Let's take a look at that. Let's think about adding some edges here.

So given all these nicely decorated nodes, we now have a rule for figuring out which ones are the bridge edges and this is the rule.

Then node 1 is connected to node 0 and node 3.

Well, for the states 1, 2, 3, presumably, I could just pass in a list of the states.

So, this is the vertical mask applied to finding horizontal edges.

You have a Warren Ellis edge.

Edge Smoothing

You can now just focus on that graph. So here's our graph.

Now, the BA edge is already in there, but the BC edge is not.

Select the rectangle tool again and start from this upper edge.

Is red bigger than black minus blue?

I'll just hold on to it for a while.

If you think about three times the number of regions, the number of edges has to be at least that big, though, we're counting each edge twice, because each edge can actually participate in two regions.

And then at the end, it's going to draw 1 extra edge to connect the 2 components together.

There are two different ways that we can add to this graph.

Now I'm in state 4, and I see a 3.

Find Edges

Detect Edges

Edge Detection

What happens if all the nodes that are not in the beginning and ending nodes have to had even degree because the path has to go in one edge and out another edge.

Here's a planar graph on five nodes.

It's easy to add bells and whistles to the adjacency matrix to accommodate parallel edges to accommodate edge weights, which is accommodate directed arcs, directed edges.

For each of the edges in the graph, we have three of these statements to rule out all possible color matches, so there's three of these for one edge, two edge, three edge there's four edges in the graph so we have four blocks of these.

So, what would change everything?

Just to illustrate this example, imagine we'd start up with some graph H. It's a little bit of a mess.

Top Edge Detection

Right Edge Detection

Bottom Edge Detection

Left Edge Detection

We start in the start state, and the character we see is a 2, so we follow this edge to state 2.

In this case there will be 16 of them.

This statement is going to get executed once per node.

Each of those actually requires two graphs of size n 4.

In that case, the formula still holds.

Okay, on each of the edges of the graph.

In this case, I then printed out the actual graph itself just so you can see what it looks like.

We've got one, two, three, four, five, six, seven, eight, nine.

The approximate diameter, it's a three inch (3 ) shell mill

It looks like b was our winner.

But that orientation distribution corresponds to edges in the image.

It'd be nice to be able to answer this question in a little bit more generality.

But the level right above it, there's n.

We'll assume, as before, that n represents the number of nodes in the graph and m represents the number of edges.

We need red bigger than the difference, no.

Here we have four nodes. That's a power of 2 2Â .