0:33

And so in terms of proving that this star is the right thing to do,

Â in the middle cost range.

Â So first of all, it's very easy to analyze the case where C is

Â less than delta- delta squared.

Â Because now, if ij are not already connected, the value that they're getting

Â from their relationship is going to be less than or equal to delta squared.

Â And if they add a direct link, they are going to get a delta

Â minus c for that relationship, possibly some extra bits too.

Â And if c is less than delta minus delta squared,

Â then delta minus c is bigger than delta squared.

Â And so, the value that they're going to get is at least what they've got before,

Â and so both of these individuals will be better off, right?

Â So ui and uj of adding this link is better off.

Â But also, in the connections model, everybody else benefits,

Â or is not hurt by adding new links.

Â So you get benefits from the shortest paths, if we add extra links,

Â that just shortens your paths.

Â So everybody else is doing at least as well as they were before,

Â plus these two individuals are doing strictly better.

Â Therefore, it's better to add the links.

Â So here, the idea that when c is less than delta- delta squared,

Â when you get the complete network, that's straightforward, okay?

Â So what we're going to do is instead, look at the case where c is bigger than delta-

Â delta squared, so we don't want to form the complete network.

Â And we're going to do two things.

Â First, we're going to show that the value of a component

Â is highest when a component is a star, okay.

Â 2:17

And then we'll actually show that if you're going to arrange people,

Â you're best off doing it in a star.

Â And then you don't want to have multiple stars,

Â you'd be better off having one star.

Â And then we can just compare whether it's better to have

Â a big star with everybody in it, or no star at all, okay?

Â And that'll be the difference between the medium cost and the really high cost, so

Â when costs get high enough, a star is not even valuable.

Â So basically what we're going to show, is that if you're going to arrange people

Â when c is bigger than delta- delta squared, you better do it in a star.

Â And then is a question of whether a star is valuable, okay?

Â So value of a star with k players, what is it?

Â Well, with k different players, then we've got 1 person in the middle,

Â k- 1 other individuals out here, so we've got k- 1 links in total.

Â And so each one of those links gives a value of delta- c

Â to each of its participants.

Â There's two people in it, k- 1 links.

Â So the direct value of connections is 2 (k- 1) [delta- c].

Â And then the indirect values that we're getting wre coming from the fact that each

Â one of these indirect people, there's k- 1 of them, right?

Â They have k- 2 other neighbors, each at a distance of 2,

Â so each of these k- 1 people have k- 2 neighbors,

Â each one of those gives a benefit of delta squared.

Â So the overall value of a star Is 2(k -1)[delta-

Â c] + (k -1)(k- 2) delta squared.

Â Okay, so that's the value of a star.

Â Now let's look at the value of some other configuration,

Â that involves k players and m links,

Â where m has to be at least k- 1 in order to connect these k players together.

Â Okay, so if you've got m links, first of all,

Â what's the value you can get out of the links directly?

Â Well again, same kind of calculation, you've got m links, 2 people in each link,

Â delta- c, so that's going to be the value there, of the direct value.

Â And the most you could be getting indirectly is,

Â you've got k players,

Â k- 1 other people that they could be connected to,

Â 2m of those connections are direct connections.

Â So the remaining connections, this is how many remaining indirect connections there

Â can be, and at most they could be with delta squared, okay?

Â So this is the maximum possible value we can imagine for

Â some other component with m links, okay?

Â So let's take the difference between these two, so let's take this,

Â we'll take the difference between these two different, right?

Â So take this expression, subtract off this expression and what do we get?

Â If you subtract this from that, well, you can do the arithmetic.

Â The difference is going to turn out to be 2(m-(k-1)[delta

Â squared- (delta- c)], okay?

Â 5:48

And if you remember correctly,

Â we're in the region where we have got

Â delta squared > delta- c, right,

Â because we're in a region where c > delta-

Â delta squared, so we've got that holding.

Â That means that this thing is positive, and

Â so if m > k- 1, then this whole difference is positive.

Â Okay, so this is more valuable, this is more valuable.

Â The first expression is more valuable than the second expression,

Â in a situation where m > k- 1.

