<< problem 219 - Skew-cost coding Tribonacci non-divisors - problem 225 >>

# Problem 222: Sphere Packing

What is the length of the shortest pipe, of internal radius 50mm, that can fully contain 21 balls of radii 30mm, 31mm, ..., 50mm?

Give your answer in micrometres (10^{-6} m) rounded to the nearest integer.

# My Algorithm

This was a strange problem ... I came up with a solution that was slow and memory hungry.
Then I saw an optimization and it became faster and even used less memory.
But then I saw that I can basically solve it instantly - and wasn't willing the adapt the code accordingly.

My solution isn't based on the fully optimized (I dare to say "trivial") algorithm.
It basically tries to put all balls in various orders into the pipe and measures which combination is the shortest.

The first observation is that the 3D objects (balls, pipe) can be reduced to their 2D equivalents (circles between two lines).
If two balls a and b are stacked on top of each other, then their centres have a distance of d = r_a + r_b.
I assumed that the pipe is vertical. Then a right triangle exists:
the hypotenuse is d = r_a + r_b, one side is x = (50 - r_a) + (50 - r_b) = 100 - (r_a + r_b) and the other side y is:
(1) d^2 = x^2 + y^2
(2) y^2 = d^2 - x^2
(3) y = sqrt{d^2 - x^2}

Substituting d and x:
(4) y = sqrt{(r_a + r_b)^2 - (100 - (r_a + r_b))^2}

→ This formula can be found in the function getDistanceY.

The function search enumerated all combinations and returns the correct result using Dynamic Programming.
As mentioned before, my initial approach was working but was slow and memory hungry.
Then it appeared to me that the two largest balls must be at the ends of the pipe (which was verified by search).
Fixing those two balls shrinks the size of the internal cache if size by factor 4 and speeds up the algorithm substantially.

The code below does exactly that.

## Alternative Approaches

But then I thought: if I remove the two biggest balls - what happens to the rest ?
Well, the same idea can be applied over and over again: pick the next two largest balls and put them next to the previously largest balls, and so on.
In the end, the balls will be placed like this: 50mm, 48mm, 46mm, 44mm, ..., 43mm, 45mm, 47mm, 49mm.
Generating that sequence can be done in a simple for-loop and eliminates the need for search (and its large cache).
The expecting computation time will be close to zero.

# 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 pipe's radius, then the radius of the smallest ball followed by the radius of the largest ball.

This is equivalent to
echo "10 5 10" | ./222

Output:

Note: the original problem's input 50 30 50 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>

// double has 64 bits => memoized cache would be twice as large
typedef float Number;

// an arbitrarily high number (more than sum of all radii)
const Number WorstCase = 999999999;

// get distance between centre of two balls along the y axis
{
}

{
// are all balls in the pipe (except for the largest) ?

// memoize
const unsigned int Shift = numBalls - 2; // ignore the two largest balls
const Number Invalid = -1;
static std::vector<Number> cache((1 << Shift) * (maxRadius - minRadius + 1), Invalid);
if (cache[id] != Invalid)
return cache[id];

Number best = WorstCase;
{
if ((mask & bit) == 0)
continue;

// process recursively
// improvement ?
if (best > newHeight)
best = newHeight;
}

cache[id] = best;
return best;
}

int main()
{
// avoid invalid input (for live test only)
return 1;
return 2;

// all bits set => all balls still available

// the two largest balls must be at the ends of the pipe, remove those bits

auto first = maxRadius - 1;
// half of the first ball (from its border to its centre) plus everything else
auto best  = first + search(mask, first);

// convert to micrometre and round to nearest integer
std::cout << (unsigned int)round(1000 * best) << std::endl;
return 0;
}


This solution contains 15 empty lines, 16 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 2.2 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 45 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 24, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 60% (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 219 - Skew-cost coding Tribonacci non-divisors - problem 225 >>
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