Ways to color a 3*N board using 4 colors

Given a 3 X n board, find the number of ways to color it using at most 4 colors such that no two adjacent boxes have the same color. Diagonal neighbors are not treated as adjacent boxes.
Output the ways%1000000007 as the answer grows quickly.

Constraints:
1<= n < 100000

Examples :

Input : 1
Output : 36
We can use either a combination of 3 colors
or 2 colors. Now, choosing 3 colors out of 
4 is {4}\choose{3} and arranging them 
in 3! ways, similarly choosing 2 colors out 
of 4 is {4}\choose{2} and while arranging
we can only choose which of them could be at 
centre, that would be 2 ways. 
Answer = {4}\choose{3}*3! + {4}\choose{2}*2 = 36

Input : 2
Output : 588



We are going to solve this using dynamic approach because when a new column is added to the board, the ways in which colors are going to be filled depends just upon the color pattern in the current column. We can only have a combination of two colors and three colors in a column. All possible new columns that can be generated is given in the image. Please consider A, B, C and D as 4 colors.

Image Containing Color Combination Generation

All possible color combinations that can be generated from current column.

From now, we will refer 3 colors combination for a Nth column of the 3*N board as W(n) and two colors as Y(n).
We can see that each W can generate 5Y and 11W, and each Y can generate 7Y and 10W. We get two equation from here
We have two equations now,

W(n+1) = 10*Y(n)+11*W(n);
Y(n+1) = 7*Y(n)+5*W(n);

C++

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// C++ program to find number of ways
// to color a 3 x n grid using 4 colors
// such that no two adjacent have same
// color
#include <iostream>
using namespace std;
  
int solve(int A)
{
      
    // When we to fill single column
    long int color3 = 24;
    long int color2 = 12;
    long int temp = 0;
      
    for (int i = 2; i <= A; i++) 
    {
        temp = color3;
        color3 = (11 * color3 + 10 *
              color2 ) % 1000000007;
                
        color2 = ( 5 * temp + 7 *
              color2 ) % 1000000007;
    }
      
    long num = (color3 + color2)
                       % 1000000007;
                         
    return (int)num;
}
  
// Driver code
int main()
{
    int num1 = 1;
    cout << solve(num1) << endl;
  
    int num2 = 2;
    cout << solve(num2) << endl;
      
    int num3 = 500;
    cout << solve(num3) << endl;
  
    int num4 = 10000;
    cout << solve(num4);
      
    return 0;
}
  
// This code is contributed by vt_m.

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Java

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// Java program to find number of ways to color
// a 3 x n grid using 4 colors such that no two
// adjacent have same color.
public class Solution {
    public static int solve(int A) {
        long color3 = 24; // When we to fill single column
        long color2 = 12;
        long temp = 0;
        for (int i = 2; i <= A; i++)        
        {
            long temp = color3;
            color3 = (11 * color3 + 10 * color2 ) % 1000000007;
            color2 = ( 5 * temp + 7 * color2 ) % 1000000007;
        }
        long num = (color3 + color2) % 1000000007;
        return (int)num;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int num1 = 1;
        System.out.println(solve(num1));
  
        int num2 = 2;
        System.out.println(solve(num2));
  
        int num3 = 500;
        System.out.println(solve(num3));
  
        int num4 = 10000;
        System.out.println(solve(num4));
    }
}

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C#

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// C# program to find number of ways
// to color a 3 x n grid using 4
// colors such that no two adjacent
// have same color.
using System;
  
public class GFG {
      
    public static int solve(int A)
    {
          
        // When we to fill single column
        long color3 = 24;
        long color2 = 12;
        long temp = 0;
          
        for (int i = 2; i <= A; i++) 
        {
            temp = color3;
            color3 = (11 * color3 + 10 
                 * color2 ) % 1000000007;
                   
            color2 = ( 5 * temp + 7 
                 * color2 ) % 1000000007;
        }
        long num = (color3 + color2) 
                            % 1000000007;
        return (int)num;
    }
  
    // Driver code
    public static void Main()
    {
        int num1 = 1;
        Console.WriteLine(solve(num1));
  
        int num2 = 2;
        Console.WriteLine(solve(num2));
  
        int num3 = 500;
        Console.WriteLine(solve(num3));
  
        int num4 = 10000;
        Console.WriteLine(solve(num4));
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP program to find number of ways
// to color a 3 x n grid using 4 colors
// such that no two adjacent have same
// color
function solve($A)
{
      
    // When we to fill single column
    $color3 = 24;
    $color2 = 12;
    $temp = 0;
      
    for ($i = 2; $i <= $A; $i++) 
    {
        $temp = $color3;
        $color3 = (11 * $color3
                   10 * $color2 ) % 
                   1000000007;
              
        $color2 = ( 5 * $temp
                    7 * $color2 ) % 
                    1000000007;
    }
      
    $num = ($color3 + $color2) % 
                     1000000007;
                          
    return (int)$num;
}
  
// Driver code
$num1 = 1;
echo solve($num1) ,"\n";
  
$num2 = 2;
echo solve($num2) ,"\n";
  
$num3 = 500;
echo solve($num3),"\n";
  
$num4 = 10000;
echo solve($num4);
  
// This code is contributed by m_kit.
?>

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Output :

36
588
178599516
540460643

This article is contributed by Panshul Garg. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : vt_m, jit_t



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