In this case, n is 3.

In this version, we're given an n x n matrix. In this case n equals 3.

So in this case, n factorial was 7.

In this case X is one dimensional which is N 1.

We could write it as n factorial it s a lower case n n factorial over n minus k factorial.

But even in this case, n would have to be, in the smallest case, it would be 7.5.

Where n is some number, and in case of the Vaucouleurs profile, n is equal 4.

This would extend to the case where N is odd in the natural way where you break it into N 2 rounded up and N 2 rounded down.

Each of these levels we're going to be generating n new edges at this top level, n 2 for this subgraph, and n 2 for this subgraph, which adds up to n, n 4 for this one, n 4 for this one, n 4 for this one, n 4 for this one, which adds up to n.

n here has to match this n here.

So, we set n to n 1, which means that in our case we would have three subsets.

The key observation is that this thing over here is n! (n k)! and to see this let's go back to the case where we had 4 heads.

In case of this galaxy n is 1. So then you have just traditional old declining exponential.

Here's our base case if n lt or to 2, return 1.

Again, in the case for n 1, it's just a single node.

In that case, n finally, let's say you want the top 100.

So this is a case where m is indeed, n minus 1. Now, for the upper bound, why can't you have more than n choose 2?

The worst possible case that you get in this algorithm is we pick the partition on the n, and then we recurse on everything but the n.

Is it n, n 1 over n, n over n 1, n divided by n 1, n 1 divided by n, or is it n squared 1 over n 1?

In the n 8 case we first generate two n 4 graphs, which we know what they look like from before.

Is the depth that we get this way log n, n, or 2 n, and are the number of leaves at the bottom log n, n, or 2 n?

This is just n 2. It's really 1 16 n 2, which is Î (n 2).

So this quantity really is somewhere between n² and 2(n²). All that is Θ(n²).

F(n) n. Okay, and let's contrast this with a logarithm.

So really, this is n² .

In this case this wasn't intended to be a super hard question as n grows there seems to be a linear relationship.

So instead of just adding N plus 1 N times, we could say that this is just N times N plus 1.

This n and this s are constant.

Θ(n²), O(n²), and O(n³).

And the final number we print should be n , which is the case here.

This only has to hold for n that are bigger than n₀.

Okay. So basically (log n)⁷ beats 9n(log n)² and n² ³ beats (log n)⁷ and 9n(log n)² beats n² ³.

And so computing the clustering coefficient for that node is n² well I guess in this case then the total running time is not going to be n³ because we don't have n² on each of those nodes.

I run through this n times.

What happens if this was n?

At this time this was an N of 20. Now there's an N of about 100.

Now, if K is n 2 that's n times n 2 where n².

So I subscript n by n is the n by n identity matrix.

This is Θ(n²). This is Θ(nlogn).

We make sure that n cannot grow beyond the size of the input, which means that in our case n would now be 8.

So altogether we're talking about n n² or n³.

n times n minus one times n minus two.

If we divide through by n, we get 2 n n and divided by n is actually n.

So essentially this would be the top of it and this would be n and this would be n.

Russian(n, n) for lots of different values of n.