Learning Stochastically Evolving Networks via Local Probing

Transcription

1 Learig Stohastially Evolvig Networks via Loal Probig Rajesh Jayaram Advisor: Eli Upfal April 25, 2017 Abstrat We osider the problem of learig the state of dyami etwork with vertex values that are perturbed over time. Our oly iterfae to the etwork omes i the form of poit probes, where we are allowed to query loal iformatio about the state of a speifi poit i the etwork. I this paper, we osider etwork models where the values of the verties are perturbed uiformly at radom at every time step, ad where we may oly query adjaet verties to determie whether or ot they have the same value. This model is draw from umerous pratial examples, where determiig the preise value of a vertex is impossible, but differetiatig betwee adjaet verties with disagreeig values is feasible. Uder this model we osider both the oiseless ad oisy ases, where i the latter our probes are subjet to a uiform oise with probability α. We first derive a lear iverse liear lower boud tradeoff betwee the umber of probes ad the fratio of errors i the etwork for either ase. I the oiseless ase, we desig a algorithm whih radomly iitializes ad the determiistially traverses the etwork to update a hypothesis state. We show that our algorithm is always withi a ostat fator of the lower boud for arbitrarily high polyomially may time steps with high probability. For the oisy ase, the problem beomes substatially more diffiult, ad performae will deped o the expasio of the graph. We show that a algorithm whih allows at least k Ω( log() 1 λ 2 r) probes o every time step ever aumulates more tha O( r ) errors at ay time for arbitrarily high polyomially may time steps, assumig that α 1 Ω(k), ad where (1 λ 2 ) is the spetral gap of the graph. A alterate aalysis shows that for quadratially may time steps we a tighte this boud to O( assumig r(1 α log() (1 λ 2 )) ) α 1 Ω(k 2 ). Furthermore, we demostrate that if the umber of errors aumulated at time t exeeds our bouds by a fator of M, the i expetatio we retur bak withi the boud i at most O( log(m)) steps. Sie our bouds for the oisy ase are ot as tight (we lose at least logarithmi fator i the probe-error tradeoff), we demostrate experimetally that our algorithm i the oisy ase appears to perform just as well as the theoretial performae of our algorithm i the oiseless ase. 1

2 1 Itrodutio The augmetatio of a etwork struture with vertex values, or labels, is a atural ehaemet ad aptures a variety of iterestig of real world pheomeo. For istae, the augmetatio of soial etwork models with ertai biary traits of the members, suh as politial lassifiatio or ay umber of other biary divisors of a target audiee. Additioally otable is the appliatio to omputer visio, where the graph i questio is the grid (or to simplify the aalysis the disrete torus: Z Z ). Here eah vertex represets a pixel i a video image, ad vertex values a the be utilized for foregroud bakgroud segmetatio. Ideed, this settig was osidered expliitly for the stati ase i [3]. Our paper a be see as a dyami versio of [3], where vertex values are perturbed over disretized time steps. This a orrespodigly be see as movig from a stati image to a dyami video feed, where the pixel labelligs, suh as foregroud ad bakgroud, are hagig as the image i the feed hages. A importat harateristi of the algorithmi framework whih we osider i this paper is that eah every time step we are oly allowed to probe a ertai, geerally assumed to be ostat or polylogarithmi, umber of edges i the graph, ad we are the told whether or ot the iidet verties have agreeig labels. This is the same model of edge observatios used i [3]. Our method of trasportig a stati algorithm to a dymi settig bears resembalae to the method of [1], where quiksort is reitrodued o a stohastially evolvig array. Now our model of allowig oly edge observatios ad ot vertex observatios aptures a importat feature of the pratial appliatios from whih our model arises. Namely, it is frequetly impratial to determie, based o lookig at oe pixel (or idividual i a soial etwork) aloe, whether or ot it is i the bakgroud (or of a partiular biary lassifier). However, by lookig at adjaet pixels (or frieds i the etwork), it may be muh easier to determie whether the two verties should have the same lassifiatio. For pixels, a simple method of doig this is by omputig the l 1 distae betwee the RGB values of adjaet pixels, ad determiig that they are alike if the distae falls uder a ertai threshold parameter. Of ourse, ay suh ompariso method will fail with some probability, whih we refer to as oise i the edge probes. I Setio 3 we itrodue a algorithm to produe good results eve with this oise. Differet models for stohastially evolvig etworks have bee osidered before. For istae, i [2], the authors osider a similar otio of a evolvig graph where the atual topology of the graph, verties ad their eighborhoods are perturbed over time. I our paper we assume that the topology of etwork is fixed, ad fous istead o perturbatios i vertex labelligs. However, it would be perhaps iterestig to examie models where both the etwork topology ad vertex labelig are perturbed over time. 1.1 Results & Tehiques The model whih we osider i this paper is as follows: we have a graph with biary vertex labels whih flip i.i.d. at every time step with probability β. Our algorithm is allowed to probe a edge ad give the XOR of the two adjaet verties. The goal is the to maitai a hypothesis vetor of the values of the verties that has lose l 1 distae to atual values at ay time step. This is a streamig variatio of the model osidered i [3]. I this settig, we first derive lower bouds for 2

