Ways to color a 3*N board using 4 colors

Difficulty Level : Hard
Last Updated : 28 Apr, 2021

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  and arranging them 

in 3! ways, similarly choosing 2 colors out 

of 4 is  and while arranging

we can only choose which of them could be at 

centre, that would be 2 ways. 

Answer = *3! + *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++

// 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.

Java

// 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));
    }
}

Python3

# Python 3 program to find number of ways
# to color a 3 x n grid using 4 colors
# such that no two adjacent have same
# color
 
def solve(A):
     
    # When we to fill single column
    color3 = 24
    color2 = 12
    temp = 0
     
    for i in range(2, A + 1, 1):
        temp = color3
        color3 = (11 * color3 + 10 * color2 ) % 1000000007
             
        color2 = ( 5 * temp + 7 * color2 ) % 1000000007
     
    num = (color3 + color2) % 1000000007
                         
    return num
 
# Driver code
if __name__ == '__main__':
    num1 = 1
    print(solve(num1))
     
    num2 = 2
    print(solve(num2))
     
    num3 = 500
    print(solve(num3))
 
    num4 = 10000
    print(solve(num4))
     
# This code is contributed by
# Shashank_Sharma

C#

// 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.

PHP

<?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.
?>

Javascript

<script>
// JavaScript 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) {
        let color3 = 24; // When we to fill single column
        let color2 = 12;
        let temp = 0;
        for (let i = 2; i <= A; i++)       
        {
            let temp = color3;
            color3 = (11 * color3 + 10 * color2 ) % 1000000007;
            color2 = ( 5 * temp + 7 * color2 ) % 1000000007;
        }
        let num = (color3 + color2) % 1000000007;
        return num;
    }
  
// Driver Code
 
        let num1 = 1;
        document.write(solve(num1) + "<br/>");
   
        let num2 = 2;
        document.write(solve(num2) + "<br/>");
   
        let num3 = 500;
        document.write(solve(num3) + "<br/>");
   
        let num4 = 10000;
        document.write(solve(num4));
                 
</script>

Output :

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