# COMP 465: Data Mining More on PageRank

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1 COMP 465: Dt Mnng Moe on PgeRnk Sldes Adpted Fo: (Mnng Mssve Dtsets) Powe Iteton: Set = 1/ 1: = 2: = Goto 1 Exple: d 1/3 1/3 5/12 9/24 6/15 = 1/3 3/6 1/3 11/24 6/15 1/3 1/6 3/12 1/6 3/15 Iteton 0, 1, 2, ½ ½ 0 ½ ½ 0 = /2 + /2 = /2 + = /2 J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 2 Powe Iteton: Set = 1/ 1: = 2: = Goto 1 Exple: d 1/3 1/3 5/12 9/24 6/15 = 1/3 3/6 1/3 11/24 6/15 1/3 1/6 3/12 1/6 3/15 Iteton 0, 1, 2, ½ ½ 0 ½ ½ 0 = /2 + /2 = /2 + = /2 Igne ndo web sufe: At n te t, sufe s on soe pge At te t + 1, the sufe follows n out-lnk fo unfol t ndo d Ends up on soe pge lnked fo out () Pocess epets ndefntel Let: p(t) vecto whose th coodnte s the pob. tht the sufe s t pge t te t So, p(t) s pobblt dstbuton ove pges J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 3 J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 4 1

2 1 2 3 Whee s the sufe t te t+1? Follows lnk unfol t ndo p t + 1 = M p(t) p( t 1) M p( t) Suppose the ndo wlk eches stte p t + 1 = M p(t) = p(t) then p(t) s stton dstbuton of ndo wlk Ou ognl nk vecto stsfes = M So, s stton dstbuton fo the ndo wlk J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 5 ( t1) ( t) d Does ths convege? Does t convege to wht we wnt? Ae esults esonble? o equvlentl M b Exple: = b Iteton 0, 1, 2, ( t1) ( t) d J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 7 J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 8 2

3 b Exple: = b Iteton 0, 1, 2, ( t1) ( t) d 2 pobles: (1) Soe pges e ded ends (hve no out-lnks) Rndo wlk hs nowhee to go to Such pges cuse potnce to lek out (2) Spde tps: (ll out-lnks e wthn the goup) Rndo wlked gets stuck n tp And eventull spde tps bsob ll potnce Ded end J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 9 J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 10 Powe Iteton: Set = 1 = d And tete Exple: 1/3 2/6 3/12 5/24 0 = 1/3 1/6 2/12 3/24 0 1/3 3/6 7/12 16/24 1 Iteton 0, 1, 2, J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 11 s spde tp All the PgeRnk scoe gets tpped n node. ½ ½ 0 ½ ½ 1 = /2 + /2 = /2 = /2 + The Google soluton fo spde tps: At ech te step, the ndo sufe hs two optons Wth pob., follow lnk t ndo Wth pob. 1-, up to soe ndo pge Coon vlues fo e n the nge 0.8 to 0.9 Sufe wll telepot out of spde tp wthn few te steps J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets,

4 Powe Iteton: Set = 1 = d And tete Exple: 1/3 2/6 3/12 5/24 0 = 1/3 1/6 2/12 3/24 0 1/3 1/6 1/12 2/24 0 Iteton 0, 1, 2, Hee the PgeRnk leks out snce the tx s not stochstc. J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 13 ½ ½ 0 ½ ½ 0 = /2 + /2 = /2 = /2 Telepots: Follow ndo telepot lnks wth pobblt 1.0 fo ded-ends Adust tx ccodngl ½ ½ 0 ½ ½ 0 ½ ½ ⅓ ½ 0 ⅓ 0 ½ ⅓ J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 14 Wh e ded-ends nd spde tps poble nd wh do telepots solve the poble? Spde-tps e not poble, but wth tps PgeRnk scoes e not wht we wnt Soluton: eve get stuck n spde tp b telepotng out of t n fnte nube of steps Ded-ends e poble The tx s not colun stochstc so ou ntl ssuptons e not et Soluton: Mke tx colun stochstc b lws telepotng when thee s nowhee else to go J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 15 Google s soluton tht does t ll: At ech step, ndo sufe hs two optons: Wth pobblt, follow lnk t ndo Wth pobblt 1-, up to soe ndo pge PgeRnk equton [Bn-Pge, 98] = β d + (1 β) 1 Ths foulton ssues tht M hs no ded ends. We cn ethe pepocess tx M to eove ll ded ends o explctl follow ndo telepot lnks wth pobblt 1.0 fo ded-ends. d out-degee of node J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets,

5 PgeRnk equton [Bn-Pge, 98] = β + (1 β) 1 d The Google Mtx A: A = β M + 1 β 1 We hve ecusve poble: = A And the Powe ethod stll woks! Wht s? In pctce =0.8,0.9 (ke 5 steps on vg., up) [1/] x b tx whee ll entes e 1/ = 7/15 1/3 1/3 1/ / /2 1/ / / M 7/15 7/15 1/15 7/15 1/15 1/15 1/15 7/15 13/15... A 7/33 5/33 21/33 [1/] x J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 17 J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 18 Ke step s tx-vecto ultplcton new = A old Es f we hve enough n eo to hold A, old, new S = 1 bllon pges We need 4 btes fo ech ent (s) 2 bllon entes fo vectos, ppox 8GB Mtx A hs 2 entes s lge nube! A = M + (1-) [1/] x ½ ½ 0 ½ ½ 1 A = /15 7/15 1/15 7/15 1/15 1/15 1/15 7/15 13/15 J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 20 = 5

6 Suppose thee e pges Consde pge, wth d out-lnks We hve M = 1/ d when nd M = 0 othewse The ndo telepot s equvlent to: Addng telepot lnk fo to eve othe pge nd settng tnston pobblt to (1-)/ Reducng the pobblt of followng ech out-lnk fo 1/ d to / d Equvlent: Tx ech pge fcton (1-) of ts scoe nd edstbute evenl J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 21 = A, whee A = β M + 1 β =1 =1 = A = β M + 1 β = β M + 1 β =1 =1 =1 snce = 1 = β M + 1 β So we get: = β M + 1 β ote: Hee we ssued M hs no ded-ends J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 22 [x] vecto of length wth ll entes x We ust enged the PgeRnk equton 1 β = βm + whee [(1-)/] s vecto wth ll entes (1-)/ M s spse tx! (wth no ded-ends) 10 lnks pe node, ppox 10 entes So n ech teton, we need to: Copute new = M old Add constnt vlue (1-)/ to ech ent n new new ote f M contns ded-ends then < 1 nd we lso hve to enolze new so tht t sus to 1 J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets, 23 Input: Gph G nd pete β Dected gph G (cn hve spde tps nd ded ends) Pete β Output: PgeRnk vecto new Set: old = 1 epet untl convegence: new old > ε : new = β old d new = 0 f n-degee of s 0 ow e-nset the leked PgeRnk: : new = new 1 S + new whee: S = old = new If the gph hs no ded-ends then the ount of leked PgeRnk s 1-β. But snce we hve ded-ends the ount of leked PgeRnk be lge. We hve to explctl ccount fo t b coputng S. J. Leskovec, A. Rn, J. Ulln: Mnng of Mssve Dtsets,

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