50 years of Linear Probing Hashing

Transcription

1 50 years of Linear Probing Hashing Alfredo Viola Universidad de la República, Uruguay AofA 2013 Dedicated to Philippe Flajolet

2 The problem. The table has m places to hash (park) from 0 to m 1, and n elements (cars). Each element is given a hash value (preferred parking lot). If place is empty, then the element is stored there. Otherwise, looks sequentially for an empty place. If no empty place up to the end of the table, the search follows at location 0. Several R.V. to study, mainly related with cost of individual searches and total construction cost. Very important special case: Parking Problem. In parking the car is lost if no empty place up to the end. Main R.V. is the number of lost cars. Important variant: each location can hold up to b (or k) cars.

3 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

4 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

5 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

6 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

7 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

8 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

9 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

10 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

11 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

12 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

13 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

14 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

15 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

16 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

17 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

18 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, / 37

19 A discrete parking problem Limiting distribution results Analysis Further research A discrete parking problem: Example Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, cars are unsuccessful 5 / 37

20 A discrete parking problem Limiting distribution results Analysis Further research Further research: Bucket parking scheme Blake and Konheim [1976]: Bucket parking scheme Each parking lots can hold up to r cars Related to analysis of bucket hashing algorithms r 32 / 37

21 A discrete parking problem Limiting distribution results Analysis Further research Further research: Bucket parking scheme Blake and Konheim [1976]: Bucket parking scheme Each parking lots can hold up to r cars Related to analysis of bucket hashing algorithms r 32 / 37

22 A discrete parking problem Limiting distribution results Analysis Further research Further research: Bucket parking scheme Blake and Konheim [1976]: Bucket parking scheme Each parking lots can hold up to r cars Related to analysis of bucket hashing algorithms r 32 / 37

23 A discrete parking problem Limiting distribution results Analysis Further research Further research: Bucket parking scheme Blake and Konheim [1976]: Bucket parking scheme Each parking lots can hold up to r cars Related to analysis of bucket hashing algorithms r 32 / 37

24 A discrete parking problem Limiting distribution results Analysis Further research Further research: Bucket parking scheme Blake and Konheim [1976]: Bucket parking scheme Each parking lots can hold up to r cars Related to analysis of bucket hashing algorithms r 32 / 37

25 A discrete parking problem Limiting distribution results Analysis Further research Further research: Bucket parking scheme Blake and Konheim [1976]: Bucket parking scheme Each parking lots can hold up to r cars Related to analysis of bucket hashing algorithms r 32 / 37

26 A discrete parking problem Limiting distribution results Analysis Further research Further research: Bucket parking scheme Blake and Konheim [1976]: Bucket parking scheme Each parking lots can hold up to r cars Related to analysis of bucket hashing algorithms r 32 / 37

27 A discrete parking problem Limiting distribution results Analysis Further research Further research: Bucket parking scheme Blake and Konheim [1976]: Bucket parking scheme Each parking lots can hold up to r cars Related to analysis of bucket hashing algorithms r 32 / 37

28 A discrete parking problem Limiting distribution results Analysis Further research Further research: Bucket parking scheme Blake and Konheim [1976]: Bucket parking scheme Each parking lots can hold up to r cars Related to analysis of bucket hashing algorithms r 32 / 37

29 A discrete parking problem Limiting distribution results Analysis Further research Further research: Bucket parking scheme Blake and Konheim [1976]: Bucket parking scheme Each parking lots can hold up to r cars Related to analysis of bucket hashing algorithms r 32 / 37

30 Linear Probing Hashing.

31 The mathematical beauty of Linear Probing!

32 Some personal motivations... Simplest collision resolution strategy for open addressing [Peterson 1957]. Works well for tables that are not too full. Because of primary clustering, its performance deteriorates when the load factor is high. Its analysis leads to nontrivial and interesting mathematical problems. There are connections with tree inversions, tree path lenghts, graph connectivity, area under excursion, etc. Equivalent formulation in terms of the parking problem. First problem that D. Knuth analyzed [Knuth 1962] with bucket size 1, and motivated the collection "The Art of Computer Programming". The analysis for general bucket size b presents very interesting challenges. For example, can symbolic methods be used?

33 ... and a box full of surprises... The study of parking sequences and their deep relations with other problems in both discrete and continuous mathematics has been carried out by dierent research communities in parallel and with little communication among them. Several related problems have been studied by experts in probability, combinatorics and computer sciences. The methodological techniques to study these problems are very diverse, and cover a wide range of research areas. As it is said in [Chassaing and Flajolet 2003], a systematic historical approach to the problem is very dicult. I will concentrate on the contributions made by people from our community as well as some related work that have inspired their research.

