Now, let's illustrate the closeness centrality computation for Cara. First, we need to find the shortest path between her and all of the other nodes in the figure. So we'll start with Cara to Ben, C to B. So from Cara to Ben, we first have link C comma A, and then we have link A comma B for this entire path, C, A, B so this path is C, A, B and the length of that path is two because there's two lengths. Second we have from, Cara to Ana, C to A and that's easy because that's just one of the links here. C comma A and that has length one. Then we have from Cara to [SOUND] Dana. Obviously, we don't have to do Cara to Cara. Cara to herself wouldn't account for anything. So from Cara to Dana we just have link, C comma D. And so link C comma D, and that has a length of one. And from Cara to Evan we have, C, comma E. Or from C to E we have, C comma E which has a length of one. So we can write C comma E here, and then from Cara to Frank, we have one of two paths, we could either go Cara to Dana to Frank, or we could go from Cara to Evan to Frank. We wouldn't want to go from Cara to Evan to Dana to Frank because that will have a length three and that is not shortest paths, we're only considering shortest paths here. So we have G comma F, or E comma F here. So, from C to F, we can say either C, D, F, or C, E, F, both of which have length two. Now we find the average of these five shortest distances for Cara. So we have from all of this, we take, we find average C, which means average for Cara. It's the average shortest path lengths to all the other nodes. So we add 2 plus 1, plus 1, plus 1, plus 2, and we divide by the total which is 5, just to find the average. This is her average shortest path length. That's 7 5ths, and we're just about there,but again as we said, just one minor detail. We want a smaller average to give higher centrality since that indicates a person's closer to the rest of the nodes, and to account for this, we simply take the reciprocal so that a higher average gives a smaller value. So, the closeness centrality for Cara [NOISE] closeness for Cara is going to be 5 over 7, which is equal to 0.714. [BLANK_AUDIO] Now let's repeat the same thing for Dana. So we'll draw a line here so we don't get confused. Now for Dana we have to repeat the same process and, start from, we can start from Dana. To Ana, the shortest path is, D, C, A so we go from Dana to Cara to Ana, D, C, A, and that has a length of two from Dana to Ben. We just add one more link for this case, it, instead of D, C, A it's going to become, D, C, A, B. So this will be D, C, A, B. And notice that's the only way to get there in the shortest path. So the shortest path here is unique. And, for, for this, that has a length of three. And for Dana to Cara, that's simple. It's just, D comma C, or C comma D. Either way, It's just, we usually start with the first one on the left and then put the other one on the right. And this has a length of one. From Dana to Evan,that's just simply D comma E, which we don't have labeled here, we can write D comma E. And D comma E just has a length of one again. And finally, for Dana to Frank, D to F, we just have D comma F. And that has length of one again. So then we just have to take the average of these, and averaging them, we get the average for Dana,. [NOISE] It's going to be equal to 2 plus 3 plus 1 plus 1 plus 1 over 5 which is 8 over 5. Then, we take the reciprocal of that to get the closeness of Dana, which is going to be 5 over 8, which is 0.625. So, little less close to centrality than Cara. And remember before we said we had Cara and Dana having the same centrality measure. And, we said that may not be the case because of how important Cara is to the connectivity. So therefore now we've separated Cara from Dana, which is a good thing. And we can repeat this computation, finding the average of the paths and then the centrality measures for your other nodes. So note that we can simply flip the denominator and the numerator in the average equation just to get the centrality, because that's all we're really doing in the end anyway, so we're just going to write it that way here. So when we have the closeness of Anna, closeness for Anna. That's just going to be five, because the denominator's always going to be five because there's, five cases over 1 plus 1, plus 2, plus 2, plus 3. So we have 1 plus 1, plus, 2 plus, 2 plus, 3, and you can figure out, which cases each of those are. Ben is one, Cara's one, each of these are 2, and then Frank is three. And that's going to be 5 over 9, which is 0.556. The closeness for Ben is going to be 5 over 1, plus 2, plus 3, plus 3, plus 4. 1 plus 2, plus 3, plus 3, plus 4, which is 5 over 13 which is 0.385 and let's draw a line here. The closeness for Evan is going to be 5 over 2, plus 3, plus 1, plus 1, plus 1, plus 1, which is 5 over 8, which is equal to 0.625. And finally the closeness for Frank, is going to be equal to 5 over 3 plus 4, plus 2, plus 1, plus 1. And where did those come from? We have, three coming from Anna, four coming from Ben, and then two coming from Cara, and then one from each Dana and Evan. Which is going to be 5 over 11, and 5 over 11 is equal to 0.455 so these are our values. We have for Cara, 0.714, for Dana we have 0.625, for Anna we have 0.556, for Ben, 0.385, for Evan 0.625, and for Frank we have 0.455. So on the graph right now we've summarized the closeness centralities we've just computed. So from this, the new ranking according to closeness centrality. Cara comes first with 0.714. It's followed by a tie between Dana and Evan together, in second place. It's followed by Ana, in third place, then Frank, and finally, Ben. [BLANK_AUDIO] So clearly this is more reasonable than degree centrality, because a lot of the ties have been broken. Cara and Anna have both been promoted, which we said was important, because they both hold the graph together. Without them, there would be partitions in the graph. But there's still some issues to this. First, if you look, Dana and Evan really don't hold the graph together. So if we remove Dana, for instance, we're going to be left with, Ben to Anna to Cara to Evan to Frank, which is one way. And if we remove, on the other hand, if we remove Evan from the graph and we keep Dana in, we're going to be left with a graph that, looks like this. So all the remainder of the nodes are still going to be connected. So why should Anna be less important than Dana and Evan? Because Anna, as we said, if we remove Anna, we're going to partition Ben away from the rest of them, but Dana and Evan aren't really doing the same thing. We're going to move them, either one of them could be removed, and the graph would still be connected. And second is that Cara should be even more central than Dana and Evan because as we said, Cara is intuitively the most important person in the graph, and 0.714 is only about 13% higher than 0.625, so really Cara should, we should try to increase the amount of importance that Cara has over Dana and Evan. And so there's another notion of centrality that's just as useful, and perhaps can fit our intuition more correctly, and we will look at that next.