France is adjacent to Spain.

In your implementation, the thing you're going to have to really watch out for is making sure that when you update the value of the neighbor, you can find it in the heap.

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

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

These glial cells become activated.

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.

And then I apply the formula.

So given a node v or computing the clustering coefficient involves

Germany borders on France.

Joins adjacent cells when they are in the same row.

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

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

h shares borders with g and e and d . gcde

f shares borders with g , b , and a

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

So, we start up v. We mark v.

And that is exactly what we found.

The first law is two colorability.

So Kv which is the number of neighbors is 4.

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

The two villages adjoin each other.

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

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

All right so if you did this right, you would realize that what you need to do to compute an expected value the sum of the overall values that it can take on the probability of that value times the value itself.

Remove it from the set because that one's now done and we actually know the distance to that node.

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.

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

In the neighbouring room on the left a painful stillness descended.

There's really no choice. It's the node over here.

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

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

Then we go through its neighbors.

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

This is the matrix of cofactors.

Take g off the open list, its neighbors are all already done, nothing is added.

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

No, said Gregor.

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

For the guys over here, 2, 3, 4, 5, 6, and 7.

Well, we're almost there.

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

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

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

We take the ending little piece off the to do list and say Oh, q that's what we have to do.

In this case, we've got Seattle, Los Angeles, Dallas Forth Worth, not Atlanta but Pittsburgh.