Let me say a little bit about that now.

Now the recurrence relation has this form..

We have a graph with 1 node, it's going to have 0 edges, by the way that this generation process happens.

Alright, so next we're going to look at some recursively generated graphs.

So, let's start off with a really simple example. So, here is some pseudo Python for generating a graph with n nodes.

N plus 1's are there?

We now know how deep. It's log n.

We're looking at the same basic setup as the last time.

Then we are going to add to that what?

And let's say there's a bunch of terms like this.

Here's why.

Otherwise, we do it recursively.

We're given a recursive formula for the Stirling numbers, which is S(n, k) kS(n 1, k) S(n 1, k 1).

So is there any easy way to write that?

We start off with a set of n nodes, then we create a graph on half of the nodes recursively.

So, question for you.

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

And then X is going to be a matrix, and it's an n by k matrix.

Now, you get a different chain network for each different possible number of nodes.

This is pretty remarkable, and you might find this to be quite counter intuitive.

Let's look right now at the analysis of the algorithm.

So we generate a graph on n 2 nodes call that G1.

What do they tell us?

In general, we're going to have 1 print statement plus 2 print statements plus 3 print statements plus etc, etc, all the way up to n 1 print statements, which equals n( n 1) 2.

Now, selection is the same.

So this now is a recurrence relation for the number of edges.

Once again, we can think about this as happening in a kind of recursion tree.

The time it takes to run Top K via partitioning on one element.

We're making a graph with n nodes, and again we're going to assume n is some power of 2.

So, let's build up to the answer in steps.

We know that if n 1, it just returns a single node.

To do this, the algorithm multiplies your scores, then takes the nth root, where n is the number of questions.

We proved that it have some, somehow must have at most log based two, n 1 total nodes.

Choose one of the following as the correction factor.

Once again, a graph with 1 node has zero edges.

So, let's fix a level j. We know how many sub problems there are, two to the j.

Which of these algorithms has the best Θ? And I'm going to give a hint.

To generate a graph with n nodes, it 1st breaks that sub problem into 2 things, where we work out the number of edges in a graph with n 2 nodes. but doing that requires working it out for n 4 nodes and so on.

Let's say that we have an unlabeled training set of M examples, and each of these examples is going to be a feature in Rn so your training set could be, feature vectors from the last

So, so far we know, that if we only count black nodes then we're good, We only have

Realize that given n yes or no questions, there are 2 to the power of n possible answer sequences.

The keys. These two cases. The metal ones.

Rn is the set of all of these possible ordered n tuples or ordered sets of n numbers.

Nacelle casings.

We're going to take n 4 of the set of nodes and fully connect them with another set of n 4 nodes. n 4 from one side, n 4 from the other.