<< problem 601 - Divisibility streaks Roman Numerals II - problem 610 >>

# Problem 607: Marsh Crossing

Frodo and Sam need to travel 100 leagues due East from point A to point B. On normal terrain, they can cover 10 leagues per day, and so the journey would take 10 days.
However, their path is crossed by a long marsh which runs exactly South-West to North-East, and walking through the marsh will slow them down.
The marsh is 50 leagues wide at all points, and the mid-point of AB is located in the middle of the marsh.

A map of the region is shown in the diagram below:

The marsh consists of 5 distinct regions, each 10 leagues across, as shown by the shading in the map.
The strip closest to point A is relatively light marsh, and can be crossed at a speed of 9 leagues per day.
However, each strip becomes progressively harder to navigate, the speeds going down to 8, 7, 6 and finally 5 leagues per day for the final region of marsh,
before it ends and the terrain becomes easier again, with the speed going back to 10 leagues per day.

If Frodo and Sam were to head directly East for point B, they would travel exactly 100 leagues, and the journey would take approximately 13.4738 days.
However, this time can be shortened if they deviate from the direct path.

Find the shortest possible time required to travel from point A to B, and give your answer in days, rounded to 10 decimal places.

# My Algorithm

Each section of the marsh is leagues wide. However, walking the direct route means they walk sqrt{2} * 10 approx 14.1421 leagues in each section.
In order to be faster they should walk less than 14.1421 leagues in each slow section and walk a longer distance on the normal terrain.

The whole map is a two-dimensional coordinate system. Initially I tried to have city A at (0,0) and B at (100,0) but things got messy.
Then I decided to rotate the whole map by 45 degrees such that the marsh is parallel to the y-axis.
City A still remains at (0,0) while B is located at (dfrac{100}{sqrt{2}}, dfrac{100}{sqrt{2}}).
The distance between both cities is still 100 leagues.

Knowing that the marsh is exactly 50 leagues wide, the marsh' smallest x and largest x are 50 units apart.
Since x_B - x_A = dfrac{100}{sqrt{2}} - 0 approx 70.7107 the marsh starts at x_1 approx (70.7107 - 50) / 2 approx 10.3553.
The next crossing will be at x_2 = x_1 + 10, then x_3 = x_2 + 10, ..., x_6 = x_5 + 10 = x_1 + 50.

The direct route is x = y and will be stored in my container named points.
The function duration() determines the distance between consecutive points and divide them by the speed of Frodo and Sam (see speed).
It returns 13.4738023615 for the direct route.

I like Monte-Carlo simulations because even though they seem to be unpredictable they produce a predictable result after some time.
This problem can be solved by randomizing the y-components of the marsh:

• mutate picks a random point out of points, except for the first and last (A and B)
• it randomly adds or subtracts a given delta
• if duration() returns a shorter time than before then keep this mutation, else undo it
My program starts with a large delta and successively reduces it.
The way I reduce delta and how often I call mutate() is not scientific at all:
I started with many more iterations and a slower decrease but saw that the coarser values still produce the correct result.

## Alternative Approaches

The Gradient Descent method is a more straightforward optimization method (see en.wikipedia.org/wiki/Gradient_descent).
The problem can be solved with Snell's law, too (see en.wikipedia.org/wiki/Snell's_law).

# Interactive test

This feature is not available for the current problem.

# 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 <iomanip>
#include <vector>
#include <cmath>

// crossing between two marches
struct Point
{
double x;
double y;

// create a new 2D point
Point(double x_, double y_)
: x(x_), y(y_)
{}

// distance between two points
double distance(const Point& other) const
{
auto diffX = x - other.x;
auto diffY = y - other.y;
return sqrt(diffX * diffX + diffY * diffY);
}
};

// how many days does it take to walk from the first to the last point, considering "speed" of the surfaces
double duration(const std::vector<Point>& points)
{
// leagues travelled per day between point[i] and point[i+1]
static const std::vector<double> speed = { 10, 9, 8, 7, 6, 5, 10 };

double result = 0;
// for each surface: determine length and divide by speed
for (size_t i = 0; i < speed.size(); i++)
{
auto way = points[i].distance(points[i+1]);
result += way / speed[i];
}
return result;
}

// a simple pseudo-random number generator
// (produces the same result no matter what compiler you have - unlike rand() from math.h)
unsigned int myrand()
{
static unsigned long long seed = 0;
seed = 6364136223846793005ULL * seed + 1;
return (unsigned int)(seed >> 30);
}

// try to move a single point's y-value by delta
// if the mutation is better, then keep the result
// if the mutation is worse,  then undo the change
void mutate(std::vector<Point>& points, double delta)
{
auto oldDuration = duration(points);

// decide at random whether to add or subtract
if (myrand() & 1)
delta = -delta;

// change one random y-coordinate (must not be the first or last)
auto id = myrand() % 6 + 1;
points[id].y += delta;

// not faster ? => undo change
if (duration(points) >= oldDuration)
points[id].y -= delta;
}

int main()
{
// A
std::vector<Point> points;
points.push_back({ 0, 0 });

// rotate the coordinate system by 45 degrees,
// so that travelling one league is equivalent to moving 1/scaling units along x-axis and 1/scaling units along y-axis
double scaling = sqrt(2);

// when travelling the shortest distance between A and B, how wide is the marsh ?
auto direct = 50 * scaling; // c^2 = a^2 + b^2 where a=b=50

// entering the marsh
auto current = ((100 - direct) / 2) / scaling; // same as 25*(sqrt(2)-1)
points.push_back({ current, current });

// all 5 zones of the marsh
for (auto i = 1; i <= 5; i++)
points.push_back({ current + i * 10, current + i * 10 });

// B
points.push_back({ 100 / scaling, 100 / scaling });

// randomly mutate the crossing
const auto NumIterations = 10000;
for (auto delta = 0.01; delta >= 0.00001; delta /= 10)
for (auto i = 0; i < NumIterations; i++)
mutate(points, delta);

// display result with 10 digits after the decimal point
std::cout << std::fixed << std::setprecision(10)
<< duration(points) << std::endl;
return 0;
}


This solution contains 18 empty lines, 23 comments and 4 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

October 18, 2017 submitted solution

# 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
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 601 - Divisibility streaks Roman Numerals II - problem 610 >>
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 !