Let me say a little bit about that now.

So, let's start off with a really simple example. So, here is some pseudo Python for generating a graph with n nodes.

We start off with a set of n nodes, then we create a graph on half of the nodes recursively.

So we generate a graph on n 2 nodes call that G1.

What do they tell us?

N plus 1's are there?

We now know how deep. It's log n.

Once again, we can think about this as happening in a kind of recursion tree.

We're looking at the same basic setup as the last time.

Now the recurrence relation has this form..

Which of these algorithms has the best Θ? And I'm going to give a hint.

We're making a graph with n nodes, and again we're going to assume n is some power of 2.

We have a graph with 1 node, it's going to have 0 edges, by the way that this generation process happens.

Alright, so next we're going to look at some recursively generated graphs.

In general, we're going to have 1 print statement plus 2 print statements plus 3 print statements plus etc, etc, all the way up to n 1 print statements, which equals n( n 1) 2.

Then we are going to add to that what?

To do this, the algorithm multiplies your scores, then takes the nth root, where n is the number of questions.

Here's why.

Otherwise, we do it recursively.

So the zero vector in Rn, if it's arbitrary, is just a vector where everything is zero.

Rn is the set of all of these possible ordered n tuples or ordered sets of n numbers.

So is there any easy way to write that?

So, let's build up to the answer in steps.

This is pretty remarkable, and you might find this to be quite counter intuitive.

But in the end, we did somewhere between say N and 2N, primitive operations to compute this first partial product.

We're going to take n 4 of the set of nodes and fully connect them with another set of n 4 nodes. n 4 from one side, n 4 from the other.

To generate a graph with n nodes, it 1st breaks that sub problem into 2 things, where we work out the number of edges in a graph with n 2 nodes. but doing that requires working it out for n 4 nodes and so on.

Here is the algorithm.

A way to define factorial isà   for any value n, we can compute the factorial, which is the number of ways of arranging n items.

We know that if n 1, it just returns a single node.

The whole thing is a heap. We made a heap. Woohoo!

And let's say there's a bunch of terms like this.

Let's look right now at the analysis of the algorithm.

Keep directory symlinks

Very initially let's guess a particle index uniformly from the set of all indices.

We have n terms here, right, where each term was a f of x times a g of y, or f1 of x times g1 of y, and then all the way to fn of x times gn of y.

Let's say you want to find the best root and items. Is it better to use selection or sorting?

Preserve hard links

Preserve group

The answer, of course, is no.

So, here's how delta debugging works first, we split the input into n subsets where initially n has a value of 2.

So, given a number, say, 'n,' it outputs how many integers are less than or equal to n that do not share any common factors with n.

Now, selection is the same.

20 different copies in 20 different locations.

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