# Tiling with Dominoes

Given a 3 x n board, find the number of ways to fill it with 2 x 1 dominoes.

Example 1
Following are all the 3 possible ways to fill up a 3 x 2 board. Example 2
Here is one possible way of filling a 3 x 8 board. You have to find all the possible ways to do so. Examples :

Input : 2
Output : 3

Input : 8
Output : 153

Input : 12
Output : 2131


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

Defining Subproblems:
At any point while filling the board, there are three possible states that the last column can be in: An =  No. of ways to completely fill a 3 x n board. (We need to find this)
Bn =  No. of ways to fill a 3 x n board with top corner in last column not filled.
Cn =  No. of ways to fill a 3 x n board with bottom corner in last column not filled.


Note: The following states are impossible to reach: Finding Reccurences
Note: Even though Bn and Cn are different states, they will be equal for same ‘n’. i.e Bn = Cn
Hence, we only need to calculate one of them.

Calculating An:   Calculating Bn:  Final Recursive Relations are:  Base Cases:  ## C++

 // C++ program to find no. of ways  // to fill a 3xn board with 2x1 dominoes.  #include  using namespace std;     int countWays(int n)  {      int A[n + 1], B[n + 1];      A = 1, A = 0, B = 0, B = 1;      for (int i = 2; i <= n; i++) {          A[i] = A[i - 2] + 2 * B[i - 1];          B[i] = A[i - 1] + B[i - 2];      }         return A[n];  }     int main()  {      int n = 8;      cout << countWays(n);      return 0;  }

## Java

 // Java program to find no. of ways  // to fill a 3xn board with 2x1 dominoes.  import java.io.*;     class GFG {         static int countWays(int n)      {          int []A = new int[n+1];          int []B = new int[n+1];          A = 1; A = 0;          B = 0; B = 1;          for (int i = 2; i <= n; i++)           {              A[i] = A[i - 2] + 2 * B[i - 1];              B[i] = A[i - 1] + B[i - 2];          }                 return A[n];      }         // Driver code      public static void main (String[] args)       {          int n = 8;          System.out.println(countWays(n));      }  }     // This code is contributed by anuj_67.

## Python 3

 # Python 3 program to find no. of ways  # to fill a 3xn board with 2x1 dominoes.     def countWays(n):         A =  * (n + 1)      B =  * (n + 1)      A = 1     A = 0     B = 0     B = 1     for i in range(2, n+1):          A[i] = A[i - 2] + 2 * B[i - 1]          B[i] = A[i - 1] + B[i - 2]             return A[n]     n = 8 print(countWays(n))     # This code is contributed by Smitha

## C#

 // C# program to find no. of ways  // to fill a 3xn board with 2x1 dominoes.  using System;     class GFG {         static int countWays(int n)      {          int []A = new int[n+1];          int []B = new int[n+1];          A = 1; A = 0;          B = 0; B = 1;          for (int i = 2; i <= n; i++)           {              A[i] = A[i - 2] + 2 * B[i - 1];              B[i] = A[i - 1] + B[i - 2];          }                 return A[n];      }         // Driver code      public static void Main ()       {          int n = 8;          Console.WriteLine(countWays(n));      }  }     // This code is contributed by anuj_67.

## PHP

 

Output :

153


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Improved By : vt_m, Smitha Dinesh Semwal