Eulerian Number

In combinatorics, the Eulerian Number A(n, m), is the number of permutations of the numbers 1 to n in which exactly m elements are greater than previous element.

For example, there are 4 permutations of the number 1 to 3 in which exactly 1 element is greater than the previous elements.

Eulerian

Examples:

Input : n = 3, m = 1
Output : 4
Please see above diagram (There
are 4 permutations where 1 no. is
greater.

Input : n = 4, m = 1
Output : 11

Eulerian Numbers are the coefficients of the Eulerian polynomials described below.

The Eulerian polynomials are defined by the exponential generating function

The Eulerian polynomials can be computed by the recurrence

An explicit formula for A(n, m) is

We can calculate A(n, m) by recurrence relation:

Example:

Suppose, n = 3 and m = 1. 
Therefore, 
A(3, 1) 
= (3 - 1) * A(2, 0) + (1 + 1) * A(2, 1) 
= 2 * A(2, 0) + 2 * A(2, 1) 
= 2 * 1 + 2 * ( (2 - 1) * A(1, 0) + (1 + 1) * A(1, 1)) 
= 2 + 2 * (1 * 1 + 2 * ((1 - 1) * A(0, 0) + (1 + 1) * A(0, 1)) 
= 2 + 2 * (1 + 2 * (0 * 1 + 2 * 0) 
= 2 + 2 * (1 + 2 * 0) 
= 2 + 2 * 1 
= 2 + 2 
= 4 
We can verify this with example shown above.

Below is the implementation of finding A(n, m):

C++

// CPP Program to find Eulerian number A(n, m)
#include <bits/stdc++.h>
using namespace std;

// Return euleriannumber A(n, m)
int eulerian(int n, int m)
{
    if (m >= n || n == 0)
        return 0;

    if (m == 0)
        return 1;

    return (n - m) * eulerian(n - 1, m - 1)
           + (m + 1) * eulerian(n - 1, m);
}

// Driven Program
int main()
{
    int n = 3, m = 1;
    cout << eulerian(n, m) << endl;
    return 0;
}

Java

// Java program to find Eulerian number A(n, m)
import java.util.*;

class Eulerian {
    // Return eulerian number A(n, m)
    public static int eulerian(int n, int m)
    {
        if (m >= n || n == 0)
            return 0;

        if (m == 0)
            return 1;

        return (n - m) * eulerian(n - 1, m - 1)
            + (m + 1) * eulerian(n - 1, m);
    }

    // driver code
    public static void main(String[] args)
    {
        int n = 3, m = 1;
        System.out.print(eulerian(n, m));
    }
}

// This code is contributed by rishabh_jain

Python3

# Python3 Program to find Eulerian number A(n, m)

# Return euleriannumber A(n, m)


def eulerian(n, m):
    if (m >= n or n == 0):
        return 0

    if (m == 0):
        return 1

    return ((n - m) * eulerian(n - 1, m - 1) +
            (m + 1) * eulerian(n - 1, m))


# Driver code
n = 3
m = 1
print(eulerian(n, m))

# This code is contributed by rishabh_jain

// C# program to find Eulerian number A(n, m)
using System;

class Eulerian {

    // Return eulerian number A(n, m)
    public static int eulerian(int n, int m)
    {
        if (m >= n || n == 0)
            return 0;

        if (m == 0)
            return 1;

        return (n - m) * eulerian(n - 1, m - 1)
            + (m + 1) * eulerian(n - 1, m);
    }

    // driver code
    public static void Main()
    {
        int n = 3, m = 1;
        Console.WriteLine(eulerian(n, m));
    }
}

// This code is contributed by vt_m

JavaScript

<script>

// JavaScript Program to find Eulerian number A(n, m)

    // Return eulerian number A(n, m)
    function eulerian(n, m)
    {
        if (m >= n || n == 0)
            return 0;
  
        if (m == 0)
            return 1;
  
        return (n - m) * eulerian(n - 1, m - 1) +
            (m + 1) * eulerian(n - 1, m);
    }

// Driver code
        
        let n = 3, m = 1;
        document.write( eulerian(n, m) );
                  
