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CS345 Data Mining

CS345 Data Mining. Link Analysis Algorithms Page Rank. Anand Rajaraman, Jeffrey D. Ullman. Link Analysis Algorithms. Page Rank Hubs and Authorities Topic-Specific Page Rank Spam Detection Algorithms Other interesting topics we won’t cover Detecting duplicates and mirrors

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CS345 Data Mining

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  1. CS345Data Mining Link Analysis Algorithms Page Rank Anand Rajaraman, Jeffrey D. Ullman

  2. Link Analysis Algorithms • Page Rank • Hubs and Authorities • Topic-Specific Page Rank • Spam Detection Algorithms • Other interesting topics we won’t cover • Detecting duplicates and mirrors • Mining for communities

  3. Ranking web pages • Web pages are not equally “important” • www.joe-schmoe.com v www.stanford.edu • Inlinks as votes • www.stanford.edu has 23,400 inlinks • www.joe-schmoe.com has 1 inlink • Are all inlinks equal? • Recursive question!

  4. Simple recursive formulation • Each link’s vote is proportional to the importance of its source page • If page P with importance x has n outlinks, each link gets x/n votes • Page P’s own importance is the sum of the votes on its inlinks

  5. Yahoo Amazon M’soft Simple “flow” model The web in 1839 y = y /2 + a /2 a = y /2 + m m = a /2 y/2 y a/2 y/2 m a/2 m a

  6. Solving the flow equations • 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 graphs

  7. Matrix formulation • Matrix M has one row and one column for each web page • Suppose page j has n outlinks • If j ! i, then Mij=1/n • Else Mij=0 • M is a columnstochastic matrix • Columns sum to 1 • Suppose r is a vector with one entry per web page • ri is the importance score of page i • Call it the rank vector • |r| = 1

  8. j i i = 1/3 M r Example Suppose page j links to 3 pages, including i r

  9. Eigenvector formulation • The flow equations can be written r = Mr • So the rank vector is an eigenvector of the stochastic web matrix • In fact, its first or principal eigenvector, with corresponding eigenvalue 1

  10. Yahoo r = Mr Amazon M’soft y 1/2 1/2 0 y a = 1/2 0 1 a m 0 1/2 0 m Example y a m y 1/2 1/2 0 a 1/2 0 1 m 0 1/2 0 y = y /2 + a /2 a = y /2 + m m = a /2

  11. Power Iteration method • Simple iterative scheme (aka relaxation) • Suppose there are N web pages • Initialize: r0 = [1/N,….,1/N]T • Iterate: rk+1 = Mrk • Stop when |rk+1 - rk|1 <  • |x|1 = 1≤i≤N|xi| is the L1 norm • Can use any other vector norm e.g., Euclidean

  12. Yahoo Amazon M’soft Power Iteration Example y a m y 1/2 1/2 0 a 1/2 0 1 m 0 1/2 0 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 . . .

  13. Random Walk Interpretation • 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 ith component is the probability that the surfer is at page i at time t • p(t) is a probability distribution on pages

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

  15. Existence and Uniqueness 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.

  16. Spider traps • A group of pages is a spider trap if there are no links from within the group to outside the group • Random surfer gets trapped • Spider traps violate the conditions needed for the random walk theorem

  17. Microsoft becomes a spider trap Yahoo y a m y 1/2 1/2 0 a 1/2 0 0 m 0 1/2 1 Amazon M’soft y a = m 1 1 1 1 1/2 3/2 3/4 1/2 7/4 5/8 3/8 2 0 0 3 . . .

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

  19. y y y 1/3 1/3 1/3 y 1/2 a 1/2 m 0 1/2 1/2 0 + 0.2* 0.8* Random teleports ( = 0.8) 0.2*1/3 1/2 Yahoo 0.8*1/2 1/2 0.2*1/3 0.8*1/2 0.2*1/3 1/2 1/2 0 1/2 0 0 0 1/2 1 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 + 0.2 Amazon M’soft 0.8 y 7/15 7/15 1/15 a 7/15 1/15 1/15 m 1/15 7/15 13/15

  20. Random teleports ( = 0.8) 1/2 1/2 0 1/2 0 0 0 1/2 1 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 + 0.2 Yahoo 0.8 y 7/15 7/15 1/15 a 7/15 1/15 1/15 m 1/15 7/15 13/15 Amazon M’soft y a = m 1 1 1 1.00 0.60 1.40 0.84 0.60 1.56 0.776 0.536 1.688 7/11 5/11 21/11 . . .

