Skip to content
Related Articles

Related Articles

Tiling with Dominoes

View Discussion
Improve Article
Save Article
Like Article
  • 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: 
 

A_{n} = A_{n-2} + B_{n-1} + C_{n-1}
A_{n} = A_{n-2} + 2*(B_{n-1})
Calculating Bn: 
 

B_{n} = A_{n-1} + B_{n-2}
Final Recursive Relations are: 
 

A_{n} = A_{n-2} + 2*(B_{n-1}) \hspace{0.5cm} B_{n} = A_{n-1} + B_{n-2}

Base Cases: 
 

A_0 = 1 , A_1 = 0 \hspace{0.5cm} B_0 = 0 , B_1 = 1

 

C++




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




<script>
    // Javascript program to find no. of ways
    // to fill a 3xn board with 2x1 dominoes.
     
    function countWays(n)
    {
        let A = new Array(n+1);
        let B = new Array(n+1);
        A[0] = 1; A[1] = 0;
        B[0] = 0; B[1] = 1;
        for (let 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];
    }
     
    let n = 8;
      document.write(countWays(n));
     
    // This code is contributed by rameshtravel07.
</script>

PHP




<?php
// PHP program to find no. of ways
// to fill a 3xn board with 2x1 dominoes.
 
function countWays($n)
{
    $A = array();
    $B = array();
    $A[0] = 1; $A[1] = 0;
    $B[0] = 0; $B[1] = 1;
    for ( $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
$n = 8;
echo countWays($n);
 
// This code is contributed by anuj_67.
?>

Output : 

153

 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!