Introduction to expander graphs

Transcription

1 Itrouctio to expaer graphs Michael A. Nielse 1, 1 School of Physical Scieces, The Uiversity of Queesla, Brisbae, Queesla 407, Australia Date: Jue, 005) I. INTRODUCTION TO EXPANDERS Expaer graphs are oe of the eepest tools of theoretical computer sciece a iscrete mathematics, poppig up i all sorts of cotexts sice their itrouctio i the 1970s. Here s a list of some of the thigs that expaer graphs ca be use to o. Do t worry if ot all the items o the list make sese: the mai thig to take away is the sheer rage of areas i which expaers ca be applie. Reuce the ee for raomess: That is, expaers ca be use to reuce the umber of raom bits eee to make a probabilistic algorithm work with some esire probability. Fi goo error-correctig coes: Expaers ca be use to costruct error-correctig coes for protectig iformatio agaist oise. Most astoishigly for iformatio theorists, expaers ca be use to fi error-correctig coes which are efficietly ecoable a ecoable, with a o-zero rate of trasmissio. This is astoishig because fiig coes with these properties was oe of the holy grails of coig theory for ecaes after Shao s pioeerig work o coig a iformatio theory back i the 1940s. A ew proof of PCP: Oe of the eepest results i computer sciece is the PCP theorem, which tells us that for all laguages L i NP there is a raomize polyoomial-time proof verifier which ee oly check a costat umber of bits i a purporte proof that x L or x L, i orer to etermie with high probability of success) whether the proof is correct or ot. This result, origially establishe i the earlier 1990s, has recetly bee give a ew proof base o expaers. What s remarkable is that oe of the topics o this list appear to be relate, a priori, to ay of the other topics, or o they appear to be relate to graph theory. Expaer graphs are oe of these powerful uifyig tools, surprisigly commo i sciece, that ca be use to gai isight ito a a astoishig rage of apparetly isparate pheomea. I m ot a expert o expaers. I m writig these otes to help myself a hopefully others) to uersta a little bit about expaers a how they ca be ielse@physics.uq.eu.au a applie. I m ot learig about expaers with ay specific itee applicatio i mi, but rather because they seem to behi some of the eepest isights we ve ha i recet years ito iformatio a computatio. What is a expaer graph? Iformally, it s a graph G = V, E) i which every subset S of vertices expas quickly, i the sese that it is coecte to may vertices i the set S of complemetary vertices. Makig this efiitio precise is the mai goal of the remaier of this sectio. Suppose G = V, E) has vertices. For a subset S of V we efie the ege bouary of S, S, to be the set of eges coectig S to its complemet, S. That is, S cosists of all those eges v, w) such that v S a w S. The expasio parameter for G is efie by hg) S mi S: /, 1) where X eotes the size of a set X. Oe staar coitio to impose o expaer graphs is that they be -regular graphs, for some costat, i.e., they are graphs i which every vertex has the same egree,. I must amit that I m ot etirely sure why this -regularity coitio is impose. Oe possible reaso is that oig this simplifies a remarkable result which we ll iscuss later, relatig the expasio parameter hg) to the eigevalues of the ajacecy matrix of G. If you o t kow what the ajacecy matrix is, we ll give a efiitio later.) Example: Suppose G is the complete graph o vertices, i.e., the graph i which every vertex is coecte to every other vertex. The for ay vertex i S, each vertex i S is coecte to all the vertices i S, a thus S = = ). It follows that the expasio parameter is give by hg) = mi =. ) S: / For reasos I o t etirely uersta, computer scietists are most itereste i the case whe the egree,, is a small costat, like =, 3 or 4, ot = 1, as is the case for the complete graph. Here s a example with costat egree. Example: Suppose G is a square lattice i imesios, with perioic bouary coitios so as to make the graph 4-regular). The if we cosier a large coecte subset of the vertices, S, it ought to be plausible that that the ege bouary set S cotais roughly oe ege for each vertex o the perimeter of the regio S. We expect there to be roughly such vertices, sice

