<< problem 571 - Super Pandigital Numbers 47-smooth triangular numbers - problem 581 >>

# Problem 577: Counting hexagons

An equilateral triangle with integer side length n >= 3 is divided into n^2 equilateral triangles with side length 1 as shown in the diagram below.
The vertices of these triangles constitute a triangular lattice with dfrac{(n+1)(n+2)}{2} lattice points.

Let H(n) be the number of all regular hexagons that can be found by connecting 6 of these points. For example, H(3)=1, H(6)=12 and H(20)=966.

Find sum_{n=3}^12345{H(n)}.

# My Algorithm

My initial brute-force algorithm works as follows:

• assume that every hexagon's center is a lattice point, too (bold assumption - but seems to works for all H(n))
• for each such lattice point, compute the distances to every other lattice point
• compute the angle as well
• keep a data structure of all distances and angles for each lattice point
• round values to avoid errors due to numeric imprecision
A hexagon can be identified by finding:
• six points with the same distance
• their angles are 60 degrees apart from each other
The bruteForce() function computes H(20) = 966 in a few milliseconds. For n > 100 it becomes pretty slow, though.

I entered the values H(3), H(4), H(5), etc. (1,3,6,12,21,33,51,75,105,145,195,255,...) in www.oeis.org and found the sequence OEIS A011779.
Unfortunately I couldn't handle the series expansion in a few lines of C++ code.
Since the "larger" hexagons appear when n mod 3 == 0 I split the sequence into three parts:
• H(3), H(6), H( 9), etc. (1,12,51,145,...) → OEIS A236770:
• H(4), H(7), H(10), etc. (3,21,75,195,...) → OEIS A228317:
• H(5), H(8), H(11), etc. (6,33,105,255,...)
Even though I found sequences with simple formulas for n mod 3 == 0 and n mod 3 == 1, there was no such sequence for n mod 3 == 2.
However, I noticed that every H(n) for n mod 3 == 2 is a multiple of 3. And voila: 2,11,35,85 is OEIS A000914.

In the end I have a simple loop from 3 to 12345 that calls A236770(), A228317() or A000914(), depending on n % 3.

## Note

Honestly, I thought that my bruteForce() results might be wrong even for small n because of numeric imprecisions.
But the code proved to be quite stable: all H(n) for n <= 100 are correct (I didn't verify larger n).

Today is the end of the year - and the end of the fiscal year, too, so I donated 10 bucks to OEIS. For tax reasons :-)

# 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: The output is the sum of H(3...n)

This is equivalent to
echo 20 | ./577

Output:

Note: the original problem's input 12345 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 <map>
#include <algorithm>
#include <cmath>

// ---------- brute force up to n=100 ----------

const auto PI = 3.14159265358979323846264;
const auto Precision = 0.00001;

// 2D point
struct Point
{
double x, y;
Point(double x_, double y_)
: x(x_), y(y_)
{ }

// distance between two points
double distance(const Point& other) const
{
auto dx = other.x - x;
auto dy = other.y - y;
return sqrt(dx*dx + dy * dy); // I could skip sqrt because the squared distance works as well
}

// angle (in degrees, always positive)
double angle(const Point& other) const
{
auto dx = other.x - x;
auto dy = other.y - y;

// convert to degrees
auto degrees = radian * 180 / PI;
if (degrees < 0)
degrees += 360;
return degrees;
}
};

// round a double with a certain precision, e.g. round(5.126, 0.01) = 5.13
double round(double a, double precision)
{
return round(a / precision) * precision;
}

// look for hexagons using polar coordinates
unsigned long long bruteForce(unsigned int size)
{
// horizontal and vertical distance between two neighbors in the lattice / grid
double dx = 1;
double dy = sqrt(3.0) / 2; // height of https://en.wikipedia.org/wiki/Equilateral_triangle

// generate all points
std::vector<Point> points;
auto numPoints = (size + 1) * (size + 2) / 2;
for (unsigned int gridY = 0; gridY <= size; gridY++)
{
// position of the left-most point in each row
double y = gridY * dy;
double x = gridY * dx / 2;

// all points in the current row
auto width = size - gridY;
for (unsigned int gridX = 0; gridX <= width; gridX++)
{
points.push_back(Point(x, y));
x += dx;
}
}

// for each points: compute polar coordinates of the vector to each other point
std::vector<std::map<double, std::vector<double>>> polar; // map: distance => angles
for (auto i : points)
{
polar.resize(polar.size() + 1);
auto& current = polar.back();
for (auto j : points)
{
auto distance = i.distance(j);
auto angle    = i.angle(j);

// avoid precision issues
distance = round(distance, Precision);
angle    = round(angle,    Precision);

if (distance > 0)
current[distance].push_back(angle);
}
}

// find hexagons amidst these polar coordinates:
// six points having the same distance where angles are (2*pi) / 6 apart
unsigned long long numFound = 0;
for (unsigned int point = 0; point < numPoints; point++)
{
for (auto& distance : polar[point])
{
auto& candidates = distance.second;
// hexagon impossible ?
if (candidates.size() < 6)
continue;

// look for subsets with 6 elements { start,     start+60,  start+120,
//                                    start+180, start+240, start+300 }
std::sort(candidates.begin(), candidates.end());
for (auto start : candidates)
{
// smallest angle must be 0 <= start < 60 (in degrees)
if (start >= 60) // hexagon has 6 sides, so 360 / 6 = 60
break;

// check if angle+60, +120, +180, ... exist
bool valid = true;
for (auto next = start + 60; next < 360; next += 60)
// note: need the same rounding as used before !
if (!std::binary_search(candidates.begin(), candidates.end(), round(next, Precision)))
{
valid = false;
break;
}

// yes, found another hexagon
if (valid)
numFound++;
}
}
}

return numFound;
}

// ---------- super-fast algorithm ----------

// sequence oeis.org/A000914
unsigned long long A000914(unsigned long long n)
{
return n * (n+1) * (n+2) * (3*n+5) / 24;
}

// sequence oeis.org/A228317
unsigned long long A228317(unsigned long long n)
{
return n * (n-1) * (n-2) * (3*n-5) /  8;
}

// sequence oeis.org/A236770
unsigned long long A236770(unsigned long long n)
{
return n * (n+1) * (3*n*n + 3*n - 2) / 8;
}

int main()
{
unsigned int size = 12345; // n
std::cin >> size;

// brute force
//std::cout << bruteForce(size) << std::endl;

// indices for A000914, A228317 and A236770
unsigned int i000914 = 1;
unsigned int i228317 = 3;
unsigned int i236770 = 1;

unsigned long long sum = 0;
// no hexagons below n = 3
for (unsigned int n = 3; n <= size; n++)
{
unsigned long long fast = 0;
// method of calculation depends on n mod 3
switch (n % 3)
{
case 0:
fast = A236770(i236770++);
break;
case 1:
fast = A228317(i228317++);
break;
case 2:
fast = 3 * A000914(i000914++);
break;
}

sum += fast;
}

// that's it !
std::cout << sum << std::endl;
return 0;
}


This solution contains 28 empty lines, 34 comments and 5 preprocessor commands.

# Changelog

December 31, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 20% (out of 100%).

# Heatmap

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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 571 - Super Pandigital Numbers 47-smooth triangular numbers - problem 581 >>
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