And then I apply the formula.

We now know that this is the final distance and then we delete that from the dist so far structure.

France is adjacent to Spain.

The first law is two colorability.

Germany borders on France.

You can try computing when a graph is all connected. You put the neighbors on a list, the open list.

h shares borders with g and e and d . gcde

f shares borders with g , b , and a

The two villages adjoin each other.

These glial cells become activated.

It actually kind of matters, but zero seems like a reasonable choice.

And that is exactly what we found.

And it can actually be all the way to one if we have a cleek.

Okay, well, let's go to j and do a mark_component from j.

We mark A as visited and we add it to the open list.

So given a node v or computing the clustering coefficient involves

We're going to average it for the n nodes in the graph.

Joins adjacent cells when they are in the same row.

We've got this one, and this one, and this one, and this one.

This is the matrix of cofactors.

The embassy is located next to the Supreme Court.

Then what we're going to do is we're going to count up the number of links between these neighboring nodes.

E is already in there but we can add in F at the tree edge.

Let me expand the first one. This one over here.

So, we start up v. We mark v.

And our instructions again say that we look at all the neighbors of v, w and z.

Then we go through its neighbors.

In particular, we're going to make each region on the map correspond to a node on the graph and we're going to connect two nodes on the graph if they share a border and therefore can't have the same color on the map. a shares a border with b and f

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So it's not that tightly connected and in fact you know that sort of what we saw as well that V is connected to a bunch of neighbors but they are not really connected to each other all that much.

Then what we do is we recursively go and mark all the things connected to that node, count up how many things got marked in that way, and then we return.

So Kv which is the number of neighbors is 4.

We're going to have to add them in separate, all right.

If you look at any node in the left side say the middle one.

ROMEO AND JULlET by William Shakespeare

And so now we've handled that node.

Then we look at all the neighbors of that node.

Letter C, B, and D then we choose one of these nodes,

Same thing in Massachusetts.

We have to go back to the beginning and do it again.

Yes, there is. There is still one more thing on the to do list.

We sum up whether or not they're connected and then we scale that by taking the number of possible connections 2 over the neighbor size of v times the neighbor size of v minus 1.

So say, V is that node and Kv is its degree and Nv of V is the number of links between the neighbors of V. So, let's

And in fact we can get zero if we've got something that's like a star.

So in this case, it's a little bit of a quarter, which is pretty clumpy.