Problem 189: Tri-colouring a triangular grid

(see projecteuler.net/problem=189)

Consider the following configuration of 64 triangles:

grid
We wish to colour the interior of each triangle with one of three colours: red, green or blue, so that no two neighbouring triangles have the same colour. Such a colouring shall be called valid. Here, two triangles are said to be neighbouring if they share an edge.
Note: if they only share a vertex, then they are not neighbours.

For example, here is a valid colouring of the above grid:

colored grid
A colouring C' which is obtained from a colouring C by rotation or reflection is considered distinct from C unless the two are identical.

How many distinct valid colourings are there for the above configuration?

My Algorithm

I solved this problem in classic Dynamic Programming way:
- generate each possible row triangle-by-triangle, beginning with the top row
- whenever I start a new row check if I already fully processed a row similar to the most previous row

The trick is to define the word "similar": I call rows similar to each other if they are followed by the same number of combinations for all rows beneath it.
For example, each color of the top-most triangle is similar to each other because I can permute the color such that I can produce the original coloring.

My function getId() returns a hash value of a row stored in triangle. If two rows share the same hash then they must be similar.
Only a row's triangles that share an edge with the row beneath it are relevant (in general: each triangle with an even index).
In the third row of the example, only the left-most green and the two blue triangles have to be considered by getId().

Each color is represented by an integer: Red = 1, Green = 2, Blue = 3.
Initially I used these codes to get a hash value
id = row * (color[0] * 3^0 + color[2] * 3^1 + color[4] * 3^2 * ...)

Then it occurred to me that the differences between those relevant triangles are sufficient (plus 3 if difference is negative):
id = row * ((color[2] - color[0]) * 3^0 + (color[4] - color[2]) * 3^1 + (color[6] - color[4]) * 3^2 + ...)

Due to symmetry, each row is similar to itself when all relevant triangles are in reverse order (see #define SYMMETRIC_ID).
So I compute two hashes and choose the smaller one because I want to have as many rows as possible to share the same hash value / ID.

Alternative Approaches

There is something called "chromatic polynomials" which can solve this problem, too (see en.wikipedia.org/wiki/Chromatic polynomial).
I haven't tried it but it seems to be a faster way.

Modifications by HackerRank

The number of colors can be larger than 3 and the maximum height of the triangle is 11.
My solution times out for about every fifth test case.

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 "4 3" | ./189

Output:

(please click 'Go !')

Note: the original problem's input 8 3 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.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

#include <iostream>
#include <vector>
#include <map>
 
#define ORIGINAL // disable for Hackerrank
#define SYMMETRIC_ID // identify rows in reverse order as similar
 
// total height
unsigned int height = 8;
unsigned int numColors = 3;
 
// "scratchpad", used while processing the colorings
typedef char Color;
std::vector<Color> triangles;
 
// cache result for completed rows, index is determined by getId()
#ifdef ORIGINAL
std::vector<unsigned long long> cache(5102 + 1, 0); // sufficient for height<=8 and numColors<=3
#else
std::map<unsigned int, unsigned long long> cache; // handle bigger input values, too
#endif
 
// compute hash value of a row
unsigned int getId(unsigned int row)
{
auto first = row * row;
auto width = 2 * row + 1;
 
// compute hash of triangles: find differences between adjacent triangles of the same orientation
unsigned int result = row;
for (auto i = first + 2; i < first + width; i += 2)
{
auto diff = triangles[i - 2] - triangles[i];
if (diff < 0)
diff += numColors;
 
result = result * numColors + diff;
}
 
#ifdef SYMMETRIC_ID
// simple optimization based on symmetry: process the same triangles in reverse order:
unsigned int reverse = row;
for (auto i = first + width - 1; i >= first + 2; i -= 2)
{
// exactly the same as above but "i" is running in reverse
auto diff = triangles[i - 2] - triangles[i];
if (diff < 0)
diff += numColors;
 
reverse = reverse * numColors + diff;
}
if (result > reverse)
result = reverse;
#endif
 
return result;
}
 
// simple enumeration, solves triangle with 5 rows in about a second
unsigned long long search(unsigned int row = 0, unsigned int column = 0)
{
// index of the first triangle in the current row
auto first = row * row;
// index of the current triangle
auto index = first + column;
// number of triangles in the current row
auto width = 2 * row + 1;
 
// "coordinates" of the next triangle
auto nextRow = row;
auto nextColumn = column + 1;
if (nextColumn == width)
{
// start a new row
nextRow++;
nextColumn = 0;
}
 
// hash value of the previous row, only needed when starting a new row
unsigned int prevId = 0;
if (column == 0)
{
// finished ?
if (row == height)
return 1; // found another valid coloring
 
if (row == 0)
cache.clear(); // only relevant when computing multiple triangle sizes (as I did while developing my solution)
else // row > 0
{
// already cached ?
prevId = getId(row - 1);
if (cache[prevId] != 0)
return cache[prevId];
}
}
 
// recursively process current position
unsigned long long result = 0;
if (column % 2 == 0) // even: no contact to triangle in row above
for (Color color = 1; color <= (Color)numColors; color++)
{
// exclude color of left neighbor (but first triangle in a row can have any color because it has no left neighbor)
if (column > 0 && triangles[index - 1] == color)
continue;
triangles[index] = color;
result += search(nextRow, nextColumn);
}
else // odd: must respect triangle in row above
for (Color color = 1; color <= (Color)numColors; color++)
{
// exclude color of left neighbor and neighbor in row above
if (triangles[index - 1] == color || triangles[index - 2 * row] == color)
continue;
triangles[index] = color;
result += search(nextRow, nextColumn);
}
 
#ifndef ORIGINAL
// Hackerrank only
result %= 1000000007;
#endif
 
// store result
if (column == 0 && row > 0)
cache[prevId] = result;
 
return result;
}
 
int main()
{
std::cin >> height >> numColors;
triangles.resize(height * height);
 
// let's go !
std::cout << search() << std::endl;
return 0;
}

This solution contains 20 empty lines, 22 comments and 12 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 0.03 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(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

September 5, 2017 submitted solution
September 5, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler189

My code solves 72 out of 91 test cases (score: 78.89%)

I failed 0 test cases due to wrong answers and 19 because of timeouts

Difficulty

70% Project Euler ranks this problem at 70% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

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

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The 270 solved problems (level 10) had an average difficulty of 31.3% at Project Euler and
I scored 13,386 points (out of 15600 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.

more about me can be found on my homepage, especially in my coding blog.
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