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1 cspringer (Original version in the Proceedings of FPSAC') A Generalized Cover Tie for Rando Walks on Graphs Cyril Banderier and Robert P. Dobrow 2 Algoriths Project. INRIA (Rocquencourt), France. Cyril.Banderier@inria.fr, 2 Clarkson University (Potsda, NY), USA. dobrowb@clarkson.edu, Abstract. Given a rando walk on a graph, the cover tie is the rst tie (nuber of steps) that every vertex has been hit (covered) by the walk. Dene the arking tie for the walk as follows. When the walk reaches vertex v i, a coin is ipped and with probability p i the vertex is arked (or colored). We study the tie that every vertex is arked. (When all the p i's are equal to, this gives the usual cover tie proble.) General forulas are given for the arking tie of a graph. Connections are ade with the generalized coupon collector's proble. Asyptotics for sall p i's are given. Techniques used include cobinatorics of rando walks, theory of deterinants, analysis and probabilistic considerations. Introduction The following proble was subitted during a supper at the eeting Analysis of Algoriths in June 999, at Barcelona: Conjecture (Supper Conjecture) Iagine guests around a table, soeone has the water carafe and decides to pour soe water in his glass with probability p. Then he gives the water carafe randoly to his right or left neighbor. This one does the sae, and so on. Call T (p) the nuber of carafe oves before everyone has got soe water. What can one say about the average tie E(T (p))? In particular, is it true that pe(t (p))! H when p!? (H is the -th haronic nuber). One will show that this conjecture is indeed true! In fact, we will tackle the proble for a slightly ore general proble: think about a dinner where soe people are ore or less not in speaking ters with soe others and so they give the carafe preferentially to their friends! The proble we consider was otivated by the gae Trivial Pursuit. Players ove pieces around a gae board answering questions on various topics (e.g., history, sports, etc.). On certain positions in the gae board, if a player answers a question correctly he gets a colored piece. And in our siplied version of the gae, when a player gets all the colored pieces he wins. Assue that for each topic there is soe probability of answering a question correctly. How long will a typical Trivial Pursuit gae take to play? Prepared with an evaluation version of PCT E. Visit

2 The reader should be able to see the connection with the following odel. Consider the usual (discrete) rando walk on a connected directed graph G with vertex set V = fv ; :::;v g. When the rando walk reaches vertex v i, that vertex is arked with probability p i, i = ;:::;. We are interested in the arking tie T (fp ; :::;p g), the rst tie that all the vertices have been arked. When all the p i 's are equal to, this is the usual cover tie proble. When the graph is the coplete graph and the rando walk oves to any vertex uniforly at rando, the proble reduces to the classical coupon collector's proble. (E.g., How any packs of Pokeon cards should you buy in order to collect the full set of Pokeon characters?) It is well know P that the expected k tie to cover a graph with vertices is H ; where H k := i= i is the kth Haronic nuber. The generalized coupon collector's proble asks for the tie to cover the coplete graph when the transition probabilities (weights) for the underlying rando walk are not constant (see the survey [2] for a lot of applications). There are classical results established either by cobinatorics (inclusion/exclusion principles [9], shue product [5]) or probability (artingales [8], tails estiates [7]). We also give soe references for the coupon collector proble for arbitrary graphs [4]; soe articles deal with asyptotic considerations [3]. Our proble (which involves two levels of randoness!) sees quite new and of course allows us to rederive/iprove previous results (in the peculiar case p=). When all the p i 's are equal (say, to p), a nice but nally predictable result is that pe(t (p)) is a rational nuber (as usual for Markovian process with rational transition probabilities). Of course, in principle all the questions concerning the rando walk can be addressed by solving appropriate linear equations using the transition atrix for the rando walk Markov chain but such an approach quickly becoes infeasible for even sall values of. On the coplete graph, in the case of equal arking probabilities (p i = p for all i), and of probabilities i for the coupons, a natural conjecture is that E[T (p)] (=p)h when p!. The reason is because for sall p, one would expect the tie to ark a particular vertex to be Geoetric with paraeter p (and thus ean =p). A quick (but erroneous) probabilistic reasoning consists in seeing the arking probability p as a change of tie-scale and thus one gets (for any graph) E(T (p)) = E(T ())=p. This is false (even asyptocally)! However, we will show this intuition to be true for regular graphs and prove a stronger result: For an arbitrary graph G, one has pe(t (p))! K as p!, where K is the expected cover tie for the generalized coupon collector's proble where the set of weights is the stationary distribution for the rando walk. This result will be proven in Section 4. Notation (Graph) Through all the paper, G is a directed connected graph with vertices v ; :::;v and with a transition atrix A (that is, a ij is the probability of a transition fro v i to v j ). The stationary distribution of this graph is noted ; :::;. The probability to ark the vertex v i whenever it is visited is In the USA, kids have a great craze for these cards. They trade the and stare at the a lot... :-) Prepared with an evaluation version of PCT E. Visit

