Mixing Times for Random Walks on Geometric Random Graphs

Transcription

1 Mixig Times for Radom Walks o Geometric Radom Graphs Stephe Boyd Arpita Ghosh Balaji Prabhakar Devavrat Shah Iformatio Systems Laboratory, Staford Uiversity Staford, CA {boyd, arpitag, balaji, devavrat}@stafordedu Abstract A geometric radom graph, G d (, r), is formed as follows: place odes uiformly at radom oto the surface of the d-dimesioal uit torus ad coect odes which are withi a distace r of each other The G d (, r) has bee of great iterest due to its success as a model for ad-hoc wireless etworks It is well kow that the coectivity of G d (, r) exhibits a threshold property: there exists a costat α d such that for ay ɛ > 0, for r d < α d (1 ɛ) log / the G d (, r) is ot coected with high probability 1 ad for r d > α d (1 + ɛ) log / the G d (, r) is coected whp I this paper, we study mixig properties of radom walks o G d (, r) for r d () = ω(log /) Specifically, we study the scalig of mixig times of the fastest-mixig reversible radom walk, ad the atural radom walk We fid that the mixig time of both of these radom walks have the same scalig laws ad scale proportioal to r (for all d) These results hold for G d (, r) whe distace is defied usig ay L p orm Though the results of this paper are ot so surprisig, they are otrivial ad require ew methods To obtai the scalig law for the fastest-mixig reversible radom walk, we first explicitly characterize the fastest-mixig reversible radom walk o a regular (grid-type) graph i d dimesios We subsequetly use this to boud the mixig time of the fastest-mixig radom walk o G d (, r) I the course of our aalysis, we obtai a tight relatio betwee the mixig time of the fastest-mixig symmetric radom walk ad the fastest-mixig reversible radom walk with a specified equilibrium distributio o a arbitrary graph Author ames appear i alphabetical order This work is supported i part by a Staford Graduate Fellowship, ad by CS, the MARCO Focus Ceter for Circuit ad System Solutio, uder MARCO cotract 003-CT I this paper, with high probability (whp) meas with probability at least 1 1/ To study the atural radom walk, we first geeralize a method of [DS91] to boud eigevalues based o Poicare s iequality ad the apply it to the G d (, r) graph We ote that the methods utilized i this paper are ovel ad geeral eough to be useful i the cotext of other graphs 1 Itroductio A d-dimesioal geometric radom graph is obtaied by placig odes uiformly at radom o the surface of a d-dimesioal uit torus, ad coectig odes withi Euclidea distace r of each other Such a graph is deoted by G d (, r)[pe03] Geometric radom graphs have bee used successfully i applicatios where the existece of a edge betwee two odes depeds o the distace betwee the odes Classically, they have bee very useful i percolatio, statistical physics, hypothesis testig, cluster aalysis, etc [Pe03] More recetly, the G (, r) graph has bee used to model the etwork coectivity graph for wireless ad-hoc etworks ad sesor etworks [GK00] I this paper, we study the mixig times of radom walks o G d (, r) I additio to beig of theoretical iterest, the mixig time o a graph is directly related to the covergece time of iterative averagig algorithms o that graph, as show recetly i [BGPS04] I particular, the mixig time of a radom walk o G d (, r) is coected to the averagig time o a wireless sesor etwork (modeled as G d (, r)) This strogly motivates the study of mixig times of radom walks o G d (, r) As oted i [BGPS04], the atural radom walk correspods to a very simple distributed averagig algorithm The goal is to compare the performace of this algorithm with the optimal averagig algorithm This is equivalet to comparig the mixig time of the atural radom walk ad the fastest-mixig reversible radom walk with a uiform statioary distributio Thus,

