Time Complexity Analysis | Tower Of Hanoi (Recursion)

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:
1) Only one disk can be moved at a time.
2) 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.
3) No disk may be placed on top of a smaller disk.

Pseudo Code

TOH(n, x, y, z)
{
   if (n >= 1)
   {
      // put (n-1) disk to z by using y
      TOH((n-1), x, z, y)
   
       // move larger disk to right place
       move:x-->y
     
      // put (n-1) disk to right place 
      TOH((n-1), z, y, x)
   }
}

Analysis of Recursion

Recursive Equation : $T(n) = 2T(n-1) + 1$ ——-equation-1

Solving it by BackSubstitution :
$T(n-1) = 2T(n-2) + 1$ ———–equation-2
$T(n-2) = 2T(n-3) + 1$ ———–equation-3

Put value of T(n-2) in equation–2 with help of equation-3
$T(n-1)= 2( 2T(n-3) + 1 ) + 1$ ——equation-4

Put value of T(n-1) in equation-1 with help of equation-4
$T(n)= 2( 2( 2T(n-3) + 1 ) + 1 ) + 1$
$T(n) = 2^3 T(n-3) + 2^2 + + 2^1 + 1$

After Generalization :
$T(n)= 2^k T(n-k) + 2^{(k-1)} + + 2^{(k-2)} + ............ +2^2 + + 2^1 + 1$

Base condition T(0) == 1
n – k = 0
n = k;
put, k = n
$T(n) = 2^n T(0) + 2^{(n-1)} + + 2^{(n-2)} + ............ +2^2 + + 2^1 + 1$

It is GP series, and sum is $2^{(n+1)} - 1$

$T(n)= O( 2^{(n+1)} - 1)$ , or you can say $O(2^n)$ which is exponentioal

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Shubham Pandey 5

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