AloisPanholzer Onadiscreteparkingproblem

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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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Outline of the talk

1 A discrete parking problem

2 Results

3 Analysis

4 Outlook

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A discrete parking problem

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A discrete parking problem:

Parking scheme

The parking scheme:

Considerone-way street m parking lotsare in a row

n drivers wish to park in these lots

Each driver has preferred parking lotto 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:

Parking scheme

The parking scheme:

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A discrete parking problem:

Parking scheme

The parking scheme:

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A discrete parking problem:

Parking scheme

The parking scheme:

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A discrete parking problem:

Parking scheme

The parking scheme:

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A discrete parking problem:

Parking scheme

The parking scheme:

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A discrete parking problem:

Parking scheme

The parking scheme:

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A discrete parking problem:

Parking scheme

The parking scheme:

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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

⇒2 cars are unsuccessful

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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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:

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:

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:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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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:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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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:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Example

1 2 3 4 5 6 7 8

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A discrete parking problem:

Unsuccessful cars

Number of unsuccessful cars:

Parking sequencea1, . . . ,an∈ {1, . . . ,m}ⁿ

⇒k unsuccessful cars (max(n−m,0)≤k ≤n−1)

Formal descriptionof k=k(m;a₁, . . . ,a_n):

b_i := #`:a_` ≥i

⇒k = max

1≤i≤m+1{b_i+i} −m−1

k independent of specific orderof cars arriving

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A discrete parking problem:

Unsuccessful cars

b_i := #`:a_` ≥i

⇒k = max

1≤i≤m+1{b_i+i} −m−1

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A discrete parking problem:

Unsuccessful cars

b_i := #`:a_` ≥i

⇒k = max

1≤i≤m+1{b_i+i} −m−1

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A discrete parking problem:

Parking functions

Parking functions: special instancek = 0

⇒all cars can be parked

Introduced byKonheim and Weiss [1966]:

in analysis oflinear probing hashing algorithm

mplaces 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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A discrete parking problem:

Parking functions

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A discrete parking problem:

Parking functions

1 2 3 m m-1

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A discrete parking problem:

Parking functions

1 2 3 m m-1

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A discrete parking problem:

Parking functions

1 2 3 m m-1

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A discrete parking problem:

Parking functions

1 2 3 m m-1

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A discrete parking problem:

Parking functions

1 2 3 m m-1

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A discrete parking problem:

Parking functions

1 2 3 m m-1

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A discrete parking problem:

Parking functions

1 2 3 m m-1

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A discrete parking problem:

Parking functions

1 2 3 m m-1

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A discrete parking problem:

Parking functions

1 2 3 m m-1

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A discrete parking problem:

Parking functions

1 2 3 m m-1

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A discrete parking problem:

Parking functions

1 2 3 m m-1

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A discrete parking problem:

Parking functions

Topic of active research in combinatorics

⇒connectionsto 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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A discrete parking problem:

Parking functions

Generalizations:

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A discrete parking problem:

Parking functions

Generalizations:

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A discrete parking problem:

Enumeration results

Enumeration result for parking sequences:

Konheim and Weiss [1966]

g(m,n): number of parking functions form parking lots andn 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:

Enumeration results

Enumeration result for parking sequences:

Konheim and Weiss [1966]

g(m,n): number of parking functions form parking lots andn 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:

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. X_m,n: counts number of unsuccessful cars for a randomly chosen parking sequence

Probability distributionof X_m,n ?

Limiting distribution results (depending on growth ofm,n) ?

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A discrete parking problem:

Enumeration results

Known results forXm,n:

Gonnet and Munro [1984]:

Xm,n studied in analysis of algorithm“linear probing sort”

Exact and asymptotic results for expectation E(X_m,n):

E(Xm,n) = 1 2

n

X

`=2

n^`

m^`, n≤m

E(Xm,m) = rπm

8 +2

3+O m⁻¹²

Analysis uses “Poisson model”

Transfer of results to “exact filling model”

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A discrete parking problem:

Enumeration results

Known results forXm,n:

Gonnet and Munro [1984]:

Xm,n studied in analysis of algorithm“linear probing sort”

Exact and asymptotic results for expectation E(X_m,n):

E(Xm,n) = 1 2

n

X

`=2

n^`

m^`, n≤m

E(Xm,m) = rπm

8 +2

3+O m⁻¹²

Analysis uses “Poisson model”

Transfer of results to “exact filling model”

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Results

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Results:

Exact enumeration formulæ

Exact enumeration results:

Cameron, Johannsen, Prellberg and Schweitzer [2007];

Panholzer [2007]

Numberg(m,n,k) of parking sequences form parking lots andn drivers such that exactlyk drivers are unsuccessful (n≤m+k):

g(m,n,k) = (m−n+k)

n−k

X

`=0

n

`

(m−n+k+`)^`−1(n−k−`)^n−`

−(m−n+k+ 1)

n−k−1

X

`=0

n

`

(m−n+k+ 1 +`)^`−1(n−k−1−`)^n−`

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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−1X

`=0

(−1)^` n

`+ 1

!