2 i this paper, we study the mixig time of the atural radom walk ad fastest-mixig reversible radom walk with uiform statioary distributio For completeess, we first state the followig defiitio: Defiitio 1 (Reversible radom walk) A radom walk o a coected graph with odes is defied by the trasitio matrix P = [P ij ], where P ij is the probability of goig from ode i to ode j Let Π = [Π i ] be the statioary distributio, that is, ΠP = Π The radom walk is called reversible iff Π i P ij = P ji Π j for all i, j Defiitio (Mixig time) The mixig time of such a radom walk is defied as follows: for ay ode i defie i (t) = 1 j=1 P t ij Π j The, the mixig time is (11) T mix (ɛ) = sup if{t : i (t ) ɛ for all t t} i It is well-kow that the mixig time of a reversible radom walk is related to the secod largest eigevalue i absolute value of P ; the followig result quatifies this relatioship (see survey [Gur00]) Lemma 11 The mixig time of a radom walk with trasitio matrix P is bouded as follows: (1) λ max (P ) log(ɛ) 1 (1 λ max (P )) T mix (ɛ) log + log ɛ 1 1 λ max (P ), where λ max (P ) is defied as follows: let 1 = λ 1 (P ) λ (P ) 1 be ordered eigevalues of reversible matrix P, the λ max (P ) = max{λ (P ), λ (P )} Later i the paper, we will sometimes refer to λ max as the mixig rate of the radom walk ad 1 λ max (P ) as the spectral gap of P The fastest-mixig or optimal (reversible) radom walk for a give statioary distributio is the oe which has the smallest λ max of all reversible radom walks with that statioary distributio [BDX04] Throughout this paper, whe we state results for the mixig time T mix, we will mea T mix (ɛ) for ɛ = 1/ α, α > 0 I order to discuss mixig time of a radom walk with uique statioary distributio, it is ecessary that the graph G d (, r) be coected The followig is a wellkow result about coectivity of G d (, r) [GK00]: For clarity we have stated the result for a uiform statioary distributio; sice we are workig with the order otatio, the results do ot chage for the statioary distributios we ecouter i this paper Theorem 11 Let r c (d) be such that r c (d) d = 4 log The G d (, r) is coected with high probability if r r c (d), ad it is ot coected with positive probability if r = o(r c (d)) For coected G d (, r), for r = Θ(r c (d)), the graph does ot have a regular structure, makig it hard to study the mixig properties of radom walks Hece, we study mixig times of radom walks for r = ω(r c (d)); for such r, the structure of the graph becomes regular, as show i Lemma 1 Throughout this paper, we oly are iterested i radom walks which are reversible 3, ad have uiform statioary distributio Also, to remove edge effects, we cosider G d (, r) o the d- dimesioal uit torus The results of this paper also hold if G d (, r) is defied o a square or a sphere The followig is the mai result of this paper: Theorem 1 For G d (, r) with r = ω(r c (d)), with high probability, (a) the mixig time of the fastest mixig reversible radom walk with uiform statioary distributio is Θ ( r log ), ad (b) the mixig time of the modified atural radom walk, where a ode jumps to ay of its eighbors (other tha itself) with equal probability, ad has a self loop of probability 1/, is also Θ ( r log ) The rest of the paper is orgaized as follows: I Sectio we obtai the regularity property of G d (, r) for r = ω(r c (d)) (Lemma 1) We prove Theorem 1(a) i Sectio 3 ad Theorem 1(b) i Sectio 4 Withi Sectio 3, we first prove the results for d = 1 ad the exted them to d To prove Theorem 1(a), give Theorem 1(b), we oly eed to show a lower boud of Ω ( r log ) 11 Related Work There is a large body of literature o properties of radom geometric graphs Coectivity properties for the G(, r) were derived i [GK00], where the G (, r) graph was used to model wireless ad-hoc etworks as well as sesor etworks The recet book by Perose [Pe03] cotais comprehesive discussios about coectivity, vertex degree distributios, percolatio, ad may other graph properties for the G(, r) graph Recetly, sharp thresholds for mootoe properties of G(, r) graphs have bee obtaied i 3 Let P = [P ij ] be the trasitio matrix correspodig to a radom walk ad Π = [Π i ] be the correspodig statioary distributio The P is reversible iff Π i P ij = Π j P ji, i, j

3 [GRK04] For some results o coverig algorithms for these graphs, see [BBFM03] Aother popular kid of radom graph is the Beroulli radom graph G(, p), which is a graph o odes formed by placig a edge betwee two odes idepedetly with probability p Such graphs have bee studied extesively; [Bol01] cotais a comprehesive treatmet of this subject Regularity of G d (, r) I this sectio, we state a (possibly kow) relatively straightforward regularity property of G d (, r), which makes the aalysis of the mixig time of radom walks tractable Lemma 1 For G d (, r) with r = ω(r c (d)), the degree of every ode is α d r d (1 + o(1)) whp, where α d = π d/ Γ(1+d/) Proof Let odes be umbered i = 1,, Cosider a particular ode, say 1 Let radom variable X j be 1 if ode j is withi distace r of ode 1 ad 0 