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On a discrete parking problem Alois Panholzer Institute of Discrete - - PowerPoint PPT Presentation

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A discrete parking problem Results Analysis Outlook On a discrete parking problem Alois Panholzer Institute of Discrete Mathematics and Geometry Vienna University of Technology Alois.Panholzer@tuwien.ac.at International Conference on the

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A discrete parking problem Results Analysis Outlook

On a discrete parking problem

Alois Panholzer

Institute of Discrete Mathematics and Geometry Vienna University of Technology Alois.Panholzer@tuwien.ac.at International Conference on the Analysis of Algorithms, 17.4.2008

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A discrete parking problem Results Analysis Outlook

Outline of the talk

1

A discrete parking problem

2

Results

3

Analysis

4

Outlook

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A discrete parking problem Results Analysis Outlook

A discrete parking problem

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking scheme

The parking scheme: Consider one-way street m parking lots are in a row n drivers wish to park in these lots Each driver has preferred parking lot to which he drives If parking lot is empty ⇒ he parks there If not, he drives on and parks in the next free parking lot if there is one If all remaining parking lots are occupied ⇒ leaves without parking

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SLIDE 5

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking scheme

The parking scheme: Consider one-way street m parking lots are in a row n drivers wish to park in these lots Each driver has preferred parking lot to which he drives If parking lot is empty ⇒ he parks there If not, he drives on and parks in the next free parking lot if there is one If all remaining parking lots are occupied ⇒ leaves without parking

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SLIDE 6

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking scheme

The parking scheme: Consider one-way street m parking lots are in a row n drivers wish to park in these lots Each driver has preferred parking lot to which he drives If parking lot is empty ⇒ he parks there If not, he drives on and parks in the next free parking lot if there is one If all remaining parking lots are occupied ⇒ leaves without parking

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SLIDE 7

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking scheme

The parking scheme: Consider one-way street m parking lots are in a row n drivers wish to park in these lots Each driver has preferred parking lot to which he drives If parking lot is empty ⇒ he parks there If not, he drives on and parks in the next free parking lot if there is one If all remaining parking lots are occupied ⇒ leaves without parking

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SLIDE 8

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking scheme

The parking scheme: Consider one-way street m parking lots are in a row n drivers wish to park in these lots Each driver has preferred parking lot to which he drives If parking lot is empty ⇒ he parks there If not, he drives on and parks in the next free parking lot if there is one If all remaining parking lots are occupied ⇒ leaves without parking

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SLIDE 9

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking scheme

The parking scheme: Consider one-way street m parking lots are in a row n drivers wish to park in these lots Each driver has preferred parking lot to which he drives If parking lot is empty ⇒ he parks there If not, he drives on and parks in the next free parking lot if there is one If all remaining parking lots are occupied ⇒ leaves without parking

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SLIDE 10

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking scheme

The parking scheme: Consider one-way street m parking lots are in a row n drivers wish to park in these lots Each driver has preferred parking lot to which he drives If parking lot is empty ⇒ he parks there If not, he drives on and parks in the next free parking lot if there is one If all remaining parking lots are occupied ⇒ leaves without parking

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SLIDE 11

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking scheme

The parking scheme: Consider one-way street m parking lots are in a row n drivers wish to park in these lots Each driver has preferred parking lot to which he drives If parking lot is empty ⇒ he parks there If not, he drives on and parks in the next free parking lot if there is one If all remaining parking lots are occupied ⇒ leaves without parking

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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SLIDE 13

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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SLIDE 14

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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SLIDE 19

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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SLIDE 20

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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SLIDE 27

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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SLIDE 29

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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SLIDE 30

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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SLIDE 31

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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SLIDE 33

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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SLIDE 34

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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SLIDE 35

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

1 2 3 4 5 6 7 8

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Example

Example: 8 parking lots, 8 cars Parking sequence: 3, 6, 3, 8, 6, 7, 4, 5

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⇒ 2 cars are unsuccessful

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Unsuccessful cars

Number of unsuccessful cars: Parking sequence a1, . . . , an ∈ {1, . . . , m}n ⇒ k unsuccessful cars (max(n − m, 0) ≤ k ≤ n − 1) Formal description of k = k(m; a1, . . . , an): bi := #ℓ : aℓ ≥ i ⇒ k = max

1≤i≤m+1{bi + i} − m − 1

k independent of specific order of cars arriving

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SLIDE 38

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Unsuccessful cars

Number of unsuccessful cars: Parking sequence a1, . . . , an ∈ {1, . . . , m}n ⇒ k unsuccessful cars (max(n − m, 0) ≤ k ≤ n − 1) Formal description of k = k(m; a1, . . . , an): bi := #ℓ : aℓ ≥ i ⇒ k = max

