Determining time complexity of a specific function - c

Question

This is a question from one of the old exams from algorithms and data structure that I recently came upon. I'm having a hard time understanding the solution.
I need to find big-O, big-ϴ and big-Ω bounds of a function:
void recursion(int n) {
int i;
if (n == 0) {
return;
}
for (i = 0; i < n; i++) {
recursion(i);
}
}
The solution is 2^n for all three and I can't understand why. I've tried writing things down and I can't even get close to the solution. I would appreciate if anyone would explain where the 2^n comes from here.

Answer 1

Let's look at a simpler recursion which is known to be O(2^n)
void fib(int n) {
if (n < 3) {
return 1;
} else {
return fib(n - 1) + fib(n - 2);
}
}
Here you can see, for the non-trivial case of n > 2, this will result in 2^(n-2) calls to itself. For example, if n = 5:
n = 5
n = 4
n = 3
n = 2
n = 1
n = 2
n = 3
n = 2
n = 1
There are 8 (2^3) recursive calls, because each call with n > 2 spawns two more recursive calls, so fib(n+1) has twice as many recursive calls as fib(n).
So for your example:
n = 3
n = 2
n = 1
n = 0
n = 0
n = 1
n = 0
n = 0
so we get 7 recursive calls when n = 3
for n = 4
n = 4
n = 3
n = 2
n = 1
n = 0
n = 0
n = 1
n = 0
n = 0
n = 2
n = 1
n = 0
n = 0
n = 1
n = 0
n = 0
Here, we have 15 calls. Looking at the execution tree above, you can see that recusrsion(4) is basically recursion(3) + recursion(3) + 1
n = 4
n = 3 // + 1
n = 2 //
n = 1 //
n = 0 // recursion(3)
n = 0 //
n = 1 //
n = 0 //
n = 0 //
n = 2 //
n = 1 //
n = 0 // recursion(3)
n = 0 //
n = 1 //
n = 0 //
n = 0 //
So in general, recursion(n + 1) will have one more recursive calls than 2 * recursion(n)....which is basically doubling for every +1 to n....which is O(2^n)

Answer 2

Let's denote the total runtime as f(n). Due to the loop in the function the f(n) is actually a sum of f(i) for i between 0 and n-1. That's a sum of n items. Let's try to simplify the expression. A standard trick in such situations is to find a complimentary equation. Let's see what is the value of f(n-1). Similary to the previous case, it's a sum of f(i) for i between 0 and n-2. So now we have 2 equations:
f(n)=f(1)+...+f(n-1)
f(n-1)=f(1)+...+f(n-2)
Let's subtract second from the first:
f(n)-f(n-1)=f(n-1)
--> f(n)=2f(n-1)
Now this is a homogeneous linear recurrence relation with constant coefficients.
The solution is immediate (see the link for more details):
f(n)=f(1)*2n=2n

Answer 3

Since this smells like a homework question, this answer is incomplete by design.
The usual trick behind these kind of problems is to create a recurrence equation. That is, the time complexity of recursion(k+1) is somehow related to the complexity of recursion(k). Just writing down the recurrence itself is not sufficient to prove the complexity, you have to demonstrate why the recurrence is true. But, for 2n, this suggests that recursion(k+1) takes twice as long as recursion(k).
Let T(k) denote the time complexity of recursion(k). Since recursion(0) returns immediately, let T(0) = 1. For k > 0, given the iterative implementation of recursion Thus You can inductively prove that T(k) = 2k.

Answer 4

r(n) = r(n-1)+r(n-2)+...+r(0) // n calls.
r(n-1) = r(n-2)+r(n-3)+...+r(0) // n-1 calls.
r(n-2) = r(n-3)+r(n-4)+...+r(0) // n-2 calls.
.
.
.
r(1) = r(0) // 1 call.
r(0) = return; // 0 call.
So,
r(n) = r(n-1)+r(n-2)+...+r(0) // n calls.
= 2 * (r(n-2)+...+r(0)) // 2 * (n - 1) calls.
= 2 * ( 2 * (r(n-3)+...+r(0)) ) // 2 * 2 * (n - 2) calls.
.
.
.
This follows that =>
2^(n-1) * (n - (n-1))
And that would be
2^n calls...