ManyThanksto Andrea forthisbeautiful posternicodeme/talks/intro_AC.pdf · -EigenvaluesofGraphs-S.Ruzieh Probabilistic Approaches -RandomTrees-B.Chauvin-UrnModels-N.Pouyanne Pierre

CIMPA Summer School

http://lipn.univ-paris13.fr/~nicodeme/nablus14/Nablus14.php

Analysis of Random StructuresAn Najah University, Nablus, Palestine

August 18–28, 2014

The school is based upon 8 courses of 6 hours

- 5 courses upon approaches of Analytic Combinatorics

- 2 courses upon Probabilistic Approaches

- 1 course upon Random Graphs

Organising Committee:Pierre Nicodeme University Paris 13Naji Qatanani An Najah University

Speakers:Cyril Banderier University Paris 13Frederique Bassino University Paris 13Brigitte Chauvin University of VersaillesHosam Mahmoud George Washington UniversityBasile Morcrette University Paris 6Pierre Nicodeme University Paris 13Nicolas Pouyanne University of VersaillesSubhi Ruzieh An Najah University

Local organisation: An Najah University

Sponsors

Many Thanks toAndrea Sportiellofor this beautifulposter

Analysis of Random Structures

Summer School, August 18-28, 2014University An Najah, Nablus

Organizers:

Pierre Nicodème, University Paris13Naji Qatanani, University An Najah

I http://www-lipn.univ-paris13.fr/~nicodeme/nablus14

http://www-lipn.univ-paris13.fr/~nicodeme/nablus14

Content of the School - (6-hours Courses)

AnalyticCombinatorics

∣∣∣∣∣∣∣∣∣∣- Introductory Course - B. Morcrette- Random Walks - C. Banderier- Random Generation - F. Bassino- Random Trees - H. Mahmoud- Statistics of Motifs - P. Nicodème

Graphs∣∣ - Eigenvalues of Graphs - S. Ruzieh

ProbabilisticApproaches

∣∣∣∣ - Random Trees - B. Chauvin- Urn Models - N. Pouyanne

Pierre Nicodeme - 3 November 21, 2013

Course I - Introduction to Analytic Combinatorics

Conrado Martinez (I slides.pdf) Basile Morcrette (lecturer)

I Keywords: symbolic method, singularity analysis


http://www.lsi.upc.edu/~conrado/docencia/udelar-course-2011.pdf

Course II - Discrete Random Walks

Cyril Banderier - U. Paris13

I Keywords: Discrete Lattices, Functional Equations, Fastenumeration schemes


Course III: Random Generation of Combinatorial Structures

Frédérique Bassino - U. Paris13

I Keywords: Large Data, Models, Numerical Validation


Course IV - Analysis of Random Trees

Hosam Mahmoud - George Washington U.

I Keywordsdata structure, algorithms, recursive decompositions


Course V -Statistics of Motifs

Pierre Nicodème - U. Paris13

I KeywordsRandom texts, counting occurrences, languages, automatas


Course VI - Eigenvalues of Graphs

Subhi Ruzieh - An Najah U.

I KeywordsAdjacency and Distance Matrix, Bounds on Eigenvalues


Course VII - Random Trees and Probability

Brigitte Chauvin - U. Versailles I course.pdf

I Keywords Branching process, branching property, martingale,Pólya urn, binary search tree


http://lipn.univ-paris13.fr/~nicodeme/nablus14/BCcourseBerlin.pdf

Course VIII - Urn models

Nicolas Pouyanne - U. Versailles

I KeywordsPólya urn, replacement rules, enumerative and probabilisticmethods


Two introductory examples to Analytic Combinatorics

1. Word Statistics

2. Derangements in Permutations


Part I

The Waiting Time

of First Occurrence of Words


Waiting time in Uniform Bernoulli Trials

Given a string of 0 and 1 bits, the waiting time of a word w isI the position of the rightmost bitI of the first occurrence of the word w

Flip repetitively a fair coin

I head gives 1 with probability1

2

I tail gives 0 with probability1

2

Question: does, in the average, (in a random string)I the words 100 and 111 have the same waiting time?I or should 100 occurs first?I or should 111 occurs first?






