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

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:

 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}) 
 B_{n} = A_{n-1} + B_{n-2} 

Base Cases:

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

C++

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// 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;
}

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Java

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// 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.

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Python 3

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# 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

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C#

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// 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.

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PHP

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<?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.
?>

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Output :

153


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Improved By : vt_m, Smitha Dinesh Semwal