otherwise The X j s are IID Beroulli with probability p d = α d r d of success (the volume of a d dimesioal sphere with radius r is α d r d ) The degree of ode 1 is (3) d 1 = X j j= By applicatio of the Cheroff boud we obtai : (4) P ( j= X j ( 1)p d δ( 1)p d ) exp( δ ( 1)p d ) If we choose δ = 8 log p d ( 1), the the right-had side i (4) becomes exp( 4 log ) = / 4 So, for p d = ω(log /)), ode 1 has degree (5) d 1 = ( 1)p d ± 8( 1)p d log p d (1 ± o(1)), wp 1 4 Usig the uio boud, we see that P (ay ode has degree p d (1 ± o(1))) 4 = 3 (6) So for large, whp, all odes i the G d (, r) have degree p d (1 ± o(1)) 3 Fastest mixig radom walk o G d (, r) I this sectio, we characterize the scalig of the fastest mixig radom walk o G d (, r) with uiform statioary distributio We first cosider the case of d = 1, ie G 1 (, r) This is much easier tha the higher dimesioal G d (, r) with d We completely characterize G 1 (, r) with the help of oe-dimesioal regular graphs For G d (, r) with d, we obtai a lower boud o the fastest-mixig reversible radom walk Note that sice we are iterested i reversible radom walks with uiform statioary distributio, the trasitio matrix correspodig to the radom walk must be symmetric (The upper boud of the same order is implied by the atural radom walk as i Theorem (b)) The remaider of the sectio is a proof of Theorem (a) 31 Fastest mixig radom walk o G 1 (, r) Let G k deote the regular graph o odes with every ode of degree k: place the odes o the circumferece of a circle, ad coect every ode to k eighbors o the left, ad k o the right From the regularity lemma, we have that whp, every ode i G 1 (, r) has degree r(1±o(1)) Also, observe that the same techique ca be used to show that whp the umber of eighbors to the right (ditto left) is r(1 ± o(1)) Hece, whpthe G 1 (, r) is a subgraph of G k for k = 4r, sice for ay mappig of the odes of G 1 (, r) to G k, a edge betwee odes i ad j i G 1 (, r) is also preset i G k Similarly, G 1 (, r) also cotais G l, for l = (1/)r Give this, we ca ow study the problem of fidig the optimal radom walk o G k with uiform statioary distributio We have the followig lemma: Lemma 31 For k, such that k /4, the mixig rate of the fastest-mixig symmetric radom walk o G k caot be smaller tha cos(πk/) Proof It ca be show usig symmetry argumets [PXBD03] that the fastest mixig radom walk o G k with uiform statioary distributio will have a symmetric ad circulat trasitio matrix (For this simple graph, this ca be easily see usig covexity of the secod eigevalue) So we ca restrict our attetio to the (circulat symmetric) trasitio matrices (37) P = p 0 p 1 p k p k p p 1 p 1 p 0 p 1 p k p k p p 1 p k p k p p 1 p 0

4 The eigevalues of this matrix are µ m = p j e πijm/ + j=0 = p 0 + p j e πi( j)m/) j=1 p j cos(πjm/), m = 0,, 1 j=1 For m = 0, µ m = 1, which is the largest eigevalue Let p = (p 0, p 1,, p k, p k,, p 1 ) We are iterested i the smallest possible secod largest eigevalue i absolute value, ie, (38) mi p max m={1,, 1} µ m subject to 1 T p = 1, p 0 We ca obtai a lower boud for the optimal value of (38) Now, µ max m={1,, 1} (µ m ) (39) mi p µ mi p max m={1,, 1} (µ m ) The right had side is the solutio of the followig liear program with a sigle total sum costrait: (310) mi p p 0 + k j=1 p j cos(πj/) st 1 T p = 1 p 0 For k such that each of the coefficiets cos(πj/) is positive, ie, for k /4, the smallest coefficiet is cos(πk/), ad so for all such k ad, the miimum value is cos(πk/), obtaied at p k = 1/, p j = 0 for all other j 4 So the fastest mixig radom walk o this graph caot have a mixig rate smaller tha cos(πk/) The above result was proved for all k /4; however, we will be iterested oly i those cases where k = o(), ie, the graph is ot too well coected For such k, the followig lemma allows us to fid a early optimal trasitio matrix: Lemma 3 For k = o(), there is a radom walk o G k for which the mixig rate is λ max = cos(πk/) + Θ(k 4 / 4 ) The proof of the lemma is icluded i the Appedix 4 Note that this is oly a lower boud: for this p, if k divides, the secod largest eigevalue is also 1, attaied at m = /k 3 Fastest mixig radom walk o G (, r) We preset the lower boud o the fastest-mixig reversible radom walk o G (, r) i this sectio The same method ca be easily exteded to d 3 First we characterize the fastest-mixig reversible radom walk o a two-dimesioal regular graph, G kk, defied as follows: form a lattice o the uit torus, where