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# Problem 317: Firecracker

A firecracker explodes at a height of 100 m above level ground.
It breaks into a large number of very small fragments, which move in every direction;
all of them have the same initial velocity of 20 m/s.

We assume that the fragments move without air resistance, in a uniform gravitational field with g=9.81 m/s^2.

Find the volume (in m^3) of the region through which the fragments move before reaching the ground.

# My Algorithm

The German Wikipedia page on trajectory (de.wikipedia.org/wiki/Wurfparabel) says that the formula of the hull is:
(1) y = dfrac{v^2}{2g} - dfrac{g x^2}{2v^2} + h_0

where g = 9.81, h_0 = 100 and v = 20
I couldn't find it on the English page (en.wikipedia.org/wiki/Trajectory_of_a_projectile), though.

Splitting the formula into constant and variable parts:
(2) add = dfrac{v^2}{2g} + h_0
(3) factor = dfrac{g}{2v^2}
(4) y = factor * x^2 + add

The volume of a rotated body is
(5) V = pi * integral{(f(x))^2} dx

Unfortunately, equation (5) applies only to rotations around the x-axis.
The fireworks are rotated around the y-axis, thus I have to find its inverse (swap x and y):
(6) x = factor * y^2 + add
(7) factor * y^2 = x - add
(8) y^2 = dfrac{x - add}{factor}

(9) y = sqrt{dfrac{x - add}{factor}}

I replace equation (9) in (5):
(10) V = pi * integral{(sqrt{dfrac{x - add}{factor}})^2} dx

(11) V = pi * integral{dfrac{x - add}{factor}} dx

(12) V = dfrac{pi}{factor} * integral(x - add) dx

Integrating:

Only "points above ground" should be taken into consideration:

• the integral starts at zero, x_{low} = 0
• the integral ends at the highest point of any firework, which is in the centre, that means x_{high} = f(0) in (1) which is x_{high} = add
Now all I have to do is computing (12) for both borders x_{low} and x_{high} of the integral:
(14) V = dfrac{pi}{factor} * (( dfrac{x_{high}^2}{2} - add * x_{high}) - (dfrac{x_{low}^2}{2} - add * x_{low} ))

It turns out that for x_{low} = 0 most of the equation is zero and simplifies to:
(15) V = dfrac{pi}{factor} * (dfrac{x_{high}^2}{2} - add * x_{high})

x_{high} = add allows even further simplification:
(17) V = dfrac{pi}{factor} * dfrac{-add^2}{2}
(18) V = -dfrac{pi}{2} * dfrac{add^2}{factor}

Equation 18 is used in my code. And in case you forgot what add and factor are:
(19) V = -dfrac{pi}{2} * dfrac{(dfrac{v^2}{2g} + h_0)^2}{dfrac{g}{2v^2}}

The result is basically found instantly without any memory consumption.

## Alternative Approaches

... I'm a software engineer and felt that something was missing !

Therefore I wrote a simple loop to integrate the initial formula (1) numerically (well, actually I used equation (4) which is the same):

• slowly step along the x-axis and compute y = factor * x * x + add
• y will be monotoneously decreasing
• each step that I take adds a small cylinder/slice to the volume (with sliceHeight = lastY - y and sliceVolume = pi * x * x * sliceHeight)
It's tricky to find the correct step which defines the distance between two consecutive x-values:
• if step is too large, then rounding artifacts falsify the result
• if step is too small, then the algorithm can be very slow and the volume because so tiny, that it can't be properly represented by double
→ another type of rounding artifacts

My code contains a dynamic adjustment of step after each step, depending on the previous slice volume.
Nevertheless, I can't achieve a sufficient precision: two decimal places are no problem but I haven't found the
optimal value for averageSliceVolume such that the correct result is displayed.

To enable the numeric integration enable #define NUMERIC_INTEGRATION.

# 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 velocity and the initial height above ground.

This is equivalent to
echo "10 0" | ./317

Output:

Note: the original problem's input 20 100 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 <iomanip>

// nore than enough digits of pi ...
const double pi = 3.141592653589793238462;
// gravition (in m/s^2)
const double g  = 9.81;

int main()
{
std::cout << std::fixed << std::setprecision(4);

// velocity in m/s
double velocity =  20;
// height above ground (in m)
double height   = 100;

std::cin >> velocity >> height;

// split formula into constant summand ...
auto add = velocity * velocity / (2 * g) + height;
// ... and factor
auto factor = - g / (2 * velocity * velocity);

// integral from 0 to "add" of pi * (0.5 * i * i - add * i) / factor
// => it's zero at 0, just compute at "add"
auto result = -0.5 * pi * add * add / factor;
std::cout << result << std::endl;

//#define NUMERIC_INTEGRATION
#ifdef NUMERIC_INTEGRATION
// accumulated volume so far
double volume = 0;
// distance between two consecutive x
double step   = 0.00001;
// current x
double x      = 0;

// y of previous iteration
double lastY  = 0;

// preferred size of a slice, adjust "step" accordingly
double averageSliceVolume = 0.0001;

while (true)
{
// evaluate formula (4)
auto y = x * x * factor + add;
// hit the ground ? ==> done
if (y <= 0)
break;

// height of cylinder
auto sliceHeight = lastY - y;
// volume of cylinder
auto sliceVolume = pi * x * x * sliceHeight;

volume += sliceVolume;

// slice was too large / too small ? adjust distance to next x
if (sliceVolume > averageSliceVolume)
step /= 2;
else
step *= 2;

// next iteration
x      += step;
lastY   = y;
}

std::cout << volume << std::endl;
#endif

return 0;
}


This solution contains 16 empty lines, 22 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

July 5, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 35% (out of 100%).

# Heatmap

Please click on a problem's number to open my solution to that problem:

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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 315 - Digital root clocks Swapping Counters - problem 321 >>
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