CS246: Mining Massive Datasets Jure Leskovec, Stanford University.

Transcription

1 CS246: Mining Massive Datasets Jure Leskovec, Stanford University

2 What is the structure of the Web? How is it organized? 2/7/2011 Jure Leskovec, Stanford C246: Mining Massive Datasets 2

3 3 What is the structure of the Web? How is it organized? Web as a directed graph

4 4 Two types of directed graphs: DAG Directed Acyclic Graph: Has no cycles: if u can reach v, then v can not reach u Strongly connected: Any node can reach any node via a directed path Any directed graph can be expressed in terms of these two types of graphs

5 Strongly connected component (SCC) is a set of nodes S: Every pair of nodes in S can reach each other There is no larger set containing S with this property Any directed graph is a DAG on its SCCs: Each SCC is a super-node Super-node A links to super-node B if a node in A links to node in B 2/7/2011 Jure Leskovec, Stanford C246: Mining Massive Datasets 5

6 6 Take a large snapshot of the Web and try to understand how it s SCCs fit as a DAG Computational issues: Say want to find SCC containing specific node v? Observation: Out(v) nodes reachable from v (via out-edges) In(v) nodes reachable from v (via in-edges) SCC containing v: = Out(v, G) In(v, G) = Out(v, G) Out(v, G) where G is G with directions of edges flipped v

7 [Broder et al., 00] 250 million webpages, 1.5 billion links [Altavista] 2/7/2011 Jure Leskovec, Stanford C246: Mining Massive Datasets 7

8 8 Out-/In- Degree Distribution: p k : fraction of nodes with k out-/in-links Histogram of p k vs. k Normalized count, p k

9 9 Plot the same data on log-log axes: Normalized count, p k log p k = βk α p k = log β α log k

10 [Broder et al., 00] 2/7/2011 Jure Leskovec, Stanford C246: Mining Massive Datasets 10

11 Random network Power-law network Degree distribution is Binomial, i.e., all nodes have similar degree 2/7/2011 Degrees are Power-law, i.e., heavily skewed Jure Leskovec, Stanford C246: Mining Massive Datasets 11

12 12 Web pages are not equally important vs. Since there is large diversity in the connectivity of the webgraph we can rank the pages by the link structure

13 13 We will cover the following Link Analysis approaches to computing importances of nodes in a graph: Page Rank Hubs and Authorities (HITS) Topic-Specific (Personalized) Page Rank Spam Detection Algorithms

14 14 First try: Page is more important if it has more links In-coming links? Out-going links? Think of in-links as votes: has 23,400 inlinks has 1 inlink Are all in-links are equal? Links from important pages count more Recursive question!

15 15 Each link s vote is proportional to the importance of its source page If page P with importance x has n out-links, each link gets x/n votes Page P s own importance is the sum of the votes on its in-links

16 16 The web in 1839 y/2 y Yahoo y = y /2 + a /2 a = y /2 + m m = a /2 a/2 y/2 Amazon a m a/2 M soft m

17 17 3 equations, 3 unknowns, no constants No unique solution All solutions equivalent modulo scale factor Additional constraint forces uniqueness y+a+m = 1 y = 2/5, a = 2/5, m = 1/5 Gaussian elimination method works for small examples, but we need a better method for large web-size graphs

18 Matrix M has one row and one column for each web page Suppose page j has n out-links If j i, then M ij = 1/n else M ij = 0 M is a column stochastic matrix Columns sum to 1 Suppose r is a vector with one entry per web page: r i is the importance score of page i Call it the rank vector r = 1 2/7/2011 Jure Leskovec, Stanford C246: Mining Massive Datasets 18

19 19 Suppose page j links to 3 pages, including i j i 1/3 = i M r r

20 20 The flow equations can be written r = M r So the rank vector is an eigenvector of the stochastic web matrix In fact, its first or principal eigenvector, with corresponding eigenvalue 1

21 21 Yahoo Y! A MS Y! ½ ½ 0 A ½ 0 1 MS 0 ½ 0 Amazon M soft y = y /2 + a /2 a = y /2 + m m = a /2 r = Mr y ½ ½ 0 y a = ½ 0 1 a m 0 ½ 0 m

22 22 Simple iterative scheme Suppose there are N web pages Initialize: r 0 = [1/N,.,1/N] T Iterate: r k+1 = M r k Stop when r k+1 - r k 1 < ε x 1 = 1 i N x i is the L1 norm Can use any other vector norm e.g., Euclidean

