef BRUTEFORCE
// brute-force
for (unsigned int i = 2; i < maxDigits; i++)
{
seven = bruteForce(seven);
eight = bruteForce(eight);
}
#endif
 
//#define POWER
#ifdef POWER
for (unsigned int numDigits = 2; numDigits <= maxDigits; numDigits++)
{
// seven = seven * seven, reduced to numDigits
seven.multiplyLow(seven, numDigits);
//std::cout << numDigits << " " << std::flush;
}
eight = findOther(seven);
#endif
 
#define DOUBLING
#ifdef DOUBLING
// use equation (3n^2 - 2n^3) mod 14^k until I have 10000 digits
seven = fastDoubling(7, maxDigits);
eight = findOther(seven); // faster than eight = fastDoubling(8, maxDigits);
#endif
 
// now I have the 10000 steady digits ending with 7 and 8, let's find their digit sum !
 
// 1 is a steady digit, too, but a strange special case: 1^2 % 14 = 1
unsigned int sum = 1;
// prepending any digit to 1 doesn't produce a steady square
 
// okay, let's proceed with ...7 and ...8
for (unsigned int i = 0; i < maxDigits; i++)
{
// last digit appears in 10000 steady squares, next to last in 9999 steady squares, ...
auto howOften = maxDigits - i;
sum += howOften * seven[i];
sum += howOften * eight[i];
 
// but don't count those numbers having a leading zero
if (seven[i] == 0)
for (unsigned int j = 0; j < i; j++) // could be more efficient ... but it's still too fast to be measurable
sum -= seven[j];
 
if (eight[i] == 0)
for (unsigned int j = 0; j < i; j++)
sum -= eight[j];
}
 
// print digit sum in base 14
BigNum sumBase14 = sum;
std::cout << sumBase14.toString() << std::endl;
return 0;
}

This solution contains 56 empty lines, 56 comments and 10 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 1.1 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

October 13, 2017 submitted solution
October 13, 2017 added comments

Difficulty

55% Project Euler ranks this problem at 55% (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
626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 642 643 644 645 646 647 648 649 650
651 652 653 654 655 656 657 658 659 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675
676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700
701 702 703 704 705 706 707 708 709 710 711 712 713 714 715 716 717 718 719 720 721 722 723 724 725
726 727 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750
751 752 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 771 772 773 774 775
776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 795 796 797 798 799 800
801 802 803 804 805 806
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.

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 !