3 the expeted umber of errors for ay algorithm ruig out algorithm i the model. Theorem (1). I the stohasti etwork model where verties are flipped by ature with probability β, ay algorithm whih uses at most k probes o every time step ever aumulates less tha β 2 4k β2 2k errors at ay time step i expetatio. For pratial purposes it makes sese to osider oly settigs where the expeted umber of flips o every time step is ostat, thus we speify to β = 1 2 i this paper, although our results hold for ay ostat multiple of this probability. Beig the ase, our lower boud demostrates a lower boud of a liear error-probe tradeoff i expetatio. We the itrodue the Trailig Walks Algorithm, whih demostrates that this lower boud a be early met with high probability (ad ot just i expetatio) for polyomially may time steps. Theorem (7). Let ɛ be the umber of iitial errors of our hypothesis at time t = 0, ad fix ay k = k (r + s), ad fix γ > 0, δ 1. The with probability at least (1 e )(1 e δ γ 6sk +log T )(1 3 k ) the (k, r, s)-trailig Walks Algorithm has o more tha 4(log(k )) 2 ɛ + (1 + δ) 2sk errors at ay time at all before t = T where T is as i Lemma 1. γ Remark. Note that T O(( 2 )(r+1)/2 1 k ) We the geeralize to the ase of edge oise. I other words, the ase of the prior model but with the additioal ompliatio that every edge probe is give the iorret XOR value with some fixed uiform probability α. For pratial purposes, we assume that probig a edge multiple times o the same time step will ot yield differet results. Note that if we allowed this, the repeatig the same algorithm from the oiseless setio but probig eah edge logarithmially may more times would yield a obvious algorithm to esure the same bouds with high probability. It turs out, however, that after disallowig this the problem beomes quikly ifeasible for ostat values of α. I order to solve the problem theoretially, our algorithm requires that α is i O( 1 k ). Give that k is geerally assumed to be at most polylogarithmi, this is as lose to ostat as α will pratially be. Give this, our algorithm probes radom paths ad, via a majority vote from our hypothesis i the prior step, deides whih labelig suh a path should reeive. I order for suh a radom path sample to be uiform, whih will be required as we will later see, the miimum legth k of the path must deped o the spetral expasio of the graph. We prove that our algorithm, kow as the Expader Samplig algorithm, performs early as well i the ase that our graph is a good expader ad the edge oise is ot too large. The followig theorem provides our mai error boud for the Expader Samplig Algorithm. Theorem (17). Suppose verties flip i.i.d. with probability 1 2. The give k = k 2560(+q) log() (1 λ 2 ) probes per time step, ad assumig α 1 60(k ) ad k o( log() ), with probability at least (1 6 ), the maximum umber of errors that the Expader Samplig algorithm eouters before time T = q is at most ( 20(+q) log() (1 λ 2 )k + 12αk ) + 2(q + + 1) log()k. Note that M eed be at most O(q log()) for the result to hold with high probability. A alterate aalysis followig the proof of the above theorem gives the followig stroger theorem for 3

4 a smaller time horizo if we have that α 1 Ω(k 2 ), alog with some additioal properties of our algorithm. We first prove a boud o the expeted umber of errors at ay give time step. Theorem (18). Assume α 1 > 2(k ) 2 r. The the Exapder Walks Sample algorithm whih uses at least k 160 log() (1 λ 2 ) probes o ay oe path has expeted umber of errors E[ɛ t] r (1 αk ) 1 for all t T. Ad the the primary theorem: Theorem (21). Let G be ay graph with spetral gap (1 λ 2 ). Fix ay itegers r, 1, ad let k = k r 2560r log() (1 λ 2 ) be the umber of probes we allow our algorithm per time step. Further assume that the edge oise satisfies α 1 > 2(k ) 2 r. The with probability at least (1 1 ), the 2 maximum umber of errors that Expader Samplig algorithm aumulates at ay time ever before ( ) T is less tha log(), where T = ( E[ɛ t] k ) 2 ( ) k 2 2 r(1 αk ) 2 Noiseless Probes 2.1 The model We first itrodue our model, whih a be see as a stremig variatio of the model examied i [3]. Our model however oly allows edge probes, ad ot vertex probes. Furthermore, we geeralize to arbitrary graphs beyod the grid, whih was the graph for whih the mai results of [3] hold. Fix a graph G with verties v 1,..., v. Eah vertex has a value i {0, 1} at every disrete time step t, whih we deote v i,t. Our goal is to maitai a hypothesis of the value of v i,t, whih we will deote by h t (v i ). We assume that at time t = 0 we start with ɛ 0 errors for some fixed 0 ɛ 0 < 1 (our hypothesis is wrog o a ɛ 0 fratio of the verties). The model is the as follows: At time t = 0, there are exatly ɛ 0 verties v i suh that v i,0 h 0 (v i ) has size ɛ 0 At the start of every time step, eah vertex is flipped by ature uiformly ad i.i.d. with probability β, ad otherwise remais uhaged with probability 1 β. At every time step we are allowed to probe k edges, for some fixed k. For eah edge that we probe, we are told whether the verties o the edge agree or disagree. I other words, we are told the xor ( ) of the two verties. The goal is to maitai a hypothesis h t (v i ) of v i,t suh that the l 1 distae h t (G) G t 1 is small. Thus at the start of ay time step t, every vertex is flipped with some probability by ature, after whih the value v i,t is determied for all v i G. The we probe k edges, ad after this we defie out hypothesis h t (v i ). Geerally we assume k to be a ostat, thus it is atural to work with probabilities for ature flippig verties that result i a ostat umber of hages per time step. 4

5 If, o the other had, the expeted umber of hages per time step is o a order larger tha k, the problem learly beomes itratable (this a be see easily from Theorem 1). We begi by provig a geeral lower boud o expeted the umber of errors aumulated by ay algorithm i this model. Theorem 1. I the above model where verties are flipped by ature with probability β, ay algorithm whih uses at most k probes o every time step ever aumulates less tha β 2 4k β2 2k errors at ay time step i expetatio. Proof. We first osider, give ay observatio set of probes, whih output hypothesis for the labels of the verties whih has the lowest expeted error. Now sie whe probig a edge we gai oly iformatio about the two verties i questio, ad sie ature flips verties idepedetly, it follows that oditioig o a edge observatio a oly hage the probabilities of the values of the two verties i questio. So the very best ay hypothetial algorithm ould do is always orretig the value of ay vertex it sees i a edge probe. Now suppose that the probes made by ay k-algorithm are ordered by ay time t. Let S i be the set of verties that are see i the i-th to last step. Note agai that eah probe shows the algorithm at most 2 verties, ay algorithm a see at most 2k verties at a give time step. The the expeted umber of errors is at least : t β S i i Γ i=1 Where Γ is the expeted umber of verties that flip twie i the sets S 1,..., S /(2k) sie the algorithm viewed them last. Regardless of Γ, this sum is miimized by havig S i = 2k for i = 1, 2,..., /(2k). Thus the expeted umber of errors is at least /(2k) 2kβ i Γ = 2kβ i=1 ( /(2k) ( /(2k) + 1) 2 ) Γ β 2 4k Γ Now ote that the probability that a vertex flips twie i 2k steps is less tha β, thus it follows Γ < 2k β2, so β 2 4k Γ β 2 4k β2 2k Corollary 2. If β 1 Θ() the the for ay algorithm a lower boud for the expeted umber of errors at ay time step is Ω(/k). I partiular, if β = 1 2, the expeted umber of errors is at least 8k 1. Proof. Follows from Theorem 1. Remark 1. Followig from Corollary 2, for the rest of this paper we will speify to β = 1 2. The bouds give i the paper a be easily geeralized however to ay other value of β that differs by a ostat fator. We itrodue ow the Trailig k walks algorithm. First we provide some justifiatio for the ostrutio of our algorithm algorithm, alog with some ituitio. 5