Â So if we're using more then m links,

Â we're doing worse than if we just did the star, okay?

Â 7:08

Well, we get the direct value of connections, and then at most,

Â we can have all but 1 relationship in the delta squared.

Â Some relationship, at least one of them has to be the distance delta cubed, okay?

Â And then if you do the difference between these, all we've done is move some

Â relationships which could be closer, to some that are further away.

Â So we get the same value in direct connections,

Â the same number of indirect connections, but some of them are at higher distances.

Â It's gotta be lower value than a star, so the star is better, okay?

Â So what we've shown is, if we're going to connect k individuals, the best way to do

Â it, when we've got c > delta- delta squared, is via a star, right?

Â 8:02

Okay, so now what we can do is say okay, well, what are the possibilities,

Â maybe 2 stars are better than 1?

Â So we can check that if 2 star components each give non-negative utility,

Â then 1 star with all those people generates a higher utility.

Â And so what we can do is just look at,

Â what's the value from the separate components, right?

Â So we've got k people in one of the stars, k' in the other star,

Â 2 stars, what are the values?

Â If you put them all together, now we've got k plus k' people together.

Â 8:39

So before, we have this value, 1 star, we have k plus k' people together.

Â Compare this expression to this expression, and it's easy to check.

Â The first parts of these are identical, right?

Â And you can just go through, and

Â check that the second expression, here, is larger than this one.

Â Okay, so just simple arithmetic,

Â you can verify this expression is bigger than this expression, and

Â therefore having 1 star is better than 2 stars, and what's the reason?

Â The reason is, if you've got 2 separate stars, you're not getting

Â the value of indirect connections, sorry, actually there's one difference here.

Â 9:33

And in terms of indirect connections,

Â we're getting more people indirectly connected than we had before.

Â And so the second expression overall is higher,

Â both in terms of the direct connections, and the indirect connections.

Â You're better off having one giant star than two separate stars, okay?

Â So what we've shown is if you're going to connect people, you want to

Â connect them in one star, rather than in some other possible configurations.

Â Put them all together in one star.

Â So then the last thing we have to do, so

Â efficient networks are going to be either collections of stars, or empty networks.

Â Again, hereâ€™s the situation where c > delta- delta squared,

Â right, so weâ€™re in this case.

Â So either a star or the empty network, and

Â now what we can do is, check the value of a star and

Â the most, So check, putting all people together in a star, when is that valuable?

Â So if putting some people together in a star is valuable,

Â putting more people together in a star is going to be even more valuable.

Â And when is a star have a positive value, with everybody's involved?

Â Look at the value of the star when bigger than 0, and

Â that's that last expression that we saw in the medium crossed-range.

Â If c is bigger than that, then it makes sense not to connect anybody,

Â just keep everybody separate.

Â If c is smaller than that, then a star is the best thing, and

Â what we've shown is a star is the unique efficient architecture in that setting.

Â And then for cheap things, it's easy to see that we've got the complete network.

Â So this does this kind of analysis, and again,

Â to emphasize that there wasn't anything special about deltas and

Â delta squared, it just mattered that there was some value.

Â When you think of some value of a distance, one relationship,

Â that's bigger than a value of distance-2 relationship,

Â bigger than a value of a distance-3 relationship.

Â You could plug those in everywhere for delta, delta squared, and so forth,

Â and exactly the same propositions would hold.

Â You'd have something where star architectures are going to be the valuable

Â things.

Â 11:47

Now, a couple things just to say about this model, the model's obviously special.

Â And the special nature of it is, that there's not sort of diminishing returns.

Â If I add ten more people to the society and I have indirect connections to them?

Â I still get ten extra indirect connection values,

Â it's not as if they're less valuable than the first ten indirect connections.

Â So if you start putting in diminishing returns,

Â then it can be that you get other kinds of architecture.

Â So we can start enriching these models, and

Â one thing to emphasize is this is a simple one, it gives us some intuitions.

Â More generally, we can embellish these, resolve them,

Â see what works, is sort of the methodology that is being emphasized here.

Â Okay, so that's the look at the efficient networks and the connections model.

Â Next, let's see whats pairwise-stable,

Â wee if there's some differences between those two things.

Â