c 2007 Society for Industrial and Applied Mathematics

Transcription

1 SIAM J. COMPUT. Vol. 37, No. 1, c 2007 Society for Industrial and Alied Mathematics A PROBABILISTIC STUDY ON COMBINATORIAL EXPANDERS AND HASHING PHILLIP G. BRADFORD AND MICHAEL N. KATEHAKIS Abstract. This aer gives a new way of showing that certain constant degree grahs are grah exanders. This is done by giving new roofs of exansion for three ermutations of the Gabber Galil 3 exander. Our results give an exansion factor of for subgrahs of these three-regular grahs with ( 1) 2 inuts for rime. The roofs are not based on eigenvalue methods or higher algebra. The same methods show the exected number of robes for unsuccessful search in double hashing is 1 bounded by, where α is the load factor. This assumes a double hashing scheme in which two hash functions are randomly and indeendently chosen from a secified uniform distribution. The result is valid regardless of the distribution of the inuts. This is analogous to Carter and Wegman s result for hashing with chaining. This aer concludes by elaborating on how any sufficiently sized subset of inuts in any distribution exands in the subgrah of the Gabber Galil grah exander of 1 focus. This is related to any key distribution having exected robes for unsuccessful search for double hashing given the initial random, indeendent, and uniform choice of two universal hash functions. Key words. exander grahs, double hashing, Gabber Galil exander, exansion factor, combinatorial exanders, airwise indeendence, hash collisions AMS subject classifications. 05C90, 68P05, 68P20, 60C05 DOI /S X 1. Introduction. Consider a biartite grah G =(I O,E), where I and O form the biartition of the nodes with n = I = O, and let r be G s maximum degree. G is an (n, r, c)-exander if the following holds: ( )) N (1+c (Î) Î 1 Î, n for every subset Î I that contains u to n/2 elements (inuts), where N (Î) = {o O:(i, o) E for some i Î}. The constant c is the exansion factor of the grah. Further, the degree r is bounded by a constant. There are many other essentially equivalent definitions for grah exanders. The hard art in designing grah exanders is roving they exand. In fact, the decision roblem of determining exansion is co-np-comlete [6]. A series of classic aers firmly established that certain grah families exand. First, Margulis [21] showed that exanders exist, without giving bounds on their exansion. However, he did show how to construct them exlicitly. Next, Gabber and Galil [12] gave an exlicit exander construction with bounds on their exansion. Received by the editors October 15, 2004; acceted for ublication (in revised form) November 1, 2006; ublished electronically March 30, This research was suorted by the Rutgers Business School Research Resources Committee. htt:// Deartment of Comuter Science, University of Alabama, Box , Tuscaloosa, AL (gb@cs.ua.edu). Deartment of Management Sciences and Information Systems, Rutgers Business School Newark and New Brunswick, Rutgers University, 180 University Ave., Newark, NJ (mnk@andromeda.rutgers.edu). 83

2 84 PHILLIP G. BRADFORD AND MICHAEL N. KATEHAKIS Finally, Alon [2] showed that a grah is an exanding grah iff its largest and secondlargest eigenvalues are well searated. See also, for examle, [5, 4, 26, 10, 18, 29] for varying deths of coverage of eigenvalue methods for grah exansion. The eigenvalue methods have been central in much research on grah exanders. Eigenvalue methods do not give the best ossible exanding grah coefficients [33]. For examle, robabilistic methods show the existence of exanders that have better exansion than is ossible to show by the searation of the largest and second-largest eigenvalues. Pinsker [27] first showed the existence of exanders using robabilistic methods. There are some other constructions of exanders. According to Alon [3], the (eigenvalue-based) construction of Jimbo and Maruoka [15] only uses elementary but rather comlicated tools from linear algebra. Ajtai [1] also gives an algorithm using linear algebra for constructing three-regular exanding grahs. This algorithm is comlex and takes O(n 3 log 3 n) time to construct an exander. The exansion factor of these exanders is unknown but ositive. Lubotzky, Phillis, and Sarnak [19] and indeendently Margulis [22] gave the best ossible exanders using the eigenvalue methods [2, 18, 19, 29]. Kahale [] gave the best exansion constant to date for Ramanujan and related grahs. Reingold, Vadhan, and Wigderson [28] give very imortant combinatorial constructions of constant degree exanders based on their new zig-zag grah roduct. By showing how the zig-zag roduct maintains the eigenvalue bounds (then breaks them), they show how to construct exanders recursively starting from a small exander. Further, Meshulam and Wigderson [25] give grou theoretic techniques whose exansion they show deends on universal hash functions. Caalbo et al. [7] give constant degree d lossless exanders. These exand by (1 ɛ)d, for ɛ>0, which is just about as much as ossible. 3 We demonstrate exansion of = for three of the five ermutations that comrise the Gabber Galil exander [12]. These results hold for three-regular subgrahs of the Gabber Galil grahs of 2 inut vertices, where is a rime. This is done without using Eigenvalue based bounds. The actual Gabber Galil exansion was shown to be (2 3)/4 orabout Suose double hashing is based on randomly, indeendently, and uniformly choosing two hash functions h 1 and h 2 from a universal set [11]. Then this aer shows the exected number of robes for unsuccessful search in double hashing is bounded by 1, where α is the load factor. This holds regardless of the distribution of the inuts. This is analogous to Carter and Wegman s result for hashing with chaining Intuitive overview. Given three ermutations of the Gabber Galil exander grah, this aer shows no matter what subset of inuts (u to half of them) an adversary chooses, then there is at least 3 exansion. This is done in two stes while trading off the local and global structure of the grah. If the adversary allows enough local exansion, then we are done. Therefore, assume the adversary focuses on sufficiently restricting local exansion. In this case, the adversary must choose inuts in certain atterns. Now, in the second ste of our main result, it is shown that these atterns cannot block much global exansion. If the elements are in the aroriate local atterns to minimize local exansion, then the adversary has freedom to choose the number of elements in the atterns as well as where these atterns start. Certain global atterns are collision sequences (see Definition 3). Collision sequences reduce the global exansion. Constraining ourselves to local inut atterns, the exected length of all of these collision sequences is at