2


2







2


2



00100

11110010111111100011000100111001110001010001101000010010111111010111111100000101000101100

1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111


00100111100

10111111100011000100111001110001010001101000010010111111010111111100000101000101100

1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111


0010011110010111111100

011000100111001110001010001101000010010111111010111111100000101000101100

1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111


001001111001011111110001100

0100111001110001010001101000010010111111010111111100000101000101100

1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111


0010011110010111111100011000100

111001110001010001101000010010111111010111111100000101000101100

1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111


0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100

1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111


0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100

100010100111

1100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111


0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100

1000101001111100010100111

100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111


0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100

10001010011111000101001111000010111

00110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111


0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100

100010100111110001010011110000101110011000011000000111

0101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111


0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100

100010100111110001010011110000101110011000011000000111010101010101000111

1101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111


0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100

1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111


1 1 10000111

2 2 00011010000. . .3 3 1001114 111011110. . .5 111011111. . .6 4 001000101. . .7 5 110010110. . .8 6 100110001. . .9 101110100

10 7 00110000. . .11 0111110010. . .12 1101011110. . .13 111010014 8 10100101. . .15 9 00100001. . .16 10 001000011117 11 00001100011111118 12 1001010. . .19 13 00010101001110. . .20 111101011. . .21 14 10101001. . .22 15 10100111110. . .23 0111010111. . .24 16 0001001. . .25 1111101110. . .26 0110111000. . .27 17 1000000. . .28 0011111101. . .29 18 1100001. . .30 0111001000. . .


1 1 100001112 2 00011010000. . .

3 3 1001114 111011110. . .5 111011111. . .6 4 001000101. . .7 5 110010110. . .8 6 100110001. . .9 101110100

10 7 00110000. . .11 0111110010. . .12 1101011110. . .13 111010014 8 10100101. . .15 9 00100001. . .16 10 001000011117 11 00001100011111118 12 1001010. . .19 13 00010101001110. . .20 111101011. . .21 14 10101001. . .22 15 10100111110. . .23 0111010111. . .24 16 0001001. . .25 1111101110. . .26 0110111000. . .27 17 1000000. . .28 0011111101. . .29 18 1100001. . .30 0111001000. . .


1 1 100001112 2 00011010000. . .3 3 100111

4 111011110. . .5 111011111. . .6 4 001000101. . .7 5 110010110. . .8 6 100110001. . .9 101110100

10 7 00110000. . .11 0111110010. . .12 1101011110. . .13 111010014 8 10100101. . .15 9 00100001. . .16 10 001000011117 11 00001100011111118 12 1001010. . .19 13 00010101001110. . .20 111101011. . .21 14 10101001. . .22 15 10100111110. . .23 0111010111. . .24 16 0001001. . .25 1111101110. . .26 0110111000. . .27 17 1000000. . .28 0011111101. . .29 18 1100001. . .30 0111001000. . .


1 1 100001112 2 00011010000. . .3 3 1001114 111011110. . .

5 111011111. . .6 4 001000101. . .7 5 110010110. . .8 6 100110001. . .9 101110100

10 7 00110000. . .11 0111110010. . .12 1101011110. . .13 111010014 8 10100101. . .15 9 00100001. . .16 10 001000011117 11 00001100011111118 12 1001010. . .19 13 00010101001110. . .20 111101011. . .21 14 10101001. . .22 15 10100111110. . .23 0111010111. . .24 16 0001001. . .25 1111101110. . .26 0110111000. . .27 17 1000000. . .28 0011111101. . .29 18 1100001. . .30 0111001000. . .


