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Number of triangles after N moves
  • Last Updated : 05 Apr, 2021
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Find the number of triangles in Nth step, 
Rules: Draw an equilateral triangle at the start. In the i-th move, take the uncolored triangles, divides each of them in 4 parts of equal areas and color the central part. Keep a count of triangles till the Nth step.

Examples:  

Input : 1
Output : 5 
Explanation: In 1st move we get

Input : 2
Output : 17 
Explanation: In 2nd move we get

Naive approach: 
The number of triangles in the nth figure are 3 times the number of triangles in the (n-1)th figure+2. We can see by observation the nth figure is made by placing 3 triangles similar to that in (n-1) figure and an inverted triangle. We also take into account the bigger triangle that has been formed. Hence the number of triangles in the nth figure becomes (number of triangles in the (n-1)th figure)*3 + 2. 



C++




// C++ program to calculate the number of equilateral
// triangles
#include <bits/stdc++.h>
using namespace std;
// function to calculate number of triangles in Nth step
int numberOfTriangles(int n)
{
    int answer[n + 1] = { 0 };
    answer[0] = 1;
    for (int i = 1; i <= n; i++)
        answer[i] = answer[i - 1] * 3 + 2;
     
    return answer[n];
}
 
// driver program
int main()
{
    int n = 2;
    cout << numberOfTriangles(n);
    return 0;
}

Java




// Java program to find middle of three
// distinct numbers to calculate the
// number of equilateral triangles
import java.util.*;
 
class Triangle
{  
    // function to calculate number of
    // traingles in Nth step
    public static int numberOfTriangles(int n)
    {
        int[] answer = new int[n+1];
        answer[0] = 1;
         
        for (int i = 1; i <= n; i++)
            answer[i] = answer[i - 1] * 3 + 2;
     
        return answer[n];
    }
     
    // driver code
    public static void main(String[] args)
    {
        int n = 2;
        System.out.println(numberOfTriangles(n));
    }
}
 
// This code is contributed by rishabh_jain

Python3




# Python3 code to calculate the
# number of equilateral triangles
 
# function to calculate number
# of traingles in Nth step
def numberOfTriangles (n) :
    answer = [None] * (n + 1);
    answer[0] = 1;
    i = 1
    while i <= n:
        answer[i] = answer[i - 1] * 3 + 2;
        i = i + 1
     
    return answer[n];
 
# Driver code
n = 2
print(numberOfTriangles(n))
 
# This code is contributed by "rishabh_jain".

C#




// C# program to find middle of three
// distinct numbers to calculate the
// number of equilateral triangles
using System;
 
class Triangle
{
    // function to calculate number of
    // traingles in Nth step
    public static int numberOfTriangles(int n)
    {
        int[] answer = new int[n+1];
        answer[0] = 1;
         
        for (int i = 1; i <= n; i++)
            answer[i] = answer[i - 1] * 3 + 2;
     
        return answer[n];
    }
     
    // Driver code
    public static void Main()
    {
        int n = 2;
        Console.WriteLine(numberOfTriangles(n));
    }
}
 
// This code is contributed by vt_m

PHP




<?php
// PHP program to calculate
// the number of equilateral
// triangles
 
// function to calculate number
// of traingles in Nth step
function numberOfTriangles($n)
{
    $answer = array();
    $answer[0] = 1;
    for ($i = 1; $i <= $n; $i++)
        $answer[$i] = $answer[$i - 1] *
                               3 + 2;
     
    return $answer[$n];
}
 
    // Driver Code
    $n = 2;
    echo numberOfTriangles($n);
 
// This code is contributed by anuj_67.
?>

Javascript




<script>
 
// Javascript program to calculate
// the number of equilateral
// triangles
 
// Function to calculate number
// of triangles in Nth step
function numberOfTriangles(n)
{
    let answer = new Uint8Array(n + 1);
    answer[0] = 1;
     
    for(let i = 1; i <= n; i++)
        answer[i] = answer[i - 1] * 3 + 2;
     
    return answer[n];
}
 
// Driver code
let n = 2;
 
document.write(numberOfTriangles(n));
     
// This code is contributed by Mayank Tyagi
     
</script>

Output:

17

Time Complexity: O(n)

An efficient solution will be to derive a formula for Nth step: 
If we follow the naive approach for every step, then we get for Nth step the number of triangles to be  

(2*(3^n))-1. 

C++




// C++ program to calculate the number of
// equilateral triangles
#include <bits/stdc++.h>
using namespace std;
 
// function to calculate number of triangles
// in Nth step
int numberOfTriangles(int n)
{
    int ans = 2 * (pow(3, n)) - 1;
    return ans;
}
 
// driver program
int main()
{
    int n = 2;
    cout << numberOfTriangles(n);
    return 0;
}

Java




// Java program to find middle of three
// distinct numbers to calculate the
// number of equilateral triangles
import java.util.*;
import static java.lang.Math.pow;
 
class Triangle
{  
    // function to calculate number
    // of triangles in Nth step
    public static double numberOfTriangles(int n)
    {
        double ans = 2 * (pow(3, n)) - 1;
        return ans;
    }
     
    // driver code
    public static void main(String[] args)
    {
        int n = 2;
        System.out.println(numberOfTriangles(n));
    }
}
 
// This code is contributed by rishabh_jain

Python3




# Python3 code to calculate the
# number of equilateral triangles
 
# function to calculate number
# of traingles in Nth step
def numberOfTriangles (n) :
    ans = 2 * (pow(3, n)) - 1;
    return ans;
 
# Driver code
n = 2
print (numberOfTriangles(n))
 
# This code is contributed by "rishabh_jain".

C#




//C# program to find middle of three
// distinct numbers to calculate the
// number of equilateral triangles
using System;
 
class Triangle
{
    // function to calculate number
    // of triangles in Nth step
    public static double numberOfTriangles(int n)
    {
        double ans = 2 * (Math.Pow(3, n)) - 1;
        return ans;
    }
     
    // Driver code
    public static void Main()
    {
        int n = 2;
        Console.WriteLine(numberOfTriangles(n));
    }
}
 
// This code is contributed by vt_m

PHP




<?php
// PHP program to calculate the
// number of equilateral triangles
 
// function to calculate
// number of triangles
// in Nth step
function numberOfTriangles($n)
{
    $ans = 2 * (pow(3, $n)) - 1;
    return $ans;
}
 
    // Driver Code
    $n = 2;
    echo numberOfTriangles($n);
 
// This code is contributed by anuj_67.
?>

Output:

17

Time Complexity: O(log n), log n is needed to compute 3^n.
 

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