3 COMBINATORIAL EXPANDERS AND HASHING 85 most 2, no matter how the adversary chooses to osition the local atterns or how many elements the adversary chooses to ut in them. It is essential to note that showing the exected collision sequence length is at most 2 uses robability theory alied to the adversary s constrained selections of inut node atterns. Our argument shows the adversary has some very restricted choices of inut nodes in the three fixed ermutations of Gabber Galil s grah; otherwise, the adversary allows lots of local exansion. At all times, the three ermutations comrising the Gabber Galil grah remain fixed. The results are given by using robabilistic methods on these fixed grahs. Further, using virtually the same methods, start by randomly, uniformly, and indeendently selecting two universal hash functions h 1 and h 2 to build a double hashing table T. All elements will be ut in T by double hashing using h 1 and h 2. In this case, let T have fixed load factor α :1>α>0. Then we show the exected number of robes for an unsuccessful search in T, still using these initially chosen 1 hash functions, is. As in the case of our exander result, we show this using robabilistic techniques on fixed grahs Structure of this aer. Section 2 gives details of the three ermutations comrising the Gabber Galil exander and sets the foundations for showing both exansion as well as our hashing result. Section 2 has five subsections. Subsection 2.1 gives the actual grah construction. Next, subsection 2.2 defines local and global exansion. Subsection 2.3 exlains the relation of double hashing to the exander grah reresentation. Next, subsection 2.4 focuses on the results of Chor and Goldreich [9] showing randomly choosing such functions and comuting their values gives airwise indeendent and uniformly distributed values. Finally, subsection 2.5 bounds functions that are necessary for our final result. Section 3 uses our methods to show that randomly indeendently and uniformly selecting two double hash functions from a strongly universal set gives a double hashing result analogous to the classical result of Carter and Wegman [8] for hashing with chaining. Section 4 comletes the exander result, showing the subgrahs exand by 3 by enunciating the trade-off of local and global exansion. Finally, in section 5 we give our conclusions and tie together the notion of exansion with the notion of double hashing with universal hash functions. 2. Combinatorial exanders. This section gives the construction and starts the analysis of exanders without using eigenvalue bounds. Without loss, always assume that n = I = O and n = 2, where is a rime. Let Î denote the elements from I that an adversary selects from I in trying to foil any exansion. The adversary foils an exansion by selecting inuts in such a way so there are relatively few oututs. This section shows that no matter what set Î the adversary chooses, there is exansion The construction. This subsection constructs three-regular biartite grahs G 3 =(V,E) with vertices V = I Odenoting the inuts and oututs, resectively. This grah is made u of ermutations σ 0,σ 2, and σ 3 used in building Gabber and Galil s exander [12]. The ermutations comrising the Gabber Galil exander are very similar to the ermutations that make u Margulis exander. Only inuts can have edges to oututs. Let Z + = {0, 1,..., 1}. Partition the inuts I and the oututs O into blocks I j and O j, for all j Z +, containing

4 86 PHILLIP G. BRADFORD AND MICHAEL N. KATEHAKIS nodes each. In articular, for any j : >j 0, I j = { (j, 0), (j, 1),..., (j, 1) }, O j = { (j, 0), (j, 1),..., (j, 1) }. For notational convenience let (j, k) denote the kth element of both lists I j and O j for all j, k Z +. As an examle, consider = 3 in Figure 1. I 0 (0,0) (0,1) (0,2) (0,0) (0,1) (0,2) O 0 I 1 (1,0) (1,1) (1,2) (1,0) (1,1) (1,2) O 1 I 2 (2,0) (2,1) (2,2) (2,0) (2,1) (2,2) O 2 Fig. 1. The nodes in G 3 where =3. Now take I as I = 1 j=0 I j. Likewise, for O, O = 1 j=0 For any inut node (j, k) I j such that j Z + and k Z +, the grah G 3 has the following edges: 1. Identity: id(j, k) (j, k). 2. Local shift: loc(j, k) (j, (j + k + 1) mod). 3. Global shift: g(j, k) ((j + k) mod, k). These edges are directed from the inuts to the oututs. This does not affect the exansion since it is measured from how the inuts exand to the oututs. Likewise, these directed edges are consonant with the hashing result given in this aer. O j.

5 COMBINATORIAL EXPANDERS AND HASHING 87 (0,0) (0,1) (0,2) (0,3) (0,4) (1,0) (1,1) (1,2) (1,3) (1,4) (2,0) (2,1) (2,2) (2,3) (2,4) (3,0) (3,1) (3,2) (3,3) (3,4) (4,0) (4,1) (4,2) (4,3) (4,4) (0,0) (0,1) (0,2) (0,3) (0,4) (1,0) (1,1) (1,2) (1,3) (1,4) (2,0) (2,1) (2,2) (2,3) (2,4) (3,0) (3,1) (3,2) (3,3) (3,4) (4,0) (4,1) (4,2) (4,3) (4,4) Fig. 2. Local edges in G 3 where =5. Figure 2 gives local edges for G 3 where = 5, and Figure 3 gives global edges for inut blocks I 0 and I 1 in G 3 where = 5. The identity edges are not shown in either of these figures. Also see Gabber and Galil [12] or, for examle, Motwani and Raghavan [26]. Note in block I 1 the local shift edges degenerate as loc( 1,k)= ( 1,k) for all k Z +. Likewise, in nodes (j, 0) the global shift edges degenerate as g(j, 0)=(j, 0) for all j Z +. Therefore, these nodes (j, 0) for g and ( 1,k) for loc do not share all of the necessary roerties for exansion. Generally, this aer assumes the adversary does not select these degenerate elements. However, after the main theorems, Theorems 8 and 9, an accounting is made assuming the adversary does select degenerate elements. These mas are well defined on sets. So id(s) loc(s) g(s) =N (S) O for any set of inuts S I. Further, g 1 (j, k), loc 1 (j, k), and id 1 (j, k) denote the first comonent of the air, while g 2 (j, k), loc 2 (j, k), and id 2 (j, k) denote the second element. An instance of this subcase of the Gabber Galil exander is G 3 =(O I, id(i) loc(i) g(i)) The analysis. An adversary, who knows G 3 s construction, selects sublists Î j from each block I j. A sublist may be emty. This aer shows that no matter what elements the adversary selects, the grah G 3 exands. This aer assumes u

6 88 PHILLIP G. BRADFORD AND MICHAEL N. KATEHAKIS (0,0) (0,1) (0,2) (0,3) (0,4) (1,0) (1,1) (1,2) (1,3) (1,4) (2,0) (2,1) (2,2) (2,3) (2,4) (3,0) (3,1) (3,2) (3,3) (3,4) (4,0) (4,1) (4,2) (4,3) (4,4) (0,0) (0,1) (0,2) (0,3) (0,4) (1,0) (1,1) (1,2) (1,3) (1,4) (2,0) (2,1) (2,2) (2,3) (2,4) (3,0) (3,1) (3,2) (3,3) (3,4) (4,0) (4,1) (4,2) (4,3) (4,4) Fig. 3. Global edges in G 3,fromI 0 and I 1, where =5. to half of the inuts to be chosen by the adversary: Let 1 Îj j=0 Î = 1 j=0 n 2. Definition 1. Given block I j, for j Z +, the local L and global G exansions of I j are Definition 1 immediately gives Î j. L(Îj) = loc(îj) id(îj), G(Îj) = g(î) O j id(îj). 1 L(Î) = L(Îj) j=0 = loc(î) id(î)

