Twisted Tower of Hanoi Problem
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 last rod then you have to move the first rod to middle rod first and then to the last one.
Approach:
- Base Case: If the number of disk is 1, then move it to the middle rod first and then move it to the last rod.
- Recursive Case: In the recursive case following steps will produce the optimal solution:(All these moves are following the rules of twisted Tower of Hanoi problem)
- We will move first n-1 disks to the last rod first.
- Then move the largest disk to the middle rod.
- Move first n-1 disks from the last rod to the first rod.
- Move the largest disk at the middle rod to the last rod.
- Move all n-1 disks from the first rode 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; } |
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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 |
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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 |
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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 |
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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).
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