Page rank computation HPC course project a.y

Transcription

1 Page rank computation HPC course project a.y Compute efficient and scalable Pagerank MPI, Multithreading, SSE 1

2 PageRank PageRank is a link analysis algorithm, named after Brin & Page [1], and used by the Google Internet search engine, which assigns a numerical weighting to each element of a hyperlinked set of documents, such as the World Wide Web, with the purpose of "measuring" its relative importance within the set [wikipedia] [1] Brin, S. and Page, L. (1998) The Anatomy of a Large-Scale Hypertextual Web Search Engine. In: Seventh International World-Wide Web Conference (WWW 1998), April 14-18, 1998, Brisbane, Australia 2

3 PageRank: the intuitive idea PageRank relies on the democratic nature of the Web by using its vast link structure as an indicator of an individual page's value or quality. PageRank interprets a hyperlink from page x to page y as a vote, by page x, for page y However, PageRank looks at more than the sheer number of votes; it also analyzes the page that casts the vote. Votes casted by important pages weigh more heavily and help to make other pages more "important" This is exactly the idea of rank prestige in social network. 3

4 More specifically A hyperlink from a page to another page is an implicit conveyance of authority to the target page. The more in-links that a page i receives, the more prestige the page i has. Pages that point to page i also have their own prestige scores. A page of a higher prestige pointing to i is more important than a page of a lower prestige pointing to i In other words, a page is important if it is pointed to by other important pages 4

5 PageRank algorithm According to rank prestige, the importance of page i (i s PageRank score) is the sum of the PageRank scores of all pages that point to i Since a page may point to many other pages, its prestige score should be shared. The Web as a directed graph G = (V, E) The PageRank score of the page i (denoted by P(i)) is defined by: P(i) = ( j,i) E P( j) O j O j is the number of out-link of j 5

6 PageRank algorithm: iterative process 1/2 P(i) = ( j,i) E P( j) O j O j is the number of out-link of j 6

7 PageRank algorithm: iterative process 1/3 1/2 1/3 1/3 P(i) = ( j,i) E P( j) O j O j is the number of out-link of j 7

8 PageRank algorithm: First iteration 1/3 1/2 1/2 1/6 P(i) = ( j,i) E P( j) O j O j is the number of out-link of j 8

9 PageRank algorithm: Second iteration 5/12 1/2 1/3 1/4 P(i) = ( j,i) E P( j) O j O j is the number of out-link of j 9

10 PageRank algorithm: Third itera*on 9/24 1/2 11/24 1/6 P(i) = ( j,i) E P( j) O j O j is the number of out-link of j 10

11 PageRank algorithm: iterative process 2/5 1/2 2/5 1/5 11

12 Matrix notation Let n = V be the total number of pages We have a system of n linear equations with n unknowns. We can use a matrix to represent them. Let P be a n-dimensional column vector of PageRank values, i.e., P = (P(1), P(2),, P(n)) T Let A be the adjacency matrix of our graph with A ij = # % $ &% 1 O i if (i, j) E 0 otherwise P( j) We can write the n equations P(i) = with P = A T P (PageRank) ( j,i) E O j 12

13 Solve the PageRank equation P = A T P This is the characteristic equation of the eigensystem, where the solution to P is an eigenvector with the corresponding eigenvalue of 1 It turns out that if some conditions are satisfied, 1 is the largest eigenvalue and the PageRank vector P is the principal eigenvector. A well known mathematical technique called power iteration can be used to find P Problem: the above Equation does not quite suffice because the Web graph does not meet the conditions. 13

14 Using Markov chain To introduce these conditions and the enhanced equation, let us derive the same above Equation based on the Markov chain. In the Markov chain, each Web page or node in the Web graph is regarded as a state. A hyperlink is a transition, which leads from one state to another state with a probability. This framework models Web surfing as a stochastic process. Random walk It models a Web surfer randomly surfing the Web as state transition. 14

15 Random surfing Recall we use O i to denote the number of out-links of a node i Each transition probability is 1/O i if we assume the Web surfer will click the hyperlinks in the page i uniformly at random. the back button on the browser is not used the surfer does not type in an URL 15

16 Transition probability matrix Let A be the state transition probability matrix: " $ $ $ A = $. $ $ $ # A 11 A A 1n % ' A 21 A A 2n '... ' '... '... ' ' A n1 A n 2... A nn & A ij represents the transition probability that the surfer in state i (page i) will move to state j (page j). Can A be the adjacency matrix previously discussed? 16

17 Let us start Given an initial probability distribution vector that a surfer is at each state (or page) p 0 = (p 0 (1), p 0 (2),, p 0 (n)) T (a column vector) an n n transition probability matrix A we have n p 0 (i) =1 i=1 n A ij =1 (1) j=1 If the matrix A satisfies Equation (1), we say that A is the stochastic matrix of a Markov chain 17

