Complexity of factorial recursive algorithm

Question

Today in class my teacher wrote on the blackboard this recursive factorial algorithm:

int factorial(int n) {    if (n == 1)         return 1;    else         return n * factorial(n-1); } 

She said that it has a cost of T(n-1) + 1.

Then with iteration method she said that T(n-1) = T(n-2) + 2 = T(n-3) + 3 ... T(n-j) + j, so the algorithm stops when n - j = 1, so j = n - 1.

After that, she substituted j in T(n-j) + j, and obtained T(1) + n-1.

She directly said that for that n-1 = 2^(log₂n-1), so the cost of the algorithm is exponential.

I really lost the last two steps. Can someone please explain them to me?

Answer 1

Let's start off with the analysis of this algorithm. We can write a recurrence relation for the total amount of work done. As a base case, you do one unit of work when the algorithm is run on an input of size 1, so

T(1) = 1

For an input of size n + 1, your algorithm does one unit of work within the function itself, then makes a call to the same function on an input of size n. Therefore

T(n + 1) = T(n) + 1

If you expand out the terms of the recurrence, you get that

• T(1) = 1
• T(2) = T(1) + 1 = 2
• T(3) = T(2) + 1 = 3
• T(4) = T(3) + 1 = 4
• ...
• T(n) = n

So in general this algorithm will require n units of work to complete (i.e. T(n) = n).

The next thing your teacher said was that

T(n) = n = 2^{log₂ n}

This statement is true, because 2^log₂x = x for any nonzero x, since exponentiation and logarithms are inverse operations of one another.

So the question is: is this polynomial time or exponential time? I'd classify this as pseudopolynomial time. If you treat the input x as a number, then the runtime is indeed a polynomial in x. However, polynomial time is formally defined such that the runtime of the algorithm must be a polynomial with respect to the number of bits used to specify the input to the problem. Here, the number x can be specified in only Θ(log x) bits, so the runtime of 2^{log x} is technically considered exponential time. I wrote about this as length in this earlier answer, and I'd recommend looking at it for a more thorough explanation.

Hope this helps!