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?

i would like to share an old thread which can be used. [Link_TO_Thread](https://stackoverflow.com/a/43991156/8395140) — DEBASIS PANDA, Oct 03 '19 at 19:39
The above written function goes into an infinite loop for factorial(0). — Asis, Aug 18 '20 at 12:32

score 26 · Accepted Answer · edited May 23 '17 at 12:00

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!

Sounds like [Quasi-polynomial time](http://en.wikipedia.org/wiki/Quasi-polynomial_time#Quasi-polynomial_time). — Nuclearman, May 06 '13 at 06:40
Reviving an old thread here, but I think it might be helpful to note that the teacher is correct. Because complexity is measured over an "efficient" encoding of the input. Factorial has a single int `n` as input, whose efficient encoding has size `log n`, not `n`! (See also my answer here: http://programmers.stackexchange.com/questions/181268/finding-the-time-complexity-of-the-following-program/247451#247451) — A.J.Rouvoet, Jul 10 '14 at 00:38
@A.J.Rouvoet sir I'm sorry for reviving an old question; but sir I m trying to get this notion efficient encoding you have used, can you briefly explain this, I'm just curious to understand — SSH, Jul 04 '15 at 23:39

Complexity of factorial recursive algorithm

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