# Problem 163: Cross-hatched triangles

Consider an equilateral triangle in which straight lines are drawn from each vertex to the middle of the opposite side,
such as in the size 1 triangle in the sketch below. Sixteen triangles of either different shape or size or orientation or location can now be observed in that triangle.
Using size 1 triangles as building blocks, larger triangles can be formed, such as the size 2 triangle in the above sketch.
One-hundred and four triangles of either different shape or size or orientation or location can now be observed in that size 2 triangle.

It can be observed that the size 2 triangle contains 4 size 1 triangle building blocks.
A size 3 triangle would contain 9 size 1 triangle building blocks and a size n triangle would thus contain n^2 size 1 triangle building blocks.

If we denote T(n) as the number of triangles present in a triangle of size n, then

T(1) = 16
T(2) = 104

Find T(36).

# My Algorithm

I failed to find an good scheme to enumerate all triangles but realized that the basic triangle of size 1 contains 6 lines;
the size 2 triangle contains 15 lines. These lines fall into six groups: they have a slope of 0, 30, 60, 90, 120 and 150 degrees.
I drew a few pictures and found that a size n triangle has 9n-3 lines. A size 36 triangle has 321 lines.

A brute-force search through all possible combinations of three lines checks:

• all three lines must not be parallel to each other
• they must not intersect at the same point
• all intersection points must be inside (or on the edge) of the outer-most triangle (the "hull")
Lines are created by converting two points (see struct Point) on the line into the line equation ax + by = c, where
a = y_1 - y_2, b = x_2 - x_1 and c = x_2 y_1 - x_1 y_2 (see class Line).
The intersection point of two lines is:
det = a_1 b_2 - a_2 b_1 → if zero, then both lines are parallel
x = (c_1 b_2 - c_2 b_1) / det
y = (a_1 c_2 - a_2 c_1) / det

The sign of the determinant of a line and a point tells me which side of the line the point is located.
When the three lines of a triangle are created in anti-clockwise order (AB, BC, CA - note: not AC !),
then the determinant must not be negative (see insideHull).

There are {{321}choose{3}} = 5461280 potential triangles.
I can prune a few by observing that if two lines are parallel I can skip building a full triangle.
Moreover, if two lines intersect outside the hull, then I can skip checking the full triangle, too.
Pruning reduces the total number of full triangle to 1631449 (program becomes three times faster).

## Alternative Approaches

I found a pretty long and weird closed formula that finds the correct answer instantly.
I can't follow the proof, though, because I'm just a software engineer and not a mathematician.

## Note

All calculations have to consider a certain error margin (see Epsilon) because double can't exactly represently sqrt{3} (see Height).

# 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):
Note: Enter the number of building blocks.

This is equivalent to
echo 2 | ./163

Output:

Note: the original problem's input 36 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 <vector>
#include <cmath>

// allowed error, I could probably add a few zeros ...
const double Epsilon = 0.0000001;

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

// return true if two points are identical or very close (allow for error Epsilon)
bool operator==(const Point& other) const
{
return fabs(x - other.x) < Epsilon &&
fabs(y - other.y) < Epsilon; // faster Manhattan metric, not a true distance
}

double x;
double y;
};

// height in a triangle of size 1
const double Height  = sqrt(3.0) / 2;
// points of the outer-most triangle, the "hull"
// note: below are the values for a size 1 triangle, they will be scaled accordingly in main()
Point A(0,   0);
Point B(1,   0);
Point C(0.5, Height);

// dummy point to indicate that two lines don't intersect (=> they are parallel)
const Point NoIntersection(9999999, 9999999);

// a line described by ax + by = c;
class Line
{
public:
// create new line through points "from" and "to"
Line(const Point& from, const Point& to)
{
// https://de.wikipedia.org/wiki/Koordinatenform
a = from.y - to.y;
b = to.x - from.x;
c = to.x*from.y - from.x*to.y;
}

// return intersection of the current line with a second line
Point intersect(const Line& other) const
{
// Cramer's Rule: https://en.wikipedia.org/wiki/Cramer%27s_rule
auto determinant = a * other.b - other.a * b;
// parallel ?
if (fabs(determinant) < Epsilon)
return NoIntersection;

auto x = (c * other.b - other.c * b) / determinant;
auto y = (a * other.c - other.a * c) / determinant;
return Point(x, y);
}

