河内的扭曲塔问题

河内塔的基本版本可以在这里找到。
它是河内问题的扭曲塔。其中，所有规则都是相同的，只是增加了一个规则：
您不能将任何磁盘直接从第一个杆移动到最后一个杆，即，如果要将磁盘从第一个杆移动到最后一个杆，则必须先将第一个杆移动到中间杆，然后再移动到最后一个杆。

方法：

• 基本案例：如果磁盘数量为1，则先将其移动到中间杆，然后再将其移动到最后一根杆。
• 递归案例：在递归案例中，以下步骤将产生最佳解决方案：(所有这些动作都遵循河内塔扭曲问题的规则)
  1. 我们将首先将n-1个磁盘移动到最后一个磁棒。
  2. 然后将最大的圆盘移至中间杆。
  3. 将前n-1个磁盘从最后一个杆移到第一个杆。
  4. 将最大的圆盘从中间杆移到最后一个杆。
  5. 将所有n-1个磁盘从第一个杆移动到最后一个杆。

下面是上述方法的实现：

C++

// C++ implementation
#include 
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

输出：

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

递归公式：

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.

通过解决该递归，时间复杂度将为O(3 ⁿ ) 。