Why is the time complexity of Tower of Hanoi with 4 pegs 0 (2^n/2)?

Question

For example if I use 4 disks, The problem should be solved in 3 steps according to the equation. However, mine takes 9 steps. Why is it so?

Consider the code here

include <stdio.h> 
void towerOfHanoi(int n, char from_rod, char to_rod,
                  char aux_rod1, char aux_rod2)
{
if (n == 0)
    return;
if (n == 1) {
    printf("\n Move disk %d from rod %c to rod %c",
                        n, from_rod, to_rod);
    return;
}

towerOfHanoi(n - 2, from_rod, aux_rod1, aux_rod2, 
                                        to_rod);
printf("\n Move disk %d from rod %c to rod %c ",
                   n - 1, from_rod, aux_rod2);
printf("\n Move disk %d from rod %c to rod %c ",
                      n, from_rod, to_rod);
printf("\n Move disk %d from rod %c to rod %c ", 
                   n - 1, aux_rod2, to_rod);
towerOfHanoi(n - 2, aux_rod1, to_rod, from_rod, 
                                    aux_rod2);
}

// driver program
int main()
{
    int n = 4; // Number of disks

    // A, B, C and D are names of rods
towerOfHanoi(n, 'A', 'D', 'B', 'C'); 
return 0;
}

Answer 1

When you have 4 pegs,time complexity to move 1 peg will require 1 comparison. time complexity to move 2 peg will require 3 comparisons. Time complexity = 2*T(n-2) + 3 ,n>=3

While solving it through substitution method . T(n) = 2*T(n-2) + 3 ,n>=3

= 2^2T(n-4) + 2*3 + 3 = 2^3T(n-6) + 2^2*3 + 2*3 + 3 and so on.

for kth substitution,

=2^(k/2)*T(nk) + 2^(k/2 -1 )*3 + ..... + 2^2*3 + 2*3 + 3 ---- eqn_(1)

take nk=2 => k=n-2 so eqn_(1) becomes = 2^(n/2 -1)*3 + 2^(n/2 - 2).3 + .....+ 2^2*3 + 2*3 + 3

= 3{1.(2^(n/2) - 1)}

= θ(2^(n/2))