<< problem 125 - Palindromic sums abc-hits - problem 127 >>

# Problem 126: Cuboid layers

The minimum number of cubes to cover every visible face on a cuboid measuring 3 x 2 x 1 is twenty-two.

If we then add a second layer to this solid it would require forty-six cubes to cover every visible face,
the third layer would require seventy-eight cubes, and the fourth layer would require one-hundred and eighteen cubes to cover every visible face.

However, the first layer on a cuboid measuring 5 x 1 x 1 also requires twenty-two cubes; similarly the first layer on cuboids measuring 5 x 3 x 1, 7 x 2 x 1, and 11 x 1 x 1 all contain forty-six cubes.

We shall define C(n) to represent the number of cuboids that contain n cubes in one of its layers. So C(22) = 2, C(46) = 4, C(78) = 5, and C(118) = 8.

It turns out that 154 is the least value of n for which C(n) = 10.

Find the least value of n for which C(n) = 1000.

# My Algorithm

The initial cuboid (without any layers) has six sides:

• front and back side, both with an area of width * height
• left and right side, both with an area of height * depth
• top and bottom side, both with an area of depth * width
The total surface area is 2 * (width * height + height * depth + depth * width).
Each part of the surface is covered by the first layer. That means:
C(1) = 2 * (width * height + height * depth + depth * width).

For the 3 x 2 x 1 cuboid: C(1) = 2 * (3 * 2 + 2 * 1 + 1 * 3) = 2 * (6 + 2 + 3) = 2 * 11 = 22

The computation of the second, third, fourth, ... layer wasn't obvious to me. My drawing skills are sub-par at best.
So I decided to write some code: start with a shape and move that shape one unit left, right, up, down, forward and backward
(start each move at the initial position). The difference between the old shape and the shape created by these six movements is the volume of the new layer.
My function naive() (last lines of my code) performs exactly these steps. Starting with a specified cuboid it finds the volume of each additional layer and prints it.

Playing around with several cuboid sizes I saw a pattern:
C(2) = C(1) + 4 * (width + height + depth)
C(3) = C(2) + 4 * (width + height + depth) + 8
C(4) = C(3) + 4 * (width + height + depth) + 16
C(5) = C(4) + 4 * (width + height + depth) + 24

A bit more general:
C(1) = 2 * (width * height + height * depth + depth * width).
C(n) = C(n-1) + 4 * (width + height + depth) + 8 * (n - 2)

You can convert the second formula into a closed form but then the equation looks a bit messy.
The C++ compiler won't generate much faster code, too. The function Cuboid::calculate evaluates these two equations.

In order to solve C(n) = x for arbitrary x (it's x = 1000 in the original problem) I wrote code based on std::map:
• each cuboids is represented by Cuboid, it stores the size of the initial cuboid and the number of layers
• Cuboid can be sorted by the size of the outermost layer (and then by it's sides)
• since std::map is an ascendingly sorted container, it's first element is the cuboid with the smallest outermost layer
Starting with the most basic cuboid (1,1,1) with only 1 layer I repeat the following:
• pick and remove the first (= smallest) cuboid from the container
• add a cuboid with width + 1 to the container
• add another cuboid with height + 1
• and a cuboid with depth + 1
• finally add a cuboid with layer + 1 to that container
Adding a cuboid is encapsulated in the add() function. It rejects cuboids where height > width and depth > height to avoid
having the same (rotated) cuboid multiple times in the container. Adding exactly the same container twice is prevented, too.
On top of that, add() keeps track of the volume of the most recent layer (in count[]).

That algorithm stops when the smallest cuboid has exactly 1000 "siblings" (→ the count of the current volume is 1000).
It shows the correct result after about 4 seconds.

However, my profiler showed that a few million cuboids where added (and removed) from the container.
The overhead of memory allocation was responsible for the majority of the execution time.

A much faster approach is to guess an upper limit of the result and then create all cuboids (and layers) that fit into that limit.
Keeping track of just the various values of C() instead of the cuboids can be done with a small std::vector.
The program becomes about 50x faster and has almost no memory consumption. See my function fast().

