Occurrences of exactly solvable PDEs in combinatorial...

PDEs and combinatorial problems Label patterns On-line selection Urn models

Occurrences of exactly solvable PDEs incombinatorial problems

Alois Panholzer

Institut fur Diskrete Mathematik und GeometrieTU Wien, Austria

[email protected]

CanaDAM, Memorial University of Newfoundland, Saint John’s,11. 6. 2013

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PDEs and combinatorial problems

Typical situations yielding multivariate recursive descriptions:

Combinatorial enumeration problemsstudy behaviour of parametersenumeration requires auxiliary quantities

Evolution of random structuresgrowth of parametersinformation on quantity at “discretetime”

Discrete stochastic processes

p = 1/2p = 1/2

p = 1/3

p = 1/3p = 1/3p = 1/3 p = 1/3

p = 1/3

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p = 1/2p = 1/2

p = 1/3

p = 1/3p = 1/3p = 1/3 p = 1/3

p = 1/3

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p = 1/2p = 1/2

p = 1/3

p = 1/3p = 1/3p = 1/3 p = 1/3

p = 1/3

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p = 1/2p = 1/2

p = 1/3

p = 1/3p = 1/3p = 1/3 p = 1/3

p = 1/3

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−→ Recursive description of bi-(multi-)variate quantity

Bivariate generating functions:

F (z , u) =∑

∑m Fn,mz

−→ functional equation for F (z , u)

Interested in problems yielding first order (quasi-)linear PDE:

a(z , u,F ) · Fz + b(z , u,F ) · Fu = c(z , u,F )

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F (z , u) =∑

∑m Fn,mz

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F (z , u) =∑

∑m Fn,mz

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F (z , u) =∑

∑m Fn,mz

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Solution method for solving such PDEs

−→ “Method of characteristics”

Way to find transformation

ξ = ξ(z , u)

η = η(z , u)

F (ξ, η) reduces to linear first order ODE

−→ solution of generating function

Difficulty: not always possible to state transformation explicitly

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ξ = ξ(z , u)

η = η(z , u)

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ξ = ξ(z , u)

η = η(z , u)

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ξ = ξ(z , u)

η = η(z , u)

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PDEs occurring in combinatorial problems

often yield explicit solutions

exact formulæasymptotics often “automatically”

if solutions “very nice”

→ think about simpler proof (bijective, etc.)

Illustrating examples:

Label quantities in trees/mappings(with Marie-Louise Bruner; Helmut Prodinger)

On-line selection strategies under uncertainty(with Ahmed Helmi and Conrado Martinez)

Urn models (with Markus Kuba and Hsien-Kuei Hwang)

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Label patterns in trees/mappings

Problems for random trees/mappings, where labels playessential role

Occurrence and avoidance of label-patterns in trees/mappings:

Records

Alternating mappings

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Records

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Records

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n-mappings: functions f from set [n] := {1, 2, . . . , n} into itself:

f : [n]→ [n]

Functional digraph Gf of f :

Gf = (Vf ,Ef ), with Vf = [n] and Ef = {(i , f (i)), i ∈ [n]}

Simple structure: connected components of Gf : cycles of trees

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f : [n]→ [n]

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f : [n]→ [n]

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Interesting example for avoiding a label patterns”:

Alternating mappings:labels on each iteration path are up-down alternating sequence:

i = f 0(i) < f 1(i) > f 2(i) < f 3(i) > · · · or

i = f 0(i) > f 1(i) < f 2(i) > f 3(i) < · · ·

Equivalently:(f 2(i)− f (i)

)· (f (i)− i) < 0 for all i

Corresponding quantity for permutations (“labelled line”):

alternating permutations

enumerated by zig-zag numbers (tangent and secant numbers)