18 Back to the Markov chain In a Markov chain, a question of common interest is: What is the probability that, after m steps/transitions (with m è ), a random process/walker reaches a state j independently of the initial state of the walk We determine the probability that the random surfer arrives at the state/page j after 1 step (1 transition) by using the following reasoning: p 1 ( j) = n i=1 A ij (1)p 0 (i) where A ij (1) is the probability of going from i to j after 1 step. At beginning p 0 (i) = 1/N for all i 18

19 State transition We can write this in matricial form: P 1 = A T P 0 In general, the probability distribution after k steps/transitions is: P k = A T P k -1 19

20 Stationary probability distribution By the Ergodic Theorem of Markov chain a finite Markov chain defined by the stochastic matrix A has a unique stationary probability distribution if A is irreducible and aperiodic The stationary probability distribution means that after a series of transitions p k will converge to a steady-state probability vector π, i.e., lim P = π k k 20

21 PageRank again When we reach the steady-state, we have P k = P k+1 =π, and thus π =A T π π is the principal eigenvector (the one with the maximum magnitude) of A T with eigenvalue of 1 In PageRank, π is used as the PageRank vector P: P = A T P 21

22 Is P = π justified? Using the stationary probability distribution π as the PageRank vector is reasonable and quite intuitive because it reflects the long-run probabilities that a random surfer will visit the pages. a page has a high prestige if the probability of visiting it is high 22

23 Back to the Web graph Now let us come back to the real Web context and see whether the above conditions are satisfied, i.e., whether A is a stochastic matrix and whether it is irreducible and aperiodic. None of them is satisfied. Hence, we need to extend the ideal-case to produce the actual PageRank model. 23

24 A is a not stochastic matrix A is the transition matrix of the Web graph A ij = # % $ &% It does not satisfy equation: 1 O i if (i, j) E 0 otherwise n j=1 A ij =1 because many Web pages have no out-links (dangling pages) This is reflected in transition matrix A by some rows of 0 s 24

25 An example Web hyperlink graph " % $ ' $ ' $ ' A = $ ' $ ' $ ' $ ' # & 25

26 Fix the problem: two possible ways 1. Remove pages with no out-links during the PageRank computation these pages do not affect the ranking of any other page directly 2. Add a complete set of outgoing links from each such page i to all the pages on the Web. Let us use the 2 nd method: " % $ ' $ ' $ ' A = $ ' $ ' $ ' $ ' # & 26

27 A is a not irreducible Irreducible means that the Web graph G is strongly connected Definition: A directed graph G = (V, E) is strongly connected if and only if, for each pair of nodes u, v V, there is a directed path from u to v. A general Web graph represented by A is not irreducible because for some pair of nodes u and v, there is no path from u to v In our example, there is no directed path from nodes 3 to 4 27

28 A is a not aperiodic A state i in a Markov chain being periodic means that there exists a directed cycle (from i to i) that a random walker traverses multiple times Definition: A state i is periodic (with period k > 1) if k is the smallest number such that all paths leading from state i back to state i have a length that is a multiple of k A Markov chain is aperiodic if all states are aperiodic. 28

29 An example: periodic This a periodic Markov chain with k = 3 If we begin from state 1, to come back to state 1 the only path is for some number of times, say h Thus any return to state 1 will take k h = 3h transitions. 29

30 Deal with irreducible and aperiodic matrices It is easy to deal with the above two problems with a single strategy. Add a link from each page to every page and give each link a small transition probability controlled by a parameter d Obviously, the augmented transition matrix becomes irreducible and aperiodic it becomes irreducible because it is strongly connected it become aperiodic because we now have paths of all the possible lengths from state i back to state i 30

31 Improved PageRank After this augmentation, at a page, the random surfer has two options With probability d, 0<d<1, she randomly chooses an outlink to follow With smaller probability 1-d, she stops clicking and jumps to a random page The following equation models the improved model: P = ((1 d) E n + dat )P where E is a n n square matrix of all 1 s n is important, since the matrix has to be stochastic 31

32 Follow our example d = 0.9 Transposed matrix (1 d) E n + dat = The matrix made stochastic, which is still: periodic (see state 3) reducible (no path from 3 to 4) " % $ ' $ ' $ ' A = $ ' $ ' $ ' $ ' # & # & % ( % ( % ( % ( % ( % ( % ( $ ' 32

33 The final PageRank algorithm (1-d)E/n + da T is a stochastic matrix (transposed). It is also irreducible and aperiodic Note that E = e e T where e is a column vector of 1 s e T P = 1 since P is the stationary probability vector π If we scale this equation: P = ((1 d) E n + dat )P = (1 d) 1 n e et P + da T P = by multiplying both sides by n, we have: e T P = n and thus: = (1 d) 1 n e + dat P P = (1 d)e + da T P 33