The main problem is to find a good guess for the upper limit.
fast() returns 0 if no result was found, that's when the initial guess has to be incremented and another call to fast() is required.
Starting with a very high guess will find the result but might waste lots of time computing cuboids with too large volumes.
Currently, I use a StepSize = 10000 ... yes, I know it's not optimal (it doesn't find the result in its first iteration)
but I deliberately keep it this way.
If StepSize = 50 then fast() isn't faster than my original code.
If StepSize = 150000 then fast() isn't faster than my original code.
That means for the original problem, fast() is only truly faster if 50 < StepSize < 150000.

## Note

Even though the code of fast() is shorter (and can be faster), I have a strange feeling about it:
somehow it needs a good guess but I haven't found a way to come up with a straightforward way to generate that guess.

I was stuck for a while because initially I stopped when I found a layer volume with at least 1000 matching cuboids (instead of exactly 1000).
There are a few volumes below the correct solution with more than 1000 cuboids.

Kristian Edlund visualized a nice way to enumerate the cubes added in each layer. His blog can found in my links.

# 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 10 | ./126

Output:

Note: the original problem's input 1000 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.

       #include <iostream>
#include <vector>
#include <set>
#include <map>

// a cuboid with 1+ added layers
struct Cuboid
{
// cuboid size
const unsigned short width;
const unsigned short height;
const unsigned short depth;
// wrapped several times
const unsigned short layers;
// compute volume with calculated()
const unsigned int   hullVolume;

// create new cuboid
Cuboid(unsigned short width_, unsigned short height_, unsigned short depth_, unsigned short layers_)
: width(width_), height(height_), depth(depth_), layers(layers_),
hullVolume(calculate())
{}

// calculate volume of the most recent / outermost layer
unsigned int calculate() const
{
// innermost hull
auto result = 2 * ((unsigned int)width  * height +
(unsigned int)height * depth  +
(unsigned int)depth  * width);

// need just one layer ?
if (layers == 1)
return result;

auto increment = 4 * (width + height + depth);
for (unsigned short current = 2; current <= layers; current++)
{
result    += increment;
increment += 8;
}

return result;
}

// for std::set
bool operator<(const Cuboid& other) const
{
// volume
if (hullVolume < other.hullVolume) return true;
if (hullVolume > other.hullVolume) return false;
// the following lines are arbitrary => they only need to be consistent
// sides
if (width  < other.width)  return true;
if (width  > other.width)  return false;
if (height < other.height) return true;
if (height > other.height) return false;
if (depth  < other.depth)  return true;
if (depth  > other.depth)  return false;
// layers
return layers < other.layers;
}
};

// list of cuboids, sorted by volume of their outermost layer
std::set<Cuboid> todo;
// count how many cuboid layers have a certain volume (looking for 1000)
std::map<unsigned int, unsigned int> count; // could be replaced by std::vector if you can estimate the upper limit
// note: this information is implicitely stored in "todo" as well: count the consecutive cuboids with the same hullVolume

// add a cuboid with a certain amount of layers, return number of cuboids with the same hull volume
bool add(unsigned short width, unsigned short height, unsigned short depth, unsigned short layers)
{
// accept only width >= height >= depth
if (width  < height)
return false;
if (height < depth)
return false;

// create new cuboid
Cuboid cuboid(width, height, depth, layers);

if (todo.count(cuboid) != 0)
return false;

todo.insert(cuboid);
count[cuboid.hullVolume]++;

return true;
}

const unsigned int Failed = 0;
// find the correct result if you know an upper bound maxVolume, if not found return Failed
unsigned int fast(unsigned int stopIf, unsigned int maxVolume)
{
// allocate enough memory
std::vector<unsigned int> count(maxVolume + 1, 0);

// just four nested loops to generate all cuboids with hulls <= maxVolume
// ensure that x >= y >= z to avoid enumerating some volume multiple times
for (unsigned short x = 1; Cuboid(x,1,1,1).hullVolume <= maxVolume; x++)
for (unsigned short y = 1; y <= x && Cuboid(x,y,1,1).hullVolume <= maxVolume; y++)
for (unsigned short z = 1; z <= y && Cuboid(x,y,z,1).hullVolume <= maxVolume; z++)
for (unsigned short layer = 1; ; layer++)
{
auto current = Cuboid(x,y,z,layer).hullVolume;
if (current > maxVolume)
break;

count[current]++;
}

// find smallest hull volume that appears exactly 1000 times
for (size_t i = 0; i < count.size(); i++)
if (count[i] == stopIf)
return i;

