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The recurrence relation capturing the optimal execution time of the $Towers \ of \ Hanoi$ problem with $n$ discs is

1. $T(n) = 2T(n − 2) + 2$
2. $T (n) = 2T(n − 1) + n$
3. $T (n) = 2T(n/2) + 1$
4. $T (n) = 2T(n − 1) + 1$

There are three towers named as Left, Middle, Right.
Tower Left contains 3 discs in decreasing size from bottom to top.
We have to move these 3 discs from Left tower to Right tower using Middle tower.
step 1: Move 2 discs from Left tower to Middle tower using Right tower. $n-1$ discs
Step 2: Move the remaining disc from Left tower to Right tower. $1-movement$
Step 3: Move 2 discs from Middle tower to Right tower using Left tower. $n-1$ discs

Recurrence relation:
$T(n) = T(n-1) + T(n-1) + 1$
$T(n) = 2T(n-1) + 1$

this should be the best answer.👍

Recurrence relation for Towers of Hanoi is

$T(1) = 1$

$T(n) = 2 T( n-1 ) +1$

For a given N number of disks, the way to accomplish the task in a minimum number of steps is:

1. Move the top N−1 disks to an intermediate tower. (Look at the first 3 steps in the figure)
2. Move the bottom disk to the destination tower.(Look at the 4th step in the figure)
3. Finally, move the N−1 disks from the intermediate peg to the destination tower.(Look at the last 3 steps in the figure)

TOH (N=3,L,M,R)
{
if(N==0)
return;
else {
TOH(N-1,L,R,M)
MOVE(L-R)
TOH(N-1,M,L,R)}
}

$T(N)=2T(N-1)+1$

Following are the steps to follow to solve Tower of Hanoi problem recursively.

Let the three pegs be A, B and C. The goal is to move n pegs from A to C.
To move n discs from peg A to peg C:
move n-1 discs from A to B. This leaves disc n alone on peg A
move disc n from A to C
move n?1 discs from B to C so they sit on disc n

The recurrence function T(n) for time complexity of the above recursive solution can be written as following.

T(n) = 2T(n-1) + 1, Hence D is the right answer.

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