3 Linear graph with reflexion at buttos (=8) Cyrclic graph (=8) Linear graph with loops at buttos (=8) Coplete graph (=4) Fig.. Soe of the oriented connected graphs considered in our exaples. noted p i (i =;:::;), and one sets q i := p i. When all the p i 's are equal, one siply notes p the probability of arking a vertex. e G is a coplete graph with transition atrix e A (related in soe sense to A, as explained later). Notation 2 (Rando variables) T (resp. e T ) is the \waiting tie" rando variable that represents the rst tie when all the vertices of the graph G (resp. eg) have been arked. The rando variable i (resp. e i ) gives the rst tie the i-th vertex of the graph G (resp. e G) is arked. Notation 3 (Operators) One notes [z n ]S the coecient of z n in a given series S. For any subset of f;:::;g, jj stands for the nuber of eleents in and one notes the substitution u i for i 2 and u i stays unchanged for i 62, the substitution u i q i for i 2 and u i for i 62. The sae notation will be used with respect to any other set of foral variables (for exaple with t ; :::;t instead of u ; :::;u ). We will also denote the identity atrix as Id and U as the diagonal atrix with diagonal eleents u ; :::;u. Exaples. To illustrate the above notations: [z 2 ] (+4z 2 + z 3 ) =4.For =4 and = f; 2; 4g, one has (3u + u u 2 + u u 4 ) = u 3 3 and (3u + u u 2 + u u 4 ) =3q + q q 2 + +q 4. In the sequel, all the graphs considered have vertices ( > ) and are directed and irreducible: there is a sequence of connected edges linking any pair of vertices. First, we establish a cobinatorial forula for the arking tie in section 2. One will explain in section 3 a probabilistic intuition which allows us to siplify the proble and thus to prove the Supper Conjecture in Section 4, another proof is given in Section 5. The last sections deal with peculiar cases. 2 Marking Tie on any Graph We rst give a generating function-based forula (based on nite dierences) for E(T (p)), the expected arking tie. For the cover tie on the coplete graph, another approach can be found in [5]. We refer the reader to Figure 4 for an exaple of our approach on two graphs of size 3. Prepared with an evaluation version of PCT E. Visit

4 Theore (General forula for average arking tie) The average tie for arking all the vertices of an arbitrary graph G is given by E(T (fp ; :::;p g)) = ( ) jj+ u (Id AU) i ; 6=; i= where A, U, the p i 's, and the 's are dened as in Notation and 3. Proof. For the transition atrix A, the entry A n i;j of A n, gives the probability of oving fro vertex v i to v j in n steps. With the atrix U dened as in Notation 3, the coecient of the onoial u k uk in (AU) n i;j gives the probability that the rando walk oves fro v i to v j in n steps such that vertex v t is visited k t ties, t =;:::;. Thus the probability generating function for the walks on the graph (beginning in v ), where z encodes the length of the walk and the u i 's encode the nuber of ties the walk visits v i is F (z;u ; :::;u ) = u z k (AU) k ;i = u (Id zau) ;i : i= k= i= Recall that p i (respectively, q i := p i ) is the probability to ark (respectively, not to ark) the vertex v i. Taking into account the fact that the walk has visited all the vertices arking the at least once leads to the substitution u n i qi n. This justi- es the introduction of the dierence operator i f(u i ) := f() f(q i ): So F + := 2 : :: F gives the probability generating function of the walks that arked all the vertices and one has in fact F + (z) = f;:::;g ( ) jj F (z;u ; :::;u ); i.e., F + is the su of F evaluated the set (dened in Notation 3). Therefore the probability generating function for T (fp ; :::;p g) is ( z)f + (z), so E(T (fp ; :::;p jz=( z)f + (z): A change of variable z = t and a local developent in t = ( z)u = (Id AU) jz= Id zau ij Note that Id A is never invertible (whereas Id za is always invertible), so one has to deal apart with the substitution ;. ut Note that setting the p i 's to gives a forula for the coupon collector proble. 3 The Probabilistic Intuition We restrict here the discussion to the case when all the p i 's are sall (equivalently, you can think p sall with p := ax p i ). One argues that the rando walk behaves, for sall values of p, like a walk on the coplete graph (with loops). We obtain a new Markov chain that is \equivalent" to the original Markov Chain, in the sense that they both have the sae (liiting) stationary distributions. Prepared with an evaluation version of PCT E. Visit

