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Time Complexity Analysis | Tower Of Hanoi (Recursion)
  • Difficulty Level : Medium
  • Last Updated : 22 Jan, 2021

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

Put the 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(1) =1 
n – k = 1 
k = n-1
put, k = n-1
T(n) =2^{(n-1)}T(0) + + 2^{(n-2)} + ............ +2^2 +2^1 + 1

It is a GP series, and the sum is 2^n - 1

T(n)= O( 2^n - 1)  , or you can say O(2^n)  which is exponential

for 5 disks i.e. n=5 It will take 2^5-1=31 moves.

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