1≤i≤m+1{bi + i} − m − 1

k independent of specific order of cars arriving

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SLIDE 39

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Unsuccessful cars

Number of unsuccessful cars: Parking sequence a1, . . . , an ∈ {1, . . . , m}n ⇒ k unsuccessful cars (max(n − m, 0) ≤ k ≤ n − 1) Formal description of k = k(m; a1, . . . , an): bi := #ℓ : aℓ ≥ i ⇒ k = max

1≤i≤m+1{bi + i} − m − 1

k independent of specific order of cars arriving

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 41

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 42

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm

1 2 3 m m-1

m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 43

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm

1 2 3 m m-1

m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 44

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm

1 2 3 m m-1

m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 45

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm

1 2 3 m m-1

m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 46

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm

1 2 3 m m-1

m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 47

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm

1 2 3 m m-1

m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 48

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm

1 2 3 m m-1

m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 49

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm

1 2 3 m m-1

m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 50

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm

1 2 3 m m-1

m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 51

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm

1 2 3 m m-1

m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 52

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Parking functions: special instance k = 0 ⇒ all cars can be parked Introduced by Konheim and Weiss [1966]: in analysis of linear probing hashing algorithm

1 2 3 m m-1

m places at a round table (∼ = memory addresses) n guests arriving sequentially at certain places (∼ = data elements) each guest goes clockwise to first empty place

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SLIDE 53

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Topic of active research in combinatorics ⇒ connections to many other objects: labelled trees, major functions, acyclic functions, Pr¨ ufer code, non-crossing partitions, hyperplane arrangements, priority queues, Tutte polynomial of graphs, linear probing hashing algorithm, invesions in trees Generalizations: multiparking functions, G-parking functions, bucket parking functions Authors working on parking functions, amongst others:

  • M. Atkinson, D. Foata, J. Francon, I. Gessel, M. Golin, D. Knuth,
  • G. Kreweras, C. Mallows, J. Pitman, A. Postnikov, J. Riordan,
  • B. Sagan, M. Sch¨

utzenberger, L. Shapiro, R. Stanley, C. Yan, . . .

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SLIDE 54

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Topic of active research in combinatorics ⇒ connections to many other objects: labelled trees, major functions, acyclic functions, Pr¨ ufer code, non-crossing partitions, hyperplane arrangements, priority queues, Tutte polynomial of graphs, linear probing hashing algorithm, invesions in trees Generalizations: multiparking functions, G-parking functions, bucket parking functions Authors working on parking functions, amongst others:

  • M. Atkinson, D. Foata, J. Francon, I. Gessel, M. Golin, D. Knuth,
  • G. Kreweras, C. Mallows, J. Pitman, A. Postnikov, J. Riordan,
  • B. Sagan, M. Sch¨

utzenberger, L. Shapiro, R. Stanley, C. Yan, . . .

8 / 37

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SLIDE 55

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Parking functions

Topic of active research in combinatorics ⇒ connections to many other objects: labelled trees, major functions, acyclic functions, Pr¨ ufer code, non-crossing partitions, hyperplane arrangements, priority queues, Tutte polynomial of graphs, linear probing hashing algorithm, invesions in trees Generalizations: multiparking functions, G-parking functions, bucket parking functions Authors working on parking functions, amongst others:

  • M. Atkinson, D. Foata, J. Francon, I. Gessel, M. Golin, D. Knuth,
  • G. Kreweras, C. Mallows, J. Pitman, A. Postnikov, J. Riordan,
  • B. Sagan, M. Sch¨

utzenberger, L. Shapiro, R. Stanley, C. Yan, . . .

8 / 37

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SLIDE 56

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Enumeration results

Enumeration result for parking sequences: Konheim and Weiss [1966] g(m, n): number of parking functions for m parking lots and n cars g(m, n) = (m − n + 1)(m + 1)n−1 Questions for general parking sequences: “Combinatorial question”: What is the number g(m, n, k) of parking sequences a1, . . . , an ∈ {1, . . . , m}n such that exactly k drivers are unsuccessful? Exact formulæ for g(m, n, k) ?

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SLIDE 57

A discrete parking problem Results Analysis Outlook

A discrete parking problem: Enumeration results

Enumeration result for parking sequences: Konheim and Weiss [1966] g(m, n): number of parking functions for m parking lots and n cars g(m, n) = (m − n + 1)(m + 1)n−1 Questions for general parking sequences: “Combinatorial question”: What is the number g(m, n, k) of parking sequences a1, . . . , an ∈ {1, . . . , m}n such that exactly k drivers are unsuccessful? Exact formulæ for g(m, n, k) ?

