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Lock and key, bro. Lock and key.

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.

She couldn't get out.

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

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.

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

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

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.

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

That's it.

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.

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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.

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.

Well, let's go through one way to do it together.

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Now, should this state be an accepting state or not?

And if I'm in state 3, there's a self loop back to state 3.

So what we're going to do is write code.

So conceptually, it might be simpler to have an edge that somehow consumes no input.

So the bank has to stay liquidate

Now, this is not a proof, but it just so happens that any non deterministic machine can be converted to a deterministic machine using exactly the same steps.

Then we see an x, and there's no outgoing edge from state 2 on an x, so we fall off the finite state machine and die.

lex is the input, takes out the parse tree using our parser, and then interprets the parse tree, returning the finite state machine that resulted.

Now we have to combine them all together.

It's just that their finite state machine is much bigger.

A regular language is a type of language.