h shares borders with g and e and d . gcde

f shares borders with g , b , and a

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

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

France is adjacent to Spain.

Joins adjacent cells when they are in the same row.

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

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

And then I apply the formula.

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

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

Germany borders on France.

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

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

So, we start up v. We mark v.

Then we go through its neighbors.

This is the matrix of cofactors.

Same thing in Massachusetts.

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.

These glial cells become activated.

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

And now, we do essentially a search just as we're doing before, consisted of A and C.

So Kv which is the number of neighbors is 4.

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

Let's say C and add all its unexpanded neighbors to the graph, but all its neighbors are expanded.

In the route finding problem, the actions are dependent on the state.

The first law is two colorability.

So, we started off with b, the only thing on the open list, we visit both the neighbors of b add them to the open list, pull c off the open list, add it's neighbors d, b, and we work on d, take it off the open list, add its neighbors e and c.

Flood fill Fill adjoining pixels with the same color with the current color

Now, we can take e off the open list, add its neighbors f and, well this is already done.

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

So given a node v or computing the clustering coefficient involves

However many nodes it marked, we're going to add that to the total number marked.

Let's say B, and we start to build a tree starting from B, and here's how we do it.

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

The two villages adjoin each other.

And that is exactly what we found.

It's going to be fast enough, so it takes that off the open list and now what we're supposed to do, we loop through the neighbors, so here we have a statement that says, for each neighbor in the list of neighbors (current),

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.

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

Right now, we've visited everything.

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

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