We see the a, self loop back to state 2. We see the 1. We're in state 3.

So it's not all bad.

If I'm in 4, fall off the world.

In state 1 on a 'q', you loop back to state 1.

Since 3 is an accepting state up there, 2 or 3 is an accepting state down here.

If you quickly look at it and think about what's going on, once you take a path down the b to state 2, you can never get to the accepting state.

Here if I was in 2, 4, 5, or 6 and I see a c,

And on b we go to state 3, and then from state 2 on c we go to 4, and state 3 on d we go to 5, and 5 on e goes back to 2, and 5 on f goes to state 6, and 5 on g goes all the way back here to 1, and our only accepting state is 6.

From state 1 on b, we go to state 2 as well a or b.

The problem that I want to talk with you about is really the problem of

All right, well, what if I see a b?

4 and I see a b I die.

How about 'aa'?

Well, one of our edges is at state 1 on 'a.' State 1 on input 'a' goes to state 2.

That's our self loop.

Finally, let's consider the string '2x'.

But let's just finish checking up on c just in case we missed something.

We need to deliver health care, not in large institutions, but close to people's homes, and we need to deliver health care using whoever is available and affordable in our local communities.

Over here, I have defined edges, just as we suggested.

Well, one possibility is that we're already at the end of the input, at which point we should just check to see if our current state is an accepting state or not.

Well, when we enter the non deterministic machine, we could only be in state 1, but after that we could see an a, and if I see an a I could be in 2, 4, or I could take the free epsilon transition to 5, or I could keep going and take the free epsilon transition to 6.

In my converted world, the state accepts if any of its corresponding original states also accept.

6 and I see a b I die.

I don't like being patient when systems are broken, and health care is broken.

I haven't made any progress, so I'm going to stop following edges from the 1 along that path.

This new finite state machine will still accept 1, 2, 3.

Now, there's some other ways to get out of 2, 4, 5, and 6, and when we do, we might be in states 2 or 3.

I must really have been in state 3, and I stay in state 3, so either on b or c we end up in just state 3, which is also an accepting state.

We see the a. We're in state 2.

Here I have the declaration for my procedure.

I'll just put them together into one big machine.

Finally, you'll notice that 1 of my states has a double circle.

If I'm in 2 or 3, on a b from 2 I go to 3, and c I'd fall off the world, so we'd end up in state 3.

The financial drain on our healthcare industry is immeasurable.

This finite state machine accepts the same language as our ba, baa, baaa regular expression from above.

From state 2 on d, we go to state 3.

We see that a finite state machine accepts a string s if there exists even one path from the start state to any accepting state that follows s.

The empty string we come in here in state 1, and we don't go anywhere because we're out of input.

One way to do that would be to take this go to start back loop here, pass go, start again.

I'll label my new state 2456 because it keeps track of everywhere that I could have my fingers if I'm simulating this non deterministic machine.

If I'm in 2 or 3 and I see a c,

We can just start at the start state.

So to go through an example, let's say we have this finite state machine.

They tell us when to move from 1 state to another.

If I'm in 2 and I see a c, I fall off the world.