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Enumeration results

“Probabilistic question”: What is the probability that for a randomly chosen parking sequence a1, . . . , an ∈ {1, . . . , m}n exactly k drivers are unsuccessful ? r.v. Xm,n: counts number of unsuccessful cars for a randomly chosen parking sequence Probability distribution of Xm,n ? Limiting distribution results (depending on growth of m, n) ?

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Enumeration results

Known results for Xm,n: Gonnet and Munro [1984]: Xm,n studied in analysis of algorithm “linear probing sort” Exact and asymptotic results for expectation E(Xm,n): E(Xm,n) = 1 2

n

  • ℓ=2

nℓ mℓ , n ≤ m E(Xm,m) = πm 8 + 2 3 + O

  • m− 1

2

Analysis uses “Poisson model” Transfer of results to “exact filling model”

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A discrete parking problem Results Analysis Outlook

A discrete parking problem: Enumeration results

Known results for Xm,n: Gonnet and Munro [1984]: Xm,n studied in analysis of algorithm “linear probing sort” Exact and asymptotic results for expectation E(Xm,n): E(Xm,n) = 1 2

n

  • ℓ=2

nℓ mℓ , n ≤ m E(Xm,m) = πm 8 + 2 3 + O

  • m− 1

2

Analysis uses “Poisson model” Transfer of results to “exact filling model”

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A discrete parking problem Results Analysis Outlook

Results

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A discrete parking problem Results Analysis Outlook

Results: Exact enumeration formulæ

Exact enumeration results: Cameron, Johannsen, Prellberg and Schweitzer [2007]; Panholzer [2007] Number g(m, n, k) of parking sequences for m parking lots and n drivers such that exactly k drivers are unsuccessful (n ≤ m + k):

g(m, n, k) = (m − n + k)

n−k

  • ℓ=0

n ℓ

  • (m − n + k + ℓ)ℓ−1(n − k − ℓ)n−ℓ

− (m − n + k + 1)

n−k−1

  • ℓ=0

n ℓ

  • (m − n + k + 1 + ℓ)ℓ−1(n − k − 1 − ℓ)n−ℓ

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A discrete parking problem Results Analysis Outlook

Results: Exact enumeration formulaæ

Alternative expression: useful for k small

g(m, n, k) = (m − n + k + 1)(m + k + 1)n−1 − (m − n + k + 1)

k−1

❳

ℓ=0

(−1)ℓ ✥ n ℓ + 1 ✦ (m + k − ℓ)n−ℓ−2(k − ℓ)ℓ+1 − (m − n + k)

k−1

❳

ℓ=0

(−1)ℓ ✥ n ℓ ✦ (m + k − ℓ)n−ℓ−1(k − ℓ)ℓ

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A discrete parking problem Results Analysis Outlook

Results: Exact enumeration formulaæ

Alternative expression: useful for k small

g(m, n, k) = (m − n + k + 1)(m + k + 1)n−1 − (m − n + k + 1)

k−1

❳

ℓ=0

(−1)ℓ ✥ n ℓ + 1 ✦ (m + k − ℓ)n−ℓ−2(k − ℓ)ℓ+1 − (m − n + k)

k−1

❳

ℓ=0

(−1)ℓ ✥ n ℓ ✦ (m + k − ℓ)n−ℓ−1(k − ℓ)ℓ

Examples for small numbers k of unsuccessful cars:

g(m, n, 0) = (m − n + 1)(m + 1)n−1 g(m, n, 1) = (m − n + 2)(m + 2)n−1 + (n2 − n − m2 − 2m − 1)(m + 1)n−2 g(m, n, 2) = (m − n + 3)(m + 3)n−1 + (2n2 − mn − m2 − 4n − 4m − 4)(m + 2)n−2 + 1 2n(−n2 − mn + 2m2 + 2n − 5m + 1)(m + 1)n−3

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A discrete parking problem Results Analysis Outlook

Results: Limiting distributions

Exact probability distribution of Xm,n: P{Xm,n = k} = g(m,n,k)

mn

Limiting distribution results for Xm,n: Panholzer [2007] Depending on growth of m, n ⇒ nine different phases m (parking lots) ≥ n (cars) n ≪ m n ∼ ρm, 0 < ρ < 1 √m ≪ ∆ := m − n ≪ m ∆ ∼ ρ√m, ρ > 0 ∆ ≪ √m m (parking lots) < n (cars) ∆ := n − m ≪ √n ∆ ∼ ρ√n, ρ > 0 √n ≪ ∆ ≪ n n ∼ ρm, ρ > 1 m ≪ n

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A discrete parking problem Results Analysis Outlook