34 The final PageRank algorithm (cont ) Given: P = (1 d ) e + da PageRank for each page i is: n j=1 that is equivalent to the formula given in the PageRank paper [BP98] The parameter d is called the damping factor which can be set to between 0 and 1. d = 0.85 was used in the PageRank paper T P P(i) = (1 d) + d A ji P( j) P(i) = (1 d) + d [BP98] Sergey Brin and Lawrence Page. The Anatomy of a Large-Scale Hypertextual Web Search Engine. WWW Int.l Conf., A ji = # % $ & % 1 O j ( j,i) E if ( j,i) E 0 otherwise P( j) O j 34

35 Compute PageRank Use the power iteration method Initialization 0 n Norm 1 less than

36 Again PageRank Without scaling the equation (w/o multiplying by n), we have e T P = 1 (i.e., the sum of all PageRanks is one), and thus: P(i) = 1 d n + d P( j) O ( j,i) E j Important pages are cited/pointed by other important ones In the example, the most important is ID=1 P(ID=1) = P(ID=1) distributes is rank among all its 5 outgoing links ID= 2, 3, 4, 5, = *

37 Again PageRank Without scaling the equation (w/o multiplying by n), we have e T P = 1 (i.e., the sum of all PageRanks is one), and thus: P(i) = 1 d n + d P( j) O ( j,i) E j The stationary probability P(ID=1) is obtained by: (1-d)/n + d ( )= (0.15)/ ( )=

38 1 st step Write a sequential code (C or C++) that implements Pagerank Compile the code with O3 option, and measure the execution times (command time) for some inputs Input graphs: Test example: P[2] = P[1] = P[0] =

39 2 nd step Given the original incidence matrix A[][], if we know which are the dangling nodes, we can avoid filling zero-rows with values 1/n A T /n * * p k = p k Dangling nodes A T /n * * * p k = p k+1 39

40 2 nd step A T * p k + 1/n * * * p k = p k+1 A T * p k + X... i2danglings X i2danglings p k [i] n p k [i] n = p k+1 40

41 2 nd step Avoid transposing matrix A[][] Still traverse A[][] in row major order for (i=0; i<n; i++) for (j=0; j<n; j++) p_new[j] = p_new[j] + a[i][j] * p[i]; Store matrix A[][] in sparse compressed form Compressed sparse row (CSR or CRS) 41

42 2 nd step Compressed sparse row (CSR or CRS) Used for traversing matrix in row major order val col_ind row_ptr Start row 0 Position where the n-th row should Start row 1 start. Note that the matrix is sparse, and thus the row could be completely zero. Start row 2 In this case row_ptr[n] = row_prt[n+1] Start row 3 Start row 4 Start row 5 Start row 6 (1 more position) 42

43 2 nd step Store big data like A[][] on a file Once we map the file to a memory region, we access it via pointers, just as you would access ordinary variables and objects You can mmap specific section/partition of the file, and share the files between more threads #include <stdio.h> #include <sys/mman.h> #include <sys/stat.h> #include <fcntl.h> #include <unistd.h> #include <stdlib.h> int main() { int i; float val; float *mmap_region; FILE *fstream; int fd; 43

44 2 nd step /* create the file */ fstream = fopen("./mmapped_file", "w+"); for (i=0; i<10; i++) { val = i ; /* write a stream of binary floats */ fwrite(&val, sizeof(float), 1, fstream); } fclose(fstream); /* map a file to the pages starting at a given address for a given length */ fd = open("./mmapped_file", O_RDONLY); mmap_region = (float *) mmap(0, 10*sizeof(float), PROT_READ, MAP_SHARED, fd, 0); if (mmap_region == MAP_FAILED) { close(fd); printf("error mmapping the file"); exit(1); } close(fd); Starting offset address in the file 44

45 2 nd step /* Print the data mmapped */ for (i=0; i<10; i++) printf("%f ", mmap_region[i]); printf("\n"); } /* free the mmapped memory */ if (munmap(mmap_region, 10*sizeof(float)) == -1) { printf("error un-mmapping the file"); exit(1); } 45

46 3rd step The goal of this assignment is to parallelize the optimized code of the 2 assignment You can use SIMD SSE, shared-memory or message passing (also hybrid) parallelization Measure speedup and efficiency as a function of processors/cores exploited (for a couple data sets) Point out the effects of the Amdahl law, concerning the serial sections that remain serial e.g., the input output phases if you are not able to parallelize Measure how the execution time changes when we increase the problem size, without changing the number of processors/cores employed This requires to consider subsets of nodes and edges of a given input graph 46

47 3rd step The issues to solve concern decision such as the right decomposition, the right granularity, and a strategy (static/dynamic) of task assignment I would only to point out that, if we don t transpose matrix A, and decomposes the problem over the input, we have: A * p k = p k+1 A * p k = p k+1 + reduce p k+1 A * p k = p k+1 47