And we exclude negative 2 by drawing an open circle at negative 2, but all the values greater than that are valid x's that would solve, that would satisfy this inequality.

But let's just make show if c₁ is equal to 1, then we need n to be less than or equal to this expression.

If the minimum value is bigger than or equal to L i that means L i the value at node i is smaller than both the kids so we're good.

Now if something has to be less than 1 and it has to be less than minus 2, well, it just has to be less than minus 2.

Let me show you an example. Here's a balanced binary tree with 20 nodes.

You will get the precise assertion that failed and you won't implant any time bombs

But then we can shade in everything below 40.

The moment the lights go down is often the best part, you know?

We're dealing with both of them.

So anything above it anything above it will work.

In each case, we'd be updating only the action we took, associated with that state, not the whole state.

If you want to know, what is the probability that I get a value less than 20?

These values scale between 0 and 255, where 255 is white and 0 is black.

Keep your eyes open!

I realize I've never been happier.

So we have minus 1, minus 2, anything less than minus 2 will also satisfy this equation by both making the numerator and denominator negative.

In particular, suppose we color all three of the nodes, black.

It'll go from 10 3, all the way down to negative infinity.

So every time like for example you add 1 up vote, we actually increment the score by some other amount.

My part of our bargain is nearly fulfilled.

Good night, good night! as sweet repose and rest Come to thy heart as that within my breast!

Now the set that I'm mapping to is called the codomain.

Those are definitely going to have an absolute value less than 12, as long as they're also and, x has to be less than 12.

So they're saying, what are all the x's, that when you subtract 5 from them, it's going to be less than 35?

So that equals 0.

(Voice over) And get ready to say cheese,

Well, it's a minimizing node, so the least it could be is 2.

We go and tick on run and if everything works well, our test should passed by now.

This hydrogen can pretend like she has that electron, and then they're happy.

Anything that's in between these two numbers is going to have an absolute value of less than 12.

And here's a somewhat non trivial quiz.

What?

So the solution is, I can either be greater than 29, not greater than or equal to, so greater than 29, that is that right there, or I could be less than negative 34.

We're just calling replace spy, but after the call the value that spy refers to has changed.

M y heart is filled with love for you

He got 100 percent in all the important things,

What happens when we run partitioning on that value v is it's going to separate all the values of L into the ones smaller than v, equal to v, and greater than v, and that happens to be position K, so the ones on the first K positioned here are exactly the top K for the list, the smallest K elements that we were looking for.

This is the best time for oysters, you know.

Now, let's verify that if we substitute y1 and its derivatives back into this differential equation, that it holds true.

All right, let's imagine that we can do that and in fact, reducing the distance of a value turns out to be, you can use up heapify to do that because the value, here let me draw.

If you have a polynomial, you could have more than one values of x that satisfy this equation.

Well, let's see where it's going to die, but it will keep trying to add nodes in their final distances even though they aren't final distances to add.

Right. 3 times 1. So 7 should be greater than 3, and it definitely is.

Well, I mean we know if you say something has to be less than minus 2 and less than 1, just saying that it's less than minus 2 will do the job.

So any of those are going to satisfy this absolute value inequality.