<< problem 92 - Square digit chains | Almost equilateral triangles - problem 94 >> |
Problem 93: Arithmetic expressions
(see projecteuler.net/problem=93)
By using each of the digits from the set, {1, 2, 3, 4}, exactly once, and making use of the four arithmetic operations (+, -, *, /) and brackets/parentheses,
it is possible to form different positive integer targets.
For example,
8 = (4 * (1 + 3)) / 2
14 = 4 * (3 + 1 / 2)
19 = 4 * (2 + 3) - 1
36 = 3 * 4 * (2 + 1)
Note that concatenations of the digits, like 12 + 34, are not allowed.
Using the set, {1, 2, 3, 4}, it is possible to obtain thirty-one different target numbers of which 36 is the maximum,
and each of the numbers 1 to 28 can be obtained before encountering the first non-expressible number.
Find the set of four distinct digits, a < b < c < d, for which the longest set of consecutive positive integers, 1 to n,
can be obtained, giving your answer as a string: abcd.
My Algorithm
No matter what brackets or operators are used, the basic pattern is always the same:
1. an operation on two out of the four numbers is executed
2. the result will become part of the equation which consists now of three numbers
3. an operation on two out of the three numbers is executed
4. the result will become part of the equation which consists now of two numbers
5. an operation on the final two numbers is executed
I generate all sets of digits {a, b, c, d} where a < b < c < d. There are 126 such sets and four nested loops produce these sets in main
.
My eval
function accepts an arbitrary set of numbers
and does the following:
- pick any two numbers and execute all operations on these numbers, thus reducing
numbers
by one element - call itself recursively
- when only one element is left in
numbers
then mark the result asused
- values may temporarily be non-integer, e.g. (1/3)*6+7 = 9, but may produce a valid integer result.
- values may temporarily be negative, e.g. (1-2)+3+4 = 6, but may produce a valid positive integer result.
- division by zero is not allowed, not even temporarily
- any order of digits must be checked: a op b and b op a
However,
double
(or float
) may have some rounding issues, that's why I added an Epsilon
to make sure that results like 3.9999
are treated as 4
.getSequenceLength
is the glue between main
and eval
:It finds the first gap, that means the first number not represented by any combination of arithmetic operations.
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho "1 " | ./93
Output:
(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.
The code contains #ifdef
s to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL
to produce the result for the original problem (default setting for most problems).
#include <vector>
#include <iostream>
#include <cmath>
const double Epsilon = 0.00001;
// try all arithmetic operations of any two elements of "numbers", set their result in "used" to true
void eval(const std::vector<double>& numbers, std::vector<bool>& used)
{
// 1. if array holds just one element, add it to the "used" list and we are done
// 2. pick any two numbers
// 3. loop through all operators
// 4. add result to the array and call eval() recursively
// step 1
if (numbers.size() == 1)
{
auto result = numbers.front() + Epsilon;
// reject non-integer result (caused by division)
if (fmod(result, 1) > 10*Epsilon)
return;
int index = int(result + Epsilon);
// reject negative and very large results
if (index >= 0 && index < (int)used.size())
used[index] = true;
return;
}
// step 2
auto next = numbers;
for (size_t i = 0; i < numbers.size(); i++)
for (size_t j = i + 1; j < numbers.size(); j++)
{
// fetch two numbers
double a = numbers[i];
double b = numbers[j];
// prepare for next recursive step
next = numbers;
next.erase(next.begin() + j); // delete the higher index first
next.erase(next.begin() + i);
// steps 3 and 4 (unrolled)
next.push_back(a + b); // add
eval(next, used);
next.back() = a - b; // subtract (I)
eval(next, used);
next.back() = b - a; // subtract (II)
eval(next, used);
next.back() = a * b; // multiply
eval(next, used);
if (b != 0)
{
next.back() = a / b; // divide (I)
eval(next, used);
}
if (a != 0)
{
next.back() = b / a; // divide (II)
eval(next, used);
}
// note: I overwrite the last element because that's simpler than
// pop_back(); push_back(...);
}
}
// evaluate all expressions and count how many numbers from 1 to x can be expressed without any gaps
unsigned int getSequenceLength(const std::vector<double>& numbers)
{
// find all results
std::vector<bool> used(1000, false);
eval(numbers, used);
// longest sequence, beginning from 1
unsigned int result = 0;
while (used[result + 1])
result++;
return result;
}
int main()
{
#define ORIGINAL
#ifdef ORIGINAL
unsigned int longestSequence = 0;
unsigned int longestDigits = 0;
// four different digits
for (unsigned int a = 1; a <= 6; a++)
for (unsigned int b = a+1; b <= 7; b++)
for (unsigned int c = b+1; c <= 8; c++)
for (unsigned int d = c+1; d <= 9; d++)
{
// evaluate all combinations
auto sequenceLength = getSequenceLength({ double(a), double(b), double(c), double(d) });
// new record ?
if (longestSequence < sequenceLength)
{
longestSequence = sequenceLength;
// build one number out of the four digits
longestDigits = a * 1000 + b * 100 + c * 10 + d;
}
}
// print result
std::cout << longestDigits << std::endl;
#else
// 1..5 digits
unsigned int numDigits;
std::cin >> numDigits;
// read those digits
std::vector<double> numbers(numDigits);
for (auto& x : numbers)
std::cin >> x;
// print length of longest sequence
std::cout << getSequenceLength(numbers);
#endif
return 0;
}
This solution contains 20 empty lines, 25 comments and 7 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in 0.02 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
March 12, 2017 submitted solution
May 8, 2017 added comments
Hackerrank
see https://www.hackerrank.com/contests/projecteuler/challenges/euler093
My code solves 25 out of 25 test cases (score: 100%)
Difficulty
Project Euler ranks this problem at 15% (out of 100%).
Hackerrank describes this problem as medium.
Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.
Links
projecteuler.net/thread=93 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
C# www.mathblog.dk/project-euler-93-longest-sequence/ (written by Kristian Edlund)
Java github.com/nayuki/Project-Euler-solutions/blob/master/java/p093.java (written by Nayuki)
Scala github.com/samskivert/euler-scala/blob/master/Euler093.scala (written by Michael Bayne)
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own. Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
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 | |
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 |
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.
Copyright
I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.
All of my solutions can be used for any purpose and I am in no way liable for any damages caused.
You can even remove my name and claim it's yours. But then you shall burn in hell.
The problems and most of the problems' images were created by Project Euler.
Thanks for all their endless effort !!!
<< problem 92 - Square digit chains | Almost equilateral triangles - problem 94 >> |