lattice poits are located at (i/, j/ ), / i, j + /, ad place the odes at these poits A edge betwee two vertices exists if the L distace betwee them is at most k/ For such G kk the fastest-mixig time scales as follows: Lemma 33 The mixig rate of the fastest-mixig reversible radom walk o G kk is o smaller tha cos (πk/ ), that is, the mixig time of the fastestmixig radom walk is such that T mix = Ω( log /k ) Proof As i the oe-dimesioal case, by symmetry, the optimal trasitio probability betwee odes i ad j will deped oly o the distace betwee these odes Usig this, we ca write the trasitio matrix correspodig to such a symmetric radom walk o G kk as the Kroecker (or tesor) product P k P k, where P k R is as i (37) This is ot difficult to visualize: for i, j = 0,, 1, a, b = 1,,, (311) (P P ) i+a,j+b = P i+1,j+1 P ab Now the eigevalues of A B are all products of eigevalues of A ad B, so that for 0 i, j 1, λ ij (P P ) = λ i (P )λ j (P ) = (p 0 + p m cos(π im ) (p 0 + m=1 m=1 p m cos(π jm ) The eigevalue 1 is obtaied by settig i = j = 0; all other eigevalues will have absolute value less equal 1 We wat to fid a lower boud for the secod largest eigevalue i absolute value, call it λ max As before, choose i = j = 1 The λ 11 max 0 λ ij mi λ 11 mi p p max λ ij, 0 so that mi p λ 11 is a lower boud for λ max Makig the assumptio agai that k /4, the miimizig

5 p is the oe with p k = 1/ ad p i = 0, i k (which correspods to trasitio probabilities of 1/4 for each of the 4 farthest diagoal odes, ad 0 everywhere else) The value of λ 11 correspodig to this distributio is cos (π k ) This is of order 1 Θ( k ), sice cos (π k ) = πk cos( ) = 1 Θ( k ) Thus, (1 λ max ) = O(k /) Hece by Lemma 11, the correspodig mixig time T mix = Ω( log /k ) 5 The G kk graph was costructed usig the L distace betwee vertices Therefore, the graph formed by placig edges betwee vertices based o distace measured i ay L p orm (for the same k) is a subgraph of G kk, ad has a mixig time lower bouded by the mixig time of G kk Thus our bouds will be valid for the G(, r) graph costructed accordig to ay L p orm Now we ll use the boud o the fastest mixig walk o G 11 to obtai a boud for G (, r) First we create a ew graph G (, r) as follows: place a square grid with squares of side r o the uit torus Usig argumets similar to that used i the proof of Lemma 1, each square of area r cotais r (1 + o(1)) odes for r = ω(r c (d)) For each of these r squares do the followig: coect every ode i a square to all the odes i the eighborig 8 squares, as well as the odes i the same square Thus, each ode is coected to 9r (1 + o(1)) odes i G (, r) By defiitio, all edges i G (, r) are preset i G (, r) ad therefore, the fastest-mixig radom walk o G (, r) is at least as fast as that of G (, r) Thus, lower-boudig the fastest-mixig radom walk o G (, r) is sufficiet to obtai lower boud o the fastest mixig radom walk o G (, r) Now, costruct a graph G of r odes as follows: for each square i the square grid used i G (, r), create a ode i G Thus, G has r odes Two odes are coected i G if the correspodig squares i the grid are adjacet Thus, each ode is coected to 8 other odes Thus, G is a regular graph G 11 with r odes I order to use the lower boud o the fastest mixig radom walk o G 11 of r odes (ieg) as a lower boud o G (, r), we eed to show that the fastest-mixig symmetric radom walk o G (, r) iduces a time-homogeeous reversible radom walk o G This will be implied by the followig Lemma Lemma 34 There exists a fastest-mixig symmetric radom walk o G (, r), whose trasitio matrix P 5 It is easy to see that a result similar to Lemma 3 ca be obtaied for d usig the same method has the followig property: for ay two odes i ad j belogig to the same square, P ik = P jk for k i, j, ad P ii = P jj Proof We prove this by cotradictio Suppose the claimed statemet is ot true, ie, there is o trasitio matrix achievig the smallest λ max with the above property Sice the optimal value of λ max must be attaied ([BDX04]), cosider such a optimizig P 1, ad let i ad j be two odes i the same square for which the above property is ot true Let A be the permutatio matrix with A ij = A ji = 1, A ii = A jj = 0, ad all other diagoal etries 1 ad all other o-diagoal etries 0 Note that A is a symmetric permutatio matrix, ad therefore A 1 = A T = A Cosider the matrix P = AP 1 A; sice A = A 1, P 1 ad P are similar, ad so have the same eigevalues Note that sice i ad