Problem Set 2. General Instructions

Transcription

1 CS224W: Analysis of Networks Fall 2017 Problem Set 2 General Instructions Due 11:59pm PDT October 26, 2017 These questions require thought, but do not require long answers. Please be as concise as possible. You are allowed to take a maximum of 1 late period (see the information sheet at the end of this document for the definition of a late period). Submission instructions: You should submit your answers via GradeScope and your code via the SNAP submission site. Submitting answers: Prepare answers to your homework in a single PDF file and submit it via GradeScope. Make sure that the answer to each sub-question is on a separate, single page. The number of the question should be at the top of each page. We recommend that you use the submission template latex file to prepare your submission. Fill out the information sheet located at the end of this problem set or at the end of the template file and sign it in order to acknowledge the Honor Code (if typesetting the homework, you may type your name instead of signing). This should be the last page of your submission. Failure to fill out the information sheet will result in a reduction of 2 points from your homework score. Starter Code: For problems 2 and 3, starter code is available at cs224w/homeworks/hw2/starter_code/. Note that the starter code is a nudge towards one way of solving the problem, amongst many. Feel free to cannibalize it in any way possible to construct your own solution. Submitting code: Upload your code at Put all the code for a single question into a single file and upload it. Questions 1 The Configuration Model [25 points Praty, Ziyi] A common method of analyzing the properties of real world networks is to analyze their behavior in comparison with a generated theoretical model referred to as a null model. We have previously discussed some null models in class (Lecture 3), such as the Erdős-Rényi model and the configuration model. While the Erdős-Rényi model has many nice theoretical properties, the configuration model is useful because it generates random networks with a specified degree sequence. In other words, given a real network, the configuration model allows you to sample from the space of networks that have the exact same sequence of degrees (i.e., the sampled random networks have the same degree distribution as the network you are studying). In this problem, we will analyze the configuration model in detail. For more background the on the configuration model, check out these lecture notes by Aaron Clauset: edu/~aaronc/courses/5352/fall2013/csci5352_2013_l11.pdf We will use the configuration model to examine the properties of the power grid of the western US (as of 1998). In this network, the nodes represent transformers, substations, and generators and the

2 CS224W: Analysis of Networks - Problem Set 2 2 (undirected) edges represent transmission lines. The dataset for this problem should be downloaded from txt. The first two columns of the file are the two nodes comprising an undirected edge; ignore the third column. 1.1 Simulating the configuration model through stub-matching [15 points] First, we will generate a configuration model by the stub matching method that was briefly mentioned during the random graphs lecture (Lecture 3). The intuition behind this algorithm is that we first break the network apart into a bunch of stubs, which are basically nodes with dangling edges; then we generate a random network by randomly pairing up these stubs and connecting them. A random network is generated from the stub-matching algorithm as follows: 1. First, calculate the degree sequence of the real world network by creating a vector k = {k 1, k 2,...k n } where k i is the degree of node i and n is the number of nodes in the graph. 2. Then create an array v and fill it by writing the index i exactly k i times for each element in the degree sequence. Each of these i s represents a stub of the i th node in the graph. 3. Next, randomly permute the elements of v. 4. Finally, create a random network by connecting adjacent pairs in v with an edge. In other words, connect up the nodes corresponding to the first and second elements, the third and fourth, and so on. Formally, v is of length 2m and the m undirected edges of the graph are (v 1, v 2 ), (v 3, v 4 ),..., (v 2m 1, v 2m ). If this was confusing, don t worry! Figure 1 provides a visual example of the algorithm that should clarify things. One import note is that this algorithm can sometimes create improper (i.e., nonsimple) networks with self-loops or multiple edges between two nodes; if this happens, then you must reject this sampled network and try again. In other words, you must make sure the sampled network is simple (no self-loops or multi-edges) after you construct it and draw a new sample if the constructed network is non-simple. (a) [10 points] Draw 100 random (simple) network samples using the stub-matching algorithm using the degree sequence of the power grid network. Calculate the average clustering coefficient of each of these samples and report the mean value across samples. Solution The stub-matching algorithm should generate a clustering coefficient around (b) [2 points] Consider node i with degree k i, and node j with degree k j. Under a random matching on the half-edges, what is the probability that node i and j are connected? What does the result imply? For simplicity, self-loops and multi-edges are allowed during stub-matching for this subproblem. Solution A closed form for the exact probability seems hard to derive. We instead derive an upper bound on the probability as follows. Note we use stubs and half-edges interchangeably