2 we are i two imesios, a so S / 1/. Sice the graph ca cotai regios S with up to O ) vertices, we expect ) 1 hg) = O 3) for this graph. I o ot kow the exact result, but am cofiet that this expressio is correct, up to costat factors a higher-orer correctios. It be a goo exercise to figure out exactly what hg) is. Note that as the lattice size is icrease, the expasio parameter ecreases, teig towar 0 as. Example: Cosier a raom -regular graph, i which each of vertices is coecte to other vertices, chose at raom. Let S be a subset of at most / vertices. The a typical vertex i S will be coecte to roughly / vertices i S, a thus we expect S /, a so S. 4) Sice has its miimum at approximately / it follows that hg) /, iepeet of the size. Exercise: Show that a iscoecte graph always has expasio parameter 0. I each of our examples, we have t costructe just a sigle graph, but rather a etire family of graphs, iexe by some parameter, with the property that as gets larger, so too oes the umber of vertices i the graph. Havig access to a etire family i this way turs out to be much more useful tha havig just a sigle graph, a fact which motivates the efiitio of expaer graphs, which we ow give. Suppose we have a family G j = V j, E j ) of -regular graphs, iexe by j, a such that V j = j for some icreasig fuctio j. The we say that the family {G j } is a family of expaer graphs if the expasio parameter is boue strictly away from 0, i.e., there is some small costat c such that hg j ) c > 0 for all G j i the family. We ll ofte abuse omeclature slightly, a just refer to the expaer {G j }, or eve just G, omittig explicit metio of the etire family of graphs. II. EXPLICIT EXAMPLES OF EXPANDERS We ve see previously that a family of -regular raom graphs o vertices efies a expaer. For applicatios it is ofte more useful to have more explicit costructios for expaers. I particular, for applicatios to algorithms it is ofte useful to costruct expaers o O ) vertices, where is some parameter escribig problem size. Just to store a escriptio of a raom graph o so may vertices requires expoetially much time a space, a so is ot feasible. Fortuately, more parsimoious costructios are possible, which we ow escribe. Example: I this example the family of graphs is iexe by a prime umber, p. The set of vertices for the graph G p is just the set of poits i Z p, the fiel of itegers moulo p. We costruct a 3-regular graph by coectig each vertex x 0 to x 1, x + 1 a x 1. The vertex x = 0 is coecte to p 1, 0 a 1. Accorig to the lecture otes by Liial a Wigerso, this was prove to be a family of expaers by Lubotsky, Phillips a Sarak i 1988, but I o t kow a lower bou o the expasio parameter. Note that eve for p = O ) we ca o basic operatios with this graph e.g., raom walkig alog its vertices), usig computatioal resources that are oly polyomial i time a space. This makes this graph potetially far more useful i applicatios tha the raom graphs cosiere earlier. Example: A similar but slightly more complex example is as follows. The vertex set is Z m Z m, where m is some positive iteger, a Z m is the aitive group of itegers moulo m. The egree is 4, a the vertex x, y) has eges to x ± y, y), a x, x ± y), where all aitio is oe moulo m. Somethig which cocers me a little about this efiitio, but which I have t resolve, is what happes whe m is eve a we choose y = m/ so that, e.g., the vertices x + y, y) a x y, y) coicie with oe aother. We woul expect this uplicatio to have some effect o the expasio parameter, but I have t thought through exactly what. III. GRAPHS AND THEIR ADJACENCY MATRICES How ca we prove that a family of graphs is a expaer? State aother way, how oes the expasio parameter hg) vary as the graph G is varie over all graphs i the family? Oe way of tacklig the problem of computig hg) is to o a brute force calculatio of the ratio S / for every subset S of vertices cotaiig o more tha half the vertices i the graph. Doig this is a time-cosumig task, sice if there are vertices i the graph, the there are expoetially may such subsets S. Problem: I geeral, how har is it to fi the subset S miimizig S /? Ca we costruct a NP- Complete variat of this problem? I o t kow the aswer to this questio, a I o t kow if ayoe else oes, either. Fortuately, there is a extraoriarily beautiful approach to the problem of etermiig hg) which is far less computatioally itesive. It ivolves the ajacecy matrix AG) of the graph G. By efiitio, the rows a colums of the ajacecy matrix are labelle by the vertices of V. For vertices v a w the etry AG) vw is efie to be 1 if v, w) is a ege, a 0 if it is ot a ege. It is a marvellous fact that properties of the eigevalue spectrum of the ajacecy matrix AG) ca be use to