23 23 Power iteration: Set r i =1/n Y! Y! A MS Y! ½ ½ 0 r i = j M ij r j And iterate A MS A ½ 0 1 MS 0 ½ 0 Example: y a = m 1/3 1/3 1/3 1/3 1/2 1/6 5/12 1/3 1/4 3/8 11/24 1/6... 2/5 2/5 1/5

24 24 Imagine a random web surfer At any time t, surfer is on some page P At time t+1, the surfer follows an outlink from P uniformly at random Ends up on some page Q linked from P Process repeats indefinitely Let p(t) be a vector whose i th component is the probability that the surfer is at page i at time t p(t) is a probability distribution on pages

25 25 Where is the surfer at time t+1? Follows a link uniformly at random p(t+1) = M p(t) Suppose the random walk reaches a state such that p(t+1) = M p(t) = p(t) Then p(t) is called a stationary distribution for the random walk Our rank vector r satisfies r = M r So it is a stationary distribution for the random surfer

26 26 A central result from the theory of random walks (aka Markov processes): For graphs that satisfy certain conditions, the stationary distribution is unique and eventually will be reached no matter what the initial probability distribution at time t = 0.

27 27 Some pages are dead ends (have no out-links) Such pages cause importance to leak out Spider traps (all out links are within the group) A group of pages is a spider trap if there are no links from within the group to the outside of the group Random surfer gets trapped And eventually spider traps absorb all importance

28 28 Power iteration: Set r i =1 Y! r i = j M ij r j And iterate Example: y 1 1 ¾ 5/8 0 a = 1 ½ ½ 3/8 0 m 1 3/2 7/4 2 3 A MS Y! A MS Y! ½ ½ 0 A ½ 0 0 MS 0 ½ 1

29 29 The Google solution for spider traps At each time step, the random surfer has two options: With probability β, follow a link at random With probability 1-β, jump to some page uniformly at random Common values for β are in the range 0.8 to 0.9 Surfer will teleport out of spider trap within a few time steps

30 30 0.2*1/3 Yahoo 1/2 0.8*1/2 1/2 0.8*1/2 0.2*1/3 0.2*1/3 Amazon M soft y y 1/2 a 1/2 m 0 0.8* y 1/2 1/2 0 1/2 1/ / / * y 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 y 7/15 7/15 1/15 a 7/15 1/15 1/15 m 1/15 7/15 13/15

31 31 Yahoo 1/2 1/ / /2 1 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 Amazon M soft y 7/15 7/15 1/15 a 7/15 1/15 1/15 m 1/15 7/15 13/15 y a = m /11 5/11 21/11

32 32 Some pages are dead ends (have no out-links) Such pages cause importance to leak out Power iteration: Set r i =1 r i = j M ij r j And iterate Example: y 1 1 ¾ 5/8 0 a = 1 ½ ½ 3/8 0 m 1 ½ ¼ ¼ 0 A Y! MS Y! A MS Y! ½ ½ 0 A ½ 0 0 MS 0 ½ 0

33 33 Teleports Follow random teleport links with probability 1.0 from dead-ends Adjust matrix accordingly

34 Suppose there are N pages Consider a page j, with set of outlinks O(j) We have M ij = 1/ O(j) when j i and M ij = 0 otherwise The random teleport is equivalent to adding a teleport link from j to every other page with probability (1-β)/N reducing the probability of following each outlink from 1/ O(j) to β/ O(j) Equivalent: tax each page a fraction (1-β) of its score and redistribute evenly 2/7/2011 Jure Leskovec, Stanford C246: Mining Massive Datasets 34

35 35 Construct the N x N matrix A as follows A ij = β M ij + (1-β)/N Verify that A is a stochastic matrix The page rank vector r is the principal eigenvector of this matrix satisfying r = A r Equivalently, r is the stationary distribution of the random walk with teleports

36 36 Key step is matrix-vector multiplication r new = A r old Easy if we have enough main memory to hold A, r old, r new Say N = 1 billion pages We need 4 bytes for each entry (say) 2 billion entries for vectors, approx 8GB Matrix A has N 2 entries is a large number!

