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Twisted Tower of Hanoi Problem

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  • Difficulty Level : Easy
  • Last Updated : 16 Aug, 2021

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) 
    1. We will move the first n-1 disks to the last rod first.
    2. Then move the largest disk to the middle rod.
    3. Move the first n-1 disk from the last rod to the first rod.
    4. Move the largest disk from the middle rod to the last rod.
    5. 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 moves
void 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 Code
int 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 moves
static 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 Code
public 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 moves
static 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 Code
public 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 moves
function 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 disks
var 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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