6 2.2 Ituitio I this model, observe the followig fat: if we kow that our hypothesis for a vertex v i at time t is orret with ertaity, the if v j is ay eighbor of v i the we a probe the edge (v i, v j ) ad, alog with the iformatio h t (v i ), we a orretly determie the value of v j. Thus if we kow the value of oe vertex with ertaity we a determie the values of all those aroud it with ertaity. Exploitig this fat, we ould take a determiisti walk where we fix the eighbors of every vertex we see o the walk with ertaity. Meaig if C is a hamiltoia yle i G (or a yle that visits all verties that has legth + O(1)), the we a just walk dow the yle i order oe step at a time fixig everythig we see. If we are allowed to probe k edges at a time, we a take k determiisti walks at a time, traversig all the verties of G i the order speified by C ad fixig everythig. There are oly two (fatal) problems: 1. If we start o a vertex v that we are wrog about, the we are i trouble sie we will hage all verties o the path to be iorret. 2. If, i the middle of the step betwee v i ad v j, either v i or v j is flipped the we will also be i trouble. To avoid this we the have to determie whether v i was flipped at every time step first. To do this we may, for istae, probe the vertex before v i to see if it agrees with what we expet. But eve i this ase: 3. Worstly, if both verties are flipped (v i ad the oe before it) o the same time step t, the eve if we hek the last edge we probed before we move o to the ew oe to see if ay hages are made, the we may otiue o makig errors thikig that we are orret ad that othig haged. Note that the probability of ase 1 depeds o how may errors there are whe we start our algorithm, ase 2 ours i expetatio oe every 2 steps of a sigle path, ad ase 3 ours i expetatio oe every 4 2 steps of a sigle path. So we eed to avoid these ases. We solve this problem by keepig trak of a trailig set of "orret" verties whih we have already fixed. At every time step, we probe the edges betwee all these veries ad reassig them the value that makes the most sese give their value i the last step. The, ad oly the, do we probe the ext vertex i the path ad set it s value to agree with the oe we last probed (whih was part of our trailig orret set). O the ext time step we add the ew vertex to the set ad subtrat the oldest oe from this set. 2.3 The k-trailig Walks Algorithm Let v i,t be the atual value of v i at time t, ad let h t (v i ) be our hypothesis. We assume we kow the startig state with probability 1 1/ɛ, thus h 0 (v i ) v i,0 with probability at most ɛ < 1 for all v i G. This meas we start with ɛ errors. Our algorithm is as follows. We fix a yle that goes through all the verties of G that we will traverse. We will later show how to geeralize to graphs whih do ot otai suh a yle (see Theorem 9). As we traverse this yle we probe the edges i it ad fix the verties alog the way to agree both the edge observatios ad our prior hypothesis. This proess was desribed 6

7 earlier. However, we make two ruial hages. Firstly, istead of traversig startig at oe plae, we traverse startig from k plaes o the yle simultaeously. Thus at every time step we move dow the yle at some uiform rate startig from k distit positios. We all eah idepedet traversig proess a thread of the algorithm (i the sese of multithreadig). Seodly, for every thread, we keep trak of a trailig set of the last r verties that that thread saw. Before attemptig to fix the ext vertex i lie for a give thread, that thread will first probe all the edges i its trailig set ad update its hypothesis for the etire trailig set i tadem. Sie probig all edges i a path fores the set of verties to take o oe of two possible assigmets (determied uiquely by whether or ot we delare ay oe vertex to be 1 or 0), we hoose to update our hypothesis to be the assigmet, out of the two, that has the least umber of hages from our prior hypothesis. So let C be a yle suh that every vertex v G appears i C at least oe. Order the verties of the yle v 0, v 2,..., v m 1, v 1... where m = C. For ow we assume m =, ad we will later show how to deal with graphs whih do ot otai a Hamiltoia yle. Our algorithm will be give as iput a parameter k, the umber of threads, positive odd iteger r, whih determies the size of our trailig set, ad a iteger s whih determies how may steps o the path we will take o every step. Thus our algorithm will use a total of k (r + s) probes every time step, where k is the umber of threads, r the size of the trailig set, ad s the umber of steps take forward o every step. First several defiitios useful i desribig our algorithm are give, ad the the algorithm is stated formally below. Defiitio 1. We all the Trailig Walks algorithm that uses k = k (r + s) probes, osistig of k threads eah with a trailig set of size r ad eah whih take s steps o every time step a (k, r, s)-trailig Walks Algorithm. Defiitio 2. We say that a thread T j is good at time t if h t (S j ) is etirely orret ad otais o errors. We say that a thread is bad if it is ot good. We say that a thread flips at time t if it goes from beig good to bad o time t. Defiitio 3. We say that a thread T j is at the positio v i at time t, deoted F t (T j ) = v i, if v i was the last edge o the path probed by Thread T j at time t. 7

8 Algorithm 1 k-trailig Walks (iitializatio t = 1) Iput: (k, r, s), yle C, ad iitial hypothesis h 0 1: Set h 1 h 0 2: for i = 1, 2,..., k do 3: Uiformly sample σ(i) U {0, 1,..., C 1}. 4: ed for 5: for j {1,..., k } do 6: Iitialize S j. 7: Set F 1 (T j ) v σ(j). 8: for i = σ(j), (σ(j) + 1),..., (σ(j) + r + s mod C ) do 9: Probe the edge e (v i, v i+1 ) ad pik x {0, 1} so that h 0 (v i ) x = e. 10: Update h 1 (v i+1 ) x. 11: Set S j S j {v i }. 12: if S j > r + 1 the 13: S j S j \ {v i (r+1) mod C } 14: ed if 15: F 1 (T j ) i 16: ed for 17: ed for Algorithm 2 k-trailig Walks (time t) 1: for j {1,..., k } do 2: Iitialize Sj t St 1 j ad E ad h t+1 h t 3: for (v i, v i+1 ) E(S j ) do 4: Probe the edge e (v i, v i+1 ). 5: Add E E {e}. 6: ed for 7: Let I 1, I 2 be the two possible assigmets to Sj t whih agree with E. 8: Set h t (Sj t) arg mi I {I 0,I 1 } h t 1 (Sj t) I 1. 9: for i = F t (T j ), F t (T j ) + 1,..., F t (T j ) + s mod C do 10: Probe the edge e (v i, v i+1 ) ad pik x {0, 1} so that h t (v i ) x = e. 11: Update h t+1 (v i+1 ) x. 12: 13: Sj t St j {v i}. if Sj t > r + 1 the 14: Sj t St j \ {v i (r+1) mod C } 15: ed if 16: F t (T j ) i 17: ed for 18: ed for 8