</script>

PHP

<?php
// PHP Program to find
// Eulerian number A(n, m)

// Return euleriannumber A(n, m)
function eulerian($n, $m)
{
    if ($m >= $n || $n == 0)
        return 0;

    if ($m == 0)
        return 1;

    return ($n - $m) * eulerian($n - 1, $m - 1) + 
                 ($m + 1) * eulerian($n - 1, $m);
}

// Driven Code
$n = 3; $m = 1;
echo eulerian($n, $m);

// This code is contributed by anuj_67.
?>

Output

Time Complexity: O(2ⁿ)
Auxiliary Space: O(log(n)), Due, to recursive call stack

Below is the implementation of finding A(n, m) using Dynamic Programming:

C++

// CPP Program to find Eulerian number A(n, m)
#include <bits/stdc++.h>
using namespace std;

// Return euleriannumber A(n, m)
int eulerian(int n, int m)
{
    int dp[n + 1][m + 1];

    memset(dp, 0, sizeof(dp));

    // For each row from 1 to n
    for (int i = 1; i <= n; i++) {

        // For each column from 0 to m
        for (int j = 0; j <= m; j++) {

            // If i is greater than j
            if (i > j) {

                // If j is 0, then make that
                // state as 1.
                if (j == 0)
                    dp[i][j] = 1;

                // basic recurrence relation.
                else
                    dp[i][j] = ((i - j) * dp[i - 1][j - 1])
                               + ((j + 1) * dp[i - 1][j]);
            }
        }
    }

    return dp[n][m];
}

// Driven Program
int main()
{
    int n = 3, m = 1;
    cout << eulerian(n, m) << endl;
    return 0;
}

Java

// Java program to find Eulerian number A(n, m)
import java.util.*;

class Eulerian {
    // Return euleriannumber A(n, m)
    public static int eulerian(int n, int m)
    {
        int[][] dp = new int[n + 1][m + 1];

        // For each row from 1 to n
        for (int i = 1; i <= n; i++) {

            // For each column from 0 to m
            for (int j = 0; j <= m; j++) {

                // If i is greater than j
                if (i > j) {

                    // If j is 0, then make
                    // that state as 1.
                    if (j == 0)
                        dp[i][j] = 1;

                    // basic recurrence relation.
                    else
                        dp[i][j]
                            = ((i - j) * dp[i - 1][j - 1])
                              + ((j + 1) * dp[i - 1][j]);
                }
            }
        }

        return dp[n][m];
    }

    // driver code
    public static void main(String[] args)
    {
        int n = 3, m = 1;
        System.out.print(eulerian(n, m));
    }
}

// This code is contributed by rishabh_jain

Python3

# Python3 Program to find Eulerian
# number A(n, m)

# Return euleriannumber A(n, m)


def eulerian(n, m):
    dp = [[0 for x in range(m+1)]
          for y in range(n+1)]

    # For each row from 1 to n
    for i in range(1, n+1):

        # For each column from 0 to m
        for j in range(0, m+1):

            # If i is greater than j
            if (i > j):
                # If j is 0, then make that
                # state as 1.

                if (j == 0):
                    dp[i][j] = 1

                # basic recurrence relation.
                else:
                    dp[i][j] = (((i - j) *
                                 dp[i - 1][j - 1]) +
                                ((j + 1) * dp[i - 1][j]))

    return dp[n][m]


# Driven Program
n = 3
m = 1
print(eulerian(n, m))

# This code is contributed by Prasad Kshirsagar

// C# program to find Eulerian number A(n, m)
using System;

class Eulerian {

    // Return euleriannumber A(n, m)
    public static int eulerian(int n, int m)
    {
        int[, ] dp = new int[n + 1, m + 1];

        // For each row from 1 to n
        for (int i = 1; i <= n; i++) {

            // For each column from 0 to m
            for (int j = 0; j <= m; j++) {

                // If i is greater than j
                if (i > j) {

                    // If j is 0, then make
                    // that state as 1.
                    if (j == 0)
                        dp[i, j] = 1;