3 3 uersta properties of the graph G. This occurs so frequetly that we refer to the spectrum of AG) as the spectrum of the graph G. It is useful because the eigevalue spectrum ca be compute quickly, a certai properties, such as the largest a smallest eigevalue, the etermiat a trace, ca be compute extremely quickly. More geerally, by recastig graphs i terms of ajacecy matrices, we ope up the possibility of usig tools from liear algebra to stuy the properties of graphs. Although we re most itereste i stuyig expaers, for the rest of this sectioi m goig to igress from the stuy of expaers, stuyig how the liear algebraic poit of view ca help us uersta graphs, without worryig about how this coects to expaers. This igressio is partially motivate by the fact that this is beautiful stuff at least i my opiio), a is partially because our later iscussio of expaers will be base o this liear algebraic poit of view, a so it s goo to get comfortable with this poit of view. The followig exercise provies a goo example of how graph properties ca be relate to the eigevalues of the graph. Exercise: Prove that if two graphs are isomorphic, the they have the same spectrum. This result is ofte useful i provig that two graphs are ot isomorphic: simply compute their eigevalues, a show that they are ifferet. A useful extesio of the exercise is to fi a example of two graphs which have the same spectra, but which are ot isomorphic. Note that the ajacecy matrix may be cosiere as a matrix over ay fiel, a the result of the exercise is true over ay fiel. I ve ofte woere if the coverse is true, but o t kow the aswer.) Noetheless, by a large, we ll cosier the ajacecy matrix as a matrix over the fiel R of real umbers. Assumig that G is a uirecte graph, we see that AG) is a real symmetric matrix, a thus ca be iagoalize. We will fi it coveiet to write the eigevalues of a graph G i oicreasig orer, as λ 1 G) λ G)... λ G). A fact we ll make a lot of use of is that whe G is -regular the largest eigevalue of G is just. To see this, ote that the vector 1 1, 1,..., 1) is a eigevector of G with eigevalue. To prove that is the largest eigevalue seems to be a little bit harer. We ll just sketch a proof. To prove this it is sufficiet to show that v T AG)v for all ormalize vectors v. From the -regularity of G it follows that AG)/ is a oubly stochastic matrix, i.e., has o-egative etries, a all rows a colums sum to oe. A theorem of Birkhoff esures that AG)/ ca be writte as a covex combiatio of permutatio matrices, so AG) = j p jp j, where p j are probabilities, a the P j are permutatio matrices. This gives v T AG)v = j p jv T P j v. But v T P j v 1 for ay permutatio P j, which gives the esire result. The followig propositio gives aother example of the relatioships oe ca fi betwee a graph a its spectrum. Propositio: A -regular graph G is coecte if a oly if λ 1 G) > λ G). Proof: The easy irectio is the reverse implicatio, for which we prove the cotrapositive, amely, that a -regular iscoecte graph has λ 1 G) = λ G). This follows by breakig G up ito iscoecte compoets G 1 a G, a observig that AG) = AG 1 ) AG ), where is the matrix irect sum. Sice both G 1 a G are -regular it follows that they both have maximal eigevalue, a so appears at least twice i the spectrum of AG). At the momet, I o t see a easy way of provig the forwar implicatio. Oe ot very satisfyig proof is to observe that AG)/ is the Markov trasitio matrix for a raom walk o the graph, a that sice the graph is coecte, the raom walk must coverge to a uique istributio, which implies that i the limit of large there ca oly be oe vector v such that G / )v = v. This