34 1962: Summer work by Don Knuth...

35 First published paper.

36 Historical note on rst published paper.

37 Original results. Let a hash table with m positions and n inserted elements. Let P m;n the probability of the last position being empty. P m;n = 1 n m. Let C m;n the R.V. for the number of successful searches of a random element. E [C m;n ] = 1 2 (1 + Q 0((m; n 1)). E [C m;m ] = with 0 < 1. p E [C n n;n ] = + O(1) (proved on may 20, 1965). 8 Let U m;n the R.V. for the number of unsuccessful searches of a random element. E [U m;n ] = 1 2 (1 + Q 1((m; n 1)). E [U m;m ] = (1 ) 2 with 0 < 1. The Ramanujan Q function is the special case Q 0 (n; n) of! nx k + r n k Q r (m; n) = k m n : k=0

38 ... question by Ramanujan to Hardy in 1913.

39 The problem...

40 ... and the Ramanujan's Q function into play!

41 Methodology...

42 ... and more methodology!

43 Collision Resolution Strategies. In open addressing, when two keys collide, either one of them may stay in that location, while the other one keeps probing. First-Come-First-Served (standard). Each collision is resolved in favor of the rst record that probed the location. Last-Come-First-Served [Poblete and Munro 1989]. Each collision is resolved in favor of the incoming record. Robin Hood [Celis, Larson and Munro 1985]. Each collision is resolved in favor of the record that is further away from its home location.

44 Original proposal of Robin Hood. "Analysis of a le addressing method" [Schay and Spruth 1962]. Modication of the algorithm proposed by Peterson in This is the Robin Hood strategy! Expected value of search cost does not change, as already noted by D. Knuth in his original note. They do an analysis based on a Poisson approximation, and nd E [C m;m ] = with 0 < 1.

45 Two models to analyze the problem. Exact lling model. A xed number of keys n, are distributed among m locations, and all m n possible arrangements are equally likely to occur. Poisson model. Each location receives a number of keys that is Poisson distributed with parameter b, and is independent of the number of keys going elsewhere. This implies that the total number of keys, N, is itself a Poisson distributed random variable with parameter bm: P r[n e bm (bm) n = n] = : n!

46 Poisson Transform. Results in one model can be transfered into the other model by the Poisson Transform: X X P m [f m;n ; b] = P r[n = n]f m;n = e bm (bm) n f m;n : n! n0 n0 Inversion Theorem: [Gonnet and Munro 1984] If P m [f m;n ; b] = X k0 a m;k (bm) k then f m;n = X k0 a m;k n k (bm) k : The Poisson model is an approximation of the exact lling model when n; m! 1 with n=m = b with 0 < 1.

47 , Diagonal Poisson Transform. [Munro, Poblete, Viola 1997] Let a hash table of size m, with n + 1 keys, and let P be a property for (chosen uniformily at random). Let f m;n be the result of applying a linear operator (e.g. an expected value) to the probability generating function of P. X f m;n = Pr[ 2 cluster of size i + 1] f i+2;i i0 X n (m i 2) n i 1 (m n 2)(i + 2) i = i m n f i+2;i : i0 Then P m [f m;n ; ] = D 2 [f n+2;n ; ] with D c [f n ; ] = (1 ) X n0 e (n+c) ((n + c))n f n : n!

48 Combinatorial interpretation. Any Linear Probing Hash table can be seen as a sequence of almost full tables (a subtable with all but the last bucket full). Example: [3-3],[4-4],[5-5],[6-2]. This interpretation can be nicely handled by Analytic Combinatorics, since for example, it implies that it is enough to study almost full tables, and then use the sequence construction.

49 Distribution of individual displacements. Complementary probabilistic and combinatorial approaches. [Janson 2005, Viola 2005]. Let P (z) be the probability generating function for the displacement of a random element wth 0 < 1 and 2 ff CF S; RHg. Then, with T (x) = ze T (x) the tree function, P F CF S (z) = (1 T (ze )) 2 (1 ) 2 2(1 z) P RH (z) = 1 e z e ze e z : LCFS is much more challenging and [Janson 2005] presents an exact expression. Exact results follow by depoissonization. :

50 Parking sequences, acyclic maps, priority queues. [Seitz 2009] surveys some of these relations. [Gilbey and Kalikow 1999]: priority queues. Presents generalizations of the parking problem. Distributional analysis of the overow. Analysis of the overow in parking with buckets. g(m; n; k): the number of defective parking functions of defect k. Let G(z; u; v) = Then X X X m0 n0 d0 G(z; u; v) = 1 g(m; n; k) zn n! um v d. T (zu) 1 zv T (zu) z 1 u ezv : v [Cameron, Johannsen, Prellberg and Schweitzer 2008]. In [Viola 2005] parking is used as a subproblem for RH.

