<< problem 308 - An amazing Prime-generating Automaton Nim Square - problem 310 >>

In the classic "Crossing Ladders" problem, we are given the lengths x and y of two ladders resting on the opposite walls of a narrow, level street.
We are also given the height h above the street where the two ladders cross and we are asked to find the width of the street (w). Here, we are only concerned with instances where all four variables are positive integers.
For example, if x = 70, y = 119 and h = 30, we can calculate that w = 56.

In fact, for integer values x, y, h and 0 < x < y < 200, there are only five triplets (x,y,h) producing integer solutions for w:
(70, 119, 30), (74, 182, 21), (87, 105, 35), (100, 116, 35) and (119, 175, 40).

For integer values x, y, h and 0 < x < y < 1 000 000, how many triplets (x,y,h) produce integer solutions for w?

# My Algorithm

Each ladders forms two right triangles:

• the first triangle consists of the ladder x (or y), the height X (or Y) where the ladder touches the wall and the width w of the lane.
• the second triangle consists of the lower part of the ladder x_1 (or y_1), the height h where both ladders cross and a part w_1 (or w_2) of the lane.

It can be seen that
(1) w = w_1 + w_2.

The angle alpha between ladder x and the lane is part of both triangle formed by the first ladder:
(2) cos alpha = dfrac{X}{w} and cos alpha = dfrac{h}{w_1}

The same can be said about an angle beta of ladder y:
(3) cos beta = dfrac{Y}{w} and cos beta = dfrac{h}{w_2}

Both equations of (2) can be merged (and the same for (3) ):
(4) dfrac{X}{w} = dfrac{h}{w_1}

(5) dfrac{Y}{w} = dfrac{h}{w_2}

Let's transform (4) and (5), merge them and solve for w_1:
(6) w * h = X * w_1
(7) w * h = Y * w_2
(8) X * w_1 = Y * w_2
(9) w_1 = frac{Y * w_2}{X}

Combining (1) and (9):
(10) w = frac{Y * w_2}{X} + w_2

Transform (7) and substitute in (10):
(11) w = frac{Y * w_2}{h}
(12) frac{Y * w_2}{h} = frac{Y * w_2}{X} + w_2

Simplify:
(13) frac{Y * w_2}{h} = frac{Y * w_2 + X * w_2}{X}
(14) frac{Y}{h} = frac{Y + X}{X}

And solve for h:
(15) h = dfrac{Y}{frac{Y + X}{X}}

(16) h = dfrac{XY}{Y + X}

That means that I have to find the height above ground where both ladders touch the wall.
If frac{XY}{Y + X} is an integer, then I have a solution.

Wikipedia has the formula in a different notation (see en.wikipedia.org/wiki/Crossed_ladders_problem , they replaced X by A and Y by B):
dfrac{1}{X} + dfrac{1}{Y} = dfrac{1}{h}

My program generates all Pythagorean triples (see en.wikipedia.org/wiki/Pythagorean_triple , my code is based on my solution for problem 75).
A container sides enumerates for each of the legs the length of the other leg (if c^2 = a^2 + b^2 then it stores a in sides[b] and b in sides[a]).
This is done for all basic triples as well as for all multiples.

In the second phase all values of sides[i] are compared to each other:
• they are triangles where one side is equal, that's the shared width w of the lane
• the other leg is the height of the point where each ladder touches a wall
• the height h of their intersection is h = dfrac{XY}{Y + X} according to equation (16)
• and the height h is only an integer if XY is a multiple of Y + X

## Note

There is probably a shorter way to derive equation (16). However, any website about the "Crossing Ladders Problem" doesn't show
how they arrive at that formula, they present it as "obviously easy". It wasn't easy for me and took quite a few attempts to get it right.

On the other hand, it's an easy task for Wolfram Alpha ... it can do all the heavy lifting (unfortunately without printing intermediate steps):
www.wolframalpha.com/input/?i=h/u=a/w and h/v=b/w and u+v=w

# 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 200 | ./309

Output:

Note: the original problem's input 1000000 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 <algorithm>

// most of the code is based on problem 75

// find greatest common divisor
template <typename T>
T gcd(T a, T b)
{
while (a != 0)
{
T c = a;
a = b % a;
b = c;
}
return b;
}

int main()
{
// 10^6
unsigned int limit = 1000000;
std::cin >> limit;

std::vector<std::vector<unsigned int>> sides(limit);

// generate all Pythagorean triples (see https://en.wikipedia.org/wiki/Pythagorean_triple )
for (unsigned int m = 2; m*m < limit; m++)
for (unsigned int n = (m & 1) + 1; n < m; n += 2)
{
// only valid m and n
if (gcd(m, n) != 1)
continue;

// compute basic triple
auto a = m*m - n*n;
auto b = 2*m*n;
auto c = m*m + n*n;

// and all multiples
for (unsigned int k = 1; k*c < limit; k++) // longest side must be < 10^6
{
// either height is shorter than width ...
sides[k*a].push_back(k*b);
// or the other way around
sides[k*b].push_back(k*a);
}
}

auto count = 0;
// process all triangles with one identical side (the width of the street)
for (const auto& current : sides)
{
// compare all potential heights
for (size_t left = 0; left < current.size(); left++)
for (size_t right = left + 1; right < current.size(); right++)
{
auto X = current[left];
auto Y = current[right];

// see "Algorithm" for an explanation of the formula
if ((X * (unsigned long long)Y) % (X + Y) == 0)
count++;
// note: need 64 bit multiplication because of potential overflows
}
}

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


This solution contains 10 empty lines, 14 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.8 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 59 MByte.

(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 27, 2017 submitted solution

# 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

I stopped working on Project Euler problems around the time they released 617.
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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 308 - An amazing Prime-generating Automaton Nim Square - problem 310 >>
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