3 CS224W: Analysis of Networks - Problem Set 2 3 A B C D Figure 1: Example run of the stub matching algorithm. (A) A toy example network where the nodes are labeled with different colors. (B) We break the toy network into stubs, where each node of degree k has k half-edges coming out of it. (C), We label the nodes and make a vector v where each node s label is repeated in the vector k times (where k is that node s degree). (D), We take a random permutation of v and then connect adjacent pairs (dashed boxes) to create a random network. below. Nodes i and j become connected when a half-edge of i and a half-edge of j are matched in the stub-matching algorithm. For s = 1,..., k i, let A s denote the event that s-th half-edge of node i is matched to one of the half-edges of node j. Then, Pr(nodes i and j are connected) = Pr(A 1 A ki ) k i Pr(A s ) = s=1 k i s=1 k j 2m 1 = k ik j 2m 1. The first inequality is by the union bound. The second equality follows from the fact that Pr(A s ) = for all s. To see this, note that s-th half-edge is randomly matched to one of 2m 1 other k j 2m 1 half-edges and k j of those belong to node j. So, the upper bound is k ik j 2m 1. Note the exact probability will be actually less than k ik j 2m 1 because the events A s are not disjoint. (c) [3 points] We artificially reject self-loops when building the graph, but for large graphs, selfloops won t be a big trouble. Consider a graph with a very large number of edges m. What

4 CS224W: Analysis of Networks - Problem Set 2 4 is the expected number of self-loops if we build the graph by stub-matching? What does this result imply about the fraction of self-loops in the graph? (Hint: write the expected number as function of k 2 and k, where k m = 1 n i km i and n is the number of nodes) Solution Let X i be the number of self-loops at node i and X = n i=1 X i be the total number of self-loops in the graph. By Linearity of Expectation, E[X] = n i=1 E[X i]. For each node i, we compute E[X i ] as follows. Let A st be the event that s-th and t-th half-edges of node i are matched to each other and I st be the indicator that is equal to 1 if A st happens, and 0 if otherwise. Then, X i = 1 s<t k i I st and E[X i ] = E[I st ] 1 s<t k i = Pr(A st ) 1 s<t k i = 2m 1 1 s<t k i 1 = k i(k i 1) 2(2m 1). We used the fact that Pr(A st ) = 1 2m 1 (the s-th half-edge is matched to one of 2m 1 other half-edges randomly and the t-th half-edge is one of them). So, E[X] = n k i (k i 1) i=1 4m 2 and we can rewrite this as E[X] = k2 k Consequently, the fraction of self-loops is E[X] m = 1 m 2 k 2 n using the fact that k = 2m n. k 2 k. As long as k 2 is bounded in terms 2 k 2 n of m, the fraction of self-loops will be small. In particular, if k 2 is bounded by some constant, i.e., O(1), then the fraction of self-loops is O(1/m). Solution There were two major issues with Parts (b) and (c). First, our original solutions were intended for simplified variants of the given problem statements. In Part (b), for example, we asked for the exact probability but we had a proof for an upper bound expression. Second, Prof. Clauset s notes (linked from the problem set) relied on some asymptotic argument in a few places and the assumptions were not always clear. We realized that these issues were causing a lot of problems and some students were being misled greatly. Once again, we apologize for the confusion and frustration this may have caused. 1.2 Simulating the configuration model through rewiring [10 points] A second (and more popular) approach to sampling from the configuration model is to do edge rewiring. The idea with this algorithm is to start with an empirical network and then randomly rewire edges until it is essentially random. Implement edge rewiring for the US power grid graph. following process: To do this we iteratively repeat the

5 CS224W: Analysis of Networks - Problem Set 2 5 Figure 2: As the iterations of rewiring increase, the clustering coefficient decreases and begins to approximate the value computed from stub-matching. This makes sense since the limiting case of rewiring is essentially totally random, just like stub-matching. 1. Randomly select two distinct edges e 1 = (a, b) and e 2 = (c, d) from the graph. We will try to re-wire these edges. 2. Randomly select one of endpoint of edge e 1 and call it u. Let v be the other endpoint in e 1. At this point, either u = a, v = b or u = b, v = a. Do the same for edge e 2. Call the randomly selected endpoint w and the other endpoint x. 3. Perform the rewiring. In the graph, replace the undirected edges e 1 = (a, b) and e 2 = (c, d) with the undirected edges (u, w) and (v, x) as long as this results in a simple network (no self-loops or multi-edges). If the result is not a simple network, reject this rewiring and return to step 1; otherwise, keep the newly swapped edges and return to step 1. Run your edge rewiring implementation for iterations on the power grid network. Every 100 iterations, calculate the average clustering coefficient of the rewired network. Then plot the average clustering coefficient as a function of the number of iterations. Briefly comment on the curve that represents the model being rewired by both explaining its shape and how it relates to the clustering coefficient generated by the stub-matching algorithm (hint: they should be pretty similar). Solution See the figure attached. What to submit Page 1: The average clustering coefficient generated from the stub-matching algorithm (averaged over 100 samples).

