Tower of Hanoi is a mathematical puzzle where we have three rods and n disks. The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules:
- Only one disk can be moved at a time.
- Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack i.e. a disk can only be moved if it is the uppermost disk on a stack.
- No disk may be placed on top of a smaller disk.
Approach :
Take an example for 2 disks : Let rod 1 = 'A', rod 2 = 'B', rod 3 = 'C'. Step 1 : Shift first disk from 'A' to 'B'. Step 2 : Shift second disk from 'A' to 'C'. Step 3 : Shift first disk from 'B' to 'C'. The pattern here is : Shift 'n-1' disks from 'A' to 'B'. Shift last disk from 'A' to 'C'. Shift 'n-1' disks from 'B' to 'C'. Image illustration for 3 disks :
Examples:
Input : 2 Output : Disk 1 moved from A to B Disk 2 moved from A to C Disk 1 moved from B to C Input : 3 Output : Disk 1 moved from A to C Disk 2 moved from A to B Disk 1 moved from C to B Disk 3 moved from A to C Disk 1 moved from B to A Disk 2 moved from B to C Disk 1 moved from A to C
C++
// C++ recursive function to // solve tower of hanoi puzzle #include <bits/stdc++.h> using namespace std; void towerOfHanoi( int n, char from_rod, char to_rod, char aux_rod) { if (n == 1) { cout << "Move disk 1 from rod " << from_rod << " to rod " << to_rod<<endl; return ; } towerOfHanoi(n - 1, from_rod, aux_rod, to_rod); cout << "Move disk " << n << " from rod " << from_rod << " to rod " << to_rod << endl; towerOfHanoi(n - 1, aux_rod, to_rod, from_rod); } // Driver code int main() { int n = 4; // Number of disks towerOfHanoi(n, 'A' , 'C' , 'B' ); // A, B and C are names of rods return 0; } // This is code is contributed by rathbhupendra |
Java
// JAVA recursive function to // solve tower of hanoi puzzle import java.util.*; import java.io.*; import java.math.*; class GFG { static void towerOfHanoi( int n, char from_rod, char to_rod, char aux_rod) { if (n == 1 ) { System.out.println( "Move disk 1 from rod " + from_rod+ " to rod " +to_rod); return ; } towerOfHanoi(n - 1 , from_rod, aux_rod, to_rod); System.out.println( "Move disk " + n + " from rod " + from_rod + " to rod " + to_rod ); towerOfHanoi(n - 1 , aux_rod, to_rod, from_rod); } // Driver code public static void main(String args[]) { int n = 4 ; // Number of disks towerOfHanoi(n, 'A' , 'C' , 'B' ); // A, B and C are names of rods } } // This code is contributed by jyoti369 |
Python3
# Recursive Python function to solve tower of hanoi def TowerOfHanoi(n , from_rod, to_rod, aux_rod): if n = = 1 : print ( "Move disk 1 from rod" ,from_rod, "to rod" ,to_rod) return TowerOfHanoi(n - 1 , from_rod, aux_rod, to_rod) print ( "Move disk" ,n, "from rod" ,from_rod, "to rod" ,to_rod) TowerOfHanoi(n - 1 , aux_rod, to_rod, from_rod) # Driver code n = 4 TowerOfHanoi(n, 'A' , 'C' , 'B' ) # A, C, B are the name of rods # Contributed By Harshit Agrawal |
C#
// C# recursive program to solve tower of hanoi puzzle using System; class GFG { static void towerOfHanoi( int n, char from_rod, char to_rod, char aux_rod) { if (n == 1) { Console.WriteLine( "Move disk 1 from rod " + from_rod + " to rod " + to_rod); return ; } towerOfHanoi(n-1, from_rod, aux_rod, to_rod); Console.WriteLine( "Move disk " + n + " from rod " + from_rod + " to rod " + to_rod); towerOfHanoi(n-1, aux_rod, to_rod, from_rod); } // Driver method public static void Main(String []args) { int n = 4; // Number of disks towerOfHanoi(n, 'A' , 'C' , 'B' ); // A, B and C are names of rods } } //This code is contributed by shivanisinghss2110 |
PHP
<?php // Tower of Hanoi (n-disk) algorithm in PHP with Display of Pole/rod // Contents the 3 poles representation $poles = array ( array (), array (), array ()); function TOH( $n , $A = "A" , $B = "B" , $C = "C" ){ if ( $n > 0){ TOH( $n -1, $A , $C , $B ); echo "Move disk from rod $A to rod $C \n" ; move( $A , $C ); dispPoles(); TOH( $n -1, $B , $A , $C ); } else { return ; } } function initPoles( $n ){ global $poles ; for ( $i = $n ; $i >=1; -- $i ){ $poles [0][] = $i ; } } function move( $source , $destination ){ global $poles ; // get source and destination pointers if ( $source == "A" ) $ptr1 =0; elseif ( $source == "B" ) $ptr1 = 1; else $ptr1 = 2; if ( $destination == "A" ) $ptr2 = 0; elseif ( $destination == "B" ) $ptr2 = 1; else $ptr2 = 2; $top = array_pop ( $poles [ $ptr1 ]); array_push ( $poles [ $ptr2 ], $top ); } function dispPoles(){ global $poles ; echo "A: [" .implode( ", " , $poles [0]). "] " ; echo "B: [" .implode( ", " , $poles [1]). "] " ; echo "C: [" .implode( ", " , $poles [2]). "] " ; echo "\n\n" ; } $numdisks = 4; initPoles( $numdisks ); echo "Tower of Hanoi Solution for $numdisks disks: \n\n" ; dispPoles(); TOH( $numdisks ); // This code is contributed by ShreyakChakraborty ?> |
Output: Tower of Hanoi Solution for 4 disks: A: [4, 3, 2, 1] B: [] C: [] Move disk from rod A to rod B A: [4, 3, 2] B: [1] C: [] Move disk from rod A to rod C A: [4, 3] B: [1] C: [2] Move disk from rod B to rod C A: [4, 3] B: [] C: [2, 1] Move disk from rod A to rod B A: [4] B: [3] C: [2, 1] Move disk from rod C to rod A A: [4, 1] B: [3] C: [2] Move disk from rod C to rod B A: [4, 1] B: [3, 2] C: [] Move disk from rod A to rod B A: [4] B: [3, 2, 1] C: [] Move disk from rod A to rod C A: [] B: [3, 2, 1] C: [4] Move disk from rod B to rod C A: [] B: [3, 2] C: [4, 1] Move disk from rod B to rod A A: [2] B: [3] C: [4, 1] Move disk from rod C to rod A A: [2, 1] B: [3] C: [4] Move disk from rod B to rod C A: [2, 1] B: [] C: [4, 3] Move disk from rod A to rod B A: [2] B: [1] C: [4, 3] Move disk from rod A to rod C A: [] B: [1] C: [4, 3, 2] Move disk from rod B to rod C A: [] B: [] C: [4, 3, 2, 1]
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References:
http://en.wikipedia.org/wiki/Tower_of_Hanoi
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