# Homework 3 - Solutions

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2 Problem 2: Suppose a directed network takes the form of a tree with all edges pointing inward towards a central vertex as shown in Figure 1(left). Assume the random walk based PageRank definition, namely that with a probability α the node follows any of the outgoing links of the current node and with probability 1 α it jumps to any node with equal probability. Question 1: If the tree has nodes and height h, show that the PageRank of its root r is given by the formula: ( ) 1 α P (r) = Σ h k=0(α k n k ), where n k is the number of nodes at level k (i.e. at distance k hops from the root) Figure 1: Question 2: Assume that we move some leaf nodes around as shown in Figure 1(right). How does the PageRank of the root change? Solution: Question 1: Below we use b k : b a to indicate that node b at level k has a link to a. Recall that edges go upwards from level k+1 to k and every node has out-degree 1; also, 0 has only the root r. Assume the first level of the tree 1 = {a 1,..., a n1 }. Then, ( ) P (r) = 1 α +α 1 α + α P (b) b 2 :b a α + α P (b) b 2 :b a n1 (1) The index sets {b 2 : b a i }, for i = 1,..., n 1 are pairwise disjoint; therefore, 2

3 [ (1 P (r) = 1 α ) α +α n 1 + α ] ( ) 1 α P (b) = [1+αn 1 ]+α 2 P (b). b 2 b 2 (2) Repeating the above manipulations on levels 2, 3, and up to level h 1, we get ( ) 1 α P (r) = Σ h 1 k=0 (αk n k ) + α h P (b). b h At the last level h all nodes are leaves, which have no incoming edges, hence the PageRank of any b h is (1 α). Then α ( ) 1 α h P (b) = α h n h, b h and the result follows. Question 2: Since the PageRank of the root only depends on the number of nodes at each level, it makes no difference if we re-arrange any nodes without changing the cardinality of each set i. Problem 3: Exercise 14.3 from Kleinberg s book (etworks, Crowds, and Markets) Solution: (a) We apply the rule, as described in Kleinberg s book. We set authorities and hub values to 1. Then update first the authorities as a (1) = M T h (0). Then h (1) = Ma (1), where M is the adjacency matrix. Running this gives us the following hub and authority scores: A = 11, B = 7 and C = 11, D = 11, E = 18, F = 7. A = 11 18, B = 7 18 and C = 11 47, D = 11 47, E = 18 47, F = 7 47 (normalized). (b) (Option 1) In this option authority and hub scores of X and Y are not changing over different rounds, as they are just connected to each other. Hence, the authority (and hub) score of X (and Y) after round 2 is simply 1 in the unnormalized case, and the normalized authority scores are: A = 11 19, B = 7 19, X = It is clear that we have not managed to make our page X look that important. (Option 2) Repeating the basic steps twice, we get the following normalized authorities. A = 23 49, B = 18 49, X = So, while in Option 1 the new page had 7 times less authority than B and 11 times less than A, now it has almost half the importance of B and 1/3 of 3

4 A. We have therefore improved the relative authority of X by 3 4 with this simple trick! The reason is that, by having Y point to other pages (and thus the important ones as well) we make it seem like an important hub pointing to our page X. Instead, before, Y was not only the only hub pointing to X, but also an uninteresting one. It is easy to check that the best policy given the introduction of one page and one hub is indeed to have Y point to all other pages, if we do not know (cannot guess) the authority scores of these other pages. (c) We will make both Y and Z hubs. ote that we cannot add links to sites we do not own here!! From the previous example, we could see that by having a hub point to other pages (besides X) it has the positive effect of raising its hub score, which it then transfer to X. But it also has the negative effect of raising the score of the other pages it points to, which doesn t help X. To improve this, we adopt the following strategy: hub the hubs Y and Z point to the more popular page among A and B, that is A, and X, as shown in Figure 2. This increases hub scores of Y and Z, but the hub score only goes to A (which is anyway the top page) and X, not B. This allows X to climb to a higher authority score than B. A = 31 54, B = 9 54, X = Figure 2: Problem 4: Consider two networks with 10 nodes, networks A and B. Question 1: Design networks A and B such that A is (around) 2 times more resilient to random vertex removal. I.e. you will need (on average) to remove 4