                    // basic recurrence relation.
                    else
                        dp[i, j]
                            = ((i - j) * dp[i - 1, j - 1])
                              + ((j + 1) * dp[i - 1, j]);
                }
            }
        }

        return dp[n, m];
    }

    // driver code
    public static void Main()
    {
        int n = 3, m = 1;
        Console.WriteLine(eulerian(n, m));
    }
}

// This code is contributed by vt_m

JavaScript

<script>

// Javascript Program to find 
// Eulerian number A(n, m)

// Return euleriannumber A(n, m)
function eulerian(n, m)
{
    var dp = Array.from(Array(n+1), 
    ()=> Array(m+1).fill(0));

    // For each row from 1 to n
    for (var i = 1; i <= n; i++) {

        // For each column from 0 to m
        for (var j = 0; j <= m; j++) {

            // If i is greater than j
            if (i > j) {

                // If j is 0, then make that 
                // state as 1.
                if (j == 0)
                    dp[i][j] = 1;

                // basic recurrence relation.
                else
                    dp[i][j] = ((i - j) *
                     dp[i - 1][j - 1]) + 
                    ((j + 1) * dp[i - 1][j]);
            }
        }
    }

    return dp[n][m];
}

// Driven Program
var n = 3, m = 1;
document.write( eulerian(n, m) );

</script>

PHP

<?php
// PHP Program to find Eulerian
// number A(n, m) 

// Return euleriannumber A(n, m) 
function eulerian($n, $m) 
{ 
    $dp = array(array());
    
    for ($i = 0; $i < $n + 1; $i++)
        for($j = 0; $j < $m + 1; $j++)
            $dp[$i][$j] = 0 ;

    // For each row from 1 to n 
    for ($i = 1; $i <= $n; $i++) 
    { 

        // For each column from 0 to m 
        for ($j = 0; $j <= $m; $j++) 
        { 

            // If i is greater than j 
            if ($i > $j) 
            { 

                // If j is 0, then make that 
                // state as 1. 
                if ($j == 0) 
                    $dp[$i][$j] = 1; 

                // basic recurrence relation. 
                else
                    $dp[$i][$j] = (($i - $j) * 
                                    $dp[$i - 1][$j - 1]) + 
                                  (($j + 1) * $dp[$i - 1][$j]); 
            } 
        } 
    } 

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

// Driver Code
$n = 3 ;
$m = 1; 
echo eulerian($n, $m) ; 

// This code is contributed by Ryuga
?>

Output

Time Complexity: O(n*m)
Auxiliar Space: O(n*m)

Efficient approach: Space optimization

In the previous approach, the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.

Implementation steps:

Create a 1D vector dp of size m+1.
Set a base case by initializing the values of DP .
Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
Now Create a variable prev to store previous computations and a temp variable to update prev at every iteration.
After every iteration assign the value of temp to prev for further iteration.
At last return and print the final answer stored in dp[m].

Implementation:

C++

// CPP Program to find Eulerian number A(n, m)
#include <bits/stdc++.h>
using namespace std;

// Return euleriannumber A(n, m)
int eulerian(int n, int m)
{
    vector<int> dp(m+1,0);
    
    // For each row from 1 to n
    for (int i = 1; i <= n; i++) {
        int prev = 0;
        
        // For each column from 0 to m
        for (int j = 0; j <= m; j++) {
            int temp = dp[j];
            
            // If i is greater than j
            if (i > j) {
                
                // If j is 0, then make that
                // state as 1.
                if (j == 0)
                    dp[j] = 1;
                
                // basic recurrence relation.
                else
                    dp[j] = ((i - j) * prev) + ((j + 1) * dp[j]);
                
                prev = temp;
            }
        }
    }
    
    return dp[m];
}

// Driven Program
int main()
{
    int n = 3, m = 1;
    cout << eulerian(n, m) << endl;
    return 0;
}

Java

import java.util.*;

public class Main {
    // Return euleriannumber A(n, m)
    static int eulerian(int n, int m) {
        int[] dp = new int[m+1];
        Arrays.fill(dp, 0);