  21. Matrix formulation • Suppose there are N pages • Consider a page j, with set of outlinks O(j) • We have Mij = 1/|O(j)| when j!i and Mij = 0 otherwise • The random teleport is equivalent to • adding a teleport link from j to every other page with probability (1-)/N • reducing the probability of following each outlink from 1/|O(j)| to /|O(j)| • Equivalent: tax each page a fraction (1-) of its score and redistribute evenly

  22. Page Rank • Construct the N£N matrix A as follows • Aij = Mij + (1-)/N • Verify that A is a stochastic matrix • The page rank vectorr is the principal eigenvector of this matrix • satisfying r = Ar • Equivalently, r is the stationary distribution of the random walk with teleports

  23. Dead ends • Pages with no outlinks are “dead ends” for the random surfer • Nowhere to go on next step

  24. Non- stochastic! Microsoft becomes a dead end 1/2 1/2 0 1/2 0 0 0 1/2 0 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 + 0.2 Yahoo 0.8 y 7/15 7/15 1/15 a 7/15 1/15 1/15 m 1/15 7/15 1/15 Amazon M’soft y a = m 1 1 1 1 0.6 0.6 0.787 0.547 0.387 0.648 0.430 0.333 0 0 0 . . .

  25. Dealing with dead-ends • Teleport • Follow random teleport links with probability 1.0 from dead-ends • Adjust matrix accordingly • Prune and propagate • Preprocess the graph to eliminate dead-ends • Might require multiple passes • Compute page rank on reduced graph • Approximate values for deadends by propagating values from reduced graph

  26. Computing page rank • Key step is matrix-vector multiplication • rnew = Arold • Easy if we have enough main memory to hold A, rold, rnew • Say N = 1 billion pages • We need 4 bytes for each entry (say) • 2 billion entries for vectors, approx 8GB • Matrix A has N2 entries • 1018 is a large number!

  27. Rearranging the equation r = Ar, where Aij = Mij + (1-)/N ri = 1≤j≤N Aij rj ri = 1≤j≤N [Mij + (1-)/N] rj =  1≤j≤N Mij rj + (1-)/N 1≤j≤N rj =  1≤j≤N Mij rj + (1-)/N, since |r| = 1 r = Mr + [(1-)/N]N where [x]N is an N-vector with all entries x

  28. Sparse matrix formulation • We can rearrange the page rank equation: • r = Mr + [(1-)/N]N • [(1-)/N]N is an N-vector with all entries (1-)/N • M is a sparse matrix! • 10 links per node, approx 10N entries • So in each iteration, we need to: • Compute rnew = Mrold • Add a constant value (1-)/N to each entry in rnew

  29. Sparse matrix encoding • 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 destination nodes

  30. Basic Algorithm • Assume we have enough RAM to fit rnew, plus some working memory • Store rold and matrix M on disk Basic Algorithm: • Initialize: rold = [1/N]N • Iterate: • Update: Perform a sequential scan of M and rold to update rnew • Write out rnew to disk as rold for next iteration • Every few iterations, compute |rnew-rold| and stop if it is below threshold • Need to read in both vectors into memory

  31. Update step Initialize all entries of rnew to (1-)/N For each page p (out-degree n): Read into memory: p, n, dest1,…,destn, rold(p) for j = 1..n: rnew(destj) += *rold(p)/n rold rnew src degree destination 0 0 1 1 2 2 3 3 4 4 5 5 6 6

  32. Analysis • In each iteration, we have to: • Read rold and M • Write rnew back to disk • IO Cost = 2|r| + |M| • What if we had enough memory to fit both rnew and rold? • What if we could not even fit rnew in memory? • 10 billion pages

  33. Block-based update algorithm rold rnew src degree destination 0 0 1 1 2 3 2 4 3 5 4 5

  34. Analysis of Block Update • Similar to nested-loop join in databases • Break rnew into k blocks that fit in memory • Scan M and rold once for each block • k scans of M and rold • k(|M| + |r|) + |r| = k|M| + (k+1)|r| • Can we do better? • Hint: M is much bigger than r (approx 10-20x), so we must avoid reading it k times per iteration

  35. Block-Stripe Update algorithm src degree destination rnew 0 rold 1 0 1 2 3 2 4 3 5 4 5

  36. Block-Stripe Analysis • Break M into stripes • Each stripe contains only destination nodes in the corresponding block of rnew • Some additional overhead per stripe • But usually worth it • Cost per iteration • |M|(1+) + (k+1)|r|

  37. Next • Topic-Specific Page Rank • Hubs and Authorities • Spam Detection

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