So, now we've illustrated a page rank calculation. But here's the next question. Does the method we just presented always lead to a unique solution? And the answer is, not quite yet. There are some modifications that we have to make in order to be guaranteed that we don't either have no solutions or multiple solutions. Each of which case isn't good because then there's no single consensus that we can come up with. So, let's consider again, this our example over here and we've augmented it now with one link point to this node V over here. Okay. So now, let's look at V for a second. V isn't pointing to anyone. It has no outgoing links. If you notice. So, what does that imply? Well, remember as we said that a node has to spread its important score to other nodes, right? So, if there was one link over here then this would be V. if there were two, then this would be V over two, and so on. Right. But that's not the case here. There's no spreading of any important score at all. So, the only logical conclusion then is that v has an importance score of 0, because it's not spreading to anyone. So, he has no importance score at all. Because that's the only way it'll stay consistent with what we just did. And now, V importance score of 0, we say that V was aggregating its importance score from Z. So, this is Z over 4 because there's four links 1, 2, 3, 4. So now, remember V has to be equal to Z over 4 by our equations because V is aggregating all of this importance from this link. So, if V is 0, then that means that Z has to be 0, so then now Z is 0. Okay. Well, if Z is 0, then that means that all these outgoing links have to be 0 because he is spreading all of his importance equally to everyone. And so, 0 divided by 4 is still 0. And well, then what's going to happen? Look at Y. Y is spreading all of its important score to Z, right? So, since Z is 0, Z has to be equal to Y plus X over 2, right? And we can't have any of them be less than 0. So, since Z is equal to 0, then the only conclusion is that both of these have to be equal to 0. So, now it's saying that Y has to be equal to 0 and X has to be equal to 0, which means that this length is 0, this length is 0. As you can see, it's going to cascade through and everyone is going to end up having an important score of 0. So, we call, the, the problem all started with this node V over here. And the problem was that V didn't spread it's important score to anyone. And we call V a dangling node because it's, it's just a node that doesn't point to any others. And once you get to that node you're kind of sunk cause you can't go anywhere else. Right? If you're following the web graph, you can't possibly get out. So, the only thing that happens is that it leads to no solution. So, here we have basically the only solution working is that all the important scores are 0 which is the trivial solution. And so, we have no, and we should say no meaningful. Because technically, 0 is a solution. Having them always 0, but there's no meaningful solution here. There's no way to rank from our solution that we've gotten. So, the way that we solve the problem of dangling nodes, is that we assume that the dangling node points to every other node in the graph. So, basically what we say is that we assume V has a link to W. We assume V has a link to Z. We assume V has a link to X. We assume V has a link to Y. And we also assume that V has a link to itself. So, it has a hyperlink, and that points back to it's own page. And by doing that, then we don't have this problem anymore because as I ran the surfer idea, once we get to V, then we can actually get out of V again. And continue with that random surfing process. So now, we fixed the dangling node problem, but there's a, there's a second issue, in the idea of a disconnected graph. So, we call a connected component a group of nodes which can reach each other, but can't reach anyone outside of their group. And, so basically in this graph right over here, we have two different connected components, right. So, if our web graph looked like this and obviously a lot of the internet might be something like this because the internet is very sparse, as we said. So, there's going to be groups of nodes that aren't connected so each other. So that we have A, B and C which are all connected on one hand and we have D and E who are connected on this side over here. So, the problem is that if we start in A go to B and then go to C, and then go to B. And then, we go to C B, C B and so on. Even if we're choosing within this group over here, we can't ever leave this sub graph. because there's, there's no, way to get out of this sub graph over here. And the same thing in here, is if we're D and E, there's no way that we can get out of this. We can't get over to this side because there's no link going over there and there's no link going this way either. This is going to lead us to actually infinitely many solution. So, there's an infinite number of values of important scores that could satisfy this because we don't have any relationship between these two over here. So, we, we have no relationship, then basically we can establish any relationship we want and it will still work. And this is going to be a problem whenever we have more than one connected component because the entire graph isn't one connected component. We have two or three sub graphs. And so, we're stuck in one of these sub graphs forever, as we said, by a random surfing philosophy. The solution to this is we add that thing that we neglected initially where we said maybe they random surfer will eventually get tired, right? So, if he's, if he's starting at A, and he comes over to B, and then he goes to C, then back to B again. Then maybe eventually he'll get bored and decide to just enter a random URL and then he'll end up on D. And then, he can go to D, to E. And he'll be stuck in that sub graph until he enters a random URL again and comes back to C. So, we're basically establishing an artificial link between each node. So, we're saying in a sense, each node is connected to every other node because there's a way to reach them just through our web browser rather than through the actual hyperlink connectivity on the pages. So we add that purely random surfing that we neglected initially into the mix.