1 1 100001112 2 00011010000. . .3 3 1001114 111011110. . .5 111011111. . .6 4 001000101. . .7 5 110010110. . .8 6 100110001. . .9 101110100

10 7 00110000. . .11 0111110010. . .12 1101011110. . .13 111010014 8 10100101. . .15 9 00100001. . .16 10 001000011117 11 00001100011111118 12 1001010. . .19 13 00010101001110. . .20 111101011. . .21 14 10101001. . .22 15 10100111110. . .23 0111010111. . .24 16 0001001. . .25 1111101110. . .26 0110111000. . .27 17 1000000. . .28 0011111101. . .29 18 1100001. . .30 0111001000. . .


1000 randomly chosen strings

w = 100 w = 111

p p

p : position of first occurrence of wPierre Nicodeme - 17 November 21, 2013

Aim of the computationW : random variable counting the first occurrence of theword w in a random string; (rightmost position of the occurrence)Probability generating function of W

R(z) =∑n≥0

P(W = n)× zn

By differentiationI Expectation of W

∂R(z)

∂z

∣∣∣∣z=1

=∑n≥0

n×P(W = n) = E(W )

I Second Moment of W∂

∂zz∂R(z)

∂z

∣∣∣∣z=1

=∑n≥0

n2 ×P(W = n) = E(W 2)

I Standard Deviation: σ(W ) =√E(W 2)−E2(W )



R(z) =∑n≥0

P(W = n)× zn


∂R(z)

∂z

∣∣∣∣z=1

=∑n≥0

n×P(W = n) = E(W )


∂zz∂R(z)

∂z

∣∣∣∣z=1

=∑n≥0

n2 ×P(W = n) = E(W 2)




R(z) =∑n≥0

P(W = n)× zn


∂R(z)

∂z

∣∣∣∣z=1

=∑n≥0

n×P(W = n) = E(W )


∂zz∂R(z)

∂z

∣∣∣∣z=1

=∑n≥0

n2 ×P(W = n) = E(W 2)




R(z) =∑n≥0

P(W = n)× zn


∂R(z)

∂z

∣∣∣∣z=1

=∑n≥0

n×P(W = n) = E(W )


∂zz∂R(z)

∂z

∣∣∣∣z=1

=∑n≥0

n2 ×P(W = n) = E(W 2)

I Standard Deviation: σ(W ) =√

E(W 2)−E2(W )


More about R(z)

R(z) =∑n≥0

P(the word w occurs first at end position n)× zn

Example: w = 01

0000 10000001 10010010 10100011 10110100 11000101 11010110 11100111 1111

P (01 occurs for the first time at position 4) =3

16Pierre Nicodeme - 19 November 21, 2013

Ordinary Generating Functions as a Counting ToolLetI A = {0, 1} (alphabet)I and S be a set of words.

Then the ordinary generating function S(z) of the set S is

S(z) =∑w∈S

P(w)× z|w|

and, when the constructions are not ambiguous,for two sets S1 and S2I if T = S1

⋃S2, then T (z) = S1(z)+S2(z)

I if T = S1•S2 (pairwise concatenation), then T (z) = S1(z)×S2(z)

We need bijections between the membersof the formal sets or languages equations


ExamplesRemarkI P(u.v) = P(u)×P(v)

I ε empty word of size 0

P(ε.w) = P(ε)×P(w) = P(w) =⇒ P(ε) = 1

Assume P(0) = P(1) =1

2

S1 = {ε, 0, 01, 11, 101}

S1(z) =∑w∈S1

P(w)z|w| = 1 +z

2+z2

2+z3

8

= P(ε)z|ε| +P(0)z|0| +P(01)z|01| +P(11)z|11| +P(101)z|101|

= 1 +(z2

)1+(z2

)2+(z2

)2+(z2

)3Pierre Nicodeme - 21 November 21, 2013

Examples

I Non ambiguous Union:

S1 = {0, 01} S1(z) =z

2+z2

4

S2 = {ε, 11, 101} S2(z) = 1 +z2

4+z3

8

T = S1⋃S2 = {ε, 0, 01, 11, 101} T (z) = 1 +

z

2+z2

2+z3

8

I Ambiguous Union:

S1 = {0, 01} S1(z) =z

2+z2

4

S2 = {01, 11} S2(z) =z2

4+z2

4

T = S1⋃S2 = {0, 01, 11} T (z) =

z

2+

2z2

46=S1(z) + S2(z)


Examples

I Non ambiguous Union:

S1 = {0, 01} S1(z) =z

2+z2

4

S2 = {ε, 11, 101} S2(z) = 1 +z2

4+z3

8

T = S1⋃S2 = {ε, 0, 01, 11, 101} T (z) = 1 +

z

2+z2

2+z3

8I Ambiguous Union:

S1 = {0, 01} S1(z) =z

2+z2

4

S2 = {01, 11} S2(z) =z2

4+z2

4

T = S1⋃S2 = {0, 01, 11} T (z) =

z

2+

2z2

46=S1(z) + S2(z)


ExamplesNon-Ambiguous UnionI Set of words of even length

E = {ε, 00, 01, 10, 11, 0000, 0001, . . . } =((0 + 1)2

)?E(z) = 1+22

(z2

)2+· · ·+22n

(z2

)2n+· · · = 1+z2+· · · = 1

1− z2I Set of words of odd length

O = {0, 1, 000, 001, 010, . . . , 111, . . . , 00000, . . . } = (0 + 1).E

O(z) = z + z3 + z5 + · · · = z

1− z2I All the words

E⋃O = (0 + 1)?

E(z)+O(z) =1

1− z2+

z

1− z2=

1

1− z= 1+z+· · ·+2n

(z2

)n+. . .


Examples: Non-Ambiguous Concatenation

U = 0? = {ε, 0, 00, 000, . . . }, V = 1.0? = {1, 10, 100, . . . }

U(z) =1

1− z

2

, V (z) =z

2× 1

1− z

2

T = U•V = 0?.1.0?

T (z) = U(z)×V (z) =z

2(1− z

2

)2 =∑n≥0

z

2×n×

(z2

)n−1= n

(z2

)nWhy?

5 words of size 5 in T : {10000, 01000, 00100, 00010, 00001}= {ε•10000, 0•1000, 00•100, 000•10, 0000•1}= {ε•10000, 0•1000, 00•100, 000•10, 0000•1}



U = 0? = {ε, 0, 00, 000, . . . }, V = 1.0? = {1, 10, 100, . . . }

U(z) =1

1− z

2

, V (z) =z

2× 1

1− z

2

T = U•V = 0?.1.0?

T (z) = U(z)×V (z) =z

2(1− z

2

)2 =∑n≥0

z

2×n×

(z2

)n−1= n

(z2

)nWhy?

5 words of size 5 in T : {10000, 01000, 00100, 00010, 00001}= {ε•10000, 0•1000, 00•100, 000•10, 0000•1}

= {ε•10000, 0•1000, 00•100, 000•10, 0000•1}



U = 0? = {ε, 0, 00, 000, . . . }, V = 1.0? = {1, 10, 100, . . . }

U(z) =1

1− z

2

, V (z) =z

2× 1

1− z

2

T = U•V = 0?.1.0?

T (z) = U(z)×V (z) =z

2(1− z

2

)2 =∑n≥0

z

2×n×

(z2

)n−1= n

(z2

)nWhy?