6 CS224W: Analysis of Networks - Problem Set 2 6 Expression for the probability that nodes i and j with degrees k i and k j are connected. 2-3 sentences explaining the expression and what it implies. The expected number of self-loops. 2-3 sentences explaining what the results implies about the fraction of self-loops in the graph? Page 2: A plot of the average clustering coefficient as a function of the iteration number for the rewiring algorithm. A brief comment on this plot (1 2 sentences).

7 CS224W: Analysis of Networks - Problem Set Signed Triad Analysis [45 points Poorvi, Anunay] In this question you will explore the notion of balance in signed networks. The first part will explore the distribution of signed triads in an empirical network. The remaining parts will consider different generative models that could potentially produce this sort of triad distribution. 2.1 Signed Triads in Epinions [15 points] Download the following files. Epinions dataset: txt Starter code: py Epinions is a consumer review site where members can decide if they trust each other or not. This leads to a web of trust which you will analyze in this question. Since a member can either trust or distrust other members, we have a special type of network called signed network where every edge has a sign. A positive sign on an edge indicates trust between the two users whereas a negative edge indicates distrust between the two users. We will consider the graph as undirected and study the various forms of triads as shown in Figure 3. (a) [5 points] Calculate the count and fraction of triads of each type in the Epinions network. (Count each triad only once and do not count self edges) Solution Triad type Count Fraction t t t t (b) [5 points] Calculate the fraction of positive and negative edges in the graph. Let the fraction of positive edges be p. Assuming that each edge of a triad will independently be assigned a positive sign with probability p and a negative sign with probability 1 p, calculate the probability of each type of triad. Solution If counting self-loops: Number of positive edges = Number of negative edges = Fraction of positive edges = p = Fraction of negative edges = If not counting self-loops: Number of positive edges = Number of negative edges =

8 CS224W: Analysis of Networks - Problem Set 2 8 Figure 3: The four different types of signed triads (a) t (b) t (c) t (d) t 3 Fraction of positive edges = p = Fraction of negative edges = Baseline probabilities: Triad type Probability t t t t (c) [5 points] Compare the probabilities from part (b) with fractions calculated in part (a). Which type of triads do you see more in data as compared to the random baseline values? Which type of triads do you see less? Provide an explanation for this observation. Solution Triads t 0, t 1, t 3 are seen more in the graph as compared to baseline whereas triads t 2 are seen significantly less than baseline. One reason for this might be that triad t 2 is intuitively unstable since B s distrust of C conflicts with B s trust of A who trusts C. 2.2 Balance in a Random Signed Network [8 points] Now that we have seen how signed triads are distributed in a real network, we will analyze a simple generative model of signed networks to see how surprising this empirical distribution really is. In particular, we will analyze how likely it is for a balanced triad to occur in this random model. Consider the following simple model for constructing random signed networks, which we will call the G + model. Start with a complete graph on n nodes. For each edge e mark its sign as positive with probability p (and thus negative with probability 1 p). All edges are undirected. Let G B denote the event that a graph G is balanced. In this question, we ll show that P (G B ) 0 as n for graphs generated according to the G + model. Assume that p = 1/2. (a) [2 points] Let T be a maximum set of disjoint-edge triangles in G. A disjoint-edge set of triangles is one in which every edge is in exactly one triangle. Give a simple lower bound for T, the number of triangles that don t share any edges in G (the bound should be an increasing function of n).