5 twice as many nodes in network A than network B, in order to disconnect the network (i.e.the remaining nodes are not all anymore in the same connected component; or equivalently, there is at least one pair of remaining nodes without a connected path between them). You are allowed to have a different number of links in A and B. Question 2: Can you achieve the same if you are also forced to have the same number of total links in both A and B (you can choose how many)? Question 3: Assume that you perform a random walk on each of A and B. Design A and B such that the time to converge to the stationary distribution in A is around 2 times higher than in B. ote: You can try different draft designs based on your intuition about resilience and convergence speed of different topologies, and then feel free to use Matlab to evaluate the actual values and to optimize your designs. Solution: Questions 1 and 2: The point of this problem is to understand that resilience is about having disjoint paths (paths that don t share the same nodes) between most node pairs in the network. If fewer pairs have such disjoint paths, then it means that some nodes are critical in the various paths, and if removed, would disconnect the network. Since we remove nodes randomly, we just need to identify the (number of) critical nodes (i.e. nodes whose removal would kill the network) and to find the expected number of steps until enough of them are removed. Figures 3 and 4 show two examples I came up with (there are others). In both cases it would take (approximately) two times more removals to disconnect the network in A than in B. You can find the exact values using a recursion and/or Matlab. Figure 3: Two example networks for Question 1 A has more links than B the dark nodes are the critical ones: the removal of any of them disconnects the network 5

6 Figure 4: Two example networks for Question 2 - A has the same number of links as B Question 3: We have learnt in class that the convergence speed of a random walk on a graph is related λ 2, the second largest eigenvalue of the probability transition matrix for a random walk on this graph. We are asked to find graphs A and B such that the convergence speed on A is twice as fast as in B. Converge speed is the time until the error, i.e. how far we are from the stationary distribution of the walk is small enough (e.g. equal to a small value ϵ). As we saw in the notes and class, the error at step n (assuming n is large) is proportional to the second eigenvalue raised to n. Denoting the current probability distribution of the random walk at step n as p (n) and the stationary distribution as π, then error = p (n) π λ n 2 We require that the time (number of steps) to reach a given error in graph A is double that for graph B. Using the above equation for the error, we thus need that: (λ A 2 ) n = (λ B 2 ) 2n λ A 2 = (λ B 2 ) 2 So, we now know how the eigenvalues of the two graphs should relate, in order for one to have twice as fast convergence. But how can we design a graph for which we only know one eigenvalue??? The trick is to remember that the 2nd eigenvalue of a graph is related to the bottleneck of that graph. Consider for example the graph of Fig 5. The red link is clearly a bottleneck. We could hope then that by doubling the bandwidth of this bottleneck, e.g. adding one more link between the two well connected clusters, will almost double the convergence speed. Indeed, calculating the eigenvalues for the two graphs, gives us that: (λ B 2 ) = (λ A 2 ) = (λ B 2 ) 2 = This is extremely useful!! It means that we can compare convergence speeds or search speeds on different graphs, by only looking at their high level structure! 6

7 Figure 5: Two example networks for Question 3 - The bottleneck in B is twice the size as that in A Problem 5: Consider a 2D finite lattice graph consisting of ( 1) total nodes ( on each side), as shown in Figure 6 (without the red links). As we have discussed in class, this graph is not small-world. We will try to make the graph small-world, by overlaying a star topology over the lattice (the red links). Specifically, from each node we create an extra link with probability p (0 < p < 1) connecting it to the center node, as shown in Figure 6. Figure 6: Question 1: What should the value of p be in order to ensure that a short path of no more than 4log() hops exists between most nodes (i.e. between at least 95% of node pairs)? Question 2: What is the percentage of extra links we end up adding as the size of the lattice () goes to infinity? (i.e. what is the ratio new links?) 7