// return determinant
double determinant(const Point& p) const
{
return a * p.x + b * p.y - c;
}

private:
// line parameters
double a;
double b;
double c;
};

// return true if P is inside the triangle ABC
bool insideHull(const Point& p)
{
// choose anti-clockwise order of points such that points on the inside have a non-negative determinant
Line bottom  (A, B);
Line topRight(B, C);
Line topLeft (C, A);

// all determinants must have a positive sign (or zero)
if (bottom  .determinant(p) < -Epsilon) return false;
if (topRight.determinant(p) < -Epsilon) return false;
if (topLeft .determinant(p) < -Epsilon) return false;

// yeah ... found a "good" one :-)
return true;
}

// return true if the three lines a,b,c create a valid triangle inside the outermost triangle
bool isValidTriangle(const Line& a, const Line& b, const Line& c)
{
// find intersections of these three lines
Point ab = a.intersect(b);
Point bc = b.intersect(c);
Point ac = a.intersect(c);

// they must not be parallel
if (ab == NoIntersection) // note: this case was actually already tested in main()
return false;
if (bc == NoIntersection)
return false;
if (ac == NoIntersection)
return false;

// degenerated case: all lines have the same intersection point
if (ab == bc) // note: no need to test ab == ac and ac == bc
return false;

// intersections points must be inside the outer-most triangle
return insideHull(ab) && insideHull(bc) && insideHull(ac);
// insideHull(ab) was already tested in main()
}

int main()
{
// number of building blocks
unsigned int size = 36;
std::cin >> size;

// middle between A,B and A,C and B,C (of the basic triangle with size=1)
Point AB((A.x+B.x)/2, (A.y+B.y)/2);
Point AC((A.x+C.x)/2, (A.y+C.y)/2);
Point BC((B.x+C.x)/2, (B.y+C.y)/2);

// create all lines
std::vector<Line> lines;
// A-B
for (unsigned int i = 0; i < size; i++)
lines.push_back(Line(Point(A.x, i * Height), Point(B.x, i * Height)));
// A-BC
for (unsigned int i = 0; i < size; i++)
{
lines.push_back(Line(Point(i, A.y), Point(BC.x + i, BC.y)));
if (i > 0)
lines.push_back(Line(Point(-(double)i, A.y), Point(BC.x - (double)i, BC.y)));
}
// A-C
for (unsigned int i = 0; i < size; i++)
lines.push_back(Line(Point(i, A.y), Point(C.x + i, C.y)));
// B-C
for (unsigned int i = 0; i < size; i++)
lines.push_back(Line(Point(i+1, B.y), Point(C.x + i, C.y)));
// B-AC
for (unsigned int i = 0; i < 2*size-1; i++)
lines.push_back(Line(Point(i+1, B.y), Point(AC.x + i, AC.y)));
// C-AB
for (unsigned int i = 1; i < 2*size; i++)
lines.push_back(Line(Point(i * C.x, 0), Point(i * C.x, Height)));

// resize hull according to user input
A.x *= size; A.y *= size;
B.x *= size; B.y *= size;
C.x *= size; C.y *= size;

unsigned int count = 0;
// generate each combination of lines
for (unsigned int i = 0; i < lines.size(); i++)
for (unsigned int j = i + 1; j < lines.size(); j++)
{
// fast check whether the intersection of two lines results in a valid point
auto first = lines[i].intersect(lines[j]);
if (first == NoIntersection || !insideHull(first))
continue;

// and the third line ...
for (unsigned int k = j + 1; k < lines.size(); k++)
// now we have a potential triangle, check whether it's valid and fully inside the hull
if (isValidTriangle(lines[i], lines[j], lines[k]))
count++;
}

// display result
std::cout << count << std::endl;
return 0;
}


This solution contains 25 empty lines, 41 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.05 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 7, 2017 submitted solution

# Hackerrank

My code solves 12 out of 12 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 70% (out of 100%).

Hackerrank describes this problem as easy.

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.

# 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
 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 642 643 644 645 646 647 648 649 650 651 652 653 654 655 656 657 658 659 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702 703 704 705 706 707 708 709 710 711 712 713 714 715 716 717 718 719 720 721 722 723 724 725 726 727 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750 751 752 753 754 755
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.

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 !