5 Let f i g i= be the stationary distribution for the rando walk on G. Note that i gives the probability that the walk is in vertex v i in stationarity (i.e., after a \long tie"). When p is sall, the tie to ark the graph is high. And the proportion of tie n that the walk is in vertex v i will be i n + O( p n) for large n, with probability close to. (This can be shown, for instance, by a central liit theore for Markov chains, or fro results on large deviations.) Thus, for \long enough" walks over G (when p is sall, all walks are \long enough"), the average length of tie to ark all the vertices should be very nearly the average length of tie to ark the vertices of the coplete graph (with loops), but with transition probabilities corresponding to the stationary distribution for the original graph. Let G e denote the coplete graph on vertices with loops. Dene its transition atrix A e with the transition probabilities ea i;j := j (for i; j =;:::;). Consider a rando walk process for the general arking proble that begins in the vertex v i with probability i. For such a process de- ne k e to be the rst tie that vertex ev k is arked. Then the k e are geoetric rando variables with paraeter k p k and P ( k e = n) = k p k ( k p k ) n, but there are not independent (as i ~ 6= j ~ for i 6= j). Observe that T = ax( ; :::; ) and T e = ax( e ; :::; e ). Consider now anew process on the graph. Instead of one \rando walker," consider particles at each of the vertices all oving siultaneously to neighboring vertices according to the sae transition echanis, but independently of each other. If we let Y i (and Yi e ) denote the rst tie that vertex v i is arked then observe that the Y i has the sae distribution as i but the Y i 's are independent and the i 's are not. Dene now Z = ax(y ; :::;Y ) and Z e = ax( Y e ; :::;e Y ). Our intuition is as follows: As the Y e i s and the Y i 's will behave siilarly for sall p, one should have that E( Z) e E(Z): As siultaneous arkings (for the Y process) occur with probability O(p 2 ), one has that E(Z) E(T ). Also E( Z) e E( T e ) and thus E(T ) E( T e ). Thus the study of the expected arking tie on the graph G e should answer our question as to the expected arking tie on the graph G. We ake the above rigorous in the next section. 4 Algebraic and Cobinatorial Proof Theore 2 (Closed for forula with stationary distribution) Let be a graph with transition probabilities ea i;j := j. The expected tie E(e T ) for arking the whole graph G e is =p ties the tie needed for visiting all the vertices E(e T )= p E(coupon collector on G); e and an inclusion-exclusion forula holds E(e T )= p 6=; where the 's are dened as in Notation 3. ( ) jj ( :::+ ) ; e G Prepared with an evaluation version of PCT E. Visit

6 Proof. Consider the i s, the q i's, and the p i 's as foral variables, then Prob(e T n) =fp > g :::fp> g( p + q + :::+ p + q ) n ; where fp >Sg (with S 2 IR[[p ; :::;p ; q ; :::;q ]]) stands for the su of onoials in p of positive exponent in S. Thus, we have Prob(e T n) = f;:::;g ( ) jj ( ( p + q + :::+ p + q )) n ; where the substitutions are dened as in Notation 3. Multiplying by z n and suing for n, when all the p i 's are equal to p, with the substitutions now taken to act on the i 's, leads to Prob( e T n)z n = f;:::;g ( ) jj z(q + p ( :::+ )) : Multiplying by z and dierentiating in z = gives the expected tie E(e T )= 6=; ( ) jj p( + ( :::+ )) : ut graph G graph G e /2 /2 3/8 3/8 /4 transition atrix A = /3 /3 /3 ea = 3/8 3/8 /4 /4 /4 /2 3/8 3/8 /4 stationary distribution (3/8, 3/8, /4) (3/8, 3/8, /4) u! u! U = u 2 U = u 2 u 3 u 3 generating atrix of the walks (Id zau) (Id z AU) e expected covering tie 95=2 ' 7:9 29=5 ' 5:8 expected arking tie (closed for) expected arking tie (asyptotics)! p + 87p p 3 75p +6p 2 + 9p 3 29=5p E(T (p)) = 29=5p + O() E(e T (p)) = 29=5p Fig. 2. Exaple of our approach on a graph of size 3. Most of the results in the literature are about the (expected) covering tie of the e G-colun (and other higher oents). In our situation (the G-colun), the lack of independence is the ain diculty. Our paper shows that in order to get the last entries in the G-colun, one can proceed as follow: Consider the rst 2 entries of the G-colun, then deal with eg-colun whose last entries gives the wanted result (for G).! Prepared with an evaluation version of PCT E. Visit