meas that G s largest eigevalue is o-egeerate, from which it follows that G s largest eigevalue is o-egeerate. This is a sketch, but it ca all be establishe rigorously with a little work a the ai of well-kow theorems o Markov chais. The proof sketche i the previous paragraph is ot really satisfactory, sice it ivolves a appeal to theorems which are i some sese less elemetary tha the result uer iscussio. Aother possibility which I ve explore but have t mae work with complete rigour is to ivestigate G / more explicitly. With a little thought oe ca prove that the etry G vw is just the umber of paths betwee v a w of legth. Sice G is coecte, we expect i the limit of large this umber woul be omiate by a term which oes ot epe o w, a woul just scale like the total umber of paths of legth startig at v which is ), ivie by the total umber of possible estiatios w, which is, givig G vw/ 1/. This woul oly be true if G has self-loops v, v).) Of course, the matrix whose etries are all 1/ has a sigle eigevalue 1, with all the rest 0, which woul suffice to establish the theorem. QED Problem: How shoul we iterpret the etermiat of a graph? What about the trace? Problem: If we cosier AG) as a matrix over the fiel Z = {0, 1}, the it is possible to efie a matrix sum G 1 + G, whose ajacecy matrix is just AG 1 ) + AG ), a a matrix prouct G 1 G whose ajacecy matrix is just AG 1 )AG ). May questios aturally suggest themseves: 1) whe is there a ege betwee v a w i G 1 + G ; ) whe is there a ege betwee v a w i G 1 G these first two questios are easy to aswer); 3) for which graphs is AG) ivertible, a thus a atural iverse graph G 1 exists; 4) how ca we iterpret the iverse graph; 5) whe o two graphs commute? Problem: Alog similar lies to the previous problem, it s possible to efie a tesor prouct of graphs. What

4 4 are the properties of the graph tesor prouct? The ieas I ve escribe i this sectioare examples of the importat geeral priciple that oce you ve efie a mathematical object, you shoul seek out alterate represetatios or eve just partial represetatios) of that object i terms of mathematical objects that you alreay uersta. By recastig graphs as matrices, we ope up the possibility of usig all the tools of liear algebra to aswer questios about graphs. This ca work i oe of two ways: we ca ask a questio about graphs, a try to see if it s possible to give a liear algebraic aswer, or we ca ask what implicatio kow results of liear algebra have for graphs what oes the Gaussia elimiatio proceure correspo to, or the spectral ecompositio, or two matrices commutig, or the wege prouct, or whatever. Explorig such coectios has the potetial to greatly erich both subjects. IV. EXPANSION AND THE EIGENVALUE GAP Let s retur our attetio to expaer graphs, a see what the eigevalues of a graph have to o with its expasio parameter. We efie the gap for the graph G to be the ifferece G) λ 1 G) λ G) betwee the largest a seco-largest eigevalues. The expasio parameter a the gap are coecte by the followig theorem: Theorem: The expasio parameter hg) for a - regular graph G is relate to the gap G) by: G) hg) G). 5) Thus, properties of the eigevalue gap ca be use to euce properties of the expasio parameter. For example, if the eigevalue gap for a family of -regular graphs is boue below by a positive costat, the the expasio parameter must also be boue below by a positive costat, a so the family is a expaer. Oe reaso for fiig the coectio betwee the gap a the expasio parameter iterestig is that it is far easier to estimate the gap of a by matrix tha it is to eumerate the expoetially may subsets S of the vertex set V, a compute S / for each oe. Proof iscussio: We alreay uersta that λ 1 G) = for this graph, with eigevector 1 = 1, 1,..., 1). So we ll cocetrate o tryig to uersta the behaviour of the seco largest eigevalue, λ G). The theorem tells us that the ifferece betwee a λ G) is cotrolle both above a below by the expasio parameter hg). How ca we get cotrol over the seco largest eigevalue of G? Oe way is to observe that λ G) is just the maximum of the expressio v T Av/v T v, where A is the ajacecy matrix of G, a we maximize over all vectors v orthogoal to the eigevector 