Results: Limiting distributions

Exact probability distribution of Xm,n: P{Xm,n = k} = g(m,n,k)

mn

Limiting distribution results for Xm,n: Panholzer [2007] Depending on growth of m, n ⇒ nine different phases m (parking lots) ≥ n (cars) n ≪ m n ∼ ρm, 0 < ρ < 1 √m ≪ ∆ := m − n ≪ m ∆ ∼ ρ√m, ρ > 0 ∆ ≪ √m m (parking lots) < n (cars) ∆ := n − m ≪ √n ∆ ∼ ρ√n, ρ > 0 √n ≪ ∆ ≪ n n ∼ ρm, ρ > 1 m ≪ n

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A discrete parking problem Results Analysis Outlook

Results: Limiting distributions

Weak convergence of Xm,n (m parking lots, n cars): n ≪ m : Xm,n

(d)

− − → X P{X = 0} = 1 degenerate limit law

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A discrete parking problem Results Analysis Outlook

Results: Limiting distributions

Weak convergence of Xm,n (m parking lots, n cars): n ∼ ρm, 0 < ρ < 1 : Xm,n

(d)

− − → Xρ P{Xρ ≤ k} = (1 − ρ)

k

  • ℓ=0

(−1)k−ℓ (ℓ + 1)k−ℓ (k − ℓ)! ρk−ℓe(ℓ+1)ρ discrete limit law

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A discrete parking problem Results Analysis Outlook

Results: Limiting distributions

Weak convergence of Xm,n (m parking lots, n cars): √m ≪ ∆ := m − n ≪ m :

∆ mXm,n (d)

− − → X

(d)

= EXP(2) survival function: P{X ≥ x} = e−2x, x ≥ 0 asymptotically exponential distributed

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A discrete parking problem Results Analysis Outlook

Results: Limiting distributions

Weak convergence of Xm,n (m parking lots, n cars): ∆ := m − n ∼ ρ√m, ρ > 0 :

1 √mXm,n (d)

− − → Xrho

(d)

= LINEXP(2, ρ) survival function: P{Xρ ≥ x} = e−2x(x+ρ), x ≥ 0 asymptotically linear-exponential distributed

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A discrete parking problem Results Analysis Outlook

Results: Limiting distributions

Weak convergence of Xm,n (m parking lots, n cars): 0 ≤ ∆ := m − n ≪ √m :

1 √mXm,n (d)

− − → X

(d)

= RAYLEIGH(2) survival function: P{X ≥ x} = e−2x2, x ≥ 0 asymptotically Rayleigh distributed

16 / 37

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A discrete parking problem Results Analysis Outlook

Results: Limiting distributions

Weak convergence of Xm,n (m parking lots, n cars): 0 ≤ ∆ := n − m ≪ √n :

Xm,n+m−n √n (d)

− − → X

(d)

= RAYLEIGH(2) survival function: P{X ≥ x} = e−2x2, x ≥ 0 asymptotically Rayleigh distributed

16 / 37

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A discrete parking problem Results Analysis Outlook

Results: Limiting distributions

Weak convergence of Xm,n (m parking lots, n cars): ∆ := n − m ∼ ρ√n, ρ > 0 :

Xm,n+m−n √n (d)

− − → Xρ

(d)

= LINEXP(2, ρ) survival function: P{X ≥ x} = e−2x(x+ρ), x ≥ 0 asymptotically linear-exponential distributed

16 / 37

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A discrete parking problem Results Analysis Outlook

Results: Limiting distributions

Weak convergence of Xm,n (m parking lots, n cars): √n ≪ ∆ := n − m ≪ n :

∆ n (Xm,n + m − n) (d)

− − → X

(d)

= EXP(2) survival function: P{X ≥ x} = e−2x, x ≥ 0 asymptotically exponential distributed

16 / 37

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A discrete parking problem Results Analysis Outlook

Results: Limiting distributions

Weak convergence of Xm,n (m parking lots, n cars): n ∼ ρm, ρ > 1 : Xm,n + m − n

(d)

− − → Xρ P{Xρ ≥ k} = ke−ρk

∞

  • ℓ=0

(ℓ + k)ℓ−1 ℓ! (ρe−ρ)ℓ, k ≥ 1 discrete limit law

16 / 37

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SLIDE 76

A discrete parking problem Results Analysis Outlook

Results: Limiting distributions

Weak convergence of Xm,n (m parking lots, n cars): n ≪ m : Xm,n + m − n

(d)

− − → X P{X = 0} = 1 degenerate limit law

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SLIDE 77

A discrete parking problem Results Analysis Outlook

Analysis

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A discrete parking problem Results Analysis Outlook

Analysis: Outline

Outline of proof Exact enumeration results: Recursive description of parameter Generating functions approach Limiting distribution results: Asymptotic evaluation of distribution function Asymptotic evaluation of positive integer moments (Method of moments)