7 COMBINATORIAL EXPANDERS AND HASHING 89 and 1 G(Î) = G(Îj) j=0 = g(î) id(î). Local and global exansion can collide in that outut nodes that give local exansion can also give global exansion. That is, there may be some Î Î, where loc(î )=g(î ). In this case, to comute the total exansion of Î just divide L(Î )+ G(Î ) by 2. Likewise, if local exansion and global exansion share outut nodes, then just consider the case that offers more exansion (if they do not offer the same exansion). For ease of exosition, when ossible we generally refer to the elements of the inuts I from here on. Each inut is directly associated with the element that it mas to by the identity maing. Definition 2. In a block Îj, for some j Z +, the element (j, k 1 ) Îj is loccontiguous iff loc(j, k 1 )=id(j, k 2 ) for k 2 =(j +k 1 +1) mod and (j, k 2 ) Îj. Aloccontiguous set is a list (j, k 1 ), (j, k 2 ),...,(j, k t ) all in Îj and loc(j, k s )=id(j, k s+1 ) for all s : t>s 1. If L(Îj) 1, for some j Z +, then the elements in Îj are loc-contiguous. Lemma 1. If there exists a fixed d :1 d>0, where d Îj id(îj) loc(îj), for all blocks I j such that j K Z +, where K and Îj, then L(Î) (1 d) j K Îj. Proof. First, since id(îj) loc(îj) d Îj so loc(îj) id(îj) (1 d) Îj, therefore it must be that L(Îj) (1 d) Îj. Since loc(îj) I j, this roof generalizes for the index set K. Definition 3. A collision sequence of length t is the maximal sequence of elements (j 1,k),...,(j t,k), where t 1, such that (j i,k) Îj i, for all i {1,...,t} and g(j 0,k) Îj 0 and (j t+1,k) Îj t+1, where g(j 0,k) (j 1,k), g(j 1,k) (j 2,k),... g(j t,k) (j t+1,k). So Length((j 1,k)) = t. For examle, if (j 1,k) Îj 1, but g(j 1,k) Ît, where t = g 1 (j 1,k), then (j 1,k)isa length 1 collision sequence starting in inut block I j1. Therefore, a collision sequence of length 1 starts and ends in the same block. Length 1 collision sequences do not diminish exansion but rather increase it. Definition 4. Suose the elements (j s,k) Îj s for all s : t s 1 form a collision sequence. The collision sequence (j 1,k) (j t,k) ends in block I jt if (j t+1,k) Îj t+1 and g(j t,k) (j t+1,k). Collision sequences revent global exansion. That is, if we have many long collision sequences, then there is not much oortunity for global exansion.

8 90 PHILLIP G. BRADFORD AND MICHAEL N. KATEHAKIS Definition 5. Consider (j, k 0 ), (j, k 1 ), and (j, k 2 ) all from I j so that (j, k 1 )= loc(j, k 0 ) and (j, k 2 )=loc(j, k 1 ), where (j, k 0 ) Îj, (j, k 1 ) Îj (j, k 2 ) Îj; so (j, k 1 ) is selected, then (j, k 1 ) is a singleton. A singleton has local exansion of 1. Definition 5 is about elements in the same inut block I j. A collision sequence has one or more selected inuts that are all in different inut blocks. In fact, a length t collision sequence containing u singletons gives total exansion of at least u + 1. The degenerate elements (j, 0) for all j Z + do not have global exansion since g(j, 0)=(j, 0). This means if an adversary can select an element (j, 0) to extend a loccontiguous set in Îj, then they should do it since it will not give any global exansion. That is, as long as (j, 0) would not be a singleton, then selecting it increases the number of elements selected but does not increase any exansion. Likewise, the degenerate elements ( 1,k) for all k Z + do not have local exansion since loc( 1,k)=( 1,k). Therefore, if selecting a ( 1,k) either extends one collision sequence or joins two collision sequences, then an adversary should select it. Selecting such an ( 1,k) will increase the number of selected elements without increasing exansion. In fact, if ( 1,k) joins two collision sequences, then it reduces the overall exansion Double hashing. Hashing with oen addressing is a storage and search technique on a table T that assumes the number of elements to be stored in the table is at most the table size: T. Elements or keys are ut directly in the table T. No ointers or data structures are used. There is a secial element NIL denoting no element in a osition it occuies. Given t elements in the table T, the load factor is α = t T and α<1. Generally, the imortant questions that have arisen for oen address hashing are related to the number of robes necessary to find elements in the table. Consider an oen addressing table T of size m and two hash functions h 1 and h 2. Given a key x, determining the (i + 1)st hash location using double hashing is done by h(i, x) =(h 1 (x)+ih 2 (x)) mod m. Double hashing is a classical data structure, and discussions of it can be found in [11, 24, 17], for examle. Inserting the element x into the table T is done by first searching for x in T. If T does not contain x, then x can be inserted into T. Likewise, to delete x from T, then it must be determined if x is in T as well as where x is located in T. Therefore, searching for an element x is the focus of studies of double hashing. The first robe to T is to osition T [h 1 (x) modm]. If T [h 1 (x) modm] = NIL, then x is not in T. Otherwise, if x is in T [h 1 (x) modm], then double hashing reorts where x is: osition h 1 (x) modm since i = 0. If x is not in T [h 1 (x) modm], then the next element robed is T [(h 1 (x)+h 2 (x)) mod m] since i = 1. If T [(h 1 (x)+h 2 (x)) mod m] is NIL, then x is not in T. Otherwise, if x = T [(h 1 (x)+h 2 (x)) mod m], then the double hashing algorithm is found where x resides and returns the value (h 1 (x)+h 2 (x)) mod m. Otherwise, x may still be