37 37 r = A r, where A ij = β M ij + (1-β)/N r i = 1 j N A ij r j r i = 1 j N [β M ij + (1-β)/N] r j = β 1 j N M ij r j + (1-β)/N 1 j N r j = β 1 j N M ij r j + (1-β)/N, since r = 1 r = βm r + [(1-β)/N] N where [x] N is an N-vector with all entries x

38 38 We can rearrange the PageRank equation r = β M r + [(1-β)/N] N [(1-β)/N] N is an N-vector with all entries (1-β)/N M is a sparse matrix! 10 links per node, approx 10N entries So in each iteration, we need to: Compute r new = β M r old Add a constant value (1-β)/N to each entry in r new

39 39 Encode sparse matrix using only nonzero entries Space proportional roughly to number of links say 10N, or 4*10*1 billion = 40GB still won t fit in memory, but will fit on disk source node degree 0 3 1, 5, , 64, 113, 117, , 23 destination nodes

40 Assume enough RAM to fit r new into memory Store r old and matrix M on disk Initialize all entries of r new to (1-β)/N For each page p (of out-degree n): Read into memory: p, n, dest 1,,dest n, r old (p) for j = 1 n: r new (dest j ) += β r old (p) / n r new src degree destination 0 3 1, 5, , 64, 113, , 23 2/7/2011 Jure Leskovec, Stanford C246: Mining Massive Datasets 40 r old

41 41 In each iteration, we have to: Read r old and M Write r new back to disk IO Cost = 2 r + M Questions: What if we had enough memory to fit both r new and r old? What if we could not even fit r new in memory? See reading:

42 46 Measures generic popularity of a page Biased against topic-specific authorities Solution: Topic-Specific PageRank (next lecture) Uses a single measure of importance Other models e.g., hubs-and-authorities Solution: Hubs-and-Authorities (next) Susceptible to Link spam Artificial link topographies created in order to boost page rank Solution: TrustRank (next lecture)

43 47 Interesting pages fall into two classes: 1. Authorities are pages containing useful information Newspaper home pages Course home pages Home pages of auto manufacturers 2. Hubs are pages that link to authorities List of newspapers Course bulletin List of US auto manufacturers NYT: 10 Ebay: 3 Yahoo: 3 CNN: 8 WSJ: 9

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47 51 A good hub links to many good authorities A good authority is linked from many good hubs Model using two scores for each node: Hub score and Authority score Represented as vectors h and a

48 52 HITS uses adjacency matrix A[i, j] = 1 if page i links to page j, 0 else A T, the transpose of A, is similar to the PageRank matrix M but A T has 1 s where M has fractions Yahoo A = y a m y a m Amazon M soft

49 53 Yahoo A = y a m y a m Amazon M soft

50 54 Notation: Vector a=(a 1,a n ), h=(h 1,h n ) Adjacency matrix (n x n): A ij =1 if i j else A ij =0 Then: h i = So: h = A a Likewise: a = j hi i j a = A T h j A ij a j

51 55 The hub score of page i is proportional to the sum of the authority scores of the pages it links to: h = λ A a Constant λ is a scaling factor, λ = 1/ h i The authority score of page i is proportional to the sum of the hub scores of the pages it is linked from: a = μ A T h Constant μ is scaling factor, μ = 1/ a i

52 56 The HITS algorithm: Initialize h, a to all 1 s Repeat: h = A a Scale h so that its sums to 1.0 a = A T h Scale a so that its sums to 1.0 Until h, a converge (i.e., change very little)

53 A = A T = Yahoo Amazon M soft a(yahoo) a(amazon) a(m soft) = = = / h(yahoo) = 1 h(amazon) = 1 h(m soft) = 1 1 2/3 1/

54 58 Algorithm: Set: a = h = 1 n Repeat: h=m a, a=m T h Normalize Then: a=m T (M a) new h new a Thus: a=(m T M) a h=(m M T ) h a is being updated (in 2 steps): M T (M a) = (M T M) a h is updated (in 2 steps): M (M T h) = (M M T ) h Repeated matrix powering

55 59 Under reasonable assumptions about A, the HITS iterative algorithm converges to vectors h* and a*: h* is the principal eigenvector of matrix AA T a* is the principal eigenvector of matrix A T A

56 60 PageRank and HITS are two solutions to the same problem: What is the value of an in-link from u to v? In the PageRank model, the value of the link depends on the links into u In the HITS model, it depends on the value of the other links out of u The destinies of PageRank and HITS post-1998 were very different