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i = f 0(i) < f 1(i) > f 2(i) < f 3(i) > · · · or

i = f 0(i) > f 1(i) < f 2(i) > f 3(i) < · · ·

)· (f (i)− i) < 0 for all i

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i = f 0(i) < f 1(i) > f 2(i) < f 3(i) > · · · or

i = f 0(i) > f 1(i) < f 2(i) > f 3(i) < · · ·

)· (f (i)− i) < 0 for all i

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i = f 0(i) < f 1(i) > f 2(i) < f 3(i) > · · · or

i = f 0(i) > f 1(i) < f 2(i) > f 3(i) < · · ·

)· (f (i)− i) < 0 for all i

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Structure: zig-zag property on each path in functional graph

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓5 17 2 5 4 4 17 15 17 19 15 10 4 19 10 17 10 1 16

local minima alternate with local maxima: avoiding { up-up, down-down }

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓5 17 2 5 4 4 17 15 17 19 15 10 4 19 10 17 10 1 16

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓5 17 2 5 4 4 17 15 17 19 15 10 4 19 10 17 10 1 16

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Combinatorial decomposition: w.r.t. largest labelled node n

orC ′

TrT ′

⇒ treatment requires auxiliary parameter: number of local minima⇒ require also treatment of quantity for rooted labelled trees

(Cayley trees)

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orC ′

TrT ′

(Cayley trees)

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orC ′

TrT ′

(Cayley trees)

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Corresponding problem for trees:Alternating trees (= intransitive trees):

enumerative studies: Postnikov [1997]

number of rooted alternating trees: Tn =1

n∑k=0

)kn−1

occurs in combinatorial studies of hyperplane arrangements:Postnikov and Stanley [2000]

Kuba and P. [2010]:

studies of alternating tree families

via combinatorial decomposition w.r.t. largest node

using auxiliary parameter “number of local minima”

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Corresponding problem for trees:Alternating trees (= intransitive trees):

enumerative studies: Postnikov [1997]

number of rooted alternating trees: Tn =1

n∑k=0

)kn−1

occurs in combinatorial studies of hyperplane arrangements:Postnikov and Stanley [2000]

Kuba and P. [2010]:

studies of alternating tree families

via combinatorial decomposition w.r.t. largest node

using auxiliary parameter “number of local minima”

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Generating functions treatment:

trees: quasi-linear first order PDE

Fz(z , u) = uFu(z , u)eF (z,u) + u

connected mappings: linear first order PDE

Cz(z , u) = uCu(z , u)eF (z,u) + uFu(z , u)eF (z,u)

mappings: set of connected mappings

M(z , u) = eC(z,u)

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M(z , u) = eC(z,u)

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M(z , u) = eC(z,u)

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M(z , u) = eC(z,u)

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Theorem (P., 2012)

Generating function solution:

M(z) =(eF + 1)2

2eF (1 + (1− F )eF ), with F = z

1 + eF

Enumeration formula: for Mn number of alternating n-mappings

n+1∑k=0

(n + 1

)(k − 1)n

Asymptotics:

Mn ∼√

2√ρ+ 2

, ρ := 2LambertW (1

e) ≈ 0.556929 . . .

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Theorem (P., 2012)

M(z) =(eF + 1)2

2eF (1 + (1− F )eF ), with F = z

1 + eF

n+1∑k=0

(n + 1

)(k − 1)n

Asymptotics:

Mn ∼√

2√ρ+ 2

e) ≈ 0.556929 . . .

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Theorem (P., 2012)

M(z) =(eF + 1)2

2eF (1 + (1− F )eF ), with F = z

1 + eF

n+1∑k=0

(n + 1

)(k − 1)n

Asymptotics:

Mn ∼√

2√ρ+ 2

e) ≈ 0.556929 . . .

