<< problem 455 - Powers With Trailing Digits Almost Pi - problem 461 >>

# Problem 458: Permutations of Project

Consider the alphabet A made out of the letters of the word "project": A= \{c,e,j,o,p,r,t\}.
Let T(n) be the number of strings of length n consisting of letters from A that do not have a substring that is one of the 5040 permutations of "project".
T(7)=7^7 - 7! = 818503.

# My Algorithm

First let's state the most important property of the word "project": all its letters are distinct.

I wrote three (!) algorithms but only the last actually solves T(10^12) in a reasonable amount of time (function fast):
bruteForce() was a quick hack to ensure I can solve T(n) for small parameters (n < 10 is okay)
It iterates over all possible 7^n strings and checks every subset of 7 consecutive letters whether they can produce the word "project".
I assign each letter P,R,O,J,E,C,T a number between 0 and 6 which in turn is used as the position in a bitmask.
Only if seven consecutive letters are distinct then the ORed bitmask of those seven letters will be 1111111 in binary (= 2^7 - 1 = 127 in decimal).

The next algorithm's implementation is the function slow. It works quite differently because it treats the whole thing as a state machine:

• I count the number of strings WITH the word "project"
• → then the result is 7^n minus that number of strings with the word "project"
• in general: if the last x letters were distinct then the state machine is currently in state x
• the state machine has 8 states (0 to 7) and starts in state 0
• once state 7 is reached it stays there
Initially, the variable state[8] = { 1,0,0,0,0,0,0,0 } because there is only one empty string and it's in state 0.
No empty string can be in state 1,2,...,7.

When the first letter is processed then there are 7 different letters. Each causes a transition from state 0 to state 1.
When the second letter is processed then there are 7 differenz letters as well but I have to be careful:
if the second letter is equal to the first letter, then I remain in state 1 else I jump to state 2.
Remember: the current state's ID indicates how many of the last letters were distinct.

When the third letter is processed then there are 3 different state transitions:
• if the new letter is equal to the first letter, then the state machine stays in state 2
• if the new letter is equal to the second letter, then the state machine must go back to state 1
• if the new letter wasn't seen before then the state machine proceeds with state 3
The same concept repeats with state 3,4,5,6.

State 7 stands out because once I reach state 7 then I stay there, no matter what future letters will arrive.
That's because I'm interesting in all strings containing the word "project" at least once - and it doesn't matter if it appears 1x, 10x, 100x, ...
I wrote a simple loop that repeatedly adds/multiplies each state by the number of different state transitions.
That O(n) algorithm can solve T(10^12) in a few hours - still too slow !

So I wrote the fast() function: the state transitions look a lot like a matrix - why not rewrite it as a Matrix ?
The matrix's number at position x,y contains the number of transitions from state x to state y.
For example, M_{4,5} = M[4,5] = 3 because 3 (out of 7) letters cause a transition from state 4 to 5.
You'll find it in the fifth column and sixth line (the upper-right corner has index (0,0) because I used zero-based indices):

M = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 7 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ 0 & 6 & 1 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 5 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 4 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 3 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 7 \end{pmatrix}

The state variable'' looks like a vector such that the computation per step is:
(1) state_{i+1} = M \bullet state_i

And the next step:
(2) state_{i+2} = M \bullet state_{i+1}
(3) state_{i+2} = M \bullet (M \bullet state_i)
(4) state_{i+2} = (M \bullet M) \bullet state_i
(5) state_{i+2} = M^2 \bullet state_i

The last equation can be generalized to:
(6) state_{final} = M^n \bullet state_0

When solving problem 137 I wrote a fast exponentiation algorithm for a 2x2 matrix.
The Matrix class written for the current problem generalizes that algorithm to larger quadratic matrices, see its powmod member function.

I was too lazy to write a Vector class to represent the state variable - and I can get away with that lazy attitude ;-)
The meaning of M^n_{x,y} was that the matrix M contains the total number of state transitions from state x to y after n steps.
Hence M^n_{0,7} is the total number of strings with the word "project" after those n = 10^12 steps.

Now I need 7^n mod 10^9 - and that number is part of the matrix, too !
The number M_{7,7} is initially 7 and multiplied by itself n times, that's exactly what I need.
However, I saw that M^n_{7,7} = 1 after n = 10^12 steps - which I found strange and I was sure I had a bug in my code.
But Wolfram Alpha confirmed that it's correct: www.wolframalpha.com/input/?i=PowerMod[7,10^12,10^9]
So the result is M^n_{7,7} - M^n_{0,7} (mod 1000000000).

