I just wanted to make sure I'm on the right track regarding this.

Here's the function that I'm dealing with:

import math

def Mystery2(A, f, l):
    if f == l:
        return A[f]
    m = math.floor((f+l)/2)
    x = Mystery2(A, f, m)
    c = 0
    i = f
    while i <= l:
        if A[i] == x:
            c = c + 1
        i = i + 1
    if c > (l-f+1)/2:
        return x
    print("reached here")
    return Mystery2(A, m+1, l)

The precondition is given as: A is an array of integers, and f and l are integers 0 <= f <= l < len(A). From what I can see by running the function, the postcondition appears to be that it outputs the element within f and l indices that occurs more than (l-f+1)/2 times or A[l]. I'm not sure how to find a closed form for the recurrence relation representing the worst case runtime. So far I figured that with T(n) being the worst runtime of the function and n = l-f, T(n) will be constant if n = 0 and T(floor(n/2)) + T(ceiling(n/2) - 1) + n*c_1 + c, if n is greater than 0. I'm not sure how to convert this into a closed form. Thanks in advance for any help!

I just wanted to make sure I'm on the right track regarding this.

Here's the function that I'm dealing with:

import math

def Mystery2(A, f, l):
    if f == l:
        return A[f]
    m = math.floor((f+l)/2)
    x = Mystery2(A, f, m)
    c = 0
    i = f
    while i <= l:
        if A[i] == x:
            c = c + 1
        i = i + 1
    if c > (l-f+1)/2:
        return x
    return Mystery2(A, m+1, l)

The precondition is given as: A is an array of integers, and $f$ and $l$ are integers $0 \leq f \leq l < len(A)$. 

From what I can see by running the function, the postcondition appears to be that it outputs the element within f and l indices that occurs more than $ \frac{l-f+1}{2}$ times or A[l].

I'm not sure how to find a closed form for the recurrence relation representing the worst case runtime. 

So far I figured that with $T(n)$ being the worst runtime of the function and $n = l-f$, $T(n)$ will be constant if $n = 0$ and $T(\left \lfloor \frac{n}{2} \rfloor \right) + T(\left \lceil \frac{n}{2} \right \rceil - 1) + n*c_1 + c$, if $n > 0$. I'm not sure how to convert this into a closed form.

Is it okay to simplify this into $T(\left \lfloor \frac{n}{2} \rfloor \right) + T(\left \lceil \frac{n}{2} \right \rceil) + n*c_1 + c$ for the purpose of comparing it to the master theorem?

Thanks in advance for any help