9 CS224W: Analysis of Networks - Problem Set 2 9 Solution A simple bound that is an increasing function of n is T n 3. This is true because you can separate the n nodes into n 3 groups of 3 (the floor is just in case n isn t divisible by 3, and there s one or two nodes left over). There is a triangle for each group of 3, and this set is disjoint-edge since none of the triangles share an edge (in fact, they don t even share any nodes, so they certainly don t share any edges). Another simple bound is T n 1 2. Fix a node v and then pair up the remaining nodes into n 1 2 pairs. A triangle is formed by node v and the nodes of each pair, and the triangles formed in this way are disjoint in edges. We can use this bound to do Parts (b) - (d). (b) [2 points] For any triangle in G, what is the probability that it is balanced? Solution The probability that a triangle is balanced is equal to the probability that it has exactly 0 or 2 negative signs. Let I be an indicator random variable that counts the number of negative signs our triangle gets. Then P ( is balanced) = P (I = 0) + P (I = 2) = 1 ( ) = 1 2 The second equality follows because there is a ( 1 2 )3 = 1 8 chance that all 3 edges are positive, and there are ( 3 2) = 3 ways that the triangle can have exactly 2 negative signs, each of which happens with probability 1 8. (c) [2 points] Using the simple lower bound from part (i), give an upper bound on the probability that all of the triangles in T are balanced. Show that this probability approaches 0 as n. Solution If i and j (i j) are two different triangles in T, then P ( i is balanced) is independent of P ( j is balanced). Thus P (all triangles in T are balanced) = which clearly approaches 0 as n. ( ) 1 T 2 ( ) 1 n 3, 2 (d) [2 points] Explain why the last part implies that P (G B ) 0 as n. Solution This implies the desired result because all triangles in T being balanced is a necessary but not sufficient condition for the entire graph G being balanced. Thus P (G B ) P (all triangles in T are balanced), and since the right hand side approaches 0 as n, so does P (G B ).

10 CS224W: Analysis of Networks - Problem Set A Dynamic Process Model of Balance [5 points] If balanced signed networks do not show up by chance, as we showed in the previous part of the question, how do they arise? One class of mechanisms that researchers have proposed and studied are dynamic processes, in which signed networks can evolve over time to become more balanced. The following describes a very simple example of such a dynamic process: (I) Pick a triad at random. (II) If it s balanced, do nothing. (III) Otherwise, choose one of the edges uniformly at random and flip its sign (so that the triad becomes balanced). Consider the following claim: in this process, the number of balanced triads can never decrease. Is this true? If so, give a proof, otherwise give a counterexample. Solution The claim is false (i.e. the number of balanced triads can decrease). Consider an unbalanced triangle ABC on nodes A, B, and C such that sign(a, B) = +, sign(b, C) = +, sign(a, C) =. Let X 1,..., X k be k other nodes such that A and B are connected to them with positive edges: sign(a, X i ) = + and sign(b, X i ) = + for all i 1,..., k. Then if our process randomly (and unluckily) chooses the triad ABC and chooses to flip the sign of the edge (A, B) from positive to negative, it balances the triad ABC but unbalances all k of the triads ABXi. Thus the total number of balanced triads decreases by k Simulation [8 points] Now let us run simulations of the dynamic process on small networks. (I) Create a complete network on 10 nodes. (II) For each edge, choose a sign (+, ) at random (p = 1/2). (III) Run the dynamic process described in the previous part for 1,000,000 iterations. Repeat this process 100 times (so that you run the dynamic process described in the previous part for 100 different graphs generated from the G + model). What fraction of the networks end up balanced? [Hint: To check whether a network is balanced or not, remember that a graph is balanced if and only if it s possible to separate the nodes into two factions such that every pair of nodes in the same faction is linked by a positive edge and every pair of nodes in different factions is linked by a negative edge. To speed things up, you can stop the process once the network is balanced, because once the network becomes balanced it will never change.] Solution You should find that 100% of the networks end up balanced. This is because these dynamics always converge to a balanced state in finite time (on finite graphs). (You could get incredibly unlucky on a single run and run out of iterations before convergence, but this is very rare for such small networks.)

