Complexity of the recursion: T(n) = T(n-1) + T(n-2) + C

Question

I want to understand how to arrive at the complexity of the below recurrence relation.

T(n) = T(n-1) + T(n-2) + C Given T(1) = C and

Answer 1

If you were also interested in finding an explicit formula for T(n) this may help.

We know that T(1) = c and T(2) = 2c and T(n) = T(n-1) + T(n-2) + c.

So just write T(n) and start expanding...

T(n) = T(n-1) + T(n-2) + c
T(n) = 2*T(n-2) + T(n-m) + 2c
T(n) = 3*T(n-3) + 2*T(n-4) + 4c
T(n) = 5*T(n-4) + 3*T(n-5) + 7c
etc ...

You see the coefficients are Fibonacci numbers themselves!

Call F(n) the nth Fibonacci number. F(n) = (phi^n + psi^n)/sqrt(5) where phi = (1+sqrt(5))/2 and psi = -1/phi, then we have:

T(n) = F(n)*2c + F(n-1)*c + (F(n+1)-1)*c

Here is some quick code to demonstrate:

def fib_gen(n):
    """generates fib numbers to avoid rounding errors"""
    fibs=[1,1]
    for i in xrange(n-2):
        fibs.append(fibs[i]+fibs[i+1])
    return fibs

F = fib_gen(50) #just an example.
c=1

def T(n):
    """the recursive definiton"""
    if n == 1:
        return c
    if n == 2:
        return 2*c
    return T(n-1) + T(n-2) + c

def our_T(n): 
    n=n-2 #just because your intials were T(1) and T(2), sorry this is ugly!
    """our found relation"""
    return F[n]*2*c + F[n-1]*c + (F[n+1]-1)*c

and

>>> T(24)
121392
>>> our_T(24)
121392