# Tiling with Dominoes

• Difficulty Level : Hard
• Last Updated : 30 Jun, 2022

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

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 Recurrences
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 = 8print(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.

## Javascript

 

## PHP

 

Output :

153

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