(m+k−`)^n−`−2(k−`)^`+1

−(m−n+k)

k−1X

`=0

(−1)^` n

`

!

(m+k−`)^n−`−1(k−`)^`

Examplesfor 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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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−1X

`=0

(−1)^` n

`+ 1

!

(m+k−`)^n−`−2(k−`)^`+1

−(m−n+k)

k−1X

`=0

(−1)^` n

`

!

(m+k−`)^n−`−1(k−`)^`

Examplesfor 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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Results:

Limiting distributions Exact probability distribution of X_m,n:

P{X_m,n=k}= ^g^(m,n,k)_mn

Limiting distribution results forX_m,n: Panholzer [2007]

Depending on growth ofm,n ⇒ nine different phases

m(parking lots) ≥ n (cars) n m

n ∼ρm, 0< ρ <1

√m∆ :=m−nm

∆∼ρ√

m, ρ >0

∆√

m

m (parking lots) <n (cars)

∆ :=n−m√ n

∆∼ρ√

n, ρ >0

√n ∆n

n∼ρm, ρ >1 mn

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Results:

Limiting distributions Exact probability distribution of X_m,n:

P{X_m,n=k}= ^g^(m,n,k)_mn

Limiting distribution results forX_m,n: Panholzer [2007]

Depending on growth ofm,n ⇒ nine different phases

m(parking lots) ≥ n (cars) n m

n ∼ρm, 0< ρ <1

√m∆ :=m−nm

∆∼ρ√

m, ρ >0

∆√

m

m (parking lots) <n (cars)

∆ :=n−m√ n

∆∼ρ√

n, ρ >0

√n ∆n

n∼ρm, ρ >1 mn

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Results:

Limiting distributions

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

nm: X_m,n−−→^(d) X

P{X = 0}= 1

degenerate limit law

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Results:

n∼ρm, 0< ρ <1 : X_m,n−−→^(d) X_ρ

P{X_ρ≤k}= (1−ρ)

k

X

`=0

(−1)^k−`(`+ 1)^k−`

(k−`)! ρ^k^−`e^(`+1)ρ discrete limit law

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Results:

√m∆ :=m−nm: ^∆_mX_m,n −−→^(d) X ^(d)= EXP(2)

survival function: P{X ≥x}=e^−2x, x ≥0

asymptotically exponential distributed

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Results:

∆ :=m−n∼ρ√

m, ρ >0 : ^√¹_mX_m,n−−→^(d) X_rho ^(d)= LINEXP(2, ρ)

survival function: P{X_ρ≥x}=e^−2x(x+ρ), x ≥0

asymptotically linear-exponential distributed

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Results:

0≤∆ :=m−n√

m: ^√¹_mX_m,n−−→^(d) X ^(d)= RAYLEIGH(2)

survival function: P{X ≥x}=e^−2x², x≥0

asymptotically Rayleigh distributed

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Results:

0≤∆ :=n−m√

n: ^X^m,n^√^+m−n_n −−→^(d) X ^(d)= RAYLEIGH(2)

survival function: P{X ≥x}=e^−2x², x≥0

asymptotically Rayleigh distributed

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Results:

∆ :=n−m∼ρ√

n, ρ >0 : ^X^m,n^√^+m−n_n −−→^(d) X_ρ^(d)= LINEXP(2, ρ)

survival function: P{X ≥x}=e^−2x(x+ρ), x≥0

asymptotically linear-exponential distributed

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Results:

√n∆ :=n−mn: ^∆_n(X_m,n+m−n)−−→^(d) X ^(d)= EXP(2)

survival function: P{X ≥x}=e^−2x, x ≥0

asymptotically exponential distributed

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Results:

n∼ρm, ρ >1 : X_m,n+m−n−−→^(d) X_ρ

P{X_ρ≥k}=ke^−ρk

∞

X

`=0

(`+k)^`−1

`! (ρe^−ρ)^`, k ≥1 discrete limit law

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Results:

nm: X_m,n+m−n −−→^(d) X

P{X = 0}= 1

degenerate limit law

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Analysis

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Analysis:

Outline

Outline of proof

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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Analysis:

Outline

Outline of proof

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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Analysis:

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

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Analysis:

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

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Analysis:

Case n<m+k: decomposition after first empty lotj:

1 j m

g(m,n,k) =

m

X

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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Analysis:

Case n<m+k: decomposition after first empty lotj:

1 j m

g(m,n,k) =

m

X

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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Analysis:

Exact enumeration results Introducing suitable generating functions:

G(z,u,v) := X

m≥0

X

n≥0

X

k≥0

g(m,n,k)zⁿ n!u^mv^k

S(u,v) := X

m≥0

X

k≥0

s(m,k) u^mv^k (m+k)!

T(z) :=X

n≥1

nⁿ⁻¹zⁿ n! =X

n≥1

f(n−1) zⁿ (n−1)!

