Delannoy Number - GeeksforGeeks

In mathematics, a Delannoy number D describes the number of paths from the southwest corner (0, 0) of a rectangular grid to the northeast corner (m, n), using only single steps north, northeast, or east.
For Example, D(3, 3) equals 63.

Delannoy Number can be calculated by:

Delannoy number can be used to find:

Counts the number of global alignments of two sequences of lengths m and n.
Number of points in an m-dimensional integer lattice that are at most n steps from the origin.
In cellular automata, the number of cells in an m-dimensional von Neumann neighborhood of radius n.
Number of cells on a surface of an m-dimensional von Neumann neighborhood of radius n.

Examples :

Input : n = 3, m = 3
Output : 63

Input : n = 4, m = 5
Output : 681

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

Below is the implementation of finding Delannoy Number:

C++

// CPP Program of finding nth Delannoy Number.
#include <bits/stdc++.h>
using namespace std;

// Return the nth Delannoy Number.
int dealnnoy(int n, int m)
{
    // Base case
    if (m == 0 || n == 0)
        return 1;

    // Recursive step.
    return dealnnoy(m - 1, n) + 
           dealnnoy(m - 1, n - 1) +
           dealnnoy(m, n - 1);
}

// Driven Program
int main()
{
    int n = 3, m = 4;
    cout << dealnnoy(n, m) << endl;
    return 0;
}

Java

// Java Program for finding nth Delannoy Number.
import java.util.*;
import java.lang.*;

public class GfG{
    
    // Return the nth Delannoy Number.
    public static int dealnnoy(int n, int m)
    {
        // Base case
        if (m == 0 || n == 0)
            return 1;

        // Recursive step.
        return dealnnoy(m - 1, n) + 
            dealnnoy(m - 1, n - 1) +
            dealnnoy(m, n - 1);
    }
    
    // driver function
    public static void main(String args[]){
        int n = 3, m = 4;
        System.out.println(dealnnoy(n, m));
    }
}

/* This code is contributed by Sagar Shukla. */

Python3

# Python3 Program for finding 
# nth Delannoy Number.

# Return the nth Delannoy Number.
def dealnnoy(n, m):
    
    # Base case
    if (m == 0 or n == 0) :
        return 1

    # Recursive step.
    return dealnnoy(m - 1, n) + dealnnoy(m - 1, n - 1) + dealnnoy(m, n - 1)

# Driven code
n = 3
m = 4;
print( dealnnoy(n, m) )

# This code is contributed by "rishabh_jain".

C#

// C# Program for finding nth Delannoy Number.
using System;

public class GfG {

    // Return the nth Delannoy Number.
    public static int dealnnoy(int n, int m)
    {

        // Base case
        if (m == 0 || n == 0)
            return 1;

        // Recursive step.
        return dealnnoy(m - 1, n) + 
               dealnnoy(m - 1, n - 1) + 
                     dealnnoy(m, n - 1);
    }

    // driver function
    public static void Main()
    {
        int n = 3, m = 4;
        Console.WriteLine(dealnnoy(n, m));
    }
}

/* This code is contributed by vt_m. */

PHP


<?php
// PHP Program of finding nth
// Delannoy Number.

// Return the nth Delannoy Number.
function dealnnoy( $n, $m)
{
    
    // Base case
    if ($m == 0 or $n == 0)
        return 1;

    // Recursive step.
    return dealnnoy($m - 1, $n) + 
           dealnnoy($m - 1, $n - 1) +
           dealnnoy($m, $n - 1);
}

    // Driver Code
    $n = 3; 
    $m = 4;
    echo dealnnoy($n, $m);

// This code is contributed by anuj_67.
?>

Output:

Below is the Dynamic Programming program to find nth Delannoy Number:

C++

// CPP Program of finding nth Delannoy Number.
#include <bits/stdc++.h>
using namespace std;

// Return the nth Delannoy Number.
int dealnnoy(int n, int m)
{
    int dp[m + 1][n + 1];

    // Base cases
    for (int i = 0; i <= m; i++) 
        dp[i][0] = 1;
    for (int i = 0; i <= m; i++) 
        dp[0][i] = 1;    

    for (int i = 1; i <= m; i++) 
        for (int j = 1; j <= n; j++) 
            dp[i][j] = dp[i - 1][j] +
                       dp[i - 1][j - 1] + 
                       dp[i][j - 1];

    return dp[m][n];
}

// Driven Program
int main()
{
    int n = 3, m = 4;
    cout << dealnnoy(n, m) << endl;
    return 0;
}

Java

// Java Program of finding nth Delannoy Number.

import java.io.*;

class GFG {
    
    // Return the nth Delannoy Number.
    static int dealnnoy(int n, int m)
    {
        int dp[][]=new int[m + 1][n + 1];
     
        // Base cases
        for (int i = 0; i <= m; i++) 
            dp[i][0] = 1;
        for (int i = 0; i < m; i++) 
            dp[0][i] = 1;    
     
        for (int i = 1; i <= m; i++) 
            for (int j = 1; j <= n; j++) 
                dp[i][j] = dp[i - 1][j] +
                           dp[i - 1][j - 1] + 
                           dp[i][j - 1];
     
        return dp[m][n];
    }
     
    // Driven Program
    public static void main(String args[])
    {
        int n = 3, m = 4;
        System.out.println(dealnnoy(n, m));
        
    }
}

// This code is contributed by Nikita Tiwari.

Python3

# Python3 Program for finding nth
# Delannoy Number.

# Return the nth Delannoy Number.
def dealnnoy (n, m):
    dp = [[0 for x in range(n+1)] for x in range(m+1)]

    # Base cases
    for i in range(m):
        dp[0][i] = 1
    
    for i in range(1, m + 1):
        dp[i][0] = 1

    for i in range(1, m + 1):
        for j in range(1, n + 1):
            dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] + dp[i][j - 1];

    return dp[m][n]

# Driven code
n = 3
m = 4
print(dealnnoy(n, m))

# This code is contributed by "rishabh_jain".

C#

// C# Program of finding nth Delannoy Number.
using System;

class GFG {

    // Return the nth Delannoy Number.
    static int dealnnoy(int n, int m)
    {
        
        int[, ] dp = new int[m + 1, n + 1];

        // Base cases
        for (int i = 0; i <= m; i++)
            dp[i, 0] = 1;
        for (int i = 0; i < m; i++)
            dp[0, i] = 1;

        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++)
                dp[i, j] = dp[i - 1, j]
                     + dp[i - 1, j - 1]
                         + dp[i, j - 1];

        return dp[m, n];
    }

    // Driven Program
    public static void Main()
    {
        int n = 3, m = 4;
        
        Console.WriteLine(dealnnoy(n, m));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP Program of finding
// nth Delannoy Number.

// Return the nth Delannoy Number.
function dealnnoy($n, $m)
{
    $dp[$m + 1][$n + 1] = 0;

    // Base cases
    for ($i = 0; $i <= $m; $i++) 
        $dp[$i][0] = 1;
    for ( $i = 0; $i <= $m; $i++) 
        $dp[0][$i] = 1; 

    for ($i = 1; $i <= $m; $i++) 
        for ($j = 1; $j <= $n; $j++) 
            $dp[$i][$j] = $dp[$i - 1][$j] +
                      $dp[$i - 1][$j - 1] + 
                           $dp[$i][$j - 1];

    return $dp[$m][$n];
}

// Driven Code
$n = 3; $m = 4;
echo dealnnoy($n, $m) ;

// This code is contributed by SanjuTomar 
?>

Output :

anuj0503

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Practice Tags :

Dynamic Programming

Mathematical

series