9 Remark 2. Note that if oe of our threads starts o a error, the util it is flipped ito a good thread we a assume that it will oly ever reate errors. We all suh a thread a bad thread (a thread that is wrog o every oe of it s trailig set hypothesis). The umber of errors suh a thread will ause is the distae betwee the thread s startig loatio i C ad the ext thread behid it that is orret, sie a orret thread trailig behid a bad thread will orret the errors it made whe it reahes them. Remark 3. I lie 8 of Algorithm 2, we pik the wrog assigmet for the trailig set S j oly whe more tha half of the S j = r verties are flipped by ature i a sigle time step. 2.4 Aalysis of Algorithm 2 Lemma 3. Whe ruig the k-walks algorithm for k = k (r + s), the probability 1 T oe thread goes from beig good to bad at ay time t r is at most 1 T k( 2 ) (r+1)/2 that at least Proof. Ay oe of the k threads goes bad oly if more tha half the verties i the trailig set flip i oe time step. Sie eah trailig set osists of r distit verties whih all flip with idepedet probabilities, the probability that this ours for ay oe thread is give by: r j=(r+1)/2 ( r j ) ( 1 2 ) j(1 1 2 )r j 2 r 1 ( 1 2 )(r+1)/2 ( 2 ) (r+1)/2 Usig the uio boud, the probability that at least oe thread flips is at most k times this above boud, whih gives the desired iequality. Now give that it is ulikely that ay of our threads flip o a give time step, we ow aalyize the probability that we begi o a error ad, if so, how may errors this may ause us. Let B(k, ɛ) be the biomial distributio (this gives the distributio over the umber of startig bad threads). Let E {0, 1, 2,..., C 1} be the subset of error verties with expeted size ɛ. Lemma 5 will demostrate that it suffies to geerate E by addig every vertex i the path to E i.i.d. with probability ɛ. Let Z k,ɛ(e) be a radom variable, as a futio of E, that is sampled i the followig way. 1. First pik k values uiformly from {0, 1, 2,..., C 1}. Call them a 1, a 2,..., a k. 2. Set d i = a (i+1) (mod C ) a i (mod C ) for i = 1, 2,..., k. 3. Defie D = {d i a i+1 E}. 4. Retur Z k,ɛ = d i D d i. Lemma 4. Suppose that before time T, o thread i our algorithm flips. The if E V (G) is the set of bad verties at time 0, the the radom variable Z k,ɛ(e) is a upper boud o the umber of errors that our algorithm makes due to bad threads at ay time t T. 9

10 Proof. If o thread has flipped yet, the we a assume that all threads that were bad iitially stay bad, ad vie-versa for good threads. Sie the oly errors that we otribute are the errors due to bad threads, we first eed oly idetify the distributio over iitial bad threads, ad the determie how may errors a be attributed to eah. A thread is bad if it starts i the set E as defied above, ad sie we hoose the startig verties of eah thread uiformly at radom the the distributio over iitial bad threads is give by steps 1. ad 2. of the algorithm for Z k,ɛ(e). Now for eah bad thread T i, the umber of errors that are due to it at ay time step is at most the umber of verties betwee it ad the ext good thread behid it. This is beause all verties that T i visits will beome errors, but they will oly remai errors util they are fixed by the ext good thread trailig behid T i. Now if the ext thread behid T i, say T j, is also bad, the we oly attribute the errors o the verties betwee T i ad T j to the thread T i. We do this sie the total umber of errors made by a sequee of oseutive bad threads is still the distae betwee the farthest thread ad the first good thread behid it, whih is equivaletly the sum of the distaes betwee eah bad thread ad the thread behid it, whih is preisely how Z k,ɛ(e) is ostruted. Note that sie verties may be repeated, we may overout a vertex that lies several times o the path betwee a good ad bad thread. Thus Z k,ɛ is a strit upper boud i this ase. Lemma 5. The distributio of Z k,ɛ(e) does ot deped o the iitial set E but oly oly the size of E. Proof. This is easy to see, sie the k iitial values were hose uiformly at radom. Lemma 6. With probability o more tha Poly( 1, 1 (k ) ) will we ever have Z k,ɛ > 4(log(k)) 2 ɛ. Furthermore, with probability at least (1 k ɛ) we have Z k,ɛ = 0. Proof. It suffies the to osider E geerated by i.i.d. addig eah vertex ito E with probability ɛ. First we examie the biomial portio of the R.V. to boud the umber of a i s that begi i the set E. This is give by the biomial distributio B(k, ɛ). Usig Cheroff bouds, it follows. [ ] Pr B(k, ɛ) > (λ + 1)k ɛ e λk ɛ 3 Now let a 1,..., a k be the radom positios i {0, 1, 2,..., m 1} that our algorithm hoses, ad radomly mark B(k, ɛ) of them, ad all that set A. We aalyze the sum a i A a i a i 1 (mod m) (mod m). A upper boud a be plaed o this by a balls ad bis argumet. First split the yle C ito k equally sized itervals of legth m k. Let X 1,..., X k be radom variables deotig whih eah iterval the a i s are plaed ito. Let f(x) be the umber of empty bis as a futio of these R.V. s. We ow ostrut the Doob martigale, givig: ] B i = E Xi+1,...,X k [f(x) X 1,..., X i Fat 1. The above ostrutio B 1,..., B k is always a martigale. Clearly B i B i 1 1. Thus, usig this bouded differee property, we a apply Azuma s iequality to yield: 10