18 / 37

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SLIDE 79

A discrete parking problem Results Analysis Outlook

Analysis: Outline

Outline of proof Exact enumeration results: Recursive description of parameter Generating functions approach Limiting distribution results: Asymptotic evaluation of distribution function Asymptotic evaluation of positive integer moments (Method of moments)

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SLIDE 80

A discrete parking problem Results Analysis Outlook

Analysis: Exact enumeration results

Exact enumeration results Quantity of interest: g(m, n, k): number of sequences ∈ {1, . . . , m}n, such that exactly k cars are unsuccessful Recursive description of g(m, n, k): Auxiliary quantities: f (n) = (n + 1)n−1: number of parking functions ∈ {1, . . . , n}n s(m, k): number of sequences ∈ {1, . . . , m}m+k, such that all parking lots are occupied ⇔ exactly k cars are unsuccessful

19 / 37

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SLIDE 81

A discrete parking problem Results Analysis Outlook

Analysis: Exact enumeration results

Exact enumeration results Quantity of interest: g(m, n, k): number of sequences ∈ {1, . . . , m}n, such that exactly k cars are unsuccessful Recursive description of g(m, n, k): Auxiliary quantities: f (n) = (n + 1)n−1: number of parking functions ∈ {1, . . . , n}n s(m, k): number of sequences ∈ {1, . . . , m}m+k, such that all parking lots are occupied ⇔ exactly k cars are unsuccessful

19 / 37

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A discrete parking problem Results Analysis Outlook

Analysis: Exact enumeration results

Case n < m + k: decomposition after first empty lot j:

j 1 m

g(m, n, k) =

m

  • j=1

n j − 1

  • f (j − 1)g(m − j, n − j + 1, k)

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SLIDE 83

A discrete parking problem Results Analysis Outlook

Analysis: Exact enumeration results

Case n < m + k: decomposition after first empty lot j:

j 1 m

g(m, n, k) =

m

  • j=1

n j − 1

  • f (j − 1)g(m − j, n − j + 1, k)

Case n = m + k: all parking lots are occupied:

1 m

g(m, n, k) = s(m, k)

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A discrete parking problem Results Analysis Outlook

Analysis: Exact enumeration results

Introducing suitable generating functions: G(z, u, v) :=

  • m≥0
  • n≥0
  • k≥0

g(m, n, k)zn n! umvk S(u, v) :=

  • m≥0
  • k≥0

s(m, k) umvk (m + k)! T(z) :=

  • n≥1

nn−1 zn n! =

  • n≥1

f (n − 1) zn (n − 1)! T(z): satisfies functional equation T(z) = zeT(z) Equation for generating functions: G(z, u, v) = S(zu, zv) 1 − T(zu)

z

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SLIDE 85

A discrete parking problem Results Analysis Outlook

Analysis: Exact enumeration results

Introducing suitable generating functions: G(z, u, v) :=

  • m≥0
  • n≥0
  • k≥0

g(m, n, k)zn n! umvk S(u, v) :=

  • m≥0
  • k≥0

s(m, k) umvk (m + k)! T(z) :=

  • n≥1

nn−1 zn n! =

  • n≥1

f (n − 1) zn (n − 1)! T(z): satisfies functional equation T(z) = zeT(z) Equation for generating functions: G(z, u, v) = S(zu, zv) 1 − T(zu)

z

21 / 37

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A discrete parking problem Results Analysis Outlook

Analysis: Exact enumeration results

Evaluating at v = 1: 1 1 − uez =

  • m≥0
  • n≥0

mn zn n! um = G(z, u, 1) = S(zu, z) 1 − T(zu)

z

⇒ S(zu, z) = 1 − T(zu)

z

1 − uez Substituting z ← zv, u ← u

v :

S(zu, zv) = S(zv · u v , zv) = 1 − T(zu)

zv

1 − u

v ezv

Exact expression for generating function: G(z, u, v) = 1 − T(zu)

zv

  • 1 − T(zu)

z

  • ·
  • 1 − u

v ezv

22 / 37

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SLIDE 87

A discrete parking problem Results Analysis Outlook

Analysis: Exact enumeration results

Evaluating at v = 1: 1 1 − uez =

  • m≥0
  • n≥0

mn zn n! um = G(z, u, 1) = S(zu, z) 1 − T(zu)

z

⇒ S(zu, z) = 1 − T(zu)

z

1 − uez Substituting z ← zv, u ← u

v :

S(zu, zv) = S(zv · u v , zv) = 1 − T(zu)

zv

1 − u

v ezv

Exact expression for generating function: G(z, u, v) = 1 − T(zu)

zv

  • 1 − T(zu)

z

  • ·
  • 1 − u

v ezv

22 / 37

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SLIDE 88

A discrete parking problem Results Analysis Outlook

Analysis: Exact enumeration results

Evaluating at v = 1: 1 1 − uez =

  • m≥0
  • n≥0

mn zn n! um = G(z, u, 1) = S(zu, z) 1 − T(zu)

z

⇒ S(zu, z) = 1 − T(zu)

z

1 − uez Substituting z ← zv, u ← u

v :