5 words of size 5 in T : {10000, 01000, 00100, 00010, 00001}= {ε•10000, 0•1000, 00•100, 000•10, 0000•1}= {ε•10000, 0•1000, 00•100, 000•10, 0000•1}


Examples: Ambiguous ConcatenationI Set of words of even length

E = {ε, 00, 01, 10, 11, 0000, 0001, . . . }

E(z) = 1 + 4(z2

)2+ · · · = 1 + z2 + z4 + · · · = 1


O = {0, 1, 000, 001, 010, 011, 100, . . . , 111, . . . , 00000, . . . }

O(z) = z + z3 + z5 + · · · = z

1− z2I Concatenate E and O

G = {w = u.v, u ∈ E , v ∈ O} = O =⇒ G(z) =z

1− z2But

E(z)×O(z) =z

(1−z2)2= z + 2z3 + 3z5 + · · ·+ nz2n−1 + . . .

Why?

00000 has been obtained as ε•00000, 00•000, 0000•0


Examples: Ambiguous ConcatenationI Set of words of even length

E = {ε, 00, 01, 10, 11, 0000, 0001, . . . }

E(z) = 1 + 4(z2

)2+ · · · = 1 + z2 + z4 + · · · = 1


O = {0, 1, 000, 001, 010, 011, 100, . . . , 111, . . . , 00000, . . . }

O(z) = z + z3 + z5 + · · · = z

1− z2I Concatenate E and O

G = {w = u.v, u ∈ E , v ∈ O} = O =⇒ G(z) =z

1− z2But

E(z)×O(z) =z

(1−z2)2= z + 2z3 + 3z5 + · · ·+ nz2n−1 + . . .

Why?00000 has been obtained as ε•00000, 00•000, 0000•0


Guibas-Odlyzko decomposition - Waiting time for a word w

I R: set of all strings with a single occurrence of w at the rightend

I N : set of all strings with no occurrence of w

w = 111

111

00100111 ∈ R 10101111 6∈ R

Right Language R

Not Language N

01010110000 ∈ N 001101110001 6∈ N





w = 111

111

00100111 ∈ R 10101111 6∈ R

Right Language R

Not Language N

01010110000 ∈ N 001101110001 6∈ N





w = 111

111

00100111 ∈ R 10101111 6∈ R

Right Language R

Not Language N

01010110000 ∈ N 001101110001 6∈ N


Autocorrelation of wordsThe word 111 is autocorrelatedThis means thatI by concatenating 1 or 11 to 111I you find an overlapping occurrence of 111

1111 11111

The word 100 has no autocorrelation

100100

autocorrelation set Cw of w:

Cw = {h : w.h = h′.w with |h| < |w|}{C111 = {ε, 1, 11}C100 = {ε}

w = 10110 −→

C10110 = {ε, 110}10110︸︷︷︸w

110︸︷︷︸h

= 101︸︷︷︸h′

10110︸︷︷︸w

=



1111 11111


100100



w = 10110 −→

C10110 = {ε, 110}10110︸︷︷︸w

110︸︷︷︸h

= 101︸︷︷︸h′

10110︸︷︷︸w

=



1111 11111


100100



w = 10110 −→

C10110 = {ε, 110}10110︸︷︷︸w

110︸︷︷︸h

= 101︸︷︷︸h′

10110︸︷︷︸w

=


Autocorrelation Polynomial

I It is the ordinary generating functionof the autocorrelation set

Example

w = 010010010 P(0) = P(1) =1

2

010010010010010010010010

010010010010010010010010010010

0100100101001001010010010010010010

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣=⇒

C = {ε, 010, 010010, 10010010}

C(z) = 1 +(z2

)3+(z2

)6+(z2

)8


Non-Ambiguous Formal Set Equations for R and NI R: set of all strings with a single occurrence of w at the right endI N : set of all strings with no occurrence of w

