<< problem 147 - Rectangles in cross-hatched grids Searching for a maximum-sum subsequence - problem 149 >>

# Problem 148: Exploring Pascal's triangle

We can easily verify that none of the entries in the first seven rows of Pascal's triangle are divisible by 7:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

However, if we check the first one hundred rows, we will find that only 2361 of the 5050 entries are not divisible by 7.

Find the number of entries which are not divisible by 7 in the first one billion (109) rows of Pascal's triangle.

# My Algorithm

I needed a few attempts to finally solve this problem.
My first idea was to iteratively create the triangle's rows, one at a time.
To find the binomial coefficient C(n+1,k+1) you only need the previous row:
C(n+1,k+1) = C(n,k) + C(n,k+1)
To avoid overflows, each value should be the original value modulo 7.
That works because (a+b) mod 7 = ((a mod 7) + (b mod 7)) mod 7 (that's no special property of 7, you can pick any integer)

Unfortunately I knew right from the start that this approach would be slow ... and memory-consuming (roughly 2 GByte).
Nevertheless I used the code (see nextRow) to show me the results for the first 1000 rows. And then a pattern appeared (my rows start at index 0):
rowfound
01
12
23
34
45
56
67
72
84
96
108
1314
143
156

If I convert the row from decimal system to a number in base 7 then
found(row) = (digit_1(row) + 1) * (digit_2(row) + 1) * ... (digit_n(row) + 1)
For example: found(15_10) = found(111_7) = (1+1) * (1+1) * (1+1) = 6

My function countNonDivisible does exactly that: it converts its parameter to base 7 and multiplies all digits plus one.
It takes 34 seconds until the correct result is displayed on my computer.

The inner-most loop of countNonDivisible consists of modulo and division operations - they are extremely slow in comparision to addition, subtractio, multiplication.
And actually there is no need to perform these divisions:
I process all numbers consecutively in ascending order.

Therefore my final algorithm stores all digits of row in base 7. To speed up the program, not the true digits but the digits plus one are stored
because when multiplying all digits I have to add one.
That means that my array base7 has up to 12 digits (7^12 > 10^9), each from 1 to 7.
Incrementing by one can cause some digits to become 7+1=8 → reset them to 1 and carry over 1.
That algorithm needs about 3.5 seconds.

## Alternative Approaches

My final algorithm is still a brute-force algorithm. You can find a closed formula, too:
The sum of the first 7^1 = 7 rows is 28. The sum of the first 7^2 = 49 rows is 28^2 = 784 ... and so on.

## Note

A substantial reason for the performance gain of my third algorithm is that base7 fits into the CPU cache and/or CPU registers (it's just 12 bytes).
After changing the data type from std::vector<unsigned char> to std::vector<unsigned int> the execution time explodes from 3.5 to 28 seconds
(and unsigned short → 3.9 seconds).

# 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 100 | ./148

Output:

(please click 'Go !')

Note: the original problem's input 1000000000 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>

typedef std::vector<unsigned char> Row;
const unsigned int Modulo = 7;

// generate next row of Pascal's triangle modulo a number (> 1)
// return count of elements that are not a multiple of modulo (in C++ speak: x % modulo != 0)
unsigned long long nextRow(Row& row)
{
// last value is always 1
row.push_back(1);
if (row.size() == 1)
return 1;

// first and last value are never a multiple of 7
unsigned long long result = 2;

// C(n+1,k+1) = C(n,k) + C(n,k+1)
for (size_t k = row.size() - 2; k > 0; k--)
{
// note: I'm processing the row back-to-front
// therefore minus 1 instead of plus 1
unsigned char current = row[k] + row[k - 1];

// subtraction is faster than modulo: current %= modulo
// all values must be 0 ... 2*(modulo-1)
if (current >= Modulo)
current -= Modulo;

// not divisible ?
if (current != 0)
result++;

row[k] = current;
}

return result;
}

// convert to base 7 and multiply all digits plus 1
unsigned long long countNonDivisible(unsigned int row)
{
unsigned long long result = 1;
while (row > 0)
{
// one more digit ...
result *= (row % Modulo) + 1;
row    /= Modulo;
}
return result;
}

int main()
{
unsigned int numRows = 1000000000;
std::cin >> numRows;

// for simple algorithm based on nextRow()
Row current = { 1 };

// for my fastest pseudo brute-force algorithm
std::vector<unsigned char> base7(12, 1); // 7^12 > 10^9

unsigned long long count = 1;
for (unsigned int row = 1; row < numRows; row++)
{
// simple algorithm (basically takes forever and needs tons of memory)
//auto found = nextRow(current);
//std::cout << row << " " << found << std::endl;

// slightly more advanced
//auto found = countNonDivisible(row);

// and my fastest (still pseudo brute-force) algorithm:
// keep all digits of row in base 7 in an array base7 with a twist:
// each digit is one higher than it should be
// => because previously I had to add 1 before multiplying

// next number
base7[0]++;
// carry over to next digits
auto carryPos = 0;
while (base7[carryPos] == Modulo + 1)
{
base7[carryPos] = 1; // remember: start at 1 instead of 0
base7[carryPos + 1]++;
carryPos++;
}

// multiply all digits
unsigned long long found = 1;
for (auto& x : base7)
found *= x;

// keep track of the sum of all rows
count += found;
}

std::cout << count << std::endl;
return 0;
}


This solution contains 19 empty lines, 27 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 3.5 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

July 12, 2017 submitted solution
July 12, 2017 added comments

# Difficulty

Project Euler ranks this problem at 50% (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
 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
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 147 - Rectangles in cross-hatched grids Searching for a maximum-sum subsequence - problem 149 >>
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 !