Fast exponentiation is truly fast and the correct result is displayed after a few milliseconds.

## Alternative Approaches

After submitting my result I looked at the official forum and noticed that pretty much everyone solved it the same way.
But some postings differed completely: they found the generating function and solved it.

# 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 7 | ./458

Output:

Note: the original problem's input 1000000000000 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 <array>

// "project" has 7 distinct letters
const auto WordSize = 7;
enum Letters { P = 0, R = 1, O = 2, J = 3, E = 4, C = 5, T = 6 };

// only the last nine digits
const unsigned int Modulo = 1000000000;

// ---------- first algorithm ----------

// super-slow brute-force algorithm, becomes unusable for numLetters > 10
unsigned long long bruteForce(unsigned int numLetters)
{
// bruteForce( 7) =    818503
// bruteForce( 8) =   5699280
// bruteForce( 9) =  39688327
// bruteForce(10) = 276386929
unsigned long long result = 0;

// 7^numLetters
unsigned long long maxId = 1;
for (unsigned long long exponent = 1; exponent <= numLetters; exponent++)
maxId *= WordSize;

// iterate over all 7^numLetters combinations
for (unsigned long long i = 0; i < maxId; i++)
{
// convert the current number to letters (it's like converting from decimal to base 7)
auto id = i;
Letters letters[20];
for (unsigned int letter = 0; letter < numLetters; letter++)
{
// I use the data type "Letters" to simplify debugging
letters[letter] = (Letters)(id % WordSize);
id /= WordSize;
}

// look at each group of 7 consecutive letters
bool isProject = false;
for (unsigned int from = 0; from + WordSize <= numLetters; from++)
{
for (unsigned int current = 0; current < WordSize; current++)
mask |= 1 << letters[from + current];

// all seven letters used ?
if (mask == 127) // 2^7 - 1 = 127
{
isProject = true;
break;
}
}

// no seven consecutive letters can be re-order to the word "PROJECT"
if (!isProject)
result++;
}

return result;
}

// ---------- second algorithm ----------

// O(n) algorithm, needs about 2 seconds for numLetters = 10^8
unsigned long long slow(unsigned long long numLetters)
{
// state x means that currently x distinct letters were observed in the last x steps
typedef std::array<unsigned long long, 7 + 1> State;
// once we reach state 7 we have the word P,R,O,J,E,C,T (in any order) and remain in that state
// no matter what follows
State state = { 0 };
state[0] = 1; // seed

unsigned long long all = 1;
for (unsigned long long i = 1; i <= numLetters; i++)
{
// total number of strings (regardless whether they contain PROJECT or not)
all *= 7;
all %= Modulo;

State next = { 0 };

// compute all state transitions
next[1] = 7 * state[0] + 1 * state[1] + 1 * state[2] + 1 * state[3] + 1 * state[4] + 1 * state[5] + 1 * state[6];
next[2] =                6 * state[1] + 1 * state[2] + 1 * state[3] + 1 * state[4] + 1 * state[5] + 1 * state[6];
next[3] =                               5 * state[2] + 1 * state[3] + 1 * state[4] + 1 * state[5] + 1 * state[6];
next[4] =                                              4 * state[3] + 1 * state[4] + 1 * state[5] + 1 * state[6];
next[5] =                                                             3 * state[4] + 1 * state[5] + 1 * state[6];
next[6] =                                                                            2 * state[5] + 1 * state[6];
next[7] =                                                                                         + 1 * state[6];

// once I'm in state 7 I stay there
next[7] += 7 * state[7];

// keep only the last 9 digits
for (auto& x : next)
x %= Modulo;

state = std::move(next);
}

// state[7] represents the number of strings WITH the word "project"
auto withProject = state[7];
// without = all - with, but avoid negative results (due to modulo)
if (all < withProject)
all += Modulo;
return all - withProject;
}

// ---------- third and final algorithm ----------

template <typename Number, unsigned int Size>
class Matrix
{
// store all elements
std::array<std::array<Number, Size>, Size> data; // same as Number data[Size][Size];

public:
// set all elements to zero
Matrix()
: data()
{
for (unsigned int i = 0; i < Size; i++)
data[i].fill(0);
}

// access a field (read/write), indices are zero-based
Number& operator()(unsigned int column, unsigned int row)
{
return data[row][column];
}
// access a field (read-only), indices are zero-based
Number get(unsigned int column, unsigned int row) const
{
return data[row][column];
}