1. A ecouragig fact is that this expressio is quite easy to eal with, because the coitio that v be orthogoal to 1 is actually equivalet to the sum of v s etries beig equal to 0, so we have λ G) = v T Av max v:trv)=0 v T v, 6) where trv) is just the sum of the etries of the vector v. We re goig to provie a lower bou o λ G) by simply guessig a goo choice of v satisfyig trv) = 0, a usig the fact that λ G) vt Av v T v. 7) To make a goo guess, it helps to have a way of thikig about expressios like v T Av, where trv) = 0. A coveiet way of thikig is to rewrite v as the ifferece of two isjoit probability istributios, p a q, i.e., v = p q, where p a q are o-egative vectors each summig to 1, a with isjoit support. This results i terms like p T Aq, which we ca thik of i terms of trasitio probabilities betwee q a p. This will allow us to apply the expasio properties of the graph. Let s make these ieas a little more cocrete. The key is to efie 1 S to be the vector whose etries are 1 o S, a 0 elsewhere, a to observe that 1 T S A 1 T = ES, T ), 8) where ES, T ) is the umber of eges betwee the vertex sets S a T. This suggests that we shoul choose p a q i terms of vectors like 1 S, sice it will eable us to relate expressios like v T Av to the sizes of various ege sets, which, i tur, ca be relate to the expasio parameter. Suppose i particular that we choose v = 1 S 1 S. 9) This satisfies the coitio trv) = 0, a gives a v T v = ) v T Av = 1 1 ES, S) + ES, S) ES, S).11) The efiitio of a expaer graph gives us cotrol over ES, S), so it is coveiet to rewrite ES, S) a ES, S) i terms of ES, S), usig the -regularity of the graph: ES, S) + ES, S) = ; ES, S) + ES, S) =.1) A straightforwar substitutio a a little algebra gives: 1 v T Av = + 1 ) ) ES, S). 13)

5 5 Comparig with the earlier expressio for the eomiator v T v, we obtai 1 λ G) + 1 ) ES, S). 14) Now choose S so that ES, S) = hg), a /, givig after a little algebra a thus λ G) hg), 15) G) hg), 16) which was the first of the two esire iequalities i the theorem. The proof of the seco iequality is a little more complicate. Ufortuately, I have t maage to boil the proof ow to a form that I m really happy with, a for this reaso I wo t escribe the etails. If you re itereste, you shoul try to prove it yourself, or refer to the otes of Liial a Wigerso. QED Problem: Ca we geeralize this result so that it applies to a geeral uirecte graph G, ot just to - regular graphs? Ca we prove a aalogous statemet for irecte graphs, perhaps i terms of sigular values? Ca we efie a a geeralize otio of expasio which ca be applie to ay symmetric matrix A with o-egative etries, a coect that otio of expasio to the eigevalue gap? Ca we geeralize eve further? What happes if we chage the fiel over which the matrix is cosiere? V. RANDOM WALKS ON EXPANDERS May applicatios of expaers ivolve oig a raom walk o the expaer. We start at some chose vertex, a the repeately move to ay oe of the eighbours, each time choosig a eighbour uiformly at raom, a iepeetly of prior choices. To escribe this raom walk, suppose at some give time we have a probability istributio p escribig the probability of beig at ay give vertex i the graph G. We the apply oe step of the raom walk proceure escribe above, i.e., selectig a eighbour of the curret vertex uiformly at raom. The upate probability istributio is easily verifie to be: p = AG) p. 17) That is, the Markov trasitio matrix escribig this raom walk is just ÂG) AG)/, i.e., up to a costat of proportioality the trasitio matrix is just the ajacecy matrix. This relatioship betwee the ajacecy matrix a raom walks opes up a whole ew worl of coectios betwee graphs a Markov chais. Oe of the most importat coectios is betwee the eigevalues of Markov trasitio matrices a the rate at which the Markov chai coverges to its statioary istributio. I particular, the followig beautiful theorem tells us that whe the uiform istributio is a statioary istributio for the chai, the the Markov chai coverges to the uiform istributio expoetially quickly, at a rate etermie by the seco largest eigevalue of M. Exercise: Show that if M is a ormal trasitio matrix for a Markov chai the 1 = λ 1 M) λ M).... Theorem: Suppose