7 Proposition The arking tie T can be approxiated by e T : E(e T ) E(T )=O(); (p! ): Proof. In order to prove that 6=;( ) jj+ i= u (Id AU) e i ( ) jj+ u (Id AU) i = O() 6=; i= where the substitutions are dened in Notation 3, it is sucient to check that [p ] Id e AU In fact, one has ( za) = ea ij [p ] (Id AU) ij = O(): () z + O(): This is established by writing A = P + R, which decoposes A as an eigenprojection on the eigensubspace related to its eigenvector, plus the projection R on the suppleentary subspace (as it is well known P n = P = e A and PR = RP = ), thus P z n (P + R) n = ea z + (Id zr) where (Id zr) is actually regular in z = (as, by Perron{ Frobenius theory, is an eigenvalue of ultiplicity ), so ( zr) = O(). So this sets the case of the equality () when = f;:::;g (siply consider z = p). Theq other cases appear as a perturbation of two noninvertible atrices (naely Id A and Id e A), which gives (Id A + B) = ea () + O() = (Id e A + e B) where B is A with its i-th colun set to whenever i 2 (so A B = P AU), and where () is the perturbation of the eigenvalue, so () = i2 i. See [6] for the analicity of the perturbations of projections, eigenvalues, etc. ut For people with probabilistic anities, we give below another approach which also proves that E(T )=E( e T )+O() as p!. 5 Analytical and Probabilistic Proof The following proposition (identical to Proposition ) is the key point Proposition 2 (Equivalence of expected ties) The arking tie for the graph G and G e have the sae rst order asyptotics, naely Proof. The ain idea is the following E(T )=E(e T )+O(): T (T without its tails) (e T without its tails ) e T ; where eans here that the expectations have the sae rst order asyptotics as p!. Note that P (T n) = p c ; =(n ;:::;n) Prepared with an evaluation version of PCT E. Visit

8 Pwhere the su is over all nonnegative -tuples = (n ; :::;n ) such that k n k = n, p is the probability that the walk takes n steps and visits each vertex v k n k ties, k =;:::;, and c Q is the probability that all vertices have been arked by such a tour, that is c = interval" I as I := pj ln pj ( ( p)nj j= ). Dene the \central : ; j ln pj p Further dene the \ultidiensional box" B as B := Y i= [n i p n ln n; n i + p n ln n]: For sall p and for large n, by a classic result in large deviation theory, one has n j p n ln n<n j < n j + p n ln n (that is, 2 B) with probability exp( c ln 2 n) (with c >, see []). As Prob(T = n) is the probability that a success occurs exactly at the n-th step, one has E(T )= n( Prob(T n)): This su can be split as follows (with a lot of abusive but natural notations!) E(T )= (2) n<i n2i;2b n2i;62b n>i;2b n>i;62b where, for exaple, n < I eans for n before the \central interval" I. The rst su is bounded by the length of the interval of suation, which is o(=p) as p!, the sus for 62 B are bounded by exp( c ln 2 n) and so is the reaining su for n>i. One now focuses on the su over I and B. By a liit theore on Markov Chains [], p follows a ultidiensional Gaussian law g(), thus E(T )= n2i 2B g()c! + o(p ): (3) The sae schee can be applied to G, e with the sae central interval I and box B since the two rando walks have the sae stationary distribution. Thus E(T )=E(e T )+o(p )=E(e T )+O(), since E( T e ) and E(T ) are rational fractions in p (see Theore and 2). ut The Theore below is the ain result of the paper and answers precisely to the Supper Conjecture. The following \integral forula" is well known (cf. [5], where it is established for the coplete graph by eans of shue products and Laplace transfor), but we give an alternate derivation here, valid for any graph. Prepared with an evaluation version of PCT E. Visit