// multiply two matrices
Matrix operator*(const Matrix& other) const
{
Matrix result; // initially all fields are zero
for (unsigned int i = 0; i < Size; i++)
for (unsigned int j = 0; j < Size; j++)
for (unsigned int k = 0; k < Size; k++)
result(i,k) += get(j,k) * other.get(i,j);
return result;
}

// fast exponentiation with modulo
Matrix powmod(unsigned long long exponent, unsigned int modulo) const
{
// more or less the same concept as powmod from my toolbox (which works on integers instead of matrices)

Matrix result;
for (unsigned int i = 0; i < Size; i++)
result(i,i) = 1;
Matrix base = *this;

while (exponent > 0)
{
// fast exponentation:
// odd exponent ? a^b = a*a^(b-1)
if (exponent & 1)
{
result = result * base;

// modulo
for (unsigned int i = 0; i < Size; i++)
for (unsigned int k = 0; k < Size; k++)
result(i,k) = result(i,k) % modulo;
}

// even exponent ? a^b = (a*a)^(b/2)
base = base * base;

// modulo
for (unsigned int i = 0; i < Size; i++)
for (unsigned int k = 0; k < Size; k++)
base(i,k) = base(i,k) % modulo;

exponent >>= 1;
}
return result;
}
};

// solve "almost instantly", too fast to measure execution time ...
unsigned long long fast(unsigned long long numLetters)
{
// same concept as slow() but rewritten with Matrix

// state transitions
Matrix<unsigned long long, 8> mat;
mat(0,0) = 0; mat(1,0) = 0; mat(2,0) = 0; mat(3,0) = 0; mat(4,0) = 0; mat(5,0) = 0; mat(6,0) = 0; mat(7,0) = 0;
mat(0,1) = 7; mat(1,1) = 1; mat(2,1) = 1; mat(3,1) = 1; mat(4,1) = 1; mat(5,1) = 1; mat(6,1) = 1; mat(7,1) = 0;
mat(0,2) = 0; mat(1,2) = 6; mat(2,2) = 1; mat(3,2) = 1; mat(4,2) = 1; mat(5,2) = 1; mat(6,2) = 1; mat(7,2) = 0;
mat(0,3) = 0; mat(1,3) = 0; mat(2,3) = 5; mat(3,3) = 1; mat(4,3) = 1; mat(5,3) = 1; mat(6,3) = 1; mat(7,3) = 0;
mat(0,4) = 0; mat(1,4) = 0; mat(2,4) = 0; mat(3,4) = 4; mat(4,4) = 1; mat(5,4) = 1; mat(6,4) = 1; mat(7,4) = 0;
mat(0,5) = 0; mat(1,5) = 0; mat(2,5) = 0; mat(3,5) = 0; mat(4,5) = 3; mat(5,5) = 1; mat(6,5) = 1; mat(7,5) = 0;
mat(0,6) = 0; mat(1,6) = 0; mat(2,6) = 0; mat(3,6) = 0; mat(4,6) = 0; mat(5,6) = 2; mat(6,6) = 1; mat(7,6) = 0;
mat(0,7) = 0; mat(1,7) = 0; mat(2,7) = 0; mat(3,7) = 0; mat(4,7) = 0; mat(5,7) = 0; mat(6,7) = 1; mat(7,7) = 7;
// note: I could have skipped all those mat(x,y) = 0 because all cell are initialized with zero anyway

// exponentiate matrix
auto superMatrix = mat.powmod(numLetters, Modulo);

// the number of strings WITH the word "project" (number of ways to transition from state 0 to 7)
auto withProject = superMatrix.get(0,7);
// total number of combinations is 7^numLetters
// => I could use powmod(7, 10^12, Modulo) from my toolbox, but there's a 7 in the lower-right corner of the matrix at 7,7
//    and since it's exponated as well, it becomes (7^numLetters) % Modulo and that's just what I need
auto all         = superMatrix.get(7,7); // and surprisingly 7^(10^12) % 10^9 = 1 !

// without = all - with, but avoid negative results (due to modulo)
if (all < withProject)
all += Modulo;
return all - withProject;
}

int main()
{
unsigned long long limit = 1000000000000ULL; // 10^12
std::cin >> limit;
//std::cout << bruteForce(limit) << std::endl;
//std::cout << slow(limit) << std::endl;
std::cout << fast(limit) << std::endl;
return 0;
}


This solution contains 37 empty lines, 54 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 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

December 18, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 30% (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
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 455 - Powers With Trailing Digits Almost Pi - problem 461 >>
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 !