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  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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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. 
?> 

Output :

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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