11 CS224W: Analysis of Networks - Problem Set The Role of New Nodes I [4 points] Another way signed networks can evolve over time is if new nodes join the network and create new signed edges to nodes already in the network. Consider the network shown in Figure 4. Is it possible to add a node D such that it forms signed edges with all existing nodes (A, B, and C), but isn t itself part of any unbalanced triangles? Justify your answer. A + + B - Figure 4: An unbalanced network with one triangle. C Solution No. A simple argument by cases does the trick: the edge (A, D) must be either positive or negative. If it s positive, then for ACD to be balanced (C, D) must be negative. But then if (B, D) is positive, then BCD is unbalanced; and if (B, D) is negative, then ABD is unbalanced. The argument for the other case (if we set (A, D) to be negative) is similar. 2.6 The Role of New Nodes II [5 points] Using your answer to the previous sub-problem, consider the following question. Take any complete signed network, on any number of nodes, that is unbalanced. When (if ever) is it possible for a new node X able to join the network and form edges to all existing nodes in such a way that it does not become involved in any unbalanced triangles? If it is possible, give an example. If it is not, give an argument explaining why. Solution No, it s not possible for a new node X to join the network and form edges to all existing nodes in such a way that it does not become involved in any unbalanced triangles. This is because any unbalanced complete signed network must have an unbalanced triangle on three nodes call them A, B, and C. ABC must have exactly 1 or 3 negative signs. If it has 1 negative sign, then by the reasoning in the previous question we know that X cannot create edges to A, B, and C without being involved in an unbalanced triangle. If ABC has 3 negative signs, then an argument similar to the one given in the previous question holds: if (A, D) is positive, then (C, X) must be negative to balance ACX, and then setting (B, X) to be positive unbalances ABX, while setting (B, X) to be negative unbalances BCX. Similar reasoning applies to the case where we set (A, X) to be negative. Thus there s no way for X to create links with all existing nodes such that it isn t involved with any unbalanced triangles, because it can t even create links with the nodes A, B, and C without becoming involved in at least one unbalanced triangle. What to submit Page 3: (part a) the count and fraction of triads in each type of opinion network;

12 CS224W: Analysis of Networks - Problem Set 2 12 (part b) the fraction of positive and negative edges in the network and the probability of each triad (part c) which triads appear more/less than the baseline values and a brief (2 3 sentence) explanation. Page 4: (part a) a simple lower bound for T. (part b) the probability that any triangle in G is balanced. (part c) an upper bound on the probability that all triangles in T are balanced, and show that the probability approaches 0 as n. (part d) 1 2 sentences explaining why P (G B ) 0 as n. Page 5: Page 6: Page 7: Page 8: Whether the number of balanced triads never decreases. A proof (if true) or a counterexample (if false). The fraction of networks that end up balanced. Whether it is possible to add a node D, and a justification. Whether it is possible for a new node X to join. An example (if possible) or argument (if impossible).

13 CS224W: Analysis of Networks - Problem Set Decision-based Cascades: A Local Election [30 points Anthony, Yokila] It s election season and two candidates, Candidate A and Candidate B, are in a hotly contested city council race in sunny New Suburb Town. You are a strategic advisor for Candidate A in charge of election forecasting and voter acquisition tactics. Based on careful modeling, you ve created two possible versions of the social graph of voters. Each graph has 10,000 nodes, where nodes are denoted by an integer ID between 0 and The edge lists of the graphs may be downloaded from: Graph 1: txt Graph 2: txt Both graphs are undirected. Given the hyper-partisan political climate of New Suburb Town, most voters have already made up their minds: 40% know they will vote for A, 40% know they will vote for B, and the remaining 20% are undecided. Each voter s support is determined by the last digit of their node id. If the last digit is 0 3, the node supports A. If the last digit is 4 7, the node supports B. And if the last digit is 8 or 9, the node is undecided. The undecided voters will go through a 10-day decision period where they choose a candidate based on the majority of their friends. The decision period works as follows: 1. The graphs are initialized with every voter s initial state (A, B, or undecided). 2. In each iteration, every undecided voter is decides on a candidate. Voters are processed in increasing order of node ID. For every undecided voter, if the majority of their friends support A, they now support A. If the majority of their friends support B, they now support B. Majority for A means that strictly more of their friends support A than the number of their friends supporting B, and vice versa for B (ignoring undecided friends). 3. If a voter has an equal number of friends supporting A and B, we assign support for A or B in alternating fashion, starting with A. In other words, as the voters are being processed in increasing order of node ID, the first tie leads to support for A, the second tie leads to support for B, the third for A, the fourth for B, and so on. This alternating assignment happens at a global level for the whole network, across all rounds. (Keep a single global variable that keeps track of whether the current alternating vote is A or B, and initialize it to A in the first round. Then as you iterate over nodes in order of increasing ID, whenever you assign a vote using this alternating variable, change its value afterwards.) 4. When processing the updates, use the values from the current iteration. For example, when updating the votes for node 10, you should use the updated votes for nodes 0 9 from the current iteration, and nodes 11 and onwards from the previous iteration. 5. There are 10 iterations of the process described above. 6. On the 11th day, it s election day, and the votes are counted.

