00:01
In this question to determine the upper asympotic upper bound of the given recurrence relation we'll write that down.
00:27
It is equal to pt ln 2j plus n.
00:49
So using a recursion tree, we can visualize the recursive calls and their respective sizes at each level.
00:58
So let us construct the recursion tree.
01:06
So here first it will be tn then it will be tn by 2.
01:24
This n divided by 2 again, it will be tn by 2 and tn divided by 2.
01:38
Now this will further break into tn divided by 4.
01:45
So let us say that this will break into tn divided by 4 and another tn divided by 4 then and so on.
02:11
So again tn divided by 4.
02:17
There are tn divided by 4 terms for each.
02:27
This will go on.
02:33
We'll write here still tn divided by 4.
02:48
So here we at each level of the recursion tree, we have three recursive cells called as tn by 2.
03:01
So the size of each subsequent level is decreased by a factor of 2 due to ln upon 2j that is given here in the equation.
03:15
So and this number of nodes at each level is multiplies by 3.
03:22
So therefore the number of nodes at each level is 3 cap 1 that is number of nodes at each level is 3 cap.
03:43
So now let us look at the next thing that is at each level of recursion tree.
03:52
We have three recursive calls.
03:54
We have three recursive calls.
04:03
So these are of the size n by 2.
04:08
So the size of each subsequent level will decrease by a factor of 2.
04:16
Okay, and the number of nodes at each level will be multiplied by 3 as we saw in the last page.
04:25
So here the depth of recursion tree that can be determined by finding the smallest value of k such that k such that n upon 2 cap k is equal to 1.
04:46
So now here we have to solving for k.
04:52
We have to solve for k...