Another way to think about that is let's say that we're in a particular state like this one

So we said that this is equal to e to the r, so p times this is equal to p.

And if you want to factor a p out of this, this is going to be equal to p times, if you take p divided p you get a 1, p square divided by p is p.

I want to draw it a little bit bigger than that, and you'll see why a second.

Rule 0 and rule 1 both apply to S, so I'm going to yield 2 things.

You can actually compute this using total probability where P(Y) equals P(Y_BAR_X) times P(X) plus P(Y_BAR_ X) times (P X).

So, to get all 3 of them together, we just multiply these by 3.

So, let me print out the resulting set of particles.

This is going to be, this term right over here is going to be p squared minus p to the third.

Starting from s, I can take rule let's do 0.

It uses this same symbol S. There is nothing before the red dot.

This one will be number 11.

The first one looks pretty good. I just apply rule 1.

If you square it you're just going to get p squared.

And then plus negative p squared.

In fact, now a code every time you call the function robot and assign it say to a particle, here the i particle, these elements p i x, y, and orientation, which is the same as heading, are initialized at random.

But it's just a scaling factor.

It will refer to this new list that has 5 elements, and at the end of this, the len of P is 5, since there are 5 elements in P.

The second element corresponds to my second grammar rule

Suppose we introduce the variable q.

P of A given B where B is the evidence and A is the variable we care about is P of B given A times P of A over P of B.

So what I want you to do is tell me for each of these sentences, whether it is valid, satisfiable but not valid, or unsatisfiable, in other words, false for all models.

The second thing we learned has to do with negation of probabilities.

You have to add one more argument and you have to change the returned function to implement a formula just like this but this using p0, p1, p2 as arguments not just the fixed numerical numbers here.

I dropped the p somewhere along the way.

How about this? Well, I'm going to apply rule one twice.

So chart state 0 includes the following parse states

We know that p is a pointer to a list let's say it's 1, 2, and 3.

The next step is now straightforward we multiply this guy out over here p 2p² p³ leaving this guy over here intact.

Now let's actually work this out.

We're going to introduce the auxiliary variable s, which we build up in three different steps.

Now, what value should E choose for P?

Computes Jacobi symbol of a and p. p should be odd and must be positive.

Compute the Legendre symbol of a and p. p should be odd and must be positive.

We can return 1 right away.

If I were to borrow it for another year, it becomes Pe to the 3r.

So e to the r.

The mean, the expected value of this distribution, is p.

P of B given A is equal to 0.2.

Using Bayes' rule, you can transform this into P of H given R comma S times P of R given S over P of H given S.

To shift the distribution to the right, U 1, we need to find in p the element 1 place to the left.

Let's look at the measurement side, and suppose we wish to know the probability of an internal state variable given a specific measurement, and that by Bayes rule becomes P of Z1 given X1 times P of X1 over P of Z1.

Once we fly from X to Y, the plane is no longer at X, so we say not at P,X the plane is no longer at X and the plane is now at Y.

But now I have the dot, I have my finger, before P.

It's that grammar of balanced parentheses.