j belog to the same square i G, they have exactly the same eighbors, ad therefore P also respects the graph structure (ie, P ab 0 oly if a ad b have a edge betwee them) Now, λ max (P ) is a covex fuctio of P for symmetric stochastic P ([BDX04]), so λ max ( P 1 + P ) 1 λ max(p 1 ) + 1 λ max(p ) = λ max (P 1 ) (31) But P = (P + P 1 )/ has the property claimed i the lemma for odes i ad j: P ik = P jk for all k i, j, P ii = P jj = (P 1ii + P 1jj )/, ad λ max (P ) λ max (P 1 ) We ca apply the above procedure recursively (eve for multiple rows) to costruct a matrix P with smallest λ max ad the property claimed i the Lemma This cotradicts our assumptio ad completes the proof From Lemma 34, we see that uder the fastest mixig radom walk, the probability of trasitig from a ode i a square, say S 1, to some eighborig square, say S, is the same for all odes i S 1 ad S Thus, essetially we ca view the radom walk evolvig over squares That is, the fastest radom walk o G (, r) iduces a radom walk o the graph G By defiitio of mixig time, the mixig time for this iduced radom walk o G (with iduced equilibrium distributio) certaily lower bouds the mixig time for the radom walk o G (, r) Further, the iduced radom walk is reversible as the radom walk was symmetric o G (, r) Therefore, we obtai that the lower boud o mixig time for the fastest-mixig radom walk o G implies a lower boud o the mixig time for the fastest-mixig radom walk o G (, r) From Lemma 33 we have a lower boud of Ω(r log ) o the mixig time of the fastest-mixig symmetric radom walk (ie with uiform statioary distributio) From Lemma 35 give

6 below, this i tur implies lower boud of Ω(r log ) o mixig time of the fastest mixig reversible radom walk o G (, r) This completes the proof of (a) for G (, r) It is easy to see that the argumets preseted above ca be readily exteded to the case of d 3 Lemma 35 Cosider a coected graph G = ({1,, }, E) Let Tmix (π) be the mixig time (with ɛ = 1/ α for some α > 0 as i defiitio 1) of the fastest mixig reversible radom walk o G with π(i) statioary distributio π Let β(π) = max π(j) C, where C is a costat The, ( (313) Tmix(π) = Ω Tmix ( )) 1 1, ie, the fastest mixig time for π is o faster tha that of the uiform distributio Proof Cosider a reversible radom walk with statioary distributio π o G ad let its trasitio matrix be R We will prove the followig claim, which i tur implies the statemet of the Lemma Claim I There exists a symmetric radom walk o graph G with trasitio matrix S such that T mix (S) = O(T mix (R)) Proof of Claim I For a reversible matrix R, by defiitio, π(i)r(i, j) = π(j)r(j, i), i, j Defie matrix P = [P (i, j)], where for i j, { R(i, j) if π(i) π(j) P (i, j) = π(i) π(j) R(j, i) if π(i) < π(j) ad P (i, i) = 1 j i P (i, j) By defiitio ad reversibility of R, P is a symmetric doubly stochastic matrix Further, for i j, P (i, j) > 0 if ad oly if R(i, j) > 0 Hece, P ca be viewed as a trasitio matrix of a symmetric radom walk o G, whose statioary distributio is uiform Defie Q R = [Q R (i, j)], where Q R (i, j) = π(i)r(i, j) = π(j)r(j, i) Similarly, defie Q P = 1 P Let φ : {1,, } R be a o-costat fuctio Defie two quadratic forms, E R ad E P, of φ, as E R (φ, φ) = 1 (φ(i) φ(j)) Q R (i, j); E P (φ, φ) = 1 (φ(i) φ(j)) Q P (i, j) Let the variace of φ with respect to two differet radom walks be V R (φ) = 1 (φ(i) φ(j)) π(i)π(j); V P (φ) = 1 (φ(i) φ(j)) 1 Let λ (P ) ad λ (R) deote the secod largest eigevalue of matrices P ad R respectively The miimax characterizatio of eigevalues ([HJ85], page 176), gives a boud o the secod largest eigevalue of a reversible matrix X(= P, R) as (314) (1 λ (X)) = if{ E X (φ, φ) V X (φ) : φ o-costat} For ay π, i π(i) = 1, hece max i π(i) 1/ ad mi j π(j) 1/ Further, by the property of π, < C Hece, for ay k, max π(i) π(j) = maxi π(i) mi j π(j) π(k) mi i π(i) max i π(i) C 1 C, π(k) max π(i) C mi π(j) C i j Thus, for ay k, This implies that E R (φ) E P (φ) ( 1 C, C ( ) π(k) 1 1/ C, C ) ; V R ( ) (φ) 1 V P (φ) C, C similarly, Hece, from (314) we obtai (1 λ (P )) = Θ(1 λ (R)) Now, we are cosiderig T mix (ɛ) for ɛ = 1/ α, α > 0 Hece, from Lemma 11, ( T mix (R) = Θ log 1 λ max (R) ) By defiitio, (1 λ max (R)) (1 λ (R)) Hece, ( ) log T mix (R) = Ω 1 λ (R) It is easy to see that radom walk o G with symmetric trasitio matrix S = (I + P )/ has mixig time give by ( ) log T mix (S) = Θ 1 λ (P ) Thus, T mix (S) = O(T mix (R)) This completes the proof of Claim I ad the proof of Lemma 35

