The basic version of the Tower of Hanoi can be found here.
It is a twisted Tower of Hanoi problem. In which, all rules are the same with an addition of a rule:
You can not move any disk directly from the first rod to last rod i.e., If you want to move a disk from the first rod to the last rod then you have to move the first rod to the middle rod first and then to the last one.
Approach:
- Base Case: If the number of disks is 1, then move it to the middle rod first and then move it to the last rod.
- Recursive Case: In the recursive case, the following steps will produce the optimal solution:(All these moves follow the rules of the twisted Tower of Hanoi problem)
- We will move the first n-1 disks to the last rod first.
- Then move the largest disk to the middle rod.
- Move the first n-1 disk from the last rod to the first rod.
- Move the largest disk from the middle rod to the last rod.
- Move all n-1 disks from the first rod to the last rod.
Below is the implementation of the above approach:
C++
// C++ implementation#include <iostream>using namespace std;// Function to print the movesvoid twistedTOH(int n, char first, char middle, char last){ // Base case if (n == 1) { cout << "Move disk " << n << " from rod " << first << " to " << middle << " and then to " << last << endl; return; } // Move n-1 disks from first to last twistedTOH(n - 1, first, middle, last); // Move largest disk from first to middle cout << "Move disk " << n << " from rod " << first << " to " << middle << endl; // Move n-1 disks from last to first twistedTOH(n - 1, last, middle, first); // Move nth disk from middle to last cout << "Move disk " << n << " from rod " << middle << " to " << last << endl; // Move n-1 disks from first to last twistedTOH(n - 1, first, middle, last);}// Driver's Codeint main(){ // Number of disks int n = 2; // Rods are in order // first(A), middle(B), last(C) twistedTOH(n, 'A', 'B', 'C'); return 0;} |
Java
// Java implementation of the approach import java.util.*;class GFG{// Function to print the movesstatic void twistedTOH(int n, char first, char middle, char last){ // Base case if (n == 1) { System.out.println("Move disk " + n + " from rod " + first + " to " + middle + " and then to " + last); return; } // Move n-1 disks from first to last twistedTOH(n - 1, first, middle, last); // Move largest disk from first to middle System.out.println("Move disk " + n + " from rod " + first + " to " + middle); // Move n-1 disks from last to first twistedTOH(n - 1, last, middle, first); // Move nth disk from middle to last System.out.println("Move disk " + n + " from rod " + middle + " to " + last); // Move n-1 disks from first to last twistedTOH(n - 1, first, middle, last);}// Driver Codepublic static void main(String[] args){ // Number of disks int n = 2; // Rods are in order // first(A), middle(B), last(C) twistedTOH(n, 'A', 'B', 'C');}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of above approach# Function to print the moves def twistedTOH(n, first, middle, last): # Base case if (n == 1): print("Move disk", n, "from rod", first, "to", middle, "and then to", last) return # Move n-1 disks from first to last twistedTOH(n - 1, first, middle, last) # Move largest disk from first to middle print("Move disk", n, "from rod", first, "to", middle) # Move n-1 disks from last to first twistedTOH(n - 1, last, middle, first) # Move nth disk from middle to last print("Move disk", n, "from rod", middle, "to", last) # Move n-1 disks from first to last twistedTOH(n - 1, first, middle, last)# Driver Code # Number of disks n = 2# Rods are in order # first(A), middle(B), last(C) twistedTOH(n, 'A', 'B', 'C') # This code is contributed by# divyamohan123 |
C#
// C# implementation of the approach using System; class GFG{// Function to print the movesstatic void twistedTOH(int n, char first, char middle, char last){ // Base case if (n == 1) { Console.WriteLine("Move disk " + n + " from rod " + first + " to " + middle + " and then to " + last); return; } // Move n-1 disks from first to last twistedTOH(n - 1, first, middle, last); // Move largest disk from first to middle Console.WriteLine("Move disk " + n + " from rod " + first + " to " + middle); // Move n-1 disks from last to first twistedTOH(n - 1, last, middle, first); // Move nth disk from middle to last Console.WriteLine("Move disk " + n + " from rod " + middle + " to " + last); // Move n-1 disks from first to last twistedTOH(n - 1, first, middle, last);}// Driver Codepublic static void Main(String[] args){ // Number of disks int n = 2; // Rods are in order // first(A), middle(B), last(C) twistedTOH(n, 'A', 'B', 'C');}} // This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript program // Function to print the movesfunction twistedTOH(n, first, middle, last){ // Base case if (n == 1) { document.write("Move disk " + n + " from rod " + first + " to " + middle + " and then to " + last + "<br>"); return; } // Move n-1 disks from first to last twistedTOH(n - 1, first, middle, last); // Move largest disk from first to middle document.write("Move disk " + n + " from rod " + first + " to " + middle + "<br>"); // Move n-1 disks from last to first twistedTOH(n - 1, last, middle, first); // Move nth disk from middle to last document.write("Move disk " + n + " from rod " + middle + " to " + last + "<br>"); // Move n-1 disks from first to last twistedTOH(n - 1, first, middle, last);}// driver code// Number of disksvar n = 2;// Rods are in order// first(A), middle(B), last(C)twistedTOH(n, 'A', 'B', 'C');// This code contributed by shivani</script> |
Output:
Move disk 1 from rod A to B and then to C Move disk 2 from rod A to B Move disk 1 from rod C to B and then to A Move disk 2 from rod B to C Move disk 1 from rod A to B and then to C
Recurrence formula:
T(n) = T(n-1) + 1 + T(n-1) + 1 + T(n-1)
= 3 * T(n-1) + 2
where n is the number of disks.
By solving this recurrence, the Time Complexity will be O(3n).
Auxiliary Space: O(n).
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