S(zu, zv) = S(zv · u v , zv) = 1 − T(zu)

zv

1 − u

v ezv

Exact expression for generating function: G(z, u, v) = 1 − T(zu)

zv

  • 1 − T(zu)

z

  • ·
  • 1 − u

v ezv

22 / 37

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SLIDE 89

A discrete parking problem Results Analysis Outlook

Analysis: Exact enumeration results

Extracting coefficients ⇒ exact formula:

g(m, n, k) = (m − n + k)

n−k

  • ℓ=0

n ℓ

  • (m − n + k + ℓ)ℓ−1(n − k − ℓ)n−ℓ

− (m − n + k + 1)

n−k−1

  • ℓ=0

n ℓ

  • (m − n + k + 1 + ℓ)ℓ−1(n − k − 1 − ℓ)n−ℓ

Exact distribution of Xm,n: P{Xm,n = k} = g(m, n, k) mn

23 / 37

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SLIDE 90

A discrete parking problem Results Analysis Outlook

Analysis: Exact enumeration results

Extracting coefficients ⇒ exact formula:

g(m, n, k) = (m − n + k)

n−k

  • ℓ=0

n ℓ

  • (m − n + k + ℓ)ℓ−1(n − k − ℓ)n−ℓ

− (m − n + k + 1)

n−k−1

  • ℓ=0

n ℓ

  • (m − n + k + 1 + ℓ)ℓ−1(n − k − 1 − ℓ)n−ℓ

Exact distribution of Xm,n: P{Xm,n = k} = g(m, n, k) mn

23 / 37

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A discrete parking problem Results Analysis Outlook

Analysis: Exact enumeration results

Abel’s generalization of the binomial theorem: (x + y)n =

n

  • ℓ=0

x(x − ℓz)ℓ−1(y + ℓz)n−ℓ ⇒ alternative expression for g(m, n, k):

g(m, n, k) = (m − n + k + 1)(m + k + 1)n−1 − (m − n + k + 1)

k−1

❳

ℓ=0

(−1)ℓ ✥ n ℓ + 1 ✦ (m + k − ℓ)n−ℓ−2(k − ℓ)ℓ+1 − (m − n + k)

k−1

❳

ℓ=0

(−1)ℓ ✥ n ℓ ✦ (m + k − ℓ)n−ℓ−1(k − ℓ)ℓ

24 / 37

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A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (I)

Limiting distribution results for Xm,n (I) Special instance: m (parking lots) = n (cars) complex-analytic techniques Generating function of diagonal: F(u, v) =

  • m≥0
  • k≥0

mmP{Xm,m = k}um m! vk Computed via contour integral: F(u, v) = 1 2πi G(t, u

t , v)

t dt = 1 2πi

  • t − T(u)

v

  • dt
  • t − T(u)
  • ·
  • t − u

v etv

25 / 37

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SLIDE 93

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (I)

Limiting distribution results for Xm,n (I) Special instance: m (parking lots) = n (cars) complex-analytic techniques Generating function of diagonal: F(u, v) =

  • m≥0
  • k≥0

mmP{Xm,m = k}um m! vk Computed via contour integral: F(u, v) = 1 2πi G(t, u

t , v)

t dt = 1 2πi

  • t − T(u)

v

  • dt
  • t − T(u)
  • ·
  • t − u

v etv

25 / 37

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SLIDE 94

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (I)

Limiting distribution results for Xm,n (I) Special instance: m (parking lots) = n (cars) complex-analytic techniques Generating function of diagonal: F(u, v) =

  • m≥0
  • k≥0

mmP{Xm,m = k}um m! vk Computed via contour integral: F(u, v) = 1 2πi G(t, u

t , v)

t dt = 1 2πi

  • t − T(u)

v

  • dt
  • t − T(u)
  • ·
  • t − u

v etv

25 / 37

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SLIDE 95

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (I)

Explicit formula: simple pole at t = T(u) computing residue F(u, v) = (v − 1)T(u) vT(u) − ueT(u)v Method of moments: E(X r

m,m) = m!

mm [um] ∂r ∂vr F(u, v)

  • v=1

Studying derivatives of F(u, v) evaluated at v = 1: local expansion around dominant singularity u = 1

e

Singularity analysis, Flajolet and Odlyzko [1990]

26 / 37

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SLIDE 96

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (I)

Explicit formula: simple pole at t = T(u) computing residue F(u, v) = (v − 1)T(u) vT(u) − ueT(u)v Method of moments: E(X r

m,m) = m!