9 COMBINATORIAL EXPANDERS AND HASHING 91 in T. Therefore, element T [(h 1 (x)+2h 2 (x)) mod m] is robed, etc. This continues using the function in the (i + 1)st robe h(i, x) until either x is found or T [(h 1 (x)+ih 2 (x)) mod m] isnil, indicating x is not in T. In summary, the robe sequence is in the following addresses of T : h 1 (x), (h 1 (x)+h 2 (x)) mod m, (h 1 (x)+2h 2 (x)) mod m,.... Assume m>2 is rime and h 1 and h 2 are based on loc and g. For a double hash table T, this aer assumes T = m as well as the key x Z m. In the case of this aer s double hashing result, the g edges are the focal oint and the loc edges are not used. In this double hashing scheme, say the air (j, k) is generated for some key x by h 1 and h 2. That is, start in osition k in inut block I j. Hash function h 1 generates the first osition j (inut block) and hash function h 2 generates the ho-size k + 1 (how to travel from inut block to inut block). So the key x is hashed into I j, starting at local osition k. In other words, the first robe is in T [j]. If necessary, the second robe is in T [(j +(k + 1)) mod m]. If necessary, the third robe is in T [(j +2(k + 1)) mod m], etc. More recisely, first, a block j 0 and a osition k are chosen by h 1 and h 2, resectively. That is, given the key x, comute j 0 = h 1 (x) and k = h 2 (x). Next, as necessary, the following blocks are comuted: j 1 = g 1 (j 0,k) and, in general, j i = g 1 (j i 1,k), for i : m 1 L i 1, giving the ermutation j 0, g 1 (j 0,k), g 1 (j 1,k),...,g 1 (j L,k). Since m is rime, g sends this ermutation exactly once through each of the inut blocks Î0, Î1,...,ÎL, where L m 1. The grah G 3, since m a rime, with the functions g reresents all ermutations used by oen addressed double hashing on a table T [0,...,m 1]. Of course, T [j] corresonds to I j. Double hashing aroximates uniform oen address hashing [26, 11, 24]. More recisely, Guibas and Szemerédi [14] showed unsuccessful searches using double hashing take asymtotically the same number of robes as idealized uniform hashing does for any fixed load factor α less than about For any fixed α<1, see Lueker and Molodowitch [20]. However, as ointed out in Schmidt and Siegel [30], these last results assume ideal randomized functions, whereas [30] utilizes more realistic k-wise indeendent and uniform functions (where k = c log n for a suitable constant c). Theorem 1 (see [20] and [30]). Suose T has load factor of any fixed α<1. The exected number of robes for an unsuccessful search in an oen addressing double 1 hashing table is + ɛ, where ɛ is a lower-order term. Lueker and Molodowitch [20] give the most straightforward method of showing this based on assumed randomized inuts. Schmidt and Siegel [30] give the tightest bound (sharest bound on ɛ) and the weakest notion of randomness to date. That is, [30] shows the result of Theorem 1 by sulying randomized hash functions, in articular randomized hash functions of degree c log n for some constant c, giving c log n-wise indeendent functions Strong universal hash functions. Given a grah G 3, where α = Î / I for fixed α :1>α>0, let Îj / = α j, such that j Z +, and each fixed α j :1> α j > 0. This means α = Î / I = α α 1.

10 92 PHILLIP G. BRADFORD AND MICHAEL N. KATEHAKIS Consider any block Îj such that L(Îj) 1. In such blocks the adversary chooses the starting oint b j for the elements of I j, as well as the total number of elements to select from I j, exressed here as α j. More recisely, since L(Îj) 1, the adversary must have chosen the inuts so that id(îj) loc(îj) 1, leaving only the number of elements selected and their starting oint to question. Definition 6 (Carter and Wegman [8]). The set of function H is strongly universal iff randomly, uniformly, and indeendently choosing h H; then for any two different keys x 1 and x 2 and any two values y 1,y 2 Z +, it must be that Pr[h(x 1 )=y 1 ]= 1 and Pr[h(x 1 )=y 1 h(x 2 )=y 2 ]= 1 2. Theorem 2 (Carter and Wegman [8]). The functions give the strongly universal set h j,b (x) =jx + b mod for all (j, b) Z + Z + H = {h j,b for all (j, b) Z + Z + }. For all h H, the range is Z +. Generally, hash functions are exressed as h j,b (i) modm, where m is the table size, but here m =, allowing the focus to be entirely on h j,b (i) Counting frequencies of selected elements. The basic rogression from this subsection to section 3 works as follows. We start with the case where an adversary selects the same number of inut elements in each osition in all I j, for j : j Z +, while maintaining loc-contiguity of the elements in each Îj. Basic bounds on the exansion are develoed in this subsection. Subsequent subsections in section 2 incrementally allow an adversary to select any elements they choose as long as they maintain loc-contiguity. This subsection alies to both universal hashing as well as exansion. Theorem 3 (Chor and Goldreich [9]). Take h j,b H uniformly at random; then the associated values h j,b (i),...,h j,b (0), for >L i 1 and L 2, are airwise indeendent and for all i 0 the elements h j,b (i),...,h j,b (0) are uniform in Z +. Chor and Goldreich resent this result for the sequences of random variables h j,b (i),...,h j,b (1), and it is straightforward that h j,b (0) can be included since h j,b (0) = b, which is uniformly and randomly chosen. Theorem 3 will be alied to g functions between different blocks. Relations in the blocks are discussed next. Recall, for each block I j, the adversary chooses each b j as well as α j, and so Theorem 3 does not aly to each block. That is, Theorem 3 assumes the air (j, b) Z + Z + is randomly and uniformly chosen. In contrast to the strongly universal set H of Theorem 2, take U Z + Z + such that, for all j Z +, there is some air (j, b) U and further, if (j, b 1 ) U and (j, b 2 ) U, then b 1 = b 2.So H = {h j,bj, for all (j, b j ) U}, where b j deends on j.

11 COMBINATORIAL EXPANDERS AND HASHING 93 Note that H = ( 1) and H = 1. So, in our situation, selecting airs from H uniformly at random does not satisfy the hyothesis of this theorem because the adversary chooses each b j in each air (j, b j ) H. Definition 7. Now, for k Z +, denote the frequency n k = 1 j=0 δ((j, k) Îj), where δ is the indicator function, and so δ(true) =1and δ(false) =0. That is, n k is the frequency of k being selected in all blocks given the adversary s choices of the b j s and the α j s. Note that if n k = 0, then k does not contribute to exansion or lack of exansion. Further, if n k = 1, then k must contribute to global exansion by one. Aggregating the frequencies gives α = n k. Lemma 2. Suose n 1 = n 2 = = n 1 and assume L(Îj) 1 for all j Z +. Take any randomly and uniformly chosen (J 1,K) Z + Z +, where (J 2,K)= g(j 1,K); then k=0 Pr[(J 1,K) ÎJ 1 (J 2,K) ÎJ 2 ] n2 k 2. Proof. Assume (J 1,K) Z + Z + considering the collision sequence, is randomly and uniformly chosen. So we are C J1,K =(J 1,K) (J 2,K), where (J 2,K)=g(J 1,K). This gives Pr[(J 1,K) ÎJ 1 (J 2,K) ÎJ 2 ] = Pr[(j, k) Îj g(j, k) Îg 1 (j,k)] j=0 k=0 = δ((j, k) Îj) Pr[g(j, k) Îg 1 (j,k) (j, k) Îj] k=0 j=0 = = 1 k=0 n2 k 2. k=0 n 2 k 3 Pr[g(j, k) Îg 1 (j,k)] by indeendence and uniformity of Theorem 3 n k 1 j=0