11 Pr[f(X 1,..., X k ) E[f(X 1,..., X k ] > ɛ)] e 2ɛ2 k Now ote that the expeted umber of empty bis is E[f] = k (1 1 k ) k whe there are k balls ad bis. Thus with probability at least 1 1 k, we ever have more tha k (1 1 k ) k + 2k log(k ) empty bis. Let E be the umber of empty bis. Now the empty bis are uiformly distributed, so the probability tha ay oe bi is empty is (1 1 k ) k + 2 log(k ) k 1 e + 2 log(k ) k 1 2 for reasoable values of k. Thus the probability that l oseutive bis are empty is at most k 2 l. Thus with probability at least 1 1 k there are o more tha 2 log(k ) oseutive empty bis, ad we have Z k,ɛ (λ + 1)ɛ log(k ) where λ is as before. Now we osider two ases. Case 1: kɛ 1. I this ase set λ = 3 log(k ), we olude with probability at least (1 1 (k ) kɛ )(1 1 k ) 2 (1 3 k ) we have Z k,ɛ log(k)(3 log(k ) + 1)ɛ 4(log(k )) 2 ɛ Case 2: kɛ << 1, the ote that P r[b(k, ɛ) > 0] (1 ɛ) k (1 ɛk ). Thus with probability at least (1 ɛk ) we have Z k,ɛ = 0. Observe that E[Z k,ɛ] = k ɛ( k ) = ɛ, sie the expeted distae is k ad we expet there to be kɛ bad (marked) threads. We ow begi the aalysis of the etire algorithm. Theorem 7. Let ɛ be the umber of iitial errors of our hypothesis at time t = 0, ad fix ay k = k (r + s), ad fix γ > 0, δ 1. The with probability at least (1 e )(1 e δ γ 6sk +log T )(1 3 k ) the (k, r, s)-trailig Walks Algorithm has o more tha 4(log(k )) 2 ɛ + (1 + δ) 2sk errors at ay time at all before t = T where T is as i Lemma 1. γ Remark 4. Note that T O(( 2 )(r+1)/2 1 k ) Proof. We prove the result by showig that with high probability o thread flips at ay time before T. Now ote that the probability that at least oe thread flips o ay time step t is idepedet of the total umber of errors. Moreover, it depeds oly o the verties that are flipped by ature durig time step t. So this probability does ot deped o the umber of errors at time t or o where those errors are. So let x t be a radom variable defied as: x t = { 1 if at least oe thread flips at time t 0 otherwise The x t is idepedat from x t for all t t. Usig the uio boud, we kow that E[x t ] 1 T where T is as i Lemma 3. Let X t = t j=1 x j. Note that E[X t ] t T. Usig Cheroff bouds we have: [ Pr X t (1 + δ) t ] [ ] ( e δ ) E[Xt] Pr X t (1 + δ)e[x t ] T (1 + δ) 1+δ 11

12 Now set δ = T t 1, we obtai: T ( e t 1 T ) E[Xt] ( e t 1 ( T t ) T t ( T t ) T t Now if we set t = 1 γ T for γ > 0, we obtai: [ ] Pr X t 1 e1 1 γ γ ) t T = e1 t T T t e γ We have prove that with probability at least 1 e, o thread flips from beig bad to good or vie γ versa at ay time step t T. This is a lower boud o the probability that we ever aumulate γ more tha Z k,ɛ errors from bad threads at time t T. Coupled with Lemma 6, we obtai the first γ term i the boud. Now we move to aalyzig the 2sk term. These are the errors made by ature. Note that a error from ature a oly our if ature flips a vertex we are right about ito oe we are wrog about. Let { 1 if vi,t v i,t 1 b i,t = 0 otherwise The these radom variables are i.i.d., so let B T = T t=1 i=1 b i,t. Cheroff bouds give: [ Pr B t (1 + δ) t ] e δt 6 2 where δ 1. Sie we see every vertex every sk steps (whether by good or bad threads), the umber of errors due to ature s flips at ay time is stritly less tha B. We have sk [ Pr B sk (1 + δ) 2sk ] e δ 6sk Thus the probability that we ever have more tha (1 + δ) 2sk errors due to ature i ay oe roud is at least 1 e δ2 6sk. I t steps we have t sk rouds. Thus, give t sk, the probability that i the first t time steps we ever have more tha (1 + δ) 2sk errors due to ature is at least ( 1 e δ ) tsk 6sk ( 1 tsk δ ) e 6sk 1 e δ 6sk +log t Settig t = T, ad puttig together this boud with the upper boud o makig more tha Z k,ɛ bad thread errors, we obtai the desired result i the theorem Corollary 8. If there are o iitial errors, meaig ɛ = 0, the settig k = 1 we have k = r + s. The for δ > 1, with probability at least (1 e δ )(1 e 6s +log(t ) ) we ever obtai more tha (1 + δ) 2s errors at ay time at all before T O((r 1)/2 ) 12

13 Now reall that durig the above aalysis we assumed C =. I other words, we assumed that the graph i questio was Hamiltoia. This, however, is a rather strit requiremet ad ot realisti i pratie. However sie ature flips verties idepedetly from their eighbors, if we partitio G ito yles ad ru the k-trailig Walks Algorithm o eah yle idepedetly, distributig the relevat probes i amouts proportioal to the respetive sizes, the we a still apply the bouds from Theorem 7. The followig Theorem formalizes that this a be doe for all graphs. Theorem 9. Let G be ay graph, ad let V (G) = C 1 Cd X be ay partitio of the verties suh that eah C i is a yle. The the errors obtaied by ruig the k-trailig Walks algorithm o eah subgraph C i admit the error bouds give by Theorem 7 applied to eah subgraph, plus at most the additioal ost X at ay time step. Proof. The errors i eah C i are idepedet of eah other, so ruig the k-trailig Walks Algorithm o eah result i idepedet error bouds, so Theorem 7 a be applied to eah as desired. Sie we do ot attempt to fix the remaiig verties X, the additioal ost X a be iurred, but learly o more due to egletig X. We ow state a simplified versio of prior results whih demostrates that the desired error boud holds i expetatio. While weaker tha Theorem 7, it aids i givig ituitio for where the asymptoti bouds i Theorem 7 ome from. Theorem 10. Let ɛ be the umber of iitial errors of our hypothesis at time t = 0, ad fix ay k = k r. The the expeted umber of errors of the trailig k-walks algorithm at ay time step is ɛ + 2k errors at ay time t T 1 k O ( r 2 ), where T is as i Lemma 1. Proof. Our algorithm starts k threads o uiformly radom verties i C, eah havig probability ɛ of startig o a error. So i expetatio, oly k ɛ threads will start i a error. For eah thread that does ot start i a error, we expet that it will ot flip ito a error thread i the first T = O ( r ) 2 steps, sie more tha half of the trailig set S j for a thread must be flipped i oe time step i order for a good thread to beome bad ad vie-versa. Similarly, we do ot expet bad threads to flip ito good oe s durig this timeframe, thus we a assume that bad threads stay bad ad good threads stay good for t T 1 k O ( r ) 2. Now fix ay bad thread, say it starts at v i. The that thread will iorretly set the hypothesis values for eah of v i, v i+1,..., v i+τ after τ time steps. However, oe the good thread that started at a vertex v j losest behid v i (of all other good threads) reahes v i, it will fix it to be good agai. Thus the umber of verties that a be iorret at ay give time beause the iitial bad thread made them so is the umber of time steps it takes for the previous good thread to reah v i ad fix it, whih i j. Now there are verties i the yle, ad we assume that the yle has legth (or is very early hamiltoia), ad we start k threads uiformly at radom i this yle. Thus we expet that i j = k, thus i expetatio a bad thread reates at most k errors. Note that it is possible that m bad threads be started oseutively, i whih ase we expet that i j = m k if the farther oe starts at v i. However this fator of m is already take ito aout sie we are multiplyig by the expeted umber of bad threads, whih is ɛk, ad this value is idepedet of 13

