Problem 491: Double pandigital number divisible by 11

(see projecteuler.net/problem=491)

We call a positive integer double pandigital if it uses all the digits 0 to 9 exactly twice (with no leading zero).
For example, 40561817703823564929 is one such number.

How many double pandigital numbers are divisible by 11?

My Algorithm

bruteForce counts all permutations which are divisible by 11.
It finds the correct result in a reasonable amount of time when the highest digit is 6.
And of course that simple approach is way to slow when the highest digit is 9.

A number is divisible by 11 if the difference of the sum of the digits at odd and the even positions is divisible by 11.
40561817703823564929 is not divisible by 11 because 4+5+1+1+7+0+2+5+4+2 = 31 and 0+6+8+7+0+8+3+6+9+9 = 56 and 56 - 31= 25 is not divisible by 11.
There is actually an astonishing variety of rules for divisibility by 11: en.wikipedia.org/wiki/Divisibility rule

All double pandigital numbers are permutations of the string "00112233445566778899" (excluding leading zeros).
Ten of these digits will be at odd positions and ten at even positions.
If I create a bitmask which digits will be at odd positions then I need a 20-bit integer where exactly 10 bits are set.

All bitmasks with exactly 10 bits set are between 00000000001111111111b (minBitmask) and 11111111110000000000b (maxBitmask).
The nice function nextNumberWithSameBits(x) creates the next number with the same number of bits set as x and was first published in HAKMEM 175 (often called "snoob").

However, there are a few bordercases:
I can choose a digit once, twice or not at all to appear at an odd position. A bitmask might be:
99887766554433221100
01001101010111000101
→ exactly 10 bits set
For each digit I must only accept the bit patterns:
01 (=> use that digit once, e.g. for 9,6,...),
11 (twice, e.g. for 7 and 3) and
00 (none, e.g. for 8 and 2)
I must not accept bit pattern 10 because it's the same as 01 (see my loop with the variable reduce).

The sum of all digits is 2 * T(9) where T is the triangular number:
digitSum = 2 * dfrac{9 * (9 + 1)}{2} = 90

The sum of all digits at even positions can be computed when I know the sum of all digits at odd positions:
sumEven = digitSum - sumOdd

And the difference of sumEven and sumOdd has to be divisible by 11:
(sumEven - sumOdd) == 0 mod 11
((digitSum - sumOdd) - sumOdd) == 0 mod 11
(digitSum - 2 * sumOdd) == 0 mod 11

Whenever I know that the bitmasks generate matching numbers I have to figure out how many different numbers are possible.
If x_i represents how often x appears at odd positions then x_i can be either 0, 1 or 2.
The total number of permutations will be:
p(x) = dfrac{10!}{x_0!x_1!x_2!...x_9!}

If y_i represents how often y appears at even positions then y_i can be either 0, 1 or 2. The formula remains the same:
p(y) = dfrac{10!}{y_0!y_1!y_2!...y_9!}

And the sum of all p(x)p(y) will be the result.
Interesting is that the number of digits that appear twice is the same for p(x) and p(y). Even though a few x_i != y_i, it's still p(x) = p(y).
There my code simplifies to result += permutationsRepeated[repeated] * permutationsRepeated[repeated];

Excluding leading zeros is simple: just multiply with frac{9}{10}.

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 3 | ./491

Output:

(please click 'Go !')

Note: the original problem's input 9 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 <algorithm>
#include <cstdlib>
 
// count all numbers, where highest digit is maxDigit
unsigned long long bruteForce(unsigned int maxDigit)
{
unsigned long long result = 0;
 
// keep only a few digits
std::string digits = "10012233445566778899"; // smallest number without a leading zero
digits.resize(maxDigit * 2 + 2);
 
// count numbers divisible by 11
do
{
if (std::stoll(digits) % 11 == 0)
result++;
} while (std::next_permutation(digits.begin(), digits.end()));
 
return result;
}
 
// return next higher number with same number of bits set
unsigned int nextNumberWithSameBits(unsigned int x)
{
// from HAKMEM 175
// see http://www.hackersdelight.org/hdcodetxt/snoob.c.txt
auto smallest = x & -x;
auto ripple = x + smallest;
auto ones = ripple ^ x;
return ((ones >> 2) / smallest) | ripple;
}
 
// count numbers by using permutations
unsigned long long fast(unsigned int maxDigit)
{
unsigned long long result = 0;
 
int digitSum = 2 * (maxDigit + 1) * maxDigit / 2; // = (maxDigit + 1) * maxDigit = 90
auto numDigits = 2 * (maxDigit + 1); // = 20
 
// precompute number of permutations with repeated elements
unsigned long long factorial = 1;
for (unsigned int i = 1; i <= maxDigit + 1; i++)
factorial *= i; // 10! = 3628800
// 10!, 10!/2!, 10!/2!2!, 10!/2!2!2!, ...
unsigned long long permutationsRepeated[10];
for (unsigned int i = 0; i <= maxDigit; i++)
permutationsRepeated[i] = factorial >> i;
 
// smallest bitmask where maxDigits bits are set
auto minBitmask = (1 << (maxDigit + 1)) - 1;
// largest bitmask where maxDigits bits are set
auto maxBitmask = minBitmask << (maxDigit + 1);
// process all bitmasks where maxDigits bits are set
for (auto bitmask = minBitmask; bitmask <= maxBitmask; bitmask = nextNumberWithSameBits(bitmask))
{
// when picking a number for odd positions:
// choose a digit once, twice or not at all
// e.g. bits represent 99887766554433221100
// and the current choice might be 01001101010111000101
// (exactly 10 bits set)
// for each digit only accept the bit patterns 01 (once, e.g. for 9,6,...),
// 11 (twice, e.g. for 7 and 3) and
// 00 (none, e.g. for 8 and 2)
// DO NOT accept 10 because it's the same as 01
// ok will be false if bit pattern 10 is found
auto reduce = bitmask;
bool ok = true;
while (reduce > 0)
{
// lowest two bits are 10b (=2 decimal) ?
if ((reduce & 3) == 2)
{
ok = false;
break;
}
// next two bits
reduce >>= 2;
}
if (!ok)
continue;
 
// add all digits at odd positions and count how many digits are repeatedly present at odd positions
int sumOdd = 0;
auto repeated = 0; // it's actually the same value for odd and even positions
for (unsigned int pos = 0; pos < numDigits; pos++)
{
// bit set ? use that digit
if (bitmask & (1 << pos))
{
sumOdd += pos / 2;
 
// use that digit twice ? (bit pattern 11b)
if (pos & 1) // same as repeated += pos & 1;
repeated++;
}
}
 
// divisible by 11 ?
if ((digitSum - 2*sumOdd) % 11 == 0)
result += permutationsRepeated[repeated] * permutationsRepeated[repeated];
}
 
// exclude leading zeros
return result * maxDigit / (maxDigit + 1);
}
 
int main()
{
unsigned int numDigits = 9;
std::cin >> numDigits;
 
//std::cout << bruteForce(numDigits) << std::endl;
std::cout << fast(numDigits) << std::endl;
return 0;
}

This solution contains 15 empty lines, 30 comments and 4 preprocessor commands.

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

August 16, 2017 submitted solution
August 16, 2017 added comments

Difficulty

20% Project Euler ranks this problem at 20% (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 !