8 Question 3: Assume that instead of adding a link from each node to the center node (with probability p), the link is to a random node (i.e. to any of the other nodes with equal probability). The probability of this shortcut link for each node is still the same, and equal to the p value found in Question 1. Do you expect path lengths in this graph to be smaller or larger than in the previous graph? Question 4 (*Extra Credit*): Can you think of any other way to make the graph small world with fewer links than the star topology? Solution: Question 1: We have nodes on each side of the grid, adding up to nodes in total. We divide this grid into K larger squares (i.e. K < ), as shown in Figure 6. log() log() Figure 7: Let s assume that the size of each such square (i.e. K ) is log(). This means that all nodes within such a square can reach each other in at most 2log() hops. Furthermore, if there exists (at least) one shortcut link inside this square, every node in the square can also reach the central node in at most 2log() hops. Similarly, if there is a shortcut from the center to the square where the destination lies, then this destination is also not more than 2log() from the center. 8

9 Putting these together, if there is (at least) one shortcut from every log(n) log() square to the center, then there exists a path between every two nodes in the network which is no longer than 2log() + 2log() = 4log() hops. The last step of our (constructive) proof is to find out the value of p that will guarantee the existence of at least one shortcut in a square of log() log() nodes with high probability (e.g. 95%). There are a total of log() log() = (log()) 2 nodes in each such square. Each on of them might have a link to the center, independently, with probability p. This defines a random variable X that has a binomial distribution with parameter (log()) 2 (the number of bernoulli trials) and probability p of success for each trial, i.e. X binomial(log 2 (), p). We are thus looking for p such that P (X 1) 95%. If is large enough, we can take the Poisson approximation of the binomial, for convenience. That is, we can assume that X, the number of shortcuts in a square, is Poisson distributed with parameter λ = log 2 ()p. Then, P (X 1) = 1 P (X = 0) = 1 e log2 ()p 1 P (X 1) e log2 ()p log 2 ()p ln(0.05) p 3 (log()) 2. Thus, for large enough networks, this value of p will guarantee that at least 95% of nodes will have a short cut within 2log() steps to the center node. Technically, since we are interested in node pairs, we would like (P (X 1)) , and hence P (X 1) We can simply plug this value above and get a new estimate for p. In any case, the important trick in this exercise is to split the grid into log() log() squares. Question 2: Since we have to pick p 3 (log()) 2, the expected number of new links will be new links = p = 3 (log()) 2. Hence, the ratio of new-links to existing links, is (there are a total of 2 links on the grid) new links 2 = 3 2(log()) 2 = 3 2(log()) 2, which means that, as goes to infinity, the percentage of additional links to make the graph small-world is negligible, going to 0. 9

10 Question 3: To answer this question, we can consider our network to be just the large squares (i.e. K total nodes). A shortcut essentially connects two blue squares (with high probability the shortcut doesn t fall in the same square). Also, on average, there are 3 shortcuts for each square, i.e. the degree is k = 3. This means that (i) the graph is connected with high probability, and (ii) paths of length log(k) = log( log 2 ()) exist between any two blue squares. This means that, starting from a square, one can reach a shortcut link with high probability after no more than 2log() steps, then traverse the higher hierarchy (the random poisson graph) in log(k) steps to reach the destination s blue square, and another 2log() steps in the last blue square. Concluding, there exist small-world paths in this graph as well, but more hops need to be traversed to connect the destination and the source blue square, while in the star topology it only takes 2 steps. Extra Credit: You could just force exactly one link to the center node for each blue square. This would be the same graph and path length, except that the extra links are now only 1 3 of the original ones. But this is a bit like cheating. If we still assume some randomness on the choice of which nodes have a shortcut, then I don t think that this graph could be made small world with fewer links than this start topology. But the jury is out! You could prove me wrong! 10

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