12 94 PHILLIP G. BRADFORD AND MICHAEL N. KATEHAKIS This comletes the roof. If all n k α, then In other words, let 1 k=0 n 2 k 3 α2. Average[n 2 k]= 1 k=0 n 2 k. If n k α, for all k : >k 0, then Average[n 2 k] α 2 2 so that for uniformly and randomly chosen (J 1,K) Z + Z + and (J 2,K)=g(J 1,K), then Pr[(J 1,K) ÎJ 1 (J 2,K) ÎJ 2 ] Average[n2 k ] 2 Furthermore, if there is a set K = {k 0,...,k r } and K δ, where δ :1>δ 0, such that α 2. n k > α( 1) for all k K, then, letting K = Z + K, this gives n k α for all k K. This means Average[n 2 k]= n 2 k + n 2 k k K k K δ 2 δ + α α2 2 ( δ ) Therefore, if K δ for any δ :1>δ 0, and for uniformly and randomly chosen (J 1,K) Z + Z + and (J 2,K)=g(J 1,K), then Pr[(J 1,K) ÎJ 1 (J 2,K) ÎJ 2 ] α 2 + O( δ 1 ) for δ :1>δ Bounding Average[n 2 k ] when n k > α for k Z +. Consider the maximal subset K Z +, where all k K are such that the frequencies n k > α when L(Î) < 3 Î. Start with the case L(Îj) 1 for all j Z +. This subsection shows for all k K Z + such that n k > α there can be a total of at most 1/α total selected elements in all collision sequences of length more than α for all k Z +. Definition 8. Let α = Î / I. If there is some k Z+ so that k s frequency n k is such that n k > α, then there are n k α excess elements selected in the kth osition of I.

13 COMBINATORIAL EXPANDERS AND HASHING The case with u to 1 excess element selected. Here k 0 denotes a single excess element. Definition 9. Let n k,l denote any n k when all j r {j 0,...,j t 1 } are all such that Îj r = L. Further, by Definition 9 it must be that α = L/ and all n k s throughout the rest of this section are associated only with the blocks indexed by {j 0,...,j t 1 }. For the next lemma recall h ji is a hash function reresenting the loc functions corresonding to Îj i. Lemma 3. Let L(Îj) 1 for all j Z +. Assume k 0 is selected in all of I j0,...,i jt 1 and Îj r = L, for j r {j 0,...,j t 1 }, t L 2, with h j0 (v) = = h jt 1 (v) =k 0 for some v Z + ; then n k (α 1 ) for all k k 0 and α = L/. Proof. The roof is by induction on the size of Îj r = L. For the moment, assume v = 0 for the induction; this will be generalized after the induction is comlete. Basis. Consider the case where Îj r =2=L for all j r {j 0,...,j t 1 }, where t L, and h j0 (0) = = h jt 1 (0) = k 0. Then all elements u r =(j r +1+k 0 )mod, for all r : t>r 0, are distinct since {j 0,...,j t 1 } are distinct and t 2. That is, if u rx = u ry, then (j x +1+k 0 )mod =(j y +1+k 0 )mod, and so j x = j y mod, and it must be that j x <and j y <, and thus x = y, a contradiction. So n k = 1, for all k k 0, and α = 2. Further, all k k 0 are such that n k (α 1 ) t, comleting the basis. Inductive hyothesis. Suose Îj r = L 2, where j r {j 0,...,j t 1 } and t L. Then n k,l (α 1 ) t for all k k 0 and α = L. Inductive ste. Suose Îj r = L+1, where j r {j 0,...,j t 1 } and t L+1 3, where α is associated with n k,l+1. Then by the inductive hyothesis the first L elements in each block share the roerty that all but one of the n k,l s are such that n k,l (α 1 ) t. Adding one element to each block and each in a unique osition gives n k,l+1 (α ) t. Each element is ut in a unique osition since t L, and no two elements among u r =(L + 1)(j r +1)+k 0 mod, for r {t 1,...,0}, are the same: If j x j y, where ((L + 1)(j x +1)+k 0 )=((L + 1)(j y +1)+k 0 )mod, then since each element of a field (mod-) has a multilicative and additive inverse, we must have j x = j y mod, a contradiction since j x <and j y <. Therefore, alying the inductive hyothesis comletes the induction. Now we show this lemma holds for any v Z +, where h j0 (v) = = h jt 1 (v) = k 0. Consider Îj r = L for all j r {j 0,...,j t 1 } and t L, and given some v 0, then break the roblem into two cases: The first case consists of all elements h j0 (i),...,h jt 1 (i) for i : L>i v. The second case consists of all elements h j0 (i ),...,h jt 1 (i ) for i : v i 0. (Note that these cases overla since they both have k 0 in common.) Now, treating these cases searately, aly the induction above with α 1 = L v to the L v elements of h j0 (i),...,h jt 1 (i) for i : L>i v. Likewise, for each h j0 (i ),...,h jt 1 (i ), where i : v i 0, aly the induction above to the v elements. Here α 2 = v+1.

14 96 PHILLIP G. BRADFORD AND MICHAEL N. KATEHAKIS Now α = α 1 + α 2 1 since k 0 was counted twice. With this in mind, take the following inequalities, where k k 0 : ( n k α 1 1 ) ( t + α 2 1 ) t ( α 1 + α 2 2 ) ( α ) ( α 1 ), comleting the roof. With a little work, Lemma 3 generalizes to Theorem 4. Assume L(Îj) 1, for all j Z + and k 0, is selected in each block I 0,...,I 2. Let L u be the number of elements selected going above and including k 0. Similarly, let L d be the number of elements selected going down from k 0 but not including k 0. (If k 0 = h(i), then h(i + c) is above for any integer c, where i + c<, and h(i c) is down, where i c 0.) The induction is about the same; the only difference is the roof of Theorem 4 assumes the relation n k (α u + α d ) t holds before the inductive ste. More recisely, suose t = c for some integer c, and by definition α u + α d = Lu+L d, meaning ( ) Lu + L d t L u + L d. c So there are a total of L u + L d elements selected er inut block, and 1 c bounds the ercent of blocks under consideration. The inductive hyothesis says Lu+L d c is an uer bound on the number of elements for each k k 0. Now consider adding one element to each block going u and one element to each block going down. That is, increase α u to α u + 1 and increase α d to α d + 1, but at the same time none of the new u elements collide with each other and none of the new elements going down collide with each other. That is, for all r {0, 1,...,t 1}, u r =(L u + 1)(j r +1)mod are all different by the uniqueness of multilicative inverses in Z + {0}. Likewise, for all r {0, 1,...,t 1}, d r =(L d + 1)(j r +1)mod are all different by the uniqueness of multilicative inverses in Z + {0}. Furthermore, by the uniqueness of multilicative inverses in Z + {0}, for each d i there can be at most one u j so that d i = u j, where i, j {0, 1,...,t 1}. Consider increasing L u to L u + 1 and increasing L d to L d + 1, and assume t = c, for some integer c, giving ( ) Lu + L d +2 t L u + L d +2, c