14 how far apart ay two oseutive threads are. Sie we expet there to be ɛk bad threads, we obtai a expeted ɛk k = ɛ errors due to bad threads. Fially, ode that i expetatio, at ay time step all verties have bee probed at least oe i the last k time steps. Sie the bad threads will oly reate errors, we eed ot aout for ature hagig the values of the bad threads (sie this would oly remove our errors). So we have at most verties whih have ot bee probed i k time steps, eah with 1 2 hae of flippig o eah time step. Thus we expet o more tha 2k additioal total hages of good verties ito bad oe s as a result of ature s radom perturbatios. Together with the errors from the bad threads, we obtai the desired ɛ + 2k expeted errors at ay time t T. 3 Noisy Probes We ow osider a geeralizatio of the previous model, amely where there is oise i our edge probes. The model is as follows. We have a graph G with ɛ 0 iitial errors, ad ature flips the value of eah vertex uiformly with probability 1 2. For the purposes of this aalysis, we assume ɛ 0 = 0, however the results of this setio a be similarly obtaied by settig ɛ 0 to be the expetatio from Theorem 18. Further suppose that o every edge probe we have a α probability of a error for 0 < α <.5. Meaig, for eah edge we probe e, with probability 1 α we reeive the orret XOR of the verties i the edge as i the last model, ad with probability α we reieve the iorret value. Thus, our etire model a be summarized as follows: At time t = 0, there are exatly ɛ 0 verties v i suh that v i,0 h 0 (v i ) has size ɛ 0 At the start of every time step, eah vertex is flipped by ature uiformly ad i.i.d. with probability 1 1 2, ad otherwise remais uhaged with probability 1 2. At every time step we are allowed to probe k edges, for some fixed k. For eah edge that we probe, we are told whether the verties o the edge agree or disagree with probability 1 α, ad we are told the opposite of this with probability α. The goal is to maitai a hypothesis h t (v i ) of v i,t suh that the l 1 distae h t (G) G t 1 is small. Remark 5. As we will see shortly, the followig aalysis does ot utilize the fat that vertex values are beig flipped uiformly at radom speifially, but oly that we a plae a boud statig that o more tha logarithmially may vertex values are flipped with high probability. Thus, the followig aalysis, ad i partiular Lemma 14, will hold for ay distributio over vertex value perturbatios suh that a high probability boud a be plaed o the umber of flips that our i a give time step. 3.1 Diffiulties with obvious algorithms I this ase, if we are to use the k-trailig Walks Algorithm, observe that eah thread will flip from beig good to bad if oly a sigle oe of it s edge probes is oisy. Thus, eve for small values of α (suh as if α 1 Ω()), our algorithm will quikly degeerate as early all threads beome bad. 14

15 A atural questio to ask the is a we simply probe radom verties? I other words, a atural algorithm is to pik a vertex v i uiformly at radom, probe all adjaet edges, ad assig our hypothesis for v i to be the value that makes the most sese give our observatios meaig the value that disagrees with less tha half of the edge observatios. The problem with this algorithm is the same problem that we disussed i the oiseless ase. The probability of iorretly ot fixig a bad vertex is the probability that more tha half the eighbors of the probed vertex are bad oditioed o the fat that we have probed a bad vertex. Now if the errors are distributed uiformly at radom, this value is easily bouded. But sie we are more likely to make a error if we are aroud other errors, as time goes o this oditioal probability beomes larger ad larger, sie errors begi to luster. This result is uaeptable, ad makes the algorithm impossible to aalyze. The seod issue is the problem of oeted ompoets of errors. If there is a large oeted ompoet of errors, the we a oly fix it by probig o the boudary. I the grid example, for istae, a oeted ompoet of errors a oly be fixed if we probe o its boudary, whih a have size proportioal to the square root of the umber of errors! 3.2 Primer o Expader Graphs The algorithm preseted i this setio will make heavy use of bouds o the size of the spetral gap of our graph. The smaller the spetral gap, the more probes we will eed to obtai reasoable bouds. Graphs with large or ostat-sized spetral gaps are kow as expaders. We assume some familiarity with spetral theory, ad will preset ow the fudametals that will be employed i the aalysis of our algorithm. Defiitio 4. Give a graph G ad ay subset S G, defie the boudary S of S to be the set S = {(u, v) E(G) u S, v G \ S}. We the defie the edge expasio, or Cheeger Costat, h(g) of G as: S h(g) = mi 0< S G S 2 Defiitio 5. We say that a graph G with verties is a expader graph if h(g) Ω(1). Defiitio 6. Give a d-regular graph G with adjaey matrix A, order the eigevalues of A λ 1 λ 2 λ. The the spetral gap of G is defied to be d λ 2. Lemma 11. A d-regular graph G is a expader if ad oly if it s spetral gap is a ostat. Proof. The proof follows from the well kow Cheeger Iequality, whih states: 1 2 (d λ 2) h(g) 2d(d λ 2 ) Thus (d λ 2 ) Ω(h(G)), ad h(g) Ω(d λ 2 ), whih ompletes the laim. The followig theorem will be ruial to our aalysis i setio