51 Parking problem: limit distributions [Panholzer 2008] Studies P rfx m;n = kg = g(m;n;k) m n. Limit distributions for X m;n. Nine regions depending on growth of m; n. n m: X m;n L! X with P rfx = 0g = 1 (degenerated law). n m; 0 < < 1: X L m;n! X, discrete limit law. p m := m n m: m X m;n L! X(d) = EXP(2). := m n p m; > 0: 0 := m n p m: 1 p m X m;n L! X (d) = LINEXP(2,). 1 p m X m;n L! X (d) = RAYLEIGH(2). 0 := n m p n: Xm;n+m p n n X L m;n! X (d) = RAY(2). := n m p n; > 0: Xm;n+m n p n X m;n L! X (d) = LE(2,). p n := n m n: m (X m;n + m n) L! X(d) = EXP(2). n m; > 1: (X m;n + m n) L! X, discrete limit law. m n: X m;n L! X with P rfx = 0g = 0 (degenerated law).

52 Total displacement (box full of surprises!).

53

54 4 Analytic Combinatorics Two basic principles 7! \dictionaries" SYMBOLIC METHODS Generating functions 7! z 11 z + z 2 + z z z z z z 8 + Analytic functions and singularities 10

55 CONSTRUCTIONS Dictionary (I) F 7! ff n g 7! f(z) = X n 1 1 f = 1 + f + f 2 + f 3 + f n z n n! : exp(f) = 1 + f + 1 2! f ! f 3 + A[B 7! A(z)+B(z) AB 7! A(z)B(z) Seq A 7! 1 1 A(z) Set A 7! exp(a(z)) Cycle A 7! 1 log 1 A(z) 11

56 COMPLEX ASYMPTOTICS Dictionary (II) Point of regularity. f(z) f(z 0 ) + f 0 (z 0 )(z z 0 ) exp(z) Point of singularity. :9 d dz f(z) z0 (1 z) 7! n 1 () p 1 z 7! n 3=2 (1 z) 3=2 7! n 5=2 12

57 Permutations: (1 z) x y Trees: 1 p 1 z y x

58 Distributional analysis (almost full tables) Record construction cost = total displacement F n (q) = n 1 X k=0 n 1 k F k (q)(1 + q + + q k )F n 1 k (q): q{calculus X n q n2 z 2 The construction cost is worst-case quadratic 17

59 Lemma. The generating functions associated with moments f r (z) = q r F (z; q) satisfy a linear ODE that involves T (z). zf 1 = 1 2 T 3 (1 T ) 2 ; zf 2 = 1 T 4 (24 11T + 2T 2 ) 12 (1 T ) 5 Theorem. [Full tables, exact form of moments] E[d n;n ] = n (Q(n) 1) 2 E[d 2 n;n] = n 12 (5n2 + 4n 1 8n Q(n)): Q(n) := 1 + n 1 (n 1)(n 2) (n 1)(n 2)(n 3) + n n n 3 Theorem. [Full tables, asymptotic form of moments] E[d n;n ] = Var[d n;n ] = p 2 4 n3=2 2 p 2 3 n + 48 n1= O(n 1 ); p n 3 + n n3=2 Knuth (1962-3); Flajolet-Poblete-Viola (1997); Knuth (1997) 20

60 6 Limit distribution A method of \pumping" moments Start from nonlinear combinatorial decomposition (BGF) [F (z; q)] = 0 Apply derivatives U@ r q to get rth moment. Expect linear operator L with Lf r = r [f 0 ; f 1 ; : : : ; f r 1 ] Solve exactly and/or or asymptotically (singularities) Method used on Quicksort, Hennequin 1989: 100 moments; nongaussian law Path length in trees, Takacs Area below walks, Louchard 1984 In situ permutation, Knuth 1972, Prodinger et alii Path length in Cayley trees: F (z; q) ze F (qz;q) = 0 21