        // For each row from 1 to n
        for (int i = 1; i <= n; i++) {
            int prev = 0;

            // For each column from 0 to m
            for (int j = 0; j <= m; j++) {
                int temp = dp[j];

                // If i is greater than j
                if (i > j) {
                    // If j is 0, then make that state as 1.
                    if (j == 0)
                        dp[j] = 1;
                    // basic recurrence relation.
                    else
                        dp[j] = ((i - j) * prev) + ((j + 1) * dp[j]);

                    prev = temp;
                }
            }
        }

        return dp[m];
    }

    // Driven Program
    public static void main(String[] args) {
        int n = 3, m = 1;
        System.out.println(eulerian(n, m));
    }
}

Python3

# Function to find Eulerian number A(n, m)
def eulerian(n, m):
    dp = [0] * (m + 1)

    # For each row from 1 to n
    for i in range(1, n + 1):
        prev = 0

        # For each column from 0 to m
        for j in range(0, m + 1):
            temp = dp[j]

            # If i is greater than j
            if i > j:

                # If j is 0, then make that
                # state as 1.
                if j == 0:
                    dp[j] = 1

                # Basic recurrence relation.
                else:
                    dp[j] = ((i - j) * prev) + ((j + 1) * dp[j])

                prev = temp

    return dp[m]

# Driver Code
n = 3
m = 1
print(eulerian(n, m))

using System;

class EulerianNumber
{
    // Function to calculate Eulerian number A(n, m)
    static int Eulerian(int n, int m)
    {
        int[] dp = new int[m + 1];

        // For each row from 1 to n
        for (int i = 1; i <= n; i++)
        {
            int prev = 0;

            // For each column from 0 to m
            for (int j = 0; j <= m; j++)
            {
                int temp = dp[j];

                // If i is greater than j
                if (i > j)
                {
                    // If j is 0, then make that state as 1.
                    if (j == 0)
                        dp[j] = 1;
                    // Basic recurrence relation.
                    else
                        dp[j] = ((i - j) * prev) + ((j + 1) * dp[j]);

                    prev = temp;
                }
            }
        }

        return dp[m];
    }

    // Main method
    static void Main()
    {
        int n = 3, m = 1;
        int result = Eulerian(n, m);
        Console.WriteLine(result);
    }
}

JavaScript

// Function to find Eulerian number A(n, m)
function eulerian(n, m) {
  let dp = new Array(m + 1).fill(0);

  // For each row from 1 to n
  for (let i = 1; i <= n; i++) {
    let prev = 0;

    // For each column from 0 to m
    for (let j = 0; j <= m; j++) {
      let temp = dp[j];

      // If i is greater than j
      if (i > j) {
        // If j is 0, then make that state as 1.
        if (j == 0) {
          dp[j] = 1;
        }
        // Basic recurrence relation.
        else {
          dp[j] = (i - j) * prev + (j + 1) * dp[j];
        }
        prev = temp;
      }
    }
  }

  return dp[m];
}

// Driver Code
let n = 3;
let m = 1;
console.log(eulerian(n, m));

PHP

<?php
// Return eulerian number A(n, m)
function eulerian($n, $m)
{
    $dp = array_fill(0, $m + 1, 0);
    
    // For each row from 1 to n
    for ($i = 1; $i <= $n; $i++) {
        $prev = 0;
        
        // For each column from 0 to m
        for ($j = 0; $j <= $m; $j++) {
            $temp = $dp[$j];
            
            // If i is greater than j
            if ($i > $j) {
                
                // If j is 0, then make that
                // state as 1.
                if ($j == 0)
                    $dp[$j] = 1;
                
                // basic recurrence relation.
                else
                    $dp[$j] = (($i - $j) * $prev) + (($j + 1) * $dp[$j]);
                
                $prev = $temp;
            }
        }
    }
    
    return $dp[$m];
}

// Driven Program
$n = 3;
$m = 1;
echo eulerian($n, $m) . PHP_EOL;
?>

Output

Time Complexity: O(n*m)
Auxiliary Space: O(m)

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