15 COMBINATORIAL EXPANDERS AND HASHING 97 which is the number of selected elements er inut block multilied by the ercent of the inut blocks. Therefore, n k (α u + α d ) t + 2 t, giving, for all k k 0, ( n k α u + α d + 2 ) (1) t, which clearly holds for t>/2 and (α u + α d ) t bounded by an integer. In the case where t</2, then since L u + L d + 2 elements were selected er inut block and their loc-continuity gives for each k k 0, by the inductive hyothesis cn k L u + L d and by the uniqueness of multilicative inverses, n k can increase no more than 2 when both L u and L d are increased by 1 each. That is, now cn k L u + L d + 2 holds, comleting the inductive ste. This gives the next theorem. Theorem 4. Let L(Îj) 1 for all j Z +. Assume k 0 is selected in all of I j0,...,i jt 1 and Îj r = L, for j r {j 0,...,j t 1 }, t L 2, with h j0 (v) = = h jt 1 (v) =k 0 for some v Z + ; then n k α t for all k k 0 and α = L/. The next lemma allows any of t blocks to have any number of elements L selected as long as t L. That is, Îj r L for all j r {j 0,...,j t 1 } and t L. Lemma 4. Let L(Îj r ) 1 for all j r {j 0,...,j t 1 }. Assume k 0 is selected in all of I j0,...,i jt 1 and Îj r L for j r {j 0,...,j t 1 } and t L 2 while h j0 (v) = = h jt 1 (v) =k 0 for some v Z + ; then n k α t for all k k 0 and α =( Îj Îj t 1 )/(t ). Proof. Consider two sets T 1 and T 2 with s : t s 0, so that and and so That is, T 1 = { Îj 0,...,Îj s 1 }, where α 1 = Îj k for all k : s 1 k 0, T 2 = { Îj s,...,îj t 1 }, where α 2 = Îj k for all k : t 1 k s. Therefore, there is a total of α t selected elements in all of the blocks { Îj 0,...,Îj t 1 }, α t= α 1 s+ α 2 (t s). α t = α 1 s + α 2 (t s). Without loss, assume α 1 <α 2 and T 1 reresents the first s inut blocks. Now, alying Theorem 4 to all t inut blocks considering only α min = min{ α 1,α 2 },it must be that n 0 k α min t for all k k 0, and each n 0 k is comuted restricting k to the α min t elements of each of all t inut blocks. Note that Theorem 4 alies to the first α min t loc-contiguous elements from each block since t L. Without loss, assume α min t is an integer. Now, letting α max = max{ α 1,α 2 }, then since t L and assuming t = c for some integer c, then alying Theorem 4 n 1 k (α max α min ) t

16 98 PHILLIP G. BRADFORD AND MICHAEL N. KATEHAKIS for all k k 0. But, not considering the α min t elements, it must be that n 1 k (α max α min )(t s). Without loss, assume that (α max α min )(t s) and α min t are integers. This means, for k k 0, and since the elements reresented by α max and α min are loc-contiguous, n 0 k + n 1 k α min t +(α max α min )(t s) α max (t s) α min (t s)+α min t α max (t s)+α min s α 1 s + α 2 (t s) α t, since α max = α 2 and α min = α 1 and α t = α 1 s + α 2 (t s), while at the same time n 0 k + n1 k > (α 1 + α 2 )(t s) only for k = k 0 ; the roof is comleted by induction on the number of sets of blocks, each set containing blocks with the same number of elements selected. Now consider combining loc-contiguous collision sequences as described by Lemma 4. In articular, now look at bounding the length of all collision sequences in G 3 by combining different loc-contiguous subblocks. The next definition generalizes Definition 2. Definition 10. Take j Z +. The set U j,s is a maximal loc-contiguous subblock of Îj if U j,s Îj. And if U j,s 2, while L(U j,s ) 1, and for any U Îj : U j,s U and U j,s U, then L(U ) > 1. Further, if U j,0 = I j (so U j,0 = ), then U j,0 is the only maximal loc-contiguous subblock of Îj. Note that a maximal loc-contiguous subblock U j,s must contain at least two elements; otherwise, it is not loc-contiguous. Further, a block Îj may have many maximal loc-contiguous subblocks. Allow maximal loc-contiguous subblocks to be emty. This means maximal loc-contiguous subblocks cannot consist of a single loccontiguous element. Definition 11. Consider the collision sequence C 1 made u of selected elements from the max loc-contiguous subblocks U 0,0,...,U 1,0. Another collision sequence C 2 overlas with C 1 iff C 2 consists of elements from at least one of the same max loc-contiguous subblocks U 0,0,...,U 1,0. So now remove from consideration all subblocks associated with any collision sequence in c 0, i.e., U 0,0,...,U 1,0. Now with the remaining elements, ut the next largest collision sequences in a set c 1. Any collision sequence C r c 1 is associated with the sets of max loc-contiguous sequences U 0,r,...,U 1,r. By Lemma 4 and Theorem 4 all elements of any other collision sequence can share no more than α 1 elements with U 0,r,...,U 1,r, where α 1 =( U 0,r + + U 1,r )/ 2. This argument extends to all collision sequences larger than α. Lemma 5. Let K be the set of indices of all s collision sequences larger than α. Suose the adversary selects no singletons and α i =( U 0,i + + U 1,i )/ 2, where n i k is n k restricted to the jth set of max loc-contiguous subblocks U j,i, for i, where i : s i 0 and all j : 1 j 0. So for any k Z + K, then ( n 0 k + + n s k α 0 1 ) ( + α 1 1 ) ( + + α s 1 ).

