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Recursive Functions
• Difficulty Level : Medium
• Last Updated : 26 May, 2021 Recursion:
In programming terms, a recursive function can be defined as a routine that calls itself directly or indirectly.
Using the recursive algorithm, certain problems can be solved quite easily. Towers of Hanoi (TOH) is one such programming exercise. Try to write an iterative algorithm for TOH. Moreover, every recursive program can be written using iterative methods.
Mathematically, recursion helps to solve a few puzzles easily.
For example, a routine interview question,
In a party of N people, each person will shake her/his hand with each other person only once. In total how many hand-shakes would happen?

Solution:
It can be solved in different ways; graphs, recursions, etc. Let us see how recursively it can be solved.
There are N persons. Each person shakes hands with each other only once. Considering N-th person, (s)he has to shake a hand with (N-1) the person. Now the problem is reduced to small instances of (N-1) persons. Assuming TN as a total shake-hands, it can be formulated recursively.
TN = (N-1) + TN-1 [T1 = 0, i.e. the last person has already shook-hand with every one]
Solving it recursively yields an arithmetic series, which can be evaluated into N(N-1)/2.
Exercise: In a party of N couples, only one gender (either male or female) can shake hands with everyone. How many shake-hands would happen?
Usually, recursive programs result in poor time complexity. An example is a Fibonacci series. The time complexity of calculating the n-th Fibonacci number using recursion is approximately 1.6n. It means the same computer takes almost 60% more time for the next Fibonacci number. The recursive Fibonacci algorithm has overlapping subproblems. There are other techniques like dynamic programming to improve such overlapped algorithms.
However, a few algorithms, (e.g. merge sort, quick sort, etc…) result in optimal time complexity using recursion.

Base Case:
One critical requirement of recursive functions is the termination point or base case. Every recursive program must have a base case to make sure that the function will terminate. Missing base case results in unexpected behavior.

Different Ways of Writing Recursive Functions
Function calling itself: (Direct way)
Most of us are aware of at least two different ways of writing recursive programs. Given below are towers of the Hanoi code. It is an example of direct calling.

## C++

 `#include ``using` `namespace` `std;` `// Assuming n-th disk is bottom disk (count down)``void` `tower(``int` `n, ``char` `sourcePole,``        ``char` `destinationPole, ``char` `auxiliaryPole)``{``    ` `// Base case (termination condition)``if``(0 == n)``    ``return``;` `// Move first n-1 disks from source pole``// to auxiliary pole using destination as``// temporary pole``tower(n - 1, sourcePole, auxiliaryPole,``    ``destinationPole);` `// Move the remaining disk from source``// pole to destination pole``cout << ``"Move the disk "``<< n << ``" from "` `<<``    ``sourcePole <<``" to "``<< destinationPole << endl;` `// Move the n-1 disks from auxiliary (now source)``// pole to destination pole using source pole as``// temporary (auxiliary) pole``tower(n - 1, auxiliaryPole, destinationPole,``    ``sourcePole);``}` `// Driver code``int` `main()``{``    ``tower(3, ``'S'``, ``'D'``, ``'A'``);``        ` `    ``return` `0;``}` `// This code is contributed by SHUBHAMSINGH10`

## C

 `#include` `// Assuming n-th disk is bottom disk (count down)``void` `tower(``int` `n, ``char` `sourcePole, ``char` `destinationPole, ``char` `auxiliaryPole)``{``   ``// Base case (termination condition)``   ``if``(0 == n)``     ``return``;` `   ``// Move first n-1 disks from source pole``   ``// to auxiliary pole using destination as``   ``// temporary pole``   ``tower(n-1, sourcePole, auxiliaryPole,``      ``destinationPole);` `    ``// Move the remaining disk from source``   ``// pole to destination pole``   ``printf``(``"Move the disk %d from %c to %c\n"``,``    ``n,sourcePole, destinationPole);` `   ``// Move the n-1 disks from auxiliary (now source)``   ``// pole to destination pole using source pole as``   ``// temporary (auxiliary) pole``   ``tower(n-1, auxiliaryPole, destinationPole,``     ``sourcePole);``}` `int` `main()``{``   ``tower(3, ``'S'``, ``'D'``, ``'A'``);``   ` `   ``return` `0;``}`

## Java

 `// Assuming n-th disk is``// bottom disk (count down)``class` `GFG {``    ` `static` `void` `tower(``int` `n, ``char` `sourcePole,``                  ``char` `destinationPole, ``char` `auxiliaryPole)``{``    ``// Base case (termination condition)``    ``if` `(``0` `== n)``    ``return``;` `    ``// Move first n-1 disks from source pole``    ``// to auxiliary pole using destination as``    ``// temporary pole``    ``tower(n - ``1``, sourcePole, auxiliaryPole,``                          ``destinationPole);` `    ``// Move the remaining disk from source``    ``// pole to destination pole``    ``System.out.printf(``"Move the disk %d from %c to %c\n"``,``                         ``n, sourcePole, destinationPole);` `    ``// Move the n-1 disks from auxiliary (now source)``    ``// pole to destination pole using source pole as``    ``// temporary (auxiliary) pole``    ``tower(n - ``1``, auxiliaryPole, destinationPole, sourcePole);``}` `public` `static` `void` `main(String[] args)``{``    ``tower(``3``, ``'S'``, ``'D'``, ``'A'``);``}``}` `// This code is contributed by Smitha Dinesh Semwal.`