9 Theore 3 (First order asyptotics. Integral for) For any graph G with a stationary distribution ( ; :::; ), the expected arking tie is Z E(T )= K p + O(); where K = ( Y j= Proof. With the sae notations as above, the starting point is c = c (n;:::;n) = where p := ln( p). Thus E(T )= n2i Y 2B ( ( p) nj ) = g() Y j= Y j= ( exp( j x))) dx: ( e p nj ); exp( p n j j ) A + O(): In the box B, n j = n j + j p n ln n (where jj j < ), so E(T )= n2i c = Y ( exp( p n j p j p n ln n)); and thus 2B g() Y j= Now, ajoring P g() by exp( c ln 2 n) yields E(T )= n2i ( exp( p n j p j p n ln n)) A + O(): exp( p n j p p j n ln n) A + O() + o(p ): j= The sus being for p j ln pj n p j ln pj, this leads to exp( p (O(n))( exp( p o(n))) ( exp( p o(n)) and copleting the tails gives p o(n) p p j ln pjo(p j ln pj) =o(p ); E(T )= n2i = n( exp( p n j ) j= Y j= A + O() + o(p ): ( exp( p n j ))) + O() + o(p ): (4) Indeed the P rst introduced part is p =j ln pj = o(p ) and the last introduced part is n>p j ln Q pj 2 exp( p in( i )) n = o(p ). Set f(x) := ( j= exp( p x j ). As f (x) has a fast enough decay to near, the Euler-Maclaurin forula gives: Prepared with an evaluation version of PCT E. Visit

10 n= f(n) Z f(x)dx = f() + f() 2 Applying this to the forula (4) leads to Z E(T )= p ( Y j= + Z Since p = p + o(p), one has E(T )=K=p + O(). (x bxc =2)f (x)dx = O(): ( exp( j x))) dx + O() when p! : The \integral forula" allows us to copute E(T ) to any precision in linear tie (as all the integrated functions have a nice behavior), whereas the forulae of Theore and 2 are ipracticable because they coprise 2 suands. 6 What about Balanced or Regular Graphs? The ost coon odel for rando walk on a graph (the unweighted case) is to suppose that at a particular vertex the walk oves with probability proportional to the degree of that vertex. It is well known that the stationary distribution for such a walk is related to the degree of the vertices. Call a graph \balanced" if the outdegree of each vertex is equal to its indegree and if each outgoing edge is equally likely. For balanced graphs, there is a siple relation between the stationary distribution (the left eigenvector associated to eigenvalue of the transition atrix of the graph) and the degrees of edges: Proposition 3 (Stationary distribution for balanced graphs) For balanced graphs, one has i = P Ni Ni (N i is the nuber of incoing edges of vertex v i ). Proof. Let N ij be the nuber of edges fro v i to v j, and note N j the nuber of incoing edges to v j and N i the nuber of outgoing edges fro v i. The left eigenvector (for eigenvalue ) satises C A = ( ; 2 ; :::; ); N =N : :: N =N ( ; 2 ; :::; )... P N =N : ::N =N hence i Nij i = j. This equation is indeed satised by Ni i := P Ni Ni when N i = N i. ut Theore 2 rewrites in these cases: P Corollary (Balanced graphs) For balanced graphs with N := N i edges, one has E(T )= N ( ) jj p (N + N 2 + :::+ N ) + O(); 6=; where the substitutions operate on the fn j g. Proof. Direct consequence of Theore 2 and Proposition 3. ut ut Prepared with an evaluation version of PCT E. Visit