M is a ormal trasitio matrix for a Markov chai o states, with the uiform istributio u = 1/ as a statioary poit, Mu = u. The for ay startig istributio p, M t p u 1 λ M) t, 18) where 1 eotes the l 1 orm. The ormality coitio i this theorem may appear a little surprisig. The reaso it s there is to esure that M ca be iagoalize. The theorem ca be mae to work for geeral M, with the seco largest eigevalue replace by the seco largest sigular value. However, i our situatio M is symmetric, a thus automatically ormal, a we prefer the statemet i terms of eigevalues, sice it allows us to make a coectio to the expasio parameter of a graph. I particular, whe M = ÂG) we obtai: ÂG)t p u 1 ) t λ G). 19) Combiig this with our earlier results coectig the gap to the expasio parameter, we euce that ÂG)t p u 1 ) t 1 hg). 0) Thus, for a family of expaer graphs, the rate of covergece of the Markov chai is expoetially fast i the umber of time steps t. Exercise: Suppose M is a trasitio matrix for a Markov chai. Show that the uiform istributio u is a statioary poit poit for the chai, i.e., Mu = u, if a oly if M is oubly stochastic, i.e., has o-zero etries, a all rows a colums of the matrix sum to 1. Proof: We start by workig with the l orm. Sice Mu = u we have M t u = u, a so: M t p u = M t p u). 1) A computatio shows that p u is orthogoal to u. But u is a eigevector of M with the maximum eigevalue, 1, a thus p u must lie i the spa of the eigespaces with eigevalues λ M), λ 3 M),.... It follows that M t p u) λ M) t p u λ M) t, ) where we use the fact that p u 1, easily establishe by observig that p u is covex i p, a thus

6 6 must be maximize at a extreme poit i the space of probability istributios; the symmetry of u esures that without loss of geerality we may take p = 1, 0,..., 0). To covert this ito a result about the l 1 orm, we use the fact that i imesios v 1 v, a thus we obtai M t p u) 1 λ M) t, 3) which was the esire result. QED What other properties o raom walks o expaers have? We ow prove aother beautiful theorem which tells us that they move arou quickly, i the sese that they are expoetially ulikely to stay for log withi a give subset of vertices, B, uless B is very large. More precisely, suppose B is a subset of vertices, a we choose some vertex X 0 uiformly at raom from the graph. Suppose we use X 0 as the startig poit for a raom walk, X 0,..., X t, where X t is the vertex after the tth step. Let Bt) be the evet that X j B for all j i the rage 0,..., t. The we will prove that: λ G) PrBt)) + B ) t 4) Provie λ G)/ + B / < 1, we get the esire expoetial ecrease i probability. For a family of expaer graphs it follows that there is some costat ɛ > 0 such that we get a expoetial ecrease for ay B such that B / < ɛ. These results are special cases of the followig more geeral theorem about Markov chais. Theorem: Let X 0 be uiformly istribute o states, a let X 0,..., X t be a time-homogeeous Markov chai with trasitio matrix M. Suppose the uiform istributio u is a statioary poit of M, i.e., Mu = u. Let B be a subset of the states, a let Bt) be the evet that X j B for all j 0,..., t. The PrBt)) λ M) + B ) t. 5) Proof: The first step i the proof is to observe that: PrBt)) = P MP ) t P u 1, 6) where the operator P projects oto the vector space spae by those basis vectors correspoig to elemets of B. This equatio is ot etirely obvious, a provig it is a goo exercise for the reaer. The ext step is to prove that P MP λ M) + B /, where the orm here is the operator orm. We will o this below, but ote first that oce this is oe, the result follows, for we have PrBt)) = P MP ) t P u 1 P MP ) t P u 7) by the staar iequality relatig l 1 a l orms, a thus PrBt)) P MP t P u, 8) by efiitio of the operator orm, a fially PrBt)) λ M) + B ) t, 9) where we use the assume iequality for the operator orm, a the observatio that P u = B / 1/. To prove the esire operator orm iequality, P MP λ M) + B /, suppose v is a ormalize state such that P MP = v T P MP v. Decompose P v = αu + βu, where u is a ormalize state orthogoal to u. Sice P v v = 1 we must have β 1. Furthermore, multiplyig P v = αu+βu o the left by u T shows that α = u T P v. It follows that α is maximize