mm [um] ∂r ∂vr F(u, v)

  • v=1

Studying derivatives of F(u, v) evaluated at v = 1: local expansion around dominant singularity u = 1

e

Singularity analysis, Flajolet and Odlyzko [1990]

26 / 37

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SLIDE 97

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (I)

Explicit formula: simple pole at t = T(u) computing residue F(u, v) = (v − 1)T(u) vT(u) − ueT(u)v Method of moments: E(X r

m,m) = m!

mm [um] ∂r ∂vr F(u, v)

  • v=1

Studying derivatives of F(u, v) evaluated at v = 1: local expansion around dominant singularity u = 1

e

Singularity analysis, Flajolet and Odlyzko [1990]

26 / 37

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SLIDE 98

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (I)

r-th moments converge to moments of Rayleigh r.v.: E Xm,m √m r → 2− r

2 Γ

r 2 + 1

  • Theorem of Fr´

echet and Shohat: Xm,m √m

(d)

− − → RAYLEIGH(2)

27 / 37

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SLIDE 99

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (I)

r-th moments converge to moments of Rayleigh r.v.: E Xm,m √m r → 2− r

2 Γ

r 2 + 1

  • Theorem of Fr´

echet and Shohat: Xm,m √m

(d)

− − → RAYLEIGH(2)

27 / 37

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A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (II)

Limiting distribution results for Xm,n (II) Instance: m (parking lots) > n (cars) Extension of previous approach Generating function for ∆ := m − n: F∆(u, v) =

  • m≥∆
  • k≥0

mm−∆P{Xm,m−∆ = k} umvk (m − ∆)! Computed via contour integral: F∆(u, v) = 1 2πi G(t, u

t , v)t∆

t dt = 1 2πi

  • t − T(u)

v

  • t∆dt
  • t − T(u)
  • ·
  • t − u

v etv

28 / 37

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SLIDE 101

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (II)

Limiting distribution results for Xm,n (II) Instance: m (parking lots) > n (cars) Extension of previous approach Generating function for ∆ := m − n: F∆(u, v) =

  • m≥∆
  • k≥0

mm−∆P{Xm,m−∆ = k} umvk (m − ∆)! Computed via contour integral: F∆(u, v) = 1 2πi G(t, u

t , v)t∆

t dt = 1 2πi

  • t − T(u)

v

  • t∆dt
  • t − T(u)
  • ·
  • t − u

v etv

28 / 37

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SLIDE 102

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (II)

Limiting distribution results for Xm,n (II) Instance: m (parking lots) > n (cars) Extension of previous approach Generating function for ∆ := m − n: F∆(u, v) =

  • m≥∆
  • k≥0

mm−∆P{Xm,m−∆ = k} umvk (m − ∆)! Computed via contour integral: F∆(u, v) = 1 2πi G(t, u

t , v)t∆

t dt = 1 2πi

  • t − T(u)

v

  • t∆dt
  • t − T(u)
  • ·
  • t − u

v etv

28 / 37

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SLIDE 103

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (II)

Explicit formula: F∆(u, v) = (v − 1)

  • T(u)

∆+1 vT(u) − ueT(u)v Exact formula for r-th factorial moments: Lagrange inversion formula E

  • Xm,m−∆

r =

r

  • q=1

γr,q

m−∆

  • ℓ=r+q

ℓ − r − 1 q − 1 (m − ∆)ℓ mℓ γr,q: certain constants sums appearing related to Ramanujan’s Q-function

29 / 37

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SLIDE 104

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (II)

Explicit formula: F∆(u, v) = (v − 1)

  • T(u)

∆+1 vT(u) − ueT(u)v Exact formula for r-th factorial moments: Lagrange inversion formula E

  • Xm,m−∆

r =

r

  • q=1

γr,q

m−∆

  • ℓ=r+q

ℓ − r − 1 q − 1 (m − ∆)ℓ mℓ γr,q: certain constants sums appearing related to Ramanujan’s Q-function

29 / 37

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SLIDE 105

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (II)

Asymptotic evaluation: dissecting summation interval Stirling’s formula tail exchange Euler’s summation formula ⇒ as. evaluation of sums via integrals Method of moments: suitably scaled r-th moments of Xm,m−∆ converge to moments of Rayleigh r.v. linear-exponential r.v. exponential r.v.

30 / 37

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SLIDE 106

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (II)

Asymptotic evaluation: dissecting summation interval Stirling’s formula tail exchange Euler’s summation formula ⇒ as. evaluation of sums via integrals Method of moments: suitably scaled r-th moments of Xm,m−∆ converge to moments of Rayleigh r.v. linear-exponential r.v. exponential r.v.