## Python3

 `# Assuming n-th disk is``# bottom disk (count down)``def` `tower(n, sourcePole, destinationPole, auxiliaryPole):` `    ``# Base case (termination condition)``    ``if``(``0` `=``=` `n):``        ``return``    ` `    ``# Move first n-1 disks``    ``# from source pole``    ``# to auxiliary pole``    ``# using destination as``    ``# temporary pole``    ``tower(n``-``1``, sourcePole, auxiliaryPole, destinationPole)``    ` `    ``# Move the remaining``    ``# disk from source``    ``# pole to destination pole``    ``print``(``"Move the disk"``,sourcePole,``"from"``,sourcePole,``"to"``,destinationPole)``    ` `    ``# Move the n-1 disks from``    ``# auxiliary (now source)``    ``# pole to destination pole``    ``# using source pole as``    ``# temporary (auxiliary) pole``    ``tower(n``-``1``, auxiliaryPole, destinationPole,sourcePole)`  `# Driver code``tower(``3``, ``'S'``, ``'D'``, ``'A'``)`

## C#

 `// Assuming n-th disk is bottom disk``// (count down)``using` `System;` `class` `GFG {``     ` `    ``static` `void` `tower(``int` `n, ``char` `sourcePole,``                        ``char` `destinationPole,``                          ``char` `auxiliaryPole)``    ``{``        ` `        ``// Base case (termination condition)``        ``if` `(0 == n)``            ``return``;``     ` `        ``// Move first n-1 disks from source``        ``// pole to auxiliary pole using``        ``// destination as temporary pole``        ``tower(n - 1, sourcePole, auxiliaryPole,``                              ``destinationPole);``     ` `        ``// Move the remaining disk from source``        ``// pole to destination pole``        ``Console.WriteLine(``"Move the disk "` `+ n``                ``+ ``"from "` `+ sourcePole + ``"to "``                           ``+ destinationPole);``     ` `        ``// Move the n-1 disks from auxiliary``        ``// (now source) pole to destination``        ``// pole using source pole as temporary``        ``// (auxiliary) pole``        ``tower(n - 1, auxiliaryPole,``                  ``destinationPole, sourcePole);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``tower(3, ``'S'``, ``'D'``, ``'A'``);``    ``}``}`` ` `// This code is contributed by Anant Agarwal.`

## PHP

 ``

## Javascript

 ``

Output:

```Move the disk 1 from S to D
Move the disk 2 from S to A
Move the disk 1 from D to A
Move the disk 3 from S to D
Move the disk 1 from A to S
Move the disk 2 from A to D
Move the disk 1 from S to D```

The time complexity of TOH can be calculated by formulating the number of moves.
We need to move the first N-1 disks from Source to Auxiliary and from Auxiliary to Destination, i.e. the first N-1 disk requires two moves. One more move of the last disk from Source to Destination. Mathematically, it can be defined recursively.
MN = 2MN-1 + 1.
We can easily solve the above recursive relation (2N-1), which is exponential.

Recursion using mutual function call: (Indirect way)
Indirect calling. Though least practical, a function [funA()] can call another function [funB()] which in turn calls [funA()] the former function. In this case, both the functions should have the base case.
Defensive Programming:
We can combine defensive coding techniques with recursion for the graceful functionality of the application. Usually, recursive programming is not allowed in safety-critical applications, such as flight controls, health monitoring, etc. However, one can use a static count technique to avoid uncontrolled calls (NOT in safety-critical systems, but may be used in soft real-time systems).

## C++

 `void` `recursive(``int` `data)``{``   ``static` `callDepth;``  ` `   ``if``(callDepth > MAX_DEPTH)``      ``return``;` `   ``// Increase call depth``   ``callDepth++;` `   ``// do other processing``   ``recursive(data);` `   ``// do other processing``   ``// Decrease call depth``   ``callDepth--;``}` `// This code is contributed by rutvik_56`

## C

 `void` `recursive(``int` `data)``{``   ``static` `callDepth;``   ``if``(callDepth > MAX_DEPTH)``      ``return``;` `   ``// Increase call depth``   ``callDepth++;` `   ``// do other processing``   ``recursive(data);` `   ``// do other processing``   ``// Decrease call depth``   ``callDepth--;``}`

## Java

 `static` `void` `recursive(``int` `data)``{``   ``static` `callDepth;``   ``if``(callDepth > MAX_DEPTH)``      ``return``;``   ` `   ``// Increase call depth``   ``callDepth++;``   ` `   ``// do other processing``   ``recursive(data);``   ` `   ``// do other processing``   ``// Decrease call depth``   ``callDepth--;``}` `// This code is contributed by divyeh072019`

## Python3

 `def` `recursive(data):` `   ``callDepth ``=` `0``   ` `   ``if``(callDepth > MAX_DEPTH):``      ``return``;`` ` `   ``# Increase call depth``   ``callDepth``+``=``1`` ` `   ``# do other processing``   ``recursive(data);`` ` `   ``# do other processing``   ``# Decrease call depth``   ``callDepth ``-``=` `1` `  ``# This code is contributed by Pratham76`

## C#

 `static` `void` `recursive(``int` `data)``{``   ``static` `callDepth;``   ``if``(callDepth > MAX_DEPTH)``      ``return``;``  ` `   ``// Increase call depth``   ``callDepth++;``  ` `   ``// do other processing``   ``recursive(data);``  ` `   ``// do other processing``   ``// Decrease call depth``   ``callDepth--;``}` `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

callDepth depth depends on function stack frame size and maximum stack size.

Recursion using function pointers: (Indirect way)
Recursion can also implemented with function pointers. An example is a signal handler in POSIX compliant systems. If the handler causes to trigger the same event due to which the handler being called, the function will reenter.

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