Down here, I have a very similar claim but with one bit change.

Every regular expression has a finite state machine, and every finite state machine has a regular expression.

Help me out.

It hasn't changed a bit.

It also has epsilon transitions, and I'll start making its deterministic equivalent down here.

We're going to go through each edge state, each entry in our dictionary.

Yeah, we understand that.

It turns out, however, that is not a problem at all.

It can actually find a live state versus a dead state, which is awesome.

We started off by introducing the concept of a language as a set of strings.

More on that in just a bit.

No epsilons, no ambiguity.

So in this class we will learn a lot of the things he have just mentioned. in various points.

And then we can repeat the bx and by afterwards as many times as we want, and we finish with a c.

It just looks like that because this is a schema.

Now that you know how to implement regular expressions, take the regular expression, make a finite state machine, make it deterministic, call FSM sim.

We just reorganized the markets.

In this problem, we have a bit of a challenge, and I've accordingly labeled it with 2 gold stars, but what we want to do here, is optimize a finite state machine.

lose the depth of the cup, the resilience, but appear to be healthy and appear to suddenly, under a threshold, be tipping over.

That's it.

This is just asking about the left side of the diagram.

And in fact, although I didn't show it, vice versa.

Briefly in lecture, Wes outlined the proof that all regular languages are context free.

If we have two entities concatenated, we're going to have to go from the current state we're at to some middle state, and then lastly, to the goal state.

Let's do one more bit of practice converting non deterministic machines to deterministic ones.

So now let's look at the problem another way, in reverse.

I give you the accepting states, just 6, and also all of the edges.

However, it's not going to solve our problems.

So let's do this together.

BJ

With all of that in mind, let's encode our finite state machine in Python.

In regular expressions, these two stars need not be the same.

That's how Python actually does it.

Each one had a regular expression. We found that.

Now let's expand it more, and there's now two options finally

States

Status

State

State...

State

Status

So this volume and that volume.

So we call that forward or progression state space search in that we're searching through the space of exact states.

So now we end up in this state.

For example, take this one here