N•{0, 1} = N ∪R \ {ε}

w = 111

10001100101 ∈ N

and 10001100101•{0, 1} ={

100011001010100011001011

}∈ N∈ N

but

100011001011 ∈ N

and 100011001011•{0, 1} ={

10001100101101000110010111

}∈ N∈ R



N•{0, 1} = N ∪R \ {ε}

w = 111

10001100101 ∈ N

and 10001100101•{0, 1} ={

100011001010100011001011

}∈ N∈ N

but

100011001011 ∈ N

and 100011001011•{0, 1} ={

10001100101101000110010111

}∈ N∈ R



N•{0, 1} = N ∪R \ {ε}

w = 111

10001100101 ∈ N

and 10001100101•{0, 1} ={

100011001010100011001011

}∈ N∈ N

but

100011001011 ∈ N

and 100011001011•{0, 1} ={

10001100101101000110010111

}∈ N∈ R



N•w = R•C

w = 111

10001100100 ∈ Nand 10001100100•111 = 10001100100111ε ∈ RC

and

10001100101 ∈ Nand 10001100101•111 = 10001100101111 ∈ RC

and also

100011001011 ∈ Nand 100011001011•111 = 100011001011111 ∈ RC

These equations are bijections [Guibas-Odlyzko 1984]



N•w = R•Cw = 111

10001100100 ∈ Nand 10001100100•111 = 10001100100111ε ∈ RC

and

10001100101 ∈ Nand 10001100101•111 = 10001100101111 ∈ RC

and also

100011001011 ∈ Nand 100011001011•111 = 100011001011111 ∈ RC




N•w = R•Cw = 111

10001100100 ∈ Nand 10001100100•111 = 10001100100111ε ∈ RC

and

10001100101 ∈ Nand 10001100101•111 = 10001100101111 ∈ RC

and also

100011001011 ∈ Nand 100011001011•111 = 100011001011111 ∈ RC




N•w = R•Cw = 111

10001100100 ∈ Nand 10001100100•111 = 10001100100111ε ∈ RC

and

10001100101 ∈ Nand 10001100101•111 = 10001100101111 ∈ RC

and also

100011001011 ∈ Nand 100011001011•111 = 100011001011111 ∈ RC




N•w = R•Cw = 111

10001100100 ∈ Nand 10001100100•111 = 10001100100111ε ∈ RC

and

10001100101 ∈ Nand 10001100101•111 = 10001100101111 ∈ RC

and also

100011001011 ∈ Nand 100011001011•111 = 100011001011111 ∈ RC

These equations are bijections [Guibas-Odlyzko 1984]Pierre Nicodeme - 30 November 21, 2013

Translating to Generating Functions

N .{0, 1} = N ∪R \ {ε}N .w = R.Ctranslate to

N(z)×z = N(z)+R(z)−1

N(z)×w(z) = R(z)×C(z)

Solving for R(z) and N(z)

R(z) =w(z)

(1− z)C(z) + w(z)=

P(w)z|w|

(1− z)C(z) +P(w)z|w|

N(z) =C(z)

(1− z)C(z) + w(z)=

C(z)

(1− z)C(z) +P(w)z|w|


Translating to Generating Functions

N .{0, 1} = N ∪R \ {ε}N .w = R.Ctranslate to

N(z)×z = N(z)+R(z)−1

N(z)×w(z) = R(z)×C(z)

Solving for R(z) and N(z)

R(z) =w(z)

(1− z)C(z) + w(z)=

P(w)z|w|

(1− z)C(z) +P(w)z|w|

N(z) =C(z)

(1− z)C(z) + w(z)=

C(z)

(1− z)C(z) +P(w)z|w|


Waiting time W for 100 and 111

R(z) =∑n≥0

P(W = n)zn =w(z)

(1− z)C(z) + w(z)

100 111

R(z)z3/8

(1−z)×1 + z3/8

z3/8

(1−z)(1+ z

2+z2

4)+z3/8

=z3

8+2z4

16+4z5

32+. . . =

z3

8+z4

16+2z5

32+. . .

