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# Problem 309: Integer Ladders

(see projecteuler.net/problem=309)

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%2Fu%3Da%2Fw+and+h%2Fv%3Db%2Fw+and+u%2Bv%3Dw

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#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.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

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)*

# Benchmark

The correct solution to the original Project Euler problem was found in 0.8 seconds on a 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=c++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

July 27, 2017 added comments

# Difficulty

Project Euler ranks this problem at **50%** (out of 100%).

# Links

projecteuler.net/thread=309 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

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Look at my progress and performance pages to get more details.

My username at Project Euler is

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# 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.

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