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Analysis of on-line selection strategies

Hiring problem:

introduced by Broder et al [SODA, 2008];independently by Krieger et al [Ann. Prob. 2007]

(unknown number of) candidates arrive sequentially

each candidate is interviewed and gets a score(absolute or relative)

decision whether to hire current candidate or not must bemade instantly, depending on scores of candidates seen so far

Important class of selection strategies:

Quality of recruited staff increases whenever candidate is hired

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Hiring problem:

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Hiring problem:

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Hiring problem:

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Hiring problem:

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“Hiring above the median”-strategy:

candidate is recruited iff its score is better than median score ofalready hired staff

Assume s1 < s2 < · · · < sk are scores of already hired candidates

⇒ new candidate is hired iff score is higher than sb k+12c (lower median)

⇒ 5 hired candidates15 / 25

3 2 5 4

3 2 5 4 1

3 2 5 4 1 7

3 2 5 4 1 7 6

Basic question:

consider random sequence of n candidates

how many candidates hn will be hired?

Krieger et al [2007]:

Expectation:E(hn)√

n→ c.

It seems impossible to determine c analytically.

Helmi and P. [2012]: analytic expression for c

Gaither and Ward [2012]: analysis of more general strategies

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Basic question:

n→ c.

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Basic question:

n→ c.

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Basic question:

n→ c.

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Evolution of median of hired staff during “hiring process”:

n candidates interviewed; median of hired staff is `-th largest

(n + 1)-th candidate with certain score arrives

Markov chain with states (n, `)odd and (n, `)even:

P = 1− `n+1

(n, `) 7→ (n + 1, `)

P = `n+1

(n, `) 7→ (n + 1, `)

P = `n+1

(n, `) 7→ (n + 1, `+ 1)

P = 1− `n+1

(n, `) 7→ (n + 1, `)

Linear first order PDE for suitable g.f. of prob. P{hn = k}:

z(1− z)Fz(z , u) +

(zu − u − u2z2

1− z

)Fu(z , u)− zF (z , u) = 0

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P = 1− `n+1

(n, `) 7→ (n + 1, `)

P = `n+1

(n, `) 7→ (n + 1, `)

P = `n+1

(n, `) 7→ (n + 1, `+ 1)

P = 1− `n+1

(n, `) 7→ (n + 1, `)

z(1− z)Fz(z , u) +

(zu − u − u2z2

1− z

)Fu(z , u)− zF (z , u) = 0

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P = 1− `n+1

(n, `) 7→ (n + 1, `)

P = `n+1

(n, `) 7→ (n + 1, `)

P = `n+1

(n, `) 7→ (n + 1, `+ 1)

P = 1− `n+1

(n, `) 7→ (n + 1, `)

z(1− z)Fz(z , u) +

(zu − u − u2z2

1− z

)Fu(z , u)− zF (z , u) = 0

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Theorem (Helmi and P., 2011)

“Hiring above lower median”: Number hn of hired candidatesis distributed as follows:

P{hn = k} =

(n−1−b k2c

d k2e−1

)( nd k2e) =

(n−``−1)(n`)

, for k = 2`− 1 odd,

(n−``−2)

( n`−1)

, for k = 2`− 2 even.

Theorem

Expectation of hn satisfies: E(hn) =√π√

n + O(1).

hn asymptotically Rayleigh distributed with parameter σ =√

i.e., hn√n

(d)−−→ R, where R has density function

f (x) =x

4 , for x > 0.

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Theorem (Helmi and P., 2011)

“Hiring above lower median”: Number hn of hired candidatesis distributed as follows:

P{hn = k} =

(n−1−b k2c

d k2e−1

)( nd k2e) =

(n−``−1)(n`)

, for k = 2`− 1 odd,

(n−``−2)

( n`−1)

, for k = 2`− 2 even.

Theorem

Expectation of hn satisfies: E(hn) =√π√

n + O(1).

hn asymptotically Rayleigh distributed with parameter σ =√

i.e., hn√n

(d)−−→ R, where R has density function

f (x) =x

4 , for x > 0.