61 Moment problem A classical theorem: if the \Moment generating function" M(z) = X r r z r r! has nonzero radius of convergence, then the law is uniquely determined by its moments. A corollary: Convergence of moments implies convergence of distributions Example. w(x) = e x, r r! =) M(z) = 1 1 z Theorem. For almost full tables, convergence to the Airy distribution, Prf d n;n 1 xg! PrfX xg (n! 1); (n=2) 3=2 where X is Airy distributed. E[X r ] = ( 1 2 ) ( 3r 1 2 ) r. X w r r = 2=3(w) r! 1=3(w) r0 (w) = 1 (4 2 1) w + (42 1)(4 2 9) w ! 24 (42 1)(4 2 9)(4 2 25) w 3 + : 3! 24 24

62 The Airy density (area under excursions). [Louchard 1984],[Takács 1991] The constants k satisfy The density!(x) has been calculated in [Takács 1991]. The constants k are the zeros of the Airy function Ai(z).

63 7 Sparse tables A table with m cells and n elements has g = m n \gaps". SparseTable := <Full> *... * <Full> Bivariate GF is: (F (z; q)) g The analysis can be \recycled" (---- g times ----) Theorem. For -sparse tables, = n, mean and variance: m E[dm;n] = n 2 (Q 0 E[d 2 m;n ] = 12 n (m; n 1) 1); (m n) 3 + (n + 3)(m n) 2 + (8n + 1)(m n) + 5n 2 + 4n 1 ((m n) 3 + 4(m n) 2 + (6n + 3)(m n) + 8n)Q 0 (m; n 1) : Q0(m; n) := 1 + n 1 (n 1)(n 2) (n 1)(n 2)(n 3) + m m 2 + m 3 + E[d m;n ] = 2(1 ) n 2(1 ) + 3 O(n 1 ); Var[d m;n ] = (1 ) 4 n Flajolet-Poblete-Viola (1997); Knuth (1997) 26

64 The limit distribution Theorem. A Gaussian law. Proof. Integral of large powers by saddle point [z n ](F (z; q)) m n = 1 2i I m n dz (F (z; q)) z n+1 1e-05 8e-06 6e-06 4e-06 2e y x Plus continuity theorem for characteristic functions q = e i Corollary. Works for large assemblies! E.g. [Mahmoud] Distribution sorts with O(n) buckets. 27

65 Linear Probing and random graphs (I).

66 Mails exchanged with Don Knuth. Date: Mon, 29 Sep :15: (PDT)... To: Subject: note from Don Knuth Dear Ph, Ordinarily I am not happy to receive , but in this case it was very touching to learn that you had decided to dedicate such a nice paper to me, just after I had (secretly) decided to dedicate reference [22] to you! But I haven't time to study it in detail now, as I'm working 150% time on the new edition of Volume Best regards, Don

67 Combinatorial approach to Linear Probing.

68 Combinatorial Analysis (FPV).

69 Solution to the fundamental recurrence (Knuth).

70 Conclusions (Knuth).

71 Properties of the parking function F n (q). Several combinatorial relations between very important combinatorial problems. [Kreweras 1980]

72 Parking, random graphs and random trees. [Spencer 1997]. BFS traversal of a random graph with vertices {0, 1,..., n}. Induces a queue (Q k ( )) 1kn and a spanning tree. BFS induces a parking sequence (A k ( )) 1kn. Ex: (A k ( )) = ff6; 8g; f2; 3g; ; f7g; f1; 4g; f5g; f9g; ; g. x k ( ) = ja k ( )j. Ex: (x k ( )) = f2; 2; 0; 1; 2; 1; 1; 0; 0g. y k ( ) = x 1 ( ) + x 2 ( ) + : : : + x k ( ) k + 1, size of queue (Q( )) before step k. Ex: (y k ( )) = f2; 3; 2; 2; 3; 3; 3; 2; 1g. y 1 ( ) + : : : y n ( ) n is the total displacement. Ex: 12. BFS induces a random walk excursion. Ex: b.

73 Parking sequences and random graphs. C n;k : # connected graphs with n vertices and n + k 1 edges. D n+1;n : RV for total displacement in parking with n cars. Theorem: C n;k C n;0 = E h Dn;n 1 i. k Sketch: How may graphs give the same with this BFS? Graph has the edges of plus some of the (y 1 ( ) 1)+ : : : + (y 1 ( ) 1) edges joining parked car with collisions. C n+1;k = X y 1( )+:::+y 1( ) n k h Dn+1;n i = E k (n + 1) n 1. (n + 1) n 1 is both the number of parking functions with n cars and labelled trees with n + 1 nodes.