7 Remark: I fact, a stroger result ca be proved, which is ( ( )) 1 Tmix(π) = Θ Tmix 1 Oe part of this has already bee proved i the Lemma The reverse directio is obtaied similarly, as follows Cosider ay symmetric radom walk with trasitio matrix P, ad suppose a statioary distributio π is C, where C is some costat The there is a reversible radom walk R with statioary distributio π, such that T mix ( R) = O(T mix (P )) R is obtaied as follows Costruct a matrix R from P as: specified, satisfyig β(π) = max π(i) π(j) R(i, j) = { P (i, j) if π(i) π(j) π(j) π(i) P (i, j) if π(i) > π(j), for i j, ad R ii = 1 j i R ij R is a stochastic reversible matrix, with statioary distributio π, sice π(i)r(i, j) = π(j)r(j, i) Followig the same steps as above, we ca coclude that 1 λ (R) = Θ(1 λ (P )) The matrix R = (I + R)/ has the same statioary distributio π ad the secod largest eigevalue is (1 + λ (R))/ Therefore, usig Lemma 11, ( ) T mix ( R) log = Θ 1 λ (R) As before, ( ) log T mix (P ) = Θ 1 λ max (P ) ( ) log = Ω 1 λ (P ) Therefore, T mix ( R) = O(T mix (P )), ad we have the stroger result as claimed i the Remark 4 Natural radom walk o G d (, r) I this sectio, we study the mixig properties of the atural radom walk o G d (, r) Recall that uder the atural radom walk, the ext ode is equally likely to be ay of the eighborig odes It is well kow that uder the statioary distributio, the probability of the walk beig at ode i is proportioal to the degree of ode i By Lemma 1, all odes have almost equal degree Hece the statioary distributio is almost uiform (it is uiform asymptotically) The rest of this sectio is the proof of Theorem 1(b) Cosider a symmetric radom walk with trasitio matrix P as follows: let d be the maximum degree of ay ode i G d (, r), the 1/d if i j are coected P ij = 0 if i j are ot coected 1 j i P ij if i = j By defiitio, P is a doubly stochastic symmetric matrix o G d (, r) Further, d = α d r d (1 + o(1)) whp It will be clear to the reader at the ed of this sectio, that usig the proof techique of this sectio, it follows that T mix (P ) = O(r log ) This will imply the upper boud for the proof of Theorem 1(a) 41 Proof of Theorem 1(b): We use a modificatio of a method developed by Diacois-Stroock [DS91] to obtai bouds o the secod largest eigevalue usig the geometry of the G d (, r) Note that for d = 1, the proof is rather straightforward The difficulty arises i the case of d For ease of expositio i the rest of the sectio, we cosider d = Exactly the same argumet ca be used for d > We begi with some iitial setup ad otatio Square Grid: Divide the uit torus ito a square grid where each square is of area r /16, ie of side legth r/4 Cosider a ode i a square By defiitio of G (, r), this ode is coected to all odes i the same square ad all 8 eighborig squares Paths ad Distributio: A path betwee two odes i ad j, deoted by γ ij, is a sequece of odes (i, v 1,, v l 1, j), l 1, such that (i, v 1 ),, (v l 1, j) are edges i G (, r) Let γ = (γ ij ) 1 i j deote a collectio of paths for all ( ) ode pairs Let Γ be the collectio of all possible γ Cosider the probability distributio iduced o Γ by selectig paths betwee all ode-pairs i the followig maer: Paths are chose idepedetly for differet ode pairs Cosider a particular ode pair (i, j) We select γ ij as follows: let i belog to square C 0 ad j belog to square C l If C 0 = C l or i ad j are i the eighborig cells the the path betwee i ad j is (i, j) Else, let C 1,, C l 1, l be other squares lyig o the straight lie joiig i ad j Select a ode v k C k, k = 1,, l 1 uiformly at radom The the path betwee i ad j is (i, v 1,, v l 1, j)

8 Uder the above setup, we claim the followig lemma: Lemma 41 Uder the probability distributio o Γ as described above, the average umber of paths passig through a edge is O(1/r 3 ) whp, where r = ω(r c (d)) Proof We will compute the average umber of paths passig through each square i the order otatio Similar to the argumets of Lemma 1, it ca be show that each of the 16/r squares cotais r (1+o(1)) 16 odes ad each ode has degree πr (1 + o(1)) whp We restrict our cosideratio to such istaces of G (, r) Now the total umber of paths are Θ( ) sice there are ( ) ode pairs Each path cotais O(1/r) edges, as O(1/r) squares ca be lyig o a straight lie joiig two odes The total umber of squares is Θ(1/r ) Hece, by symmetry ad regularity, the umber of paths passig through each square is