30 / 37

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A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (III)

Limiting distribution results for Xm,n (III) Instance: m (parking lots) < n (cars) consider Xm,n + m − n: number of empty parking lots Asymptotic evaluation of exact formula for survival function Exact formula of survival function for ∆ := n − m:

P

  • Xn−∆,n−∆ ≥ k
  • =

n−∆−k

  • ℓ=0

k ℓ + k n ℓ (ℓ + k)ℓ(n − ∆ − k − ℓ)n−ℓ (n − ∆)n

31 / 37

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SLIDE 108

A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (III)

Limiting distribution results for Xm,n (III) Instance: m (parking lots) < n (cars) consider Xm,n + m − n: number of empty parking lots Asymptotic evaluation of exact formula for survival function Exact formula of survival function for ∆ := n − m:

P

  • Xn−∆,n−∆ ≥ k
  • =

n−∆−k

  • ℓ=0

k ℓ + k n ℓ (ℓ + k)ℓ(n − ∆ − k − ℓ)n−ℓ (n − ∆)n

31 / 37

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A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (III)

Asymptotic evaluation: dissecting summation interval Stirling’s formula tail exchange inequalities, uniform estimates Euler’s summation formula ⇒ as. evaluation of sums via integrals

32 / 37

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A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (III)

Example: ∆ ∼ ρ√n, 0 < ρ < ∞, k ∼ x√n, 0 < x < ∞ Pointwise convergence for all 0 < x < ∞: P

  • Xn−∆,n − ∆ ≥ k
  • →

1 xe

ρ2 2

√ 2πt

3 2 √1 − t

e− x2

2t − (x+ρ)2 2(1−t) dt

Evaluation of the integral: 1 xe

ρ2 2

√ 2πt

3 2 √1 − t

e− x2

2t − (x+ρ)2 2(1−t) dt = e−2x(x+ρ)

Characterization of the limiting distribution: P Xn−∆,n − ∆ √n ≥ x

  • → e−2x(x+ρ)

33 / 37

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A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (III)

Example: ∆ ∼ ρ√n, 0 < ρ < ∞, k ∼ x√n, 0 < x < ∞ Pointwise convergence for all 0 < x < ∞: P

  • Xn−∆,n − ∆ ≥ k
  • →

1 xe

ρ2 2

√ 2πt

3 2 √1 − t

e− x2

2t − (x+ρ)2 2(1−t) dt

Evaluation of the integral: 1 xe

ρ2 2

√ 2πt

3 2 √1 − t

e− x2

2t − (x+ρ)2 2(1−t) dt = e−2x(x+ρ)

Characterization of the limiting distribution: P Xn−∆,n − ∆ √n ≥ x

  • → e−2x(x+ρ)

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A discrete parking problem Results Analysis Outlook

Analysis: Limiting distribution results (III)

Example: ∆ ∼ ρ√n, 0 < ρ < ∞, k ∼ x√n, 0 < x < ∞ Pointwise convergence for all 0 < x < ∞: P

  • Xn−∆,n − ∆ ≥ k
  • →

1 xe

ρ2 2

√ 2πt

3 2 √1 − t

e− x2

2t − (x+ρ)2 2(1−t) dt

Evaluation of the integral: 1 xe

ρ2 2

√ 2πt

3 2 √1 − t

e− x2

2t − (x+ρ)2 2(1−t) dt = e−2x(x+ρ)

Characterization of the limiting distribution: P Xn−∆,n − ∆ √n ≥ x

  • → e−2x(x+ρ)

33 / 37

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A discrete parking problem Results Analysis Outlook

Outlook

34 / 37

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A discrete parking problem Results Analysis Outlook

Outlook

Possible further research directions Refined analysis: Local limit laws case n > m: convergence of moments Extensions to related problems: Analysis of “number of insertion steps” Bucket parking functions Multiparking functions

35 / 37

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A discrete parking problem Results Analysis Outlook

Outlook

Possible further research directions Refined analysis: Local limit laws case n > m: convergence of moments Extensions to related problems: Analysis of “number of insertion steps” Bucket parking functions Multiparking functions

35 / 37

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A discrete parking problem Results Analysis Outlook

Bucket parking scheme

Bucket parking scheme Blake and Konheim [1976]: Each parking lots can hold up to b cars Related to analysis of bucket hashing algorithms b

36 / 37

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A discrete parking problem Results Analysis Outlook

Generating functions approach works: G(z, u, v) = 1 1 − u

vb ezv

(1 − b

zv T(zu1/b)) · (1 − b zv T(ωzu1/b)) · · · (1 − b zv T(ωb−1zu1/b))

(1 − b

z T(zu1/b)) · (1 − b z T(ωzu1/b)) · · · (1 − b z T(ωb−1zu1/b))

37 / 37