// failed
return Failed;
}

// forward declaration of my old helper code
//unsigned int naive(unsigned int x, unsigned int y, unsigned int z, unsigned int layers);

int main()
{
// start old code to analyze a single cuboid
//naive(3,2,1, 5); return 0;

// look for that number of cuboids with the same volume
unsigned int stopIf = 1000;
std::cin >> stopIf;

// ---------- fast enumeration ----------

// find solution if an upper limit of the result is known
#define FAST_ENUMERATION
#ifdef  FAST_ENUMERATION
const unsigned int StepSize = 10000; // I deliberately choose not the optimal value !
// assume that the result is below 10000, then 20000, ... until fast() returns > 0
for (unsigned int maxVolume = StepSize; ; maxVolume += StepSize)
{
// found ?
if (current != Failed)
{
std::cout << current << std::endl;
return 0;
}
}
#endif

// ---------- slower enumeration ----------

// note: I found the correct result with this approach but later created the faster enumeration you find above

// initial cuboid: one unit wide, high and deep with one layer around it
unsigned int volume = 0;
while (count[volume] != stopIf)
{
// pick cuboid with smallest volume and remove it
auto current = *(todo.begin());
todo.erase(todo.begin());

volume = current.hullVolume;

// enlarge cuboid
add(current.width + 1, current.height, current.depth, current.layers);
add(current.width, current.height + 1, current.depth, current.layers);
add(current.width, current.height, current.depth + 1, current.layers);
add(current.width, current.height, current.depth, current.layers + 1);
}

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

// ---------- compute hull volume, slow approach ----------

// a single cube that is part of an cuboid (more or less a coordinate)
struct Cube
{
// coordinates
const int x, y, z;

// create new cube
Cube(int xx, int yy, int zz)
: x(xx), y(yy), z(zz)
{}

// for std::set
bool operator<(const Cube& other) const
{
if (x < other.x) return true;
if (x > other.x) return false;
if (y < other.y) return true;
if (y > other.y) return false;
return z < other.z;
}
};

unsigned int naive(unsigned int x, unsigned int y, unsigned int z, unsigned int layers)
{
// keep track of all cubes belonging to the shape
std::set<Cube> cuboid;
// cubes added in the most recent layer
std::vector<Cube> hull;

// initial shape
for (int xx = 0; xx < (int)x; xx++)
for (int yy = 0; yy < (int)y; yy++)
for (int zz = 0; zz < (int)z; zz++)
{
Cube current(xx, yy, zz);
cuboid.insert(current);
hull.push_back(current);
}

// add desired numbers of layers
for (unsigned int layer = 1; layer <= layers; layer++)
{
std::vector<Cube> next;
for (auto current : hull)
{
// grow in each direction x,y,z
Cube left (current.x - 1, current.y, current.z);
Cube right(current.x + 1, current.y, current.z);
Cube down (current.x, current.y - 1, current.z);
Cube up   (current.x, current.y + 1, current.z);
Cube back (current.x, current.y, current.z - 1);
Cube front(current.x, current.y, current.z + 1);

if (cuboid.count(left)  == 0)
{
cuboid.insert(left);
next.push_back(left);
}
if (cuboid.count(right) == 0)
{
cuboid.insert(right);
next.push_back(right);
}
if (cuboid.count(down)  == 0)
{
cuboid.insert(down);
next.push_back(down);
}
if (cuboid.count(up)    == 0)
{
cuboid.insert(up);
next.push_back(up);
}
if (cuboid.count(back)  == 0)
{
cuboid.insert(back);
next.push_back(back);
}
if (cuboid.count(front) == 0)
{
cuboid.insert(front);
next.push_back(front);
}
}

std::cout << "layer " << layer << ": "
<< "C(" << x << "," << y << "," << z << ")=" << next.size();
if (layer > 1)
std::cout << " delta=" << next.size() - hull.size();
std::cout << std::endl;

// next iteration
hull = std::move(next);
}

return hull.size();
}


This solution contains 40 empty lines, 57 comments and 7 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.04 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

August 18, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 55% (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.

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