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Understanding Google’s PageRank™

Understanding Google’s PageRank™. Review: The Search Engine. Goals and assumptions. The results from the query module are still excessively large sets, despite the boolean operations and content index operations. We still don’t know which pages should be ranked highest.

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Understanding Google’s PageRank™

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  1. Understanding Google’s PageRank™

  2. Review: The Search Engine

  3. Goals and assumptions The results from the query module are still excessively large sets, despite the boolean operations and content index operations. We still don’t know which pages should be ranked highest. Assume the pages with the most in-bound links are the best; a link is a vote.

  4. An Elegant Formula πT= π(k)T (aS + (1-a) E) Google’s (Brin & Page) PageRank™ equation. US Patent #6285999, filed 1998, granted 2001 This formula resolves the world’s largest matrix calculation.

  5. πT= π(k)T (aS + (1-a) E) Derived from a formula B&P worked out in graduate school (itself derived from traditional bibliometrics research literature). r(Pi) = Essential characteristic: high-ranking pages associate with high-ranking pages S r(Pj) _____ |Pj| PjÎ BPi

  6. πT= π(k)T (aS + (1-a) E) S r(Pi) : the rank of a given page PjÎ Bpi : the ranks of the set of back- linking pages r(Pj) : the rank of a given page |Pj| : the number of out-links on a page r(Pj) r(Pi) = Must be applied to a set of linked pages, or a graph. To do this we analyze the graph to see it’s out-links and back-links. Therefore. . . _____ |Pj| PjÎ BPi

  7. πT= π(k)T (aS + (1-a) E) 1 A site diagram like this: 2 3 5 4 6

  8. πT= π(k)T (aS + (1-a) E) 1 2 becomes a directed graph like this: 3 4 5 6

  9. But there’s a problem 1 2 S r(Pj) 3 Nothing’s ranked! _____ • r(Pi) = |Pj| PjÎ BPi 4 5 r(Pi) : the rank of a given page PjÎ Bpi : the ranks of the set of back- linking pages r(Pj) : the rank of a given page |Pj| : the number of out-links on a page 6

  10. The solution. . . sort of 1 2 Start by assuming all the ranks are equal. In this example each page is just 1 of 6, so the initial rank is expressed as 1/6 Then, you keep feeding the number through the formula until you get a ranking. This results in a rank , but you have to calculate these ranks one page at a time. That’s slow. 3 4 5 6

  11. Directed graph iterative node values 1 2 r0 r1 r2 Rank(i2) P1 1/6 1/18 1/36 5 P2 1/6 5/36 1/18 4 P3 1/6 1/12 1/36 5 P4 1/6 1/4 17/72 1 P5 1/6 5/36 11/72 3 P6 1/6 1/6 14/72 2 3 4 5 6

  12. This can’t go on forever CMS matrix 1 2 In the interest of speed and efficiency, we need to know if the ranks converge—that is, we need to know if there are clear rankings, or can we keep doing this indefinitely and never have a decisive ranking? To determine this, the formula must be transformed using binary adjacency transformation, and Markov chain theory. 3 4 5 6

  13. Convert the iterative calculation to a matrix calculation using binary adjacency transformation for a 1Xn matrix [ ] P1 P2 P3 P4 P5 P6 P1 0 ½ ½ 0 0 0 P2 0 0 0 0 0 0 P3 1/3 1/3 0 0 1/3 0 P4 0 0 0 0 ½ ½ P5 0 0 0 ½ 0 ½ P6 0 0 0 1 0 0

  14. Now, you can treat a row as a vector, or set of values [ ] P1 P2 P3 P4 P5 P6 P1 0 ½ ½ 0 0 0 P2 0 0 0 0 0 0 P3 1/3 1/3 0 0 1/3 0 P4 0 0 0 0 ½ ½ P5 0 0 0 ½ 0 ½ P6 0 0 0 1 0 0 pT

  15. This is a sparse matrix. That’s good. We store only the non-zero elements and represent the entire matrix as H [ ] P1 P2 P3 P4 P5 P6 P1 0 ½ ½ 0 0 0 P2 0 0 0 0 0 0 P3 1/3 1/3 0 0 1/3 0 P4 0 0 0 0 ½ ½ P5 0 0 0 ½ 0 ½ P6 0 0 0 1 0 0

  16. πT= π(k)T (aS + (1-a) E) So now this: Has become this: πT= π(k)T (H) through the transformation and reduction of the form (power method transform, eigenvector computation, stochasticity adjustment). We only need a couple more adjustments. S r(Pj) _____ • r(Pi) = |Pj| PjÎ BPi

  17. πT= πT (aS + (1-a) E) Sometimes, people teleport to a page. They just enter the URL and go. And just as easily, they can teleport out. To account for this, B&P added two adjustments: aSaccounts for people who reach a dead end and jump to another page within a site. a is a weighted probability that someone will leave. S is a matrix of probable page destinations.

  18. πT= πT (aS + (1-a) E) What about people who jump out to a completely new destination? To account for this, B&P added the final adjustments: 1- ais the inverted weighted probability that someone will leave and go to a completely new site. E is a random teleportation matrix of probable page destinations.

  19. Summary πT= π(k)T (aS + (1-a) E) A page’s rank is equal to the summed and transformed ranks of the referring pages, tempered by the ability (a weighted probability) to teleport within a site, plus the inverted probability to teleport out of a site, multiplied by the probability matrix of teleporting to a popular site.

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