74 Back to Knuth (1997). P.g.f. for the total displacement in parking with n cars: F n (q) F n (1) : Moreover and "!# Dn+1;n E = 1 F n (k) (1) k k! F n (1) ; F (1 + q) = X k0 1 k! F n (k) (1)q k : So, if F n (q) enumerates total displacements in parking with n cars, then F n (1 + q) enumerates connected graphs with n + 1 nodes discriminated by their excess!.

75 Inversions in Cayley trees. [Gessel and Wang 1979] The inversions are f4; 3g; f4; 2g; f6; 2g, and f6; 5g. Bijection with parking problem. Same functional equation. BFS starting at 1, visiting the greatest unvisited node rst. How many graphs share the same spanning tree?: Inversions! I n (t): # of inversions of a tree rooted at 1. C n (t) = t n 1 I n (1 + t).

76 Parking sequences, random excursions,... [Spencer 1997]. (A k ( )) = ff6; 8g; f2; 3g; ; f7g; f1; 4g; f5g; f9g; ; g. x k ( ) = ja k ( )j. Ex: (x k ( )) = f2; 2; 0; 1; 2; 1; 1; 0; 0g. y k ( ) = x 1 ( ) + x 2 ( ) + : : : + x k ( ) k + 1, size of queue (A( )) before step k. Ex: (y k ( )) = f2; 3; 2; 2; 3; 3; 3; 2; 1g. n! There are x 1!:::x n! parkings seqs. Q associated to (y 1 ; : : : ; y k ), n e with probability proportional to 1 i=1 x i, (n Po(1) i.i.d.).! Poisson R.W. with y 0 = 0, steps x i 1, conditioned to be positive on time 1; 2; : : : ; n and zero on time n + 1.

77 ... and the ineable Brownian excursion! (x k ( )) 0kn are Po(1) i.i.d. conditioned as above. Corresponding unlabelled tree is a Galton-Watson tree with Po(1) progeny, constrained to have n + 1 nodes, and x k is the progeny of k th node visited by the BFS. ybntc It is known that pn L! (e(t)) 0t1, ( L! denotes 0t1 convergence in law) with (e(t)) the Brownian excursion. R As a consequence Dn+1;n n p L 1 n! 0 e(t)dt: max k y k pn L! m = max e(t): 0t1 [Chassaing and Marckert 2001]. Convergence of moments. Coupling labeled trees-empirical processes using parking functions. Alternative probabilistic proof to [Flajolet, Poblete, Viola 1998] for the convergence of moments towards moments of the Airy law.

78 More asymptotic distributions: new cases. [Janson 2001] Four dierent methods used in the proof! There is a phase transition around m n p m. Let U m;n the average cost of an unsuccessful search beginning at a random cell h (averaged over h). Then

79 "Monkey Saddle" as the ALGO project's logo.

80 Distribution of lengths of parking blocks. [Pittel 1985]. Table with ` = m n empty places. B m;` k length of k th largest cluster. When `=m! with > 0 then B m;` 2 log m 3 log log m+ m 1 = ; where 2( 1 log ) m converges weakly to an extreme-value distribution. Moreover, from [Chassaing and Louchard 2002] we have Theorem: For n; m jointly to +1, 1 if p m = o(`); B m;` 1 =m L! 0; 2 if ` = o( p m); B m;` 1 =m L! 1: Phase transition occurs when ` = ( p m). Largest block reaches O(m) after p m cars arrive at the critical region. So, we have a coalescence phenomena. Studied with the help of Brownian motion theory.

81 Size of the rst created cluster. [Chassaing and Louchard 2002]. R m;n, size of the cluster of the rst arrived car. P r[r m;n = k] = n 1 (k+1) k 1 (m k 1) n k 1 (m n 1) k 1 m n. 1 P r[r m;n = k] (k + 1) k 1 e (k+1) k 1 (1 ), (0 < < 1). P r[r m;n = k] = 1 m f ; k m where f is the density of N 2 2 +N 2 (k 1)! f (; x) = 2 x 1=2 (1 x) 3=2 exp + O(m 3=2 ); m n = p m, (N standard Gaussian), with 1 ]0;1[ (x). 2 x 2(1 x) In other words, when m n = p m then Rm;n m L! : N 2 Limit cases: Rm;n m 2 +N 2. P Rm;n! 0 (! 1) and m P! 1 (! 0).

