Count of Equilateral Triangles of unit length possible from a given Hexagon

Given an array S[] consisting of the lengths of the 6 sides of a Hexagon, the task is to calculate the number of equilateral triangles of unit length that can be made from the given hexagon.

Examples:

Input: S = {1, 1, 1, 1, 1, 1} 
Output: 6 
Explanation:

Input: S = {2, 2, 1, 3, 1, 2} 
Output: 19 
Explanation:

Approach: The following observations need to be made to solve the given problem: 

• Consider an equilateral triangle of ‘X’ side length. It has been divided into smaller triangles of unit length each, by drawing lines parallel to its sides.
• Below are the images of three such equilateral triangles: 
  

Example 1: X = 2

Example 2: X = 3

Example 3: X = 5

• In each of the above three examples, the count of unit length equilateral triangles possible are: 
  

1. X = 2: 4 equilateral triangles of 1 unit length side.
2. X = 3: 9 equilateral triangles of 1 unit length side.
3. X = 5: 25 equilateral triangles of 1 unit length side.
   

• By observation, it is clear that, for an equilateral triangle of side length X, X²equilateral triangles of unit length are possible.
• Extending this observation to Hexagons, inscribe Hexagons inside the equilateral triangles, as shown below:

A regular Hexagon inscribed from equilateral triangle of side X = 3, has 6 mini triangles inside it.

An irregular Hexagon inscribed from the equilateral triangle of side X = 5, has 19 mini triangles inside it.

• It can be observed that by removing a certain number of mini triangles from the bigger triangle, the hexagon with given dimensions can be found.

The formula for counting the number of triangles of unit length can be generalized for a Hexagon having six sides S₁ , S₂ , S₃ , S₄ , S₅ , S₆ as:

Number of triangles that can be formed = ( S₁ + S₂ + S₃ )² – S₁² – S₃² – S₅²

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate the
// the number of Triangles possible
int calculateTriangles(int sides[])
{
    double count = pow(sides[0] + sides[1] +
                       sides[2], 2);
    count -= pow(sides[0], 2);
    count -= pow(sides[2], 2);
    count -= pow(sides[4], 2);

    return (int)(count);
}

// Driver Code
int main()
{

    // Regular Hexagon
    int sides[] = { 1, 1, 1, 1, 1, 1 };
    cout << (calculateTriangles(sides)) << endl;

    // Irregular Hexagon
    int sides1[] = { 2, 2, 1, 3, 1, 2 };
    cout << (calculateTriangles(sides1)) << endl;

    return 0;
}

// This code is contributed by 29AjayKumar

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Function to calculate the
// the number of Triangles possible
static int calculateTriangles(int sides[])
{
    double count = Math.pow(sides[0] + sides[1] +
                            sides[2], 2);
    count -= Math.pow(sides[0], 2);
    count -= Math.pow(sides[2], 2);
    count -= Math.pow(sides[4], 2);

    return (int)(count);
}

// Driver Code
public static void main(String[] args)
{

    // Regular Hexagon
    int sides[] = { 1, 1, 1, 1, 1, 1 };
    System.out.print((calculateTriangles(sides)) + "\n");

    // Irregular Hexagon
    int sides1[] = { 2, 2, 1, 3, 1, 2 };
    System.out.print((calculateTriangles(sides1)) + "\n");
}
}

// This code is contributed by amal kumar choubey

Python3

# Python3 Program to implement
# the above approach

# Function to calculate the
# the number of Triangles possible
def calculateTriangles(sides):
    count = pow( sides[0] + sides[1] + sides[2], 2)
    count -= pow( sides[0], 2)
    count -= pow( sides[2], 2)
    count -= pow( sides[4], 2)

    return int(count)

# Driver Code

# Regular Hexagon
sides = [1, 1, 1, 1, 1, 1]
print(calculateTriangles(sides))

# Irregular Hexagon
sides = [2, 2, 1, 3, 1, 2]
print(calculateTriangles(sides))

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to calculate the
// the number of Triangles possible
static int calculateTriangles(int []sides)
{
    double count = Math.Pow(sides[0] + sides[1] +
                            sides[2], 2);
    count -= Math.Pow(sides[0], 2);
    count -= Math.Pow(sides[2], 2);
    count -= Math.Pow(sides[4], 2);

    return (int)(count);
}

// Driver Code
public static void Main(String[] args)
{

    // Regular Hexagon
    int []sides = { 1, 1, 1, 1, 1, 1 };
    Console.Write((calculateTriangles(sides)) + "\n");

    // Irregular Hexagon
    int []sides1 = { 2, 2, 1, 3, 1, 2 };
    Console.Write((calculateTriangles(sides1)) + "\n");
}
}

// This code is contributed by amal kumar choubey

Javascript

<script>

// JavaScript program to implement
// the above approach

// Function to calculate the
// the number of Triangles possible
function calculateTriangles(sides)
{
    let count = Math.pow(sides[0] + sides[1] +
                            sides[2], 2);
    count -= Math.pow(sides[0], 2);
    count -= Math.pow(sides[2], 2);
    count -= Math.pow(sides[4], 2);

    return (count);
}

// Driver Code

    // Regular Hexagon
    let sides = [ 1, 1, 1, 1, 1, 1 ];
    document.write((calculateTriangles(sides)) + "<br/>");

    // Irregular Hexagon
    let sides1 = [ 2, 2, 1, 3, 1, 2 ];
    document.write((calculateTriangles(sides1)) + "<br/>");

</script>

Output:

6
19

Time Complexity: O(1)
Auxiliary Space: O(1)