CS683: calculating the effective resistances

Transcription

1 CS683: clculting the effective resistnces Lecturer: John Hopcroft Note tkers: June Andrews nd Jen-Bptiste Jennin Mrch 7th, 2008 On Ferury 29th we sw tht, given grph in which ech edge is lelled with resistnce (nlogy with n electricl circuit), nd given two vertices of this grph nd, the proility of reching efore returning to (considering pth tht strts from ) is P escpe = C eff C = R R eff. The purpose of this lecture will e to clculte C eff, nd thus P escpe, in some specil cses. We re minly interested in clculting the proility of escpe from the center of volume, first in one dimension, then in 2 dimension nd 3 dimensions. In ll the lecture we will keep the nottion for our strting node. Throughout, C = x C x, where C x = R x nd x is neighor of in the grph. In one dimension In one dimension our volume looks like: Bending in hlf, this cn lso e viewed s: 0 v = Therefore R eff = n+ 2, so P escpe = C eff n +. 2 In two dimensions n n v = C = C R eff = 2R eff = n+ 0 when Here lso, we wnt to clculte the resistnce. Here we wnt lower ound on the resistnce, so we put some resistnces to zero (the ones tht re old in the

2 scheme nd correspond to seril edges ); the effect of doing this is to reduce R eff nd give us n upper ound on P escpe : This cn lso e viewed s: Therefore R eff = ( ) = inf 4 i= Θ(log n). Therefore the proility of escping 0 P escpe 4Θ(log n) P escpe 0 when n. 3 In three dimensions 3. Lower ound on P escpe 2 i = nd so Let us present the technique in 2 dimensions, then pply it to the 3-dimensionl cse. As efore, to get lower ound on P escpe, we need n upper ound on R eff, therefore we will put some resistnces to infinity, i.e., just cut the wires(on prllel edges ). We will do tht in the following wy: first drw in dshes the lines of eqution x + y = 2 n (this will e x + y + z = 2 n in 3 dimensions); then from the origin, follow pth up nd pth right until hitting dotted line; when hitting dotted line, split ech pth reching the dotted line into pth up nd pth right gin, etc. Doing tht, ll the resistnces tht re not 2

3 on pth re considered to e +, nd the others re considered to e. Here is picture illustrting this process: Now, this cn lso e viewed s(p, r), where p is the numer of pths nd r is the resistnce of ech pth: Now, since the resistnce of ech pth etween dotted lines grows t rte of 2 i nd the numer of prllel pths etween dotted lines lso grows t rte of 2 i, these fctors lnce ech other nd therefore we would hve R eff But we re relly in 3 dimensions, therefore we get tht the numer of pths grows fster thn 2 i, infct it grows t the rte of 3 i, while the length of ech pth grows t the sme rte: 3

4 R eff ( ) k k=0 3 ( 2 ) 3 R eff Now for P escpe P escpe = C R eff P escpe C P escpe Upper ound on P escpe Now to get P escpe s upper ound we find lower ound on R eff using the sme principles s with the 2D lower ound, ie using cues insted of squres: R eff numerofnodesonsideofcue + R eff ( ) 25 + (( ) ( )) 36 + (( ) ( )) + R eff π2 6 Hence getting to the upper ound on P escpe : 4 Pge Rnk R P escpe R eff P escpe 8 3π 2 We now riefly discuss specil grphs with respect to pge rnk nd leve detiled discussion for the next lecture. 4

5 When crwling the we it is possile to lnd node, or group of nodes, such tht there re no out edges. This presents prolem when ending crwl or in unduly lnding t certin nodes too often for pge rnk to e ccurte. An effective wy round this is to rndomly restrt we crwl, for some reson restrting with proility 0.5 works well. Consider the cse in which we crwl from pge restrts with proility 0.5, leves the pge ltogether with proility nd follows self-loop with proility Allowing tht the pge s rnk is x, the pge rnk will eventully converge to the following formul: x = xx =.74 Even if the person seeks to rtificlly increse their pge rnk y dding n infinite numer of self loops the pge rnk will still converge to every out possiility going to the pge nd with proility 0.5 the we crwl will leve giving: x = xx =