What happens inside INC is an assignment statement.

And if we have N features then rather than summing up over our four features, we would have a sum over our N features.

Now, if K is n 2 that's n times n 2 where n².

The parameters of this model are theta0 through theta n, but instead of thinking of this as n separate parameters, which is valid, I'm instead going to think of the parameters as theta where theta here is a n 1 dimensional vector.

I won't lift a finger against Geum Jandi.

I'm untouchable.

As easy as that?

The running time for that approach is going to be big Θ of n, the number of nodes times the time that it takes to do a shortest path search.

We can add 2 n to both sides, subtract n from both sides, and we get that.

This loop runs over all pairs of nodes n², but it does this from each value from 0 to n 1.

of course, there are alternatives. there are things instead.

And in order to do that, I'm going to use modulo n, because i 1 when we're at the end of the string i is n 1.

So it's Θ(n²). It's order of n², as well as order of n³.

So, going back to the picture, what are c and n0? Well, c, of course, is just going to be two. And n0 is the crossing point.

And obviously, you know, there's no point at all to try to memorize all these commands.

(Laughter)

Once we eliminate public funding for the artists,

And that everyone's a perfect logician.

Let me ask a question now.

So, unless...

Mu, sigma squared.

And if this is actually a densely connected graph like a clique, we're talking about n² for each of the nodes in the graph which totals up to n³ and n³ for something like the Marvel Comics graph where you have 6000 nodes, 6000³ is a pretty substantial number even by current computing standards.

Choose one of the following as the correction factor.

Well, if you try to tile it, it doesn't work too well, does it.

I can not allow you to eat lunch that Ha Ni packed for you.

Okay. So basically (log n)⁷ beats 9n(log n)² and n² ³ beats (log n)⁷ and 9n(log n)² beats n² ³.

Is it Θ of n² by itsef?

So what is the logorithm? Well, for our purposes, we can just think of a logorithm as follows, okay?

And I don't know if you can see what's going to happen here, but we get n edge contributions at each of these levels, and you may recall that there are log n levels to this tree.

With the very long list that were wanting to tap values of, you might as well just sort the whole thing. n, well what happens with n? n where look we're comparing n log n to n( n) which is n³ ². n³ ² is asymptotically larger than n log n so we're still better off just sorting the whole list.

And finally, we bring N back to the left side to use number 8, 1 N

And since the situation is going from bad to worse I decide to change my First Minister, since he's become unpopular

But could you go a little further?

And if you do that, you do that responsibly,

More coffee?

Would you like another Extra BigAss Fries?

I really need you

In exchange for...?

Hey, want more soup?

So instead of just adding N plus 1 N times, we could say that this is just N times N plus 1.

However, the number of edges is always going to be between 0 and n² roughly.

Breathe deep down. If I could only suffer in her place.

F(n) is bigger than or equal to g(n).

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