82 Distribution of the length of the clusters. B m;` k length of the k th largest cluster and B m;` = B m;` k Let e(t) the normalized Brownian excursion. sup Let e(x) = e(x) x 0yx (e(y) y). B() = (B k ()) k1 decreasing widths of e(x) excusrions. Theorem: If lim ` p m = > 0, then Bm;l m L! B(). k1. As a consequence, from [Pavlov 1977] in random forest (and its relation with the parking problem), Joint distribution of B() out of reach.

83 Asymptotic distributions of block lenghts.

84 Size-biased permutations R(). R() constructed from B(). 1 Choose R 1 () with P r(r 1 () = B k ()jb()) = B k (). 2 Same for R k () but with the terms that did not appear before. Let R m;` = R() appears in the -valued fragmentation process derived from the continuum random tree (CRT) in the standard additive coalescence. [Aldous and Pitman 1998]. R m;` k k1, the sequence of block sizes ordered by Rm;` k date of birth, and R k () = m with ` = n m = dp me. p ` m = > 0, then Rm;l L m! R(). Theorem: If lim

85 Coalescence: emergence of a giant block. Up to now parking frozen at xed : n() = m b p mc. To understand coalescence: understand the dependence between parking schemes at times n( 1 ) < : : : < n( k ). Joint distribution: (size(), initial position()) 0 of each block. Coalescence in the discrete model, translates into coalescence in the continuous model at the limit. Two constructions of the additive coalescent. [Aldous and Pitman 1998] CRT as a limit of a discrete model of coalescence-fragmentation process that starts with a random unrooted labelled tree (reverse of process of deleting edges at random). [ Bertoin 2000] based on xcursions of the family of stochastic processes ( e) 0. [Chassaing and Louchard 2002] prove that asymptotically parking schemes lead to this construction of the additive coalescent.

86 Parking schemes and Pavlov's forests.

87 Parking with buckets: distribution of overow. Let g b (m; n; d) the number of parking functions of defect d, with buckets of size b. Let G b (z; u; v) = Then [Seitz 2009] G b (z; u; v) = X X X m0 n0 d0 1 1 u v k e zv g b (m; n; d) zn n! um v d.! b Y1 j=1 b Y1 j=1 1 b 1 b zv T (!j u 1 b z b z T (!j u 1 b z b ) : ) Let w m;b;k be the probability of having k cars going to overow in a b-full table with m buckets of size b and < 1, and m (b; z) = P k0 w m;b;kz k. Then [Viola 2010] Q b 1 Q j=1 z b e b(z 1) b 1 b(1 )(z 1) m (b; z) = j=1 z 1 T (! j e ) T (! j e ) + O bm :

88 Linear Probing with Buckets: bucket occupancy. Let T d (b) be the probability that a given bucket has more than d empty places, when n; m! 1, 0 n=bm = < 1. From [Viola 2010], inspired in [Blake and Konheim 1977]: Theorem Q [u d b 1 ] i=1 T d (b) = b(1 ) Q b 1 j=1 1 1 u T (!i e ) ; 0 d b 1; T (! j e ) where T is the Tree function and! is a b-th root of unity. The sequence T k;d;b = k![ k ]T d (); 0 d < b is sequence EIS A Neil Sloane's Encyclopedia of Integer Sequences.

89 Linear Probing and Paking problem with Buckets. From [Viola 2010], following [Viola and Poblete 1998] and the pioneering work by [Blake and Konheim 1977]: Theorem Let m;b be the random variable for the cost of searching a random element in a b-full table with m buckets of size b and < 1, using the Robin Hood linear probing hashing algorithm, and let m (b; z) be its probability generating function. Then m (b; z) = z b C m (b; z) = b X1 d=0 C m b; e 2id X b z 1=b b 1 p=0 Q 1)) b 1 b(1 )(1 eb(z b z b e b(z 1) Q j=1 b 1 j=1 e 2id b z 1=b p ; with z 1 T (r j e ) T (r j e ) :

90 Some nal considerations. Problem with a very rich history. Paradigm of a problem that nicely integrates analytical, combinatorial and probabilistic approaches. This integration (together with the use of symbolic methods!) has allowed the understanding of deep relations with other important problem. Problem that it is continuously evolving, and new important results are still to come. Some ongoing work: Total displacement with buckets. Relation with other problems as for b=1? FCFS with buckets. Study more properties of sequence T d (b) and nd other applications. Number of movements in deletion algorithm. Worst case Robin Hood.