11 There are several classes of graphs for which the forula can be siplied. The ore interesting one, the class of regular graphs (graphs whose all vertices have the sae nuber of incoing and outgoing edges), is the object of the following corollary. Corollary 2 (Regular graphs) When all the i 's are equal (in particular if G is a regular graph and outgoing edges are chosen uniforly at rando), one has E(T )= H p + O(). Proof. ( ) i i p( + i ) = ( ) i+ p i i = H : p E(e T )= i= The last equality follows as iteration of forward nite dierence operators and Euler's R transfor. Q Equivalently, one could use Theore 3. Here the integral becoes ( ( exp( x=))) dx, which siplies to H j=. A third proof of this forula is probabilistic and considers the axiu of i.i.d. geoetric rando variables. Thus when i = = (for i =;:::;), one has Prob(e Z = n) = Prob( i e Y = n) Prob( j e < n) j6=i i i= = p=( p=) n ( ( p=) n ) : Since E(e T )= Pn nprob( e T = n), one is interested by the coecient of the lowest ter in the Laurent developent at p =of n nz( z )n ( ( z )n ) = k= ( ) k k z ( ( z=) k+ ) 2 : As ( ( z=) k+ ) 2 = O(z 2 ) as z!, the valuation of the developent in Laurent series of z ( z=) k+ 2 is. Thus the coecient of the lowest ter of F (z) is the su of the residues of the rational fractions, then k Res(z =)= ( ) k k= (k +) 2 = H : 7 Exaples The forula for the expected tie in the coupon collector proble reduces to for the cyclic graph C, to for the line graph L (with reection) and to ( )H + for the coplete graph without loops K. For L, there is an equivalence between the coupon collector proble and the rando walks (with jups +; ) fro to of height, the last ones being linked to continued fraction theory and hence to a quotient of Chebyshev polynoials. We apply below the forula of Theore for all nonisoorphic unoriented connected graphs without ultiplicity having at ost 4 vertices. We give for each graph, the stationary distribution, the average tie for the classical coupon Prepared with an evaluation version of PCT E. Visit

12 collector proble and, in the last colun, the coecient of p (the leading coecient) in the asyptotics of the average tie for arking the whole graph: size graph stationary expected expected distributions cover tie arking tie = K () =2 K 2 (/2, /2) 2 3 =3 K 3 (/3, /3, /3) 4 3H 3 =/2 =3 L 3 (/4, /2, /4) 5 9/3 =4 L 4 (/6, /3, /3, /6) 99/ =4 K 4 (/4, /4, /4, /4) 3/2 4H 4 =25/3 =4 C 4 (/4, /4, /4, /4) 7 4H 4 =25/3 =4 Y 4 (/5, 2/5, /5, /5) / =4 T 4 (/8, 3/8, /4, /4) 63/5 369/35 =4 Q 4 (3/, /5, /5, 3/) 69/ 62/7 The graphs K, L and C are dened as above whereas T 4 is the triangle with a tail, Y 4 is the Y shape, Q 4 is the square with one diagonal. We recall that, as seen in previous sections, for the coplete graph with loops, the coupon collector tie is H and the arking tie is H p. This is nally the behavior of all regular graphs, those for which the Supper Conjecture holds. Acknowledgeent. The rst of the two authors would like to thank Philippe Flajolet for a lot of helpful rearks and does not resist quoting one of his pseudo-reforulations of our proble \You are a PhD student and you attept to prove a theore; your chance of success at each attept is p (p very close to, of course). Let T be the tie it takes you to get the theore right..." References. Aldous (David J.) and Fill (Jaes). { Reversible Markov Chains and Rando Walks on Graphs. { Book in preparation. Available at 2. Boneh (Arnon) and Hofri (Micha). { The coupon-collector proble revisited. Co. Statist. Stochastic Models, vol. 3, n, 997, pp. 39{ Csorg}o (Sandor). { A rate of convergence for coupon collectors. Acta Sci. Math. (Szeged), vol. 57, n-4, 993, pp. 337{ Feige (Uriel). { Collecting coupons on trees, and the cover tie of rando walks. Coput. Coplexity, vol. 6, n4, 996=97, pp. 34{ Flajolet (Philippe), Gardy (Daniele), and Thionier (Loys). { Birthday paradox, coupon collectors, caching algoriths and self-organizing search. Discrete Applied Matheatics, vol. 39, n3, 992, pp. 27{ Kato (Tosio). { Perturbation theory for linear operators. { Springer-Verlag, Nath (Harindar B.). { Waiting tie in the coupon-collector's proble. Austral. J. Statist., vol. 5, 973, pp. 32{ Sen (Pranab Kuar). { Invariance principles for the coupon collector's proble: a artingale approach. Ann. Statist., vol. 7, n2, 979, pp. 372{ von Schelling (Herann). { Coupon collecting for unequal probabilities. Aer. Math. Monthly, vol. 6, 954, pp. 36{3. Prepared with an evaluation version of PCT E. Visit