17 COMBINATORIAL EXPANDERS AND HASHING 99 Proof. Without loss, let C 0, C 1,...,C s be collision sequences of lengths larger than α. Assume these are listed largest (C 0 ) to smallest (C s ). No two collision sequences C i and C j, for i j, are made u of elements from the same max loccontiguous subblocks U j0,i,...,u j k 1,i for some i, by Lemma 4 and Theorem 4. In the case where two collision sequences share a max loc-contiguous subblock, then this max loc-contiguous subblock can be cut into two loc-contiguous subblocks. Using this fact, starting with the largest collision sequences first, each collision sequence is uniquely associated with a set of 1 loc-contiguous subblocks, one for each inut block I j, for j Z +. (Some of these loc-contiguous subblocks may be emty.) This means each collision sequence is associated with one loc-contiguous subblock from each inut block, U 0,0,...,U 1,0, U 0,1,...,U 1,1,..., U 0,s,...,U 1,s, and in some cases U j,i = { }. Let K = {k 0,...,k s } Z + be such that k K means n k α, and now takeitasn k = 1. If k Z + K and n i k is restricted to U 0,i,...,U 1,i, where α i =( U 0,i + + U 1,i )/( 1), then n i k ( α i 1 ), which gives the lemma. The next lemma deals with the case when the number of elements in each block is larger than the total number of such blocks, or L>t. Lemma 6. Let L(Îj) 1 for all j Z +. Given t blocks I j0,...,i jt 1, where Îj r L, for j r {j 0,...,j t 1 } so that α =( Îj Îj t 1 )/(t ), and letting S = id 2 (Îj 0 ) id 2 (Îj t 1 ), now let T = S ; t then at most T of the n k s are such that n k > α t. Proof. Take the L elements in each block and consider them in T = S t loccontiguous sets of inuts of size t each, in every block. If T = 1, then we are done. Next, consider each set of t selected loc-contiguous inut elements among the t blocks; then assume there is an inut set {i 0,...,i t 1 }, so that h j0 (i 0 )= = h jt 1 (i t 1 ). If not, then consider the largest such inut set for each of the t selected loc-contiguous sets of elements. By Theorem 4 and since there are u to t elements in each set of loc-contiguous elements, then each set of loc-contiguous blocks alone has at most one n k such that n k > α s t, for α s = α /T, for some k. This means, among all T size t loc-contiguous sets among the t inut blocks, there will be at most T elements n k i > α s t for α s = t L α and i : T 1 i 0. Let K = { k 0,...,k T 1 } be the set of n ki elements so that n ki > α s t. Let n l k denote n k restricted to the lth set of loc-contiguous blocks. In other words, n l k is the number of times k occurs in the lth set of loc-contiguous blocks.

18 100 PHILLIP G. BRADFORD AND MICHAEL N. KATEHAKIS Next, without loss, suose that α s t = Lt is an integer, and considering all loc-contiguous sets at the same time gives, for any k i K, n 0 k i + + n T 1 k i (α s + + α s ) t, where there are T total α s α t, by Lemma 5. This comletes the roof. Suose there is a collision sequence of length t made of one selected element from each of Îj 0,...,Îj t 1.Ift α, then Lemma 6 indicates that S T α 1, α since S, and where S = id 2 (Îj 0 ) id 2 (Îj t 1 ). That is, suose this occurs in a set of t inut blocks where S>tand the largest collision sequences are of length at least α. The only selected elements in excess of α in T = S α large collision sequences can be larger than α. Since T 1/α, if all of these T large collision sequences consist of elements each, there is a total of at most α α α excess elements out of a total of 2 ossible elements and α 2 selected elements. That is, the uniform random robability of selecting an element that extends a collision 1 sequence to one of these excess elements is at most α. It is also ossible that the adversary chooses t> α blocks that have more elements selected in each block than α 2, where α = Î / I. Take the case where α >α, where α = Î / I, and take α = L for aroriate L, but say α L+d, for some integer d, for all Îj 0,...,Îj t 1. This means L + d T =, α but since α = L, it must be that α= L, and therefore d T =1+. α Assuming d> α, then T 1+ 1 α, since d<, and thus say d = c for some number c 1; then d = α c α 1 cα 1 α since 1 >αand c 1. So now we discard any case where L>tby this discussion and Lemma 6.

19 COMBINATORIAL EXPANDERS AND HASHING When more than one excess element is selected in Î. The next theorem assumes no more than one loc-contiguous subblock is selected er block I j for j Z +. Therefore, L(Îj) 1 for all j Z +, which means at most total elements of all of the n k s are such that n k > (α 1 ). Definition 12. Given the frequency n k, then B[n k ] Z + is the set of block indices that have element k selected in each of them. That is, u B[n k ] iff δ((u, k) Î u ). Theorem 5. Suose L(Îj) 1 for all blocks j Z +. Then the total number of excess elements in G 3 is at most 1/α. Proof. Consider the frequencies, n k0 n k1 n ks > α. By Lemma 5, this roof must consider only the elements {k 0,...,k s } Z +. Further, by Lemma 6, at most T = 1/α frequencies are such that B[n ki ] B[n kj ] α for k i k j. Now, if all k, k {k 0,...,k s } are such that then clearly B[n k ] B[n k ], s n ki, i=0 which would comlete the roof. Furthermore, if any subset {u 0,...,u t } {k 0,...,k s } is such that for all k i {k 0,...,k s } and for all u i {u 0,...,u t }, then we need only consider the set B[n ui ] B[n ki ]=, K = {k 0,...,k s } {u 0,...,u t }. Given two distinct collision sequences, then by Lemma 4 these collision sequences can share no more than α elements as long as L(Îj) 1. This means B[n k0 ] (B[n k1 ] B[n ks ]) α, and thus removing the block indices B[n k0 ] and by alying Lemma 4 again gives B[n k1 ] (B[n k2 ] B[n ks ]) α. The roof is comleted by induction on the remaining blocks B[n k2 ],...,B[n ks ]. That is, if L(Îj) 1, for all j Z +, then the total number of excess selected elements is 1/α. This means the robability of uniformly and randomly selecting one of these u to excess elements is at most α = 1 2 α. Next, this is generalized to the case where L(Î) 3 Î. In this case, this subsection concludes by showing for all k Z + that Average[n 2 k ] is bounded so that randomly, indeendently, and uniformly choosing (J 1,K) from Z + Z +, where (J 2,K)=g(J 1,K), gives Pr[(J 1,K) ÎJ 1 (J 2,K) ÎJ 2 ] α 2 + ɛ,

20 102 PHILLIP G. BRADFORD AND MICHAEL N. KATEHAKIS where ɛ = O( 1 ). As before, start assuming L(Îj) 1 for all blocks j Z +. Take the relations, n k0 n k1 n ks > α, so that n k α for all k Z + K, where K = { k 0,...,k s }. Theorem 6. Suose L(Îj) 1 for all j Z + and α = Î / I. Take any randomly and uniformly chosen (J 1,K) Z + Z +, where (J 2,K)=g(J 1,K); then Pr[(J 1,K) ÎJ 1 (J 2,K) ÎJ 2 ] α 2 + ɛ, where ɛ is of lower-order terms. Proof. If there is at most one frequency n k0 > α, then for all of the 2 or more nonexcess elements in G 3, it must be that Pr[(J 1,K) ÎJ 1 (J 2,K) ÎJ 2 ] α 2 holds by Lemma 2. Consider the frequencies n k0 n k1 n ks > α, where s 1. The case of the u to elements in n k0,...,n ks, the robability of them extending a one-element collision sequence with excess elements by Theorem 5, is α 2 = 1 α. Thus, for all the elements in G 3 it must be that Pr[(J 1,K) ÎJ 1 (J 2,K) ÎJ 2 ] α 2 + O ( ) 1 holds. This comletes the roof. If L(Îj) > 1 for some j Z +, then there may be many collision sequences that must be considered. Start with G 3 where the adversary has selected Î inut elements. Then consider any set of at most 3 Î sets of associated loc-contiguous elements, where each associated set contains a common collision sequence. By alying Theorem 6 to each collision sequence created by increasing local exansion to decrease global exansion (by extending or joining collision sequences) gives the next theorem. Note that each extended collision sequence has some associated α i <α= Î / I and α 1+ +α u = α, and so α αu 2 α 2. This is because α α 2 u (α α u ) 2 α 2. 3 Theorem 7. Let α = Î / I. Suose L(Î) < Î and all selected elements areinaloc-contiguous subblocks and there are no singletons. Take any randomly and uniformly chosen (J 1,K) Z + Z +, where (J 2,K)=g(J 1,K); then where ɛ is of lower-order terms. Pr[(J 1,K) ÎJ 1 (J 2,K) ÎJ 2 ] α 2 + ɛ,