T(z): satisfies functional equationT(z) =ze^T^(z)

Equation for generating functions:

G(z,u,v) = S(zu,zv) 1− ^T(zu)_z

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Analysis:

Exact enumeration results Introducing suitable generating functions:

G(z,u,v) := X

m≥0

X

n≥0

X

k≥0

g(m,n,k)zⁿ n!u^mv^k

S(u,v) := X

m≥0

X

k≥0

s(m,k) u^mv^k (m+k)!

T(z) :=X

n≥1

nⁿ⁻¹zⁿ n! =X

n≥1

f(n−1) zⁿ (n−1)!

T(z): satisfies functional equationT(z) =ze^T^(z)

Equation for generating functions:

G(z,u,v) = S(zu,zv) 1− ^T(zu)_z

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Analysis:

Evaluating at v = 1:

1

1−ue^z =X

m≥0

X

n≥0

mⁿzⁿ

n!u^m =G(z,u,1) = S(zu,z) 1− ^T(zu)_z

⇒S(zu,z) = 1−^T(zu)_z 1−ue^z Substitutingz ←zv, u← ^u_v:

S(zu,zv) =S(zv·u

v,zv) = 1−^T_zv^(zu) 1−^u_ve^zv Exact expression for generating function:

G(z,u,v) = 1− ^T(zu)_zv 1−^T(zu)_z

· 1−^u_ve^zv

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Analysis:

1

1−ue^z =X

m≥0

X

n≥0

mⁿzⁿ

n!u^m =G(z,u,1) = S(zu,z) 1− ^T(zu)_z

S(zu,zv) =S(zv·u

G(z,u,v) = 1− ^T(zu)_zv 1−^T(zu)_z

· 1−^u_ve^zv

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Analysis:

1

1−ue^z =X

m≥0

X

n≥0

mⁿzⁿ

n!u^m =G(z,u,1) = S(zu,z) 1− ^T(zu)_z

S(zu,zv) =S(zv·u

G(z,u,v) = 1− ^T(zu)_zv 1−^T(zu)_z

· 1−^u_ve^zv

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Analysis:

Exact enumeration results Extracting coefficients⇒ exact formula:

g(m,n,k) = (m−n+k)

n−k

X

`=0

n

`

(m−n+k+`)^`−1(n−k−`)^n−`

−(m−n+k+ 1)

n−k−1

X

`=0

n

`

(m−n+k+ 1 +`)^`−1(n−k−1−`)^n−`

Exact distribution ofXm,n:

P{X_m,n=k}= g(m,n,k) mⁿ

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Analysis:

Exact enumeration results Extracting coefficients⇒ exact formula:

g(m,n,k) = (m−n+k)

n−k

X

`=0

n

`

(m−n+k+`)^`−1(n−k−`)^n−`

−(m−n+k+ 1)

n−k−1

X

`=0

n

`

(m−n+k+ 1 +`)^`−1(n−k−1−`)^n−`

Exact distribution ofXm,n:

P{X_m,n=k}= g(m,n,k) mⁿ

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Analysis:

Abel’s generalization of the binomial theorem:

(x+y)ⁿ=

n

X

`=0

x(x−`z)^`−1(y+`z)^n−`

⇒ 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−1X

`=0

(−1)^` n

`+ 1

!

(m+k−`)^n−`−2(k−`)^`+1

−(m−n+k)

k−1X

`=0

(−1)^` n

`

!

(m+k−`)^n−`−1(k−`)^`

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Analysis:

Limiting distribution results (I)

Limiting distribution results forXm,n (I)

Special instance: m (parking lots) = n (cars) complex-analytic techniques

Generating function of diagonal:

F(u,v) =X

m≥0

X

k≥0

m^mP{X_m,m=k}u^m m!v^k

Computed via contour integral:

F(u,v) = 1 2πi

I G(t,^u_t,v)

t dt = 1 2πi

I t−^T_v^(u) dt t−T(u)

· t−^u_ve^tv

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Analysis:

F(u,v) =X

m≥0

X

k≥0

F(u,v) = 1 2πi

I G(t,^u_t,v)

t dt = 1 2πi

· t−^u_ve^tv

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Analysis:

F(u,v) =X

m≥0

X

k≥0

F(u,v) = 1 2πi

I G(t,^u_t,v)

t dt = 1 2πi

· t−^u_ve^tv

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Analysis:

Explicit formula:

simple pole att =T(u) computing residue

F(u,v) = (v−1)T(u) vT(u)−ue^T(u)v

Method of moments:

E(X_m,m^r ) = m!

m^m[u^m] ∂^r

∂v^rF(u,v) v=1

Studying derivatives ofF(u,v) evaluated at v = 1:

local expansion around dominant singularityu = ¹_e Singularity analysis, Flajolet and Odlyzko [1990]

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Analysis:

Explicit formula:

simple pole att =T(u) computing residue

F(u,v) = (v−1)T(u) vT(u)−ue^T(u)v

Method of moments:

E(X_m,m^r ) = m!

m^m[u^m] ∂^r

∂v^rF(u,v) v=1

Studying derivatives ofF(u,v) evaluated at v = 1:

local expansion around dominant singularityu = ¹_e Singularity analysis, Flajolet and Odlyzko [1990]

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