16 Theorem 12 (Expader Walk Samplig 25). Let G be a graph with ormalized seod largest eigevalue λ 2. Let f : V (G) {0, 1} be ay futio, ad let µ = 1 v i V (G) f(v i) be its mea. If Y 0, Y 1,..., Y k is a k -step radom walk startig at a radom vertex Y 0, the we have for all γ > 0: ad [ 1 k ] Pr k f(y i ) µ > γ e γ2 (1 λ 2 )k 10 = p i=0 [ 1 k ] Pr k f(y i ) µ < γ e γ2 (1 λ 2 )k 10 i=0 Example 1. The 4-regular disrete torus (i.e. the Cayley Graph of the fiite group Z Z ) with 2 verties has spetral gap (1 λ 2 ) O( 1 ) = O( 1 Z Z ). Example 2. The ( 1)-regular omplete graph K o verties has spetral gap (( 1) λ 2 ) O( 1 ) 3.3 Expader Walks Algorithm I Setio 3.1 we remarked that i order to reliably fix errors i the oisy model we aot rely o ever beig very sure of the orretess of our hypothesis for a give vertex at ay time step. This is a diret result of a o-trivial probability of oise o every probe. Thus, we must somehow obtai a uiform sample of the verties i the graph of a suffiietly large size suh that with high probability at least half of suh verties are ot errors. This is a eessarily oditio i order to update our hypothesis orretly, sie eve i a o-oisy sample the set of verties observed i ay set of probes may still take o oe of two possible sets of values. Eve to efore this, suh a sequee of probes must be made i path i order to relate the values of suessive verties together. But by the very ature of a path, the verties o it are fored to be lose to eah other, ad therefore ot uiformly hose! Thus we must take a loger path if we wat the verties o it to be, o average, uiformly distributed. Lukily, there are lasses of graphs for whih it suffies to take logarithmially log paths to obtai suh a result. The property of the graph whih will determie how log suh a path must be i order to be uiform is kow as the expasio of the graph. Before we itrodue this otio, we will first preset out algorithm. Defiitio 7. At time step t, let ɛ t be the fratio of bad verties (errors) i the graph. I other words, at time t there are ɛ verties v i suh that v i,t h t (v i ). Our algorithm uses k = k r probes o every time step, where k will be the legth of eah radom path sampled ad r will be the umber of suh paths sampled o eah time step. 1. O eah time step, for j = 1, 2,..., r: Expader Walk Samplig Algorithm: 2. Pik a radom vertex v i, ad take a k step radom walk startig from v i, provig eah edge alog the way. 16

17 3. Let Y = Y 0, Y 1,..., Y k be the (ot eessarily uique) verties o the walk. Assumig that all edge probes are orret, assig h t ( Y ) to be the value (out of the two possible) whih miimizes the l 1 distae from our prior hypothesis h t 1 ( Y ). Algorithm 3 Expader Walk Samplig Iput: r, k, G 1: Iitialize paths P 1,..., P r = 2: for j = 1, 2,..., r do 3: Uiformly sample v 1,j U V (G). 4: for i = 2,..., k do 5: v i,j N(V i 1,j ) U 6: ed for 7: P j {v 1,j, v 2,j,..., v k,j } 8: ed for Output: P 1, P 2,..., P r Algorithm 4 Expader Hypothesis Update (time t) Iput: Paths P 1, P 2,..., P r ad hypothesis h t 1 (G) 1: h t h t 1 2: for i {1, 2,..., r} do 3: Probe all edges E i P i 4: Let I 1, I 2 be the valid assigmets to P i give E i. 5: h t (P i ) arg mi I {I0,I 1 } h t 1 (P i ) I 1 1 6: ed for Output: h t (G) The above algorithm uses k = rk probes per time step. Suppose we have a d-regular graph G with ormalized seod largest eigevalue λ 2. We will show that if G is a good expader the k O( log() (1 λ 2 ) ) is suffiiet to obtai a expeted umber of errors E[ɛ t] 2r (1 α) k, ad that ɛ t is highly oetrated aroud it s mea for t O( 2 ). 3.4 Mai Aalysis I the followig two setios we will odut the priiple aalysis of our algorithm that will justify our mai theorems, Theorem 17 ad Theorem 21. First we prove two simple tehial Lemmas. Afterwards, we itrodue Lemma 15, whih will allow us i the aalysis that follows to assume that at every time step we have fewer tha ɛ t 1/4 errors i the graph with high probability. Usig similar tehiques as i this Lemma, we prove a boud o the umber of errors made by our algorithm for polyomially may time steps i Theorem

18 Afterwards, we osider looser bouds i the ase that we allow fewer probes. I Theorem 18, we will prove that the expeted umber of errors of our algorithm at ay time step before T = 2 is at most r (1 αk ) 1. Usig this fat, we will the observe that if we have more tha this expeted umber of errors, the i expetatio the umber of errors will derease o the ext time step. Tehially, this meas that the umber of errors we have exeedig this expeted value is a supermartigale. Usig this fat, we apply Azuma s iequality for supermartigales to obtai our mai Theorems: Theorem 20 ad Theorem 21. Lemma 13. Suppose we have 0 1 radom variables x 1,..., x that are ot i.i.d. but P r[x i = 1 X] p where X is ay set of observatios of the values x j s ot iludig x i. The we a apply the stadard Cheroff bouds, usig Pr[x i = 1] = p, to boud the probability that x i is large. Proof. We prove P r( t T x t > ) P r( t T Y t > ) where Y i are Beroulli(p). But this is easy to see, sie P r( t T x t > ) = I P r[x i = 1, i I] I p I (1 p I ) = P r( t T Y t > ), whih ompletes the laim. Lemma 14. With probability at least 1 1, ature ever flips more tha 4( + q) log() at ay time before T = q. Proof. Let E t be the umber of verties flipped by ature at time t. Note that E 1,..., E T idepedet with expetatio 1 2. The by Cheroff bouds, usig δ = 6( + q) log(): are Pr [E t 1 ] 2 (1 + 6( + q) log()) e (+q) log() = 1 +q Sie eah E T are idepedet, ad sie 1 2 (1 + 6( + q) log()) < 4( + q) log(), it follows via the uio boud that with probability at least 1 1 the maximum, take over all t T = q, umber of verties that ature flips i a give time step is less 4( + q) log(). Lemma 15. Give k 2560( + q) log() 1 (1 λ 2 ) probes i every radom walk, ad assumig α < 240k ad k o( log() ), the with probability at least (1 2 ) at o time before T = q will we ever have more tha a ɛ t = 1 4 fratio of errors. Proof. First ote that via Lemma 14 bouds the probability that ature flips more tha 4( + q) log()) verties at ay time step before T = q is at most 1 whih is a boud we will ow use. Supposig ɛ t = 1/8, we will boud the probability that we obtai 4 errors before droppig below 8. Now we have piked our k large eough so that by Theorem 25, the probability that we see more tha ɛ t + 1/16th errors o our path is at most 1. Similarly, the probability that we see less tha ɛ t th errors is at most 1. Thus the probability that either of these our o ay oe path is at least 1 2. Thus with probability at least 1 2, sie k 2560( + q) log() (1 λ 2 ), if there is o oise ad if t T, the we fix at least 1 16 k 4( + q) log() = 160( + q) log() log() 4( + q) log() > 156( + q) (1 λ 2 ) (1 λ 2 ) k 20 18