21 COMBINATORIAL EXPANDERS AND HASHING Variations on hashing. Given an oen addressing hash table T with a fixed load factor of α :1>α>0, assume T is filled using double hashing to load factor α. As discussed in subsection 2.3, double hashing uses two hash functions h 1 and h 2. The goal of this section is to show that if both hash functions h 1 and h 2 are randomly, uniformly, and indeendently chosen from the strongly universal hash functions H (Definition 6), then the exected cost of an unsuccessful search using double hashing 1 is table accesses. This question was suggested by Carter and Wegman [8]. Another imortant form of hashing is hashing with chaining; see, for examle, [11, 13, 24]. Carter and Wegman [8] showed that given any strongly universal set of hash functions H, then randomly and uniformly selecting a hash function h H gives an exected chain length of at most 1 + α for fixed load factor α > 0. For instance, taking the set of hash functions H with domain and range Z + as in Definition 6, Carter and Wegman s result is imortant since the strongly universal set H (of size O( 2 )) behaves as if randomly selecting a function from the set of all functions from Z + to Z + (of size O( )). See Mehlhorn [24, 23] for lower bounds on the sizes of universal hash sets. As future research they suggest extending such an analysis to double or oen hashing. Schmidt and Siegel [30] and Siegel [31] answer this, giving c log n-indeendent functions that are comutable in constant time for a standard word model random access machine. Their results are quite general; see also [32]. Next, we focus on another answer to Carter and Wegman s question using the standard set H of strongly universal hash functions, see Definition 6, as they are reresented in the G 3 grah. Although this aer uses a different model, the g-edges in G 3 make selecting entire blocks simulate twowise indeendent functions; see Theorem 3. The G 3 grah can reresent a double hashing configuration if all elements in each inut block are either all selected or all unselected. That is, say each block Îj / = α j is such that either α j =1orα j = 0. This gives a fixed load factor α = Î / I, where 1 >α>0. Each entire inut block corresonds to a cell in the hash table T, where T is of size. So, if α j = 1, then T [j] is full; and if α j = 0, then T [j] is emty. If one wants to build a double hashing table, do this by making two indeendent and uniform random choices h 1,h 2 H, where H is the strongly universal set described in subsection 2.4. So, given a key x, the value j 0 = h 1 (x) is the first table element T [j 0 ] to robe. Now, if T [j 0 ] is full and T [j 0 ] x, then robe the values T [(j 0 + ih 2 (x)) mod ] for i =1,..., 1, until encountering the key x or an emty table element (NIL). Next, this aer shows that building a hash table by double hashing with the initial uniform and indeendent random choices h 1,h 2 H gives an oen addressing 1 table of load factor α with exected number of robes for an unsuccessful search, regardless of the inut distribution. When searching through a hash table for x, a collision sequence equates to a robe sequence (j 0 + ih 2 (x)) mod, given j 0 h 1 (x) and h 1,h 2 both randomly, uniformly, and indeendently chosen from a strongly universal set H. In this context, consider the collision sequence C J1,K starting in osition (J 1,K). So let J 1 and K be randomly, uniformly, and indeendently chosen. Since J 1 is indeendent of J 2 and (J 2,K)=g(J 1,K), then Pr[Length(C J1,K) 2] = Pr[Length(C J1,K) =1 Length(C J2,K) 1 (J 2,K)=g(J 1,K)]. In the next roof, lower-order terms that would aear if the adversary selected

22 104 PHILLIP G. BRADFORD AND MICHAEL N. KATEHAKIS degenerate elements are ignored. Proosition 1 (hashing collision sequence). Say T is an oen address hash table with any configuration of elements built by initially randomly, uniformly, and indeendently choosing h 1,h 2 from H and then erforming double hashing to insert elements into T. Let α = Î / I. Assume each block I j is such that Îj / = α j and either α j =1or α j =0. Then the exected collision sequence length in T is E[Length(C)] α. Proof. For any fixed S = {α i0,...,α is } {α 0,...,α 2 } let s< 2and α ir =1 for all i r {i 0,...,i s } and α j = 0 for all j {i 0,...,i s }. This means α = s+1. In this case, since each full block (α ir = 1) has exactly one edge going to all other blocks, then the fraction 1 s+1 of all collision sequences starting in any full blocks are of length exactly 1. Now the claim that E[Length(C)] α is shown by induction. Let C J1,K be a otential collision sequence that asses through at any randomly and uniformly chosen (J 1,K) Z + Z + ; then E[Pr[Length(C J1,K) 1]] = α since α is the robability of randomly and uniformly selecting an inut element. Basis. Since α = s+1, then randomly and uniformly choosing (J 1,K) Z + Z +, where (J 2,K)=g(J 1,K), gives ( s +1 Pr[(J 1,K) ÎJ 1 (J 2,K) ÎJ 2 ]= α 2 )( ) s by Lemma 2 noting that since α ir = 1, for all α ir S, then n 0 = = n 1 = α. Further, if T [J 1 ] is full, then Pr[(J 2,K) ÎJ 2 ]=α 1 = s and s <α. Inductive hyothesis. For some c 2, and for all i<c, assume for any collision sequences C J1,K that Pr[Length(C J1,K) i] α i. Inductive ste. Take c 2 and consider t c; then we claim for all otential collision sequences C J1,K, where (J 2,K)=g(J 1,K), Pr[Length(C J1,K) t +1]=Pr[Length(C J1,K) =1]Pr[Length(C J2,K) t]. To substantiate this claim, take a otential length t + 1 collision sequence C J1,K and suose the first robe starts in block I J1, and so C J1,K =(J 1,K) (J 2,K) (J t,k) and J i = g 1 (J i 1,K), where c t i>1. It must be that Pr[Length(C J1,K) t +1]=Pr[Length(C J1,K) =1]Pr[Length(C J2,K) t] α Pr[Length(C J2,K) t], which holds by airwise indeendence from Theorem 3 and, further, by the inductive hyothesis Pr[Length(C J2,K) t] α t + ɛ, comleting the induction. Now, for any t 1, since the random variable Length(C J1,K) is nonnegative, this means E[Length(C)] = Pr[Length(C) t] t 1 t 1 α 1 α. α t