by choosig v to be uiformly istribute over B, from which it follows that α B. A little algebra shows that v T P MP v = α u T Mu + β u T Mu. 30) Applyig α B, u T Mu = u T u = 1/, β 1, a u T Mu λ M) gives v T P MP v B + λ M), 31) which completes the proof. QED VI. REDUCING THE NUMBER OF RANDOM BITS REQUIRED BY AN ALGORITHM Oe surprisig applicatio of expaers is that they ca be use to reuce the umber of raom bits eee by a raomize algorithm i orer to achieve a esire success probability. Suppose, for example, that we are tryig to compute a fuctio fx) that ca take the values fx) = 0 or fx) = 1. Suppose we have a raomize algorithm Ax, Y ) which takes as iput x a a m-bit uiformly istribute raom variable Y, a outputs either 0 or 1. We assume that: fx) = 0 implies Ax, Y ) = 0 with certaity. fx) = 1 implies Ax, Y ) = 1 with probability at least 1 p f. That is, p f is the maximal probability that the algorithm fails, i the case whe fx) = 1, but Ax, Y ) = 0 is output by the algorithm. A algorithm of this type is calle a oe-sie raomize algorithm, sice it ca oly fail whe fx) = 1, ot whe fx) = 0. I wo t give ay cocrete examples of oe-sie raomize algorithms here, but the reaer ufamiliar with them shoul rest assure that they are useful a importat see, e.g., the book of Motwai a Raghava Cambrige Uiversity Press, 1995) for examples.

7 7 As a asie, the iscussio of oe-sie algorithms i this sectioca be extee to the case of raomize algorithms which ca fail whe either fx) = 0 or fx) = 1. The etails are a little more complicate, but the basic ieas are the same. This is escribe i Liial a Wigerso s lecture otes. Alterately, exteig the iscussio to this case is a goo problem. How ca we escrease the probability of failure for a oe-sie raomize algoerithm? Oe obvious way of ecreasig the failure probability is to ru the algorithm k times, computig Ax, Y 0 ), Ax, Y 1 ),..., Ax, Y k 1 ). If we get Ax, Y j ) = 0 for all j the we output 0, while if Ax, Y j ) = 1 for at least oe value of J, the we output fx) = 1. This algorithm makes use of km bits, a reuces the failure probability to at most p k f. Expaers ca be use to substatially ecrease the umber of raom bits require to achieve such a reuctio i the failure probability. We efie a ew algorithm A as follows. It requires a -regular expaer graph G whose vertex set V cotais m vertices, each of which ca represet a possible m-bit iput y to Ax, y). The moifie algorithm A works as follows: Iput x. Sample uiformly at raom from V to geerate Y 0. Now o a k 1 step raom walk o the expaer, geeratig raom variables Y 1,..., Y k 1. Compute Ax, Y 0 ),..., Ax, Y k 1 ). If ay of these are 1, output 1, otherwise output 0. We see that the basic iea of the algorithm is similar to the earlier proposal for ruig Ax, Y ) repeately, but the sequece of iepeet a uiformly istribute samples Y 0,..., Y k 1 is replace by a raom walk o the expaer. The avatage of oig this is that oly m + k log) raom bits are require m to sample from the iitial uiform istributio, a the log) for each step i the raom walk. Whe is a small costat this is far fewer tha the km bits use whe we simply repeately ru the algorithm Ax, Y j ) with uiform a iepeetly geerate raom bits Y j. With what probability oes this algorithm fail? Defie B x to be the set of values of y such that Ax, y) = 0, yet fx) = 1. This is the ba set, which we hope our algorithm will avoi. The algorithm will fail oly if the steps i the raom walk Y 0, Y 1,..., Y k 1 all fall withi B x. From our earlier theorem we see that this occurs with probability at most: Bx m + λ ) k 1 G). 3) But we kow that B x / m p f, a so the failure probability is at most p f + λ ) k 1 G). 33) Thus, provie p f + λ G)/ < 1, we agai get a expoetial ecrease i the failure probability as the umber of repetitios k is icrease. VII. CONCLUSION These otes have give a pretty basic itrouctio to expaers, a there s much we have t covere. More etail a more applicatios ca be fou i the olie otes of Liial a Wigerso, or i the research literature. Still, I hope that these otes have give some iea of why these families of graphs are useful, a of some of the powerful coectios betwee graph theory, liear algebra, a raom walks.