19 errors after probig ay oe path. For ay path where we fix at least this may errors, we all suh a path a suessful path. Note that we have subtrated off the maximum umber of errors that ature a reate o ay time step (oditioed o the fat that it ever makes more). Thus if we sample multiple paths o ay time step we are subtratig ature s error quatity multiple times withi a give time step, makig our boud slightly worse, but evertheless suffiiet. Observe that after a usuessful path there a be at most k + 7( + q) log() 2k ew errors, thus at most 40 suessful paths fully removes the errors from oe usuessful path. This fat will be shortly useful. Now usig the uio boud, the probability that there is o oise o a path is at least (1 k α). So we fix at least k 20 may errors with a path sample with probability at least (1 2 )(1 k α) (1 2k α) (assumig (1 2 ) > (1 k α), whih if did ot hold would oly give a tighter aalysis), ad reate at most 2k errors with probability at most 2k α. Thus, startig at ay time t, we fix 2k errors before ay usuessful path with probability at least (1 80k α). Now divide the rage (0, ) ito itervals of size 2k, ad set X t = { Xt if the umber of errors goes up oe iterval sie the last hage i iterval X t 1 1 if the umber of errors goes dow oe iterval sie the last hage i iterval Note that we a ever skip a iterval i oe time step sie they have size 2k. The X t is a bias radom walk with deset probability at least (1 80k α), sie by the uio boud this is a lower boud o the probability that we have 40 suessful paths i a row. Now suppose at time t we have 8 errors. The it would take at the very least 1 8 2k = 16k ireases i itervals before we ould attai 4 may errors, i other words whe X t hits Applyig Lemma 24, if we start with X 0 = 0, the the probability that X t hits at most ( 80αk ) 1 80αk 16k. Sie α < 1 240k dereasig below 8 is at least ( 1 2 ) 16k 16k. before 1 is by assumptio, the probability that we hit 4 errors before 16k, so the probability that this ever happes before time T = q, where there are r q path samples, is at most q r( 1 2 ) 16k. Coditioed o the fat that ature ever makes too may errors o oe time step before q (whih we showed ours with probability 1 1 ), ad give r q( ) 1 16k 2 < 1 2 whih holds as log as k o( log() ), it follows that the probability that we ever exeed 4 errors is at least (1 rq( ) 1 16k 2 )(1 1 ) (1 2 ), whih is the desired boud. Here, utilizig similar tehiques as i Lemma 15, we prove a reliable error boud for our algorithm. This error boud holds for muh more geeral distributios, where all we require is a high probability boud o the umber for flips made at ay time step. I the subsequet setio we will preset a stroger boud for the i.i.d. ase whih holds for a muh smaller time horizo whih works for smaller α, suh as α 1 Ω(k 2 ). The boud we preset i this setio holds for a muh loger time horizo. Theorem 16. Suppose verties flip aordig to ay distributio D suh that the ature ever flips more tha ζ D k verties at ay time step before q with probability at least 1 1. The give 19

20 k = k 2560( + q) log() (1 λ 2 ) probes i every radom walk, settig the parameter r = 1, ad assumig α 1 60(k ) ad k o( log() ), the with probability at least (1 5 )(1 q 2 M ) the maximum umber of errors that the Expader Samplig algorithm eouters before time T = q, ruig o a graph where verties flip aordig to D, is at most ( 10(+q) log() (1 λ 2 )k + 12αk + ζ D k ) + 2Mk. Proof. First apply Lemma 15 to oditio o the fat that we ever have more tha ɛ t 1/4 errors, whih ours with probability at least 1 2. The proof will proeed muh as i the proof of Lemma 15. Agai, for ay path where there is o oise, all suh a path a suessful path. Usig 10 log() the otatio of Theorem 25, suppose that we have a ɛ t µ = (1 λ 2 )k + ζ D+12α(k ) 2 k fratio of errors at time t, ad let Y 0,..., Y k be the verties o the radom path sampled by our algorithm. Let f(y i ) a biary futio evaluatig to 1 if ad oly if f(y i ) is a error, ad 0 otherwise. Note that sie by assumptio we have α 1 60(k ), it follows that ɛ t < 1/4, so we are ot violatig the assumptio obtaied by oditioig o Lemma 15 to begi with. By Theorem 25 we have [ k Pr f(y i ) < ζ D + 12α(k ) 2] [ 1 k 10 log() ] Pr k f(y i ) ɛ t < (1 λ 2 )k 1 i=0 Now utilizig the lower boud o k we establish a similar boud: i=0 Pr[ 1 k k f(y i ) ɛ t > 1 4 ] 1 i=0 The above is a upper boud o the probability that we iorretly assig our hypothesis for the verties o a suessful path, sie this ours oly if 1 k k i=0 f(y i) > 1/2, ad we kow ɛ t < 1/4. Thus we have established the followig fat: with probability at least 1 2 we fix at least ζ D + 12α(k ) 2 10 log() errors if we have 1/4 > ɛ t (1 λ 2 )k + ζ D+12α(k ) 2 k ad there is o oise o the path. Sie with probability at least 1 1 ature ever reates more tha ζ D errors o ay time step before q, it follows that with probability at least 1 3 the et derease i errors is at least 12α(k ) 2 after every suessful path probe whe ɛ t ( 10 log() (1 λ 2 )k + ζ D+12α(k ) 2 k, 1/4 ). Now we proeed as i the proof of Lemma 15. First break the iterval (0, ) up ito sub-itervals of size 2k, ad let X t = { Xt if the umber of errors goes up oe iterval sie last hage i iterval X t 1 1 if the umber of errors goes dow oe iterval sie last hage i iterval Now ote that (k + ζ D ) 2k is a upper boud o the differee betwee errors o subsequet time steps, thus it is impossible to skip over itervals withi oe time step. Furthermore ote that 2k it less tha = 1 12α(k ) 2 6αk suessful paths to go from oe iterval to the oe below it. The if we have ɛ t ( 10 log() (1 λ 2 )k + ζ D+12α(k ) 2 k, 1/4 ) we kow that the probability that X t = X t 1 1 is at 20