14 CS224W: Analysis of Networks - Problem Set 2 14 Note that only the undecided voters go through the decision process. The decision process does not change the loyalties of those voters who have already made up their minds. 3.1 Basic setup and forecasting [8 points] Read in the two graphs and assign the initial vote configurations to the network. Then, perform the 10 iterations of the voting process. Which candidate wins in Graph 1, and by how many votes? Which candidate wins in Graph 2, and by how many votes? Solution In graph 1, candidate B wins by 96 votes. In graph 2, candidate B wins by 256 votes. 3.2 TV Advertising [8 points] You have amassed a substantial war chest of $9000, and you have decided to spend this money by showing ads on the local news. Unfortunately, only 100 New Suburb Townians watch the local news those with ids However, your ads are extremely persuasive, so anyone who sees the ad is immediately swayed to vote for candidate A regardless of his/her previous decision. You may spend $1000 at a time on ads. The first $1,000 reaches voters , the second $1000 reaches voters , and so on. In other words, the total of $k in advertising would reach voters with ids from 3000 to k This advertising happens before the decision period. After voters are persuaded by your ads, they never change their minds again. Simulate the effect of advertising spending on the two possible social graphs. First, read in the two graphs again and assign the initial configurations as before. Now, before the decision process, you purchase $k of ads and go through the decision process of counting votes. For each of the two social graphs, plot $k (the amount you spend) on the x-axis (for values k = 1000, 2000,..., 9000) and the number of votes you win by on the y-axis (that is, the number of votes for A less the number of votes for B). Put these on the same plot. What s the minimum amount you can spend to win the election in each of the two social graphs? Note that the TV advertising affects all of the voters who see the ads and not just those who are undecided. Solution On graph 1, the minimum amount to spend is $5,000 dollars. On graph 2, it is $7,000. See Figure Wining and Dining the High Rollers [8 points] TV advertising is only one way to spend your campaign war chest. You have another idea to have a very classy $1000 per plate event for the high rollers of New Suburb Town (the people with the highest degree in the social graph). You invite high rollers in order of how many people they know, and everyone that comes to your dinner is instantly persuaded to vote for candidate A regardless of his/her previous decision. This event will happen before the decision period. When there are ties between voters with the same degree, the high roller with lowest node ID get chosen first. Simulate the effect of the high roller dinner on the two graphs. First, read in the graphs and assign the initial configuration as before. Now, before the decision process, you spend $k on the fancy

15 CS224W: Analysis of Networks - Problem Set 2 15 #votes for A - #votes for B TV Advertising Graph 1 Graph Amount spent ($) Figure 5: Election outcomes with TV advertising dinner and then go through the decision process of counting votes. For each of the two social graphs, plot $k (the amount you spend) on the x-axis (for values k = 1000, 2000,..., 9000) and the number of votes you win by on the y-axis (that is, the number of votes for A less the number of votes for B). What s the minimum amount you can spend to win the election in each of the two social graphs? Note that wining and dining sways all the voters to vote for A and not just those who are undecided. Solution For graph 1, no amount up to $9,000 is sufficient to win the election. For Graph 2, the minimum amount is 6,000 dollars. See Figure 6. #votes for A - #votes for B Wining and Dining Graph 1 Graph Amount spent ($) Figure 6: Election outcomes with wining and dining. 3.4 Analysis [6 points] Plot the degree distributions on a log-log scale of the two graphs on the same plot (as in Question 1.1 on Problem Set 1). Although both graphs have roughly the same number of edges, one was

16 CS224W: Analysis of Networks - Problem Set 2 16 generated from an Erdős-Rényi Random graph model and the other was generated from a preferential attachment model. Which is which? In 1 2 sentences, briefly summarize how this information explains the results of Question 3.3. Solution The degree distributions are in Figure 7. Graph 1 was generated from the Erdős-Rényi Random (ER) graph model and Graph 2 was generated from a preferential attachment (PA) model. The graphs have roughly the same number of edges, but with the PA model, there are nodes with much higher degrees than in the ER model and that s why in PA model Wining and Dining is more effective Graph 1 Graph 2 p k k Figure 7: Degree distributions of the social graphs What to submit Page 9: Page 10: Page 11: Page 12: Which candidate wins and by how many votes in each graph. Plot of winning margin in each graph as a function of $k (on the same plot) The minimum amount you can spend to win the election in each graph Plot of winning margin in each graph as a function of $k (on the same plot) The minimum amount you can spend to win the election in each graph Log-log plot of the degree distributions (on the same plot), which graph corresponds to which random graph model 1 2 sentences on why the random graph models explain the results of Question 3.3.