Θ( r) Cosider a particular square C For C, at least 1 Θ(r )( 1) fractio of paths passig through it have edpoits lyig i squares other tha C That is, most of the paths passig through C have C as a itermediate square, ad ot a origiatig square Such paths are equally likely to select ay of the odes i C Hece the average umber of paths cotaiig a ode, say 1, i C, is Θ( r/r ) = Θ(/r) The umber of edges betwee 1 ad eighborig squares is Θ(r ) By symmetry, the average load o a edge icidet o 1 will be Θ(1/r 3 ) This is true for all odes Hece, the average load o a edge is O(1/r 3 ) Next we will use this setup ad Lemma 41 to obtai a boud o the secod largest eigevalue usig a modified versio of Poicare s iequality stated below Lemma 4 Cosider a atural radom walk o a graph G = ({1,, }, E) with Γ as the set of all possible paths o all ode pairs Let γ be the maximum path legth (amog all paths ad over all ode pairs), d be the maximum ode degree, ad E be total umber of edges Let, accordig to some probability distributio o Γ, the maximum average load o ay of the edges be b, ie o average o edge belogs to more tha b paths The, the secod largest eigevalue, λ, is bouded above as (415) ( ) E λ 1 d γ b The proof of this Lemma ca be foud i the Appedix From Lemmas 1, 41, 4 ad the fact that all paths are of legth at most Θ(1/r), we obtai that the secod largest eigevalue correspodig to the atural radom walk o G (, r) is bouded above as: ( r ) λ 1 Θ r 4 r 4 (416) = 1 Θ(r ) We would like to ote that, for mixig time, we eed to show that the smallest eigevalue (which ca be egative), is also Θ(r ) away from 1 Oe wellkow way to avoid this difficulty is the followig: modify trasitio probabilities as Q = 1 (I + P ) Q ad P have the same statioary distributio By defiitio, Q has all o-egative eigevalues, ad λ (Q) = 1 (1 + λ (P )) Thus, the mixig time of the radom walk correspodig to Q is govered by λ (P ), ad is therefore Θ(r log ) This radom walk Q is the modified atural radom walk i Theorem 1(b) Thus, from Lemma 11 ad (416), the proof of Theorem 1(b) for G (, r) follows I geeral, the above argumet ca be carried out similarly for d > completig the proof of Theorem 1(b) 5 Coclusio We studied the scalig of mixig times for the fastest mixig reversible radom walk ad modified atural radom walks for G d (, r) We foud that both of them have mixig time of the same order, Θ(r log ) for all d I fact these scalig results apply ot just for G d (, r) costructed with the Euclidea orm, but for ay L p orm The methods used i this paper to compute mixig times are ovel ad we strogly believe that they will be useful i other cotexts Refereces [BBFM03] L Booth, J Bruck, M Fraceschetti, ad R Meester Coverig algorithms, cotiuum percolatio, ad the geometry of wireless etworks The Aals of Applied Probability, 13():7 41, 003 [BDX04] S Boyd, P Diacois, ad L Xiao Fastest mixig Markov chai o a graph To appear i SIAM Review, problems ad techiques sectio, 004 Available at wwwstafordedu/~boyd/fmmchtml [BGPS04] S Boyd, A Ghosh, B Prabhakar, ad D Shah Gossip algorithms: Desig, aalysis ad applicatios 004 Submitted, available at wwwstafordedu/~devavrat/ifocom05ps [Bol01] B Bollobas Radom Graphs Cambridge Uiversity Press, 001

9 [DS91] P Diacois ad D Stroock Geometric bouds for eigevalues of Markov chais The Aals of Applied Probability, 1(1):36 61, 1991 [GK00] P Gupta ad P R Kumar The capacity of wireless etworks IEEE Trasactios o Iformatio Theory, 46(): , March 000 [GRK04] A Goel, S Rai, ad B Krishamachari Sharp threshold for mootoe properties i geometric radom graphs I Proc Coferece o Symposium o Theory of Computatio ACM, 004 [Gur00] V Guruswami Rapidly mixig markov chais: A compariso of techiques May 000 Available at cswashigtoedu/homes/vekat/ pubs/papershtml [HJ85] R Hor ad C Johso Matrix Aalysis Cambridge Uiversity Press, Cambridge, UK, 1985 [Pe03] M Perose Radom geometric graphs Oxford studies i probability Oxford Uiversity Press, Oxford, 003 [PXBD03] P A Parrilo, L Xiao, S Boyd, ad P Diacois Symmetry aalysis of reversible markov chais Techical report No , Departmet of Statistics, Staford Uiversity, Appedix Here we preset the proofs of Lemma 3 ad 4 Lemma 61 For k = o(), there is a radom walk o G k for which the mixig rate is λ max = cos(πk/) + Θ(k 4 / 4 ) Proof For simplicity let us assume that k divides ; it is ot difficult to obtai the same results whe this is ot the case Cosider the Markov chai with trasitio