This one over here, we have a constant cost per cluster is new.

K means is a very simple almost binary algorithm that allows you to find cluster centers.

Cluster ordered dither

This is a really useful thing to do if the exact answer doesn't matter very much which often when you're doing social network analysis it doesn't really matter that much exactly what something like the clustering coefficient is.

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And the answer is 2.

Next I want to focus for a little while on the notion of a clustering coefficient.

It was clear that they're not purely randomly distributed.

Suppose you are given data like this, and you wish to learn that there's 2 clusters a cluster over here and a cluster over here.

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So here's my second quiz.

I'm going to do. The first step is to randomly initialize two points, called the cluster centroids.

But so, Ci is going to, I'm going to minimize over my values of k and find the value of K that minimizes this distance between Xi and the cluster centroid, and then, you know, the value of k that minimizes this, that's what gets set in

So a new customer will pay this minus 3.

So there's a fair amount of interconnection between a node's neighbors.

So just getting it correct to a couple of digits is probably sufficient.

The point is to always advertise in a very popular way to get them to our website.

What if it's a book that's been made into a movie?

And then the outcome over here would be 3P 4 times 1 P.

For this problem we have 6 graphs, and we want you to order them with a clustering coefficient of the red node of each graph.

In particular, you're going to minimize the negative log likelihood of your data given the model plus a constant penalty per cluster.

The last question asked, what is the maximum clustering coefficient for a node in B?

Next is lower case k.

The reason being that these aren't clusters defined by a center of data points, but they're clusters define by affinity, which means they're defined by the presence of nearby points.

K of the cluster centroid closest to Xl.

So, these two crosses here, these are called the Cluster Centroids and I have two of them because I want to group my data into two clusters.

We were just happy with the number of cluster centers, and using more will make the situation worse.

So we'd like to have a way of getting a pretty good answer in time a lot less than this.

These are Hubble Space Telescope images of a particular cluster, that serves as a gravitational lens.

You've got the red hat triathletes over there.

Is it civilized?

So as a concrete example, let's say that one of my cluster centroids, let's say cluster centroid two, has training examples, you know, 1, 5, 6, and 10 assigned to it.

And if you do that, you end up with K minus one clusters instead of k clusters.

Step 1 Guess cluster centers at random, as shown over here with the 2 stars.

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