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Suppose we compute PageRank with a β of 0.7, and we introduce the additional constraint that the sum of the PageRanks of the three pages must be 3, to handle the problem that otherwise any multiple of a solution will also be a solution. Compute the PageRanks a, b, and c of the three pages A, B, and C, respectively.

A = \sum_{i\rightarrow{j}} \beta r_i/d_i + (1 - \beta ) e / n

My R code is this.

M = matrix(c(0,1/2,1/2,0,0,1,0,0,1),ncol=3)
e = matrix(c(1,1,1),ncol=1)
v1 = matrix(c(1,1,1),ncol=1) 
v1 = v1 / 3
for( i in 1:5){

  v1 =  ((0.7 * M ) %*% v1 ) + (((1 - 0.7 ) * e ) /3 )
   v1 = v1 * 3
[1,] 0.300
[2,] 0.405
[3,] 2.295

2 Responses to PageRank

  1. Eric says:

    Hi,there. The node C is a dead end, so additional links to A and B should be added, therefor matrix M should be M = matrix(c(0,1/2,1/2,0,0,1,1/3,1/3,1/3),ncol=3). Besides, ‘e’ should be a n*n matrix, what do you think?

  2. I remember the code generated values that were accepted by Coursera but I will check this.

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