probabilities p 0 = 0, p i = δ, i = 1,, k 1, p k = 1/ (k 1)δ We will show that for a certai δ, small eough, µ 1 is ideed λ max, ad is away from cos(πk/) by Θ(k 4 / 4 ) For the trasitio matrix P correspodig to these probabilities, the eigevalues are, for m = 0,, 1, µ m = k 1 = cos( πkm δ cos( πim ) + δ k 1 k 1 ) + ( 1 πkm (k 1)δ) cos( ) (63) cos(πim/) = cos(πiq/k) 1 = 1 πim πkm (cos( ) cos( )) sice k cos(πiq/k) = 0 (sum of real parts of the kth roots of uity) (617) We wat to fid the smallest positive δ such that µ 1 is λ max (this is ot true, for example, for δ = 0) However, we eed δ to be small eough so that the residual term, δ k 1 (cos(πi/) cos(πk/)), is small compared to cos(πk/) Sice k = o() ad we hope that δ is small (o(1)), we see that the values of m for which µ m is comparable to µ 1 are those values of m for which cos(πkm/) = 1 This happes for m = k, k, 3 k,, (We oly eed cosider values of m util /, sice λ i = λ i ) At all odd multiples of /k, cos(πkm/) = 1, ad for the eve multiples, cos(πkm/) = 1 For δ to satisfy µ m µ 1, we must have for m a eve multiple of /k, (618) 1 + δ k 1 cos( πk πim (cos( ) 1) ) + δ k 1 ad for m a odd multiple of /k 1 + δ k 1 (619) 1 δ k 1 (cos( πim δ k 1 (cos( πim δ k 1 (cos πi cos πk ); πk ) + 1) cos( ) + πk cos ), πk ) + 1) cos( ) + πk cos ) (cos πi (cos πi From (618), we see that δ must satisfy 1 πk (1 cos( ) πi cos( δ (k 1)(1 cos( πk )) + k 1 (60) for m a odd multiple of /k, ad from (619), 1 πk (1 cos( ) πi cos( ) + cos( πim ) δ (k 1)(1 cos( πk )) + k 1 πim ) cos( )) (61) for m a multiple of /k So δ ca be oly as small as the maximum over the specified m of all of these right-had sides Note that the oly term depedet o m i each of these expressios is k 1 cos(πim/) For m = p/k, p odd, (6) k 1 k 1 cos(πim/) = cos(πip/k) = 0, sice cos(πip/k) = cos(π(k i)p/k) for odd p, ad if k is eve, cos(πkp/k) = 0 also For m = q/k, So δ = Θ(k/ ), ad returig to (617), we see that the residual term i µ 1 is of order (k/ )(k 3 / ), ie, k 4 / 4, while cos(πk/) 1 π k / So the differece betwee λ max ad cos(πk/) is Θ(k 4 / 4 ) Lemma 6 Cosider a atural radom walk o a graph G = ({1,, }, E) with Γ as the set of all

10 possible paths o all ode pairs Let γ be the maximum path legth (amog all paths ad over all ode pairs), d be the maximum ode degree, ad E be total umber of edges Let, accordig to some probability distributio o Γ, the maximum average load o ay of the edges be b, ie o average o edge belogs to more tha b paths The, the secod largest eigevalue, λ, is bouded above as (64) ( ) E λ 1 d γ b Proof The proof follows from a modificatio of Poicare s iequality (Propositio 1 [DS91]) Before proceedig to the proof, we itroduce some otatio Let φ : {1,, } R be a real valued fuctio o the odes Let π = (π(i)) {1 i } deote the equilibrium distributio of the radom walk Let d i be the degree of ode i, the it is well kow that π(i) = For ode pair (i, j), let di E d E Q(i, j) = π(i)p ij = π(j)p ji = 1/ E Defie the quadratic form of fuctio φ as E(φ, φ) = 1 (φ(i) φ(j)) Q(i, j) Let the variace of φ with respect to π be V (φ) = 1 (φ(i) φ(j)) π(i)π(j) For a directed edge e from i j, defie φ(e) = φ(i) φ(j) ad Q(e) = Q(i, j) First, cosider oe collectio of path γ = (γ ij ) Defie γ ij Q = e γ ij Q(e) 1 The, uder the atural radom walk, (66) (c) = (d) ( ) d 1 Q(e)φ(e) γ ij Q E e γ ij e ( ) d 1 Q(e)φ(e) γ ij E e γ ij e ( ) d 1 γ Q(e)φ(e) b(γ, e), E e where b(γ, e) deotes the umber of paths passig through edge e uder γ = (γ ij ) (a) follows by usig π(i) 1/ for all i ad addig as well as subtractig values of φ o the odes of path γ ij for all ode pair (i, j) for a give path-set γ = (γ ij ) (b) follows by Cauchy- Schwartz iequality (c) follows from (65), ad (d) follows from the fact that all path legth are smaller tha γ Note that i (66), b(γ, e) is the oly path depedet term Hece uder a probability distributio o Γ (ie set of all paths), i (66), b(γ, e) ca be replaced by b(e) where Let b = max e b(e) The, (67) V (φ) ( d E (68) = ( d γ b E b(e) = γ Γ Pr(γ)b(γ, e) ) 1 γ Q(e)φ(e) b, e ) E(φ, φ) The miimax characterizatio of eigevalue ([HJ85], page 176), gives a boud o the secod largest eigevalue as (69) λ = sup{1 E(φ, φ) V (φ) : φ a o-costat} From (68) ad (69), the statemet of the Lemma follows (65) γ ij Q = γ ij ( E ), where γ ij is the legth of the path γ ij V (φ) = 1 (φ(i) φ(j)) π(i)π(j) (a) = 1 ( ) 1/ Q(e) φ(e) π(i)π(j) Q(e) (b) 1 e γ ij γ ij Q π(i)π(j) e γ ij Q(e)φ(e)