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Polya-Eggenberger urn models

Polya-Eggenberger urn model:

Urn with 2 type of balls: n white balls, m black balls

Transition matrix: M =(a bc d

)Urn evolution process:

choose ball at random

examine color

put ball back into urn

insert balls according transition matrix:

if white ball chosen: add a white and b black balls

if black ball chosen: add c white and d black balls

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examine color

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examine color

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Example:

ball replacement matrix M =(2 11 −1

)initial configuration:

n = 7 yellow (white) balls and m = 6 black balls

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Example:

pyellow = 7/13

pblack = 6/13

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Example:

Inspected color:yellow

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Example:

2 x 1 x

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Example:

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Example:

pyellow = 9/16

pblack = 7/16

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Example:

Inspected color:black

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Example:

1 x -1 x

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Example:

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Diminishing urn models:

Polya-Eggenberger urn model with ball replacement matrix M

in addition: set of absorbing states A ⊂ N0 × N0.

urn evolves according to matrix M until absorbing state(i , j) ∈ A is reached

consider only well defined urns:urn always ends in absorbing state of A

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The pill’s problem: proposed by Knuth and McCarthy [1991]

in a bottle there are n small pills and m large pills

a large pill is equivalent to two small pills

every day a person chooses a pill at random

if a small pill is chosen, it is eaten up

if a large pill is chosen it is broken into two halves: one half iseaten and the other half, which is now considered to be asmall pill, is returned to the bottle

Main question:

What is the number of small pills Xm,n remaining whenall large pills have been consumed?

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Main question:

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Main question:

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The pill’s problem:

ball replacement matrix M =(−1 0

1 −1)

absorbing states A = {(0, n)|n ∈ N0}start with 6 large pills and one small pill

⇒ The state (0, 2) ∈ A is reached.

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1 −1)

(5,2)6/7

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1 −1)

6/72/7

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1 −1)

6/72/7

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1 −1)

(4,1)6/7

1/65/5

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1 −1)

6/72/7

1/65/51/5

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1 −1)

6/72/7

1/65/51/5

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1 −1)

(2,2)6/7

1/65/51/5

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1 −1)

6/72/7

1/65/51/5

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1 −1)

(1,2) 6/72/7

1/65/51/5

2/33/4

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1 −1)

6/72/7

1/65/51/5

2/33/4

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1 −1)

6/72/7

1/65/51/5

2/33/4

2/31/2

⇒ the state (0, 2) ∈ A is reached

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Analysis of pill’s problem urn M =(−1 0

1 −1)

Decompose path in urn evolution according to first step

−→ recurrence for probabilities P{Xm,n = k}

Introducing suitable g.f. −→ linear first order PDE

(z − z2 − u)Hz(z , u, v) + u(1− z)Hu(z , u, v)− zH(z , u, v) =uv

(1− vz)2

Explicit formula for probability generating functionhm,n(v) :=

∑k P{Xm,n = k}vk :

hm,n(v) = mv

0(1 + (v − 1)q)n(1− q − (v − 1)q log q)m−1dq

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1 −1)

(1− vz)2

hm,n(v) = mv

0(1 + (v − 1)q)n(1− q − (v − 1)q log q)m−1dq

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1 −1)

(1− vz)2

hm,n(v) = mv

0(1 + (v − 1)q)n(1− q − (v − 1)q log q)m−1dq

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1 −1)

(1− vz)2

hm,n(v) = mv

0(1 + (v − 1)q)n(1− q − (v − 1)q log q)m−1dq

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Results: Pills problem

Theorem (Hwang, Kuba and P., 2007)

If m→∞ the r.v. Xm,n converges, after suitable scaling, indistribution to an exponentially distributed r.v. X withparameter λ = 1, i.e.,

Xm,nnm + log m

(d)−−→ X(d)= Exp(1),

where X has density f (x) = e−x , x ≥ 0.

If m is fixed and n→∞ the r.v. Xm,n converges, aftersuitable scaling, in distribution to a beta distributed r.v. Bm,i.e.,

(d)−−→ Bm(d)= Beta(1,m),

where Bm has density f (x) = m(1− x)m−1, 0 ≤ x ≤ 1.

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