E(W )=∂R(z)

∂z

∣∣∣∣z=1

8 14

σ(W )√24 ≈ 4.90

√142 ≈ 11.92


1000 randomly chosen strings

w = 100

E(W ) = 8

σ(W ) ≈ 4.90

w = 111

E(W ) = 14

σ(W ) ≈ 11.92

p p

p : position of first occurrence of wPierre Nicodeme - 33 November 21, 2013

Back to the Paradox

I At each position of a random string,I the probability of getting one occurrence

I is1

8for 111 and 100.

Why does on the average 100 occur before 111?


Statistics of Words(Some tools that you can learn during my course)

AimsI Counting simultaneously occurrences of several wordsI Counting occurrences of Motifs (Regular Expressions)I Hidden Motifs in textsI Limit laws for the counts

MethodsI Combinatorics of LanguagesI Automata constructions

ApplicationsI Linguistic AnalysisI Bioinformatics (exceptional motifs in DNA, modellingevolution)

I Analysis of Random TreesPierre Nicodeme - 35 November 21, 2013

Part II

Derangements in Permutations


When nobody gets back its own hat!

“A number n of people go to the opera, leave their hats on hooksin the cloakroom and grab them at random when leaving.”

“The probability that nobody gets back his own hat is asymptoticto 1/e ≈ 37%.”

(Comtet, L. “Advanced Combinatorics”, Reidel 1974)

Usual approach: Inclusion-Exclusion

What about an approach by Analytic Combinatorics?


Permutations as set of cycles

P =

(1 2 3 4 5 65 6 1 4 3 2

)1

5 3

4 2

6

Hat problemI Give to each person a different number from 1 to nI Give to each hat the number (or label) of its ownerI If at return time, the resulting permutation

has no cycle of size 1, nobody gets back his hat.(Such a permutation is called a derangement)


Permutations as set of cycles

P =

(1 2 3 4 5 65 6 1 4 3 2

)1

5 3

4 2

6

Hat problemI Give to each person a different number from 1 to nI Give to each hat the number (or label) of its ownerI If at return time, the resulting permutation

has no cycle of size 1, nobody gets back his hat.(Such a permutation is called a derangement)


Exponential Generating Function

I S set of labelled objectsI sn number of elements of S of size nI exponential generating function S(z) of S:

S(z) =∑n≥0

sn ×zn

n!=∑α∈S

z|α|

|α|!

Example: permutationsI Sn: symmetric group of permutations of size nI P =

⋃n≥0

Sn

P (z) =∑n≥0|Sn| ×

zn

n!=

1

1− z


Admissible Constructions - Unions

Disjoint Union P = P1⋃P2

P (z) =∑α∈P

z|α|

|α|!=∑β∈P1

z|β|

|β|!+∑γ∈P2

z|γ|

|γ|!= P1(z) + P2(z)

Example

P1 ={(

1 2 33 1 2

),

(11

)}, P2 =

{(1 2 33 2 1

)}

P1(z) = z+z3

3!, P2(z) =

z3

3!, P (z) = z+

2z3

3!= P1(z)+P2(z)


Admissible Construction - Labelled Product

π1 =

(1 2 33 1 2

), π2 =

(1 22 1

)I How can we make

a labelled product π = π1 ? π2 of π1 and π2?

I It is notthe product of elements in the Symmetric Group!!!

I The labelled product of π1 and π2 will have 5 labelsHow can we distribute them?



π1 =

(1 2 33 1 2

), π2 =

(1 22 1

)I How can we make






π1 =

(1 2 33 1 2

), π2 =

(1 22 1

)I How can we make





Labelled Product - Example

P1 =

(1 2 32 3 1

)P2 =

(1 22 1

)Compute P1?P2

I for each “subset” S = (s1, s2, s3) of size 3 of (1, 2, 3, 4, 5)I apply the permutation P1 to SI (respect each pairwise order in P1(S) with respect to P1)

12345◦•◦••

S = 2 45

∣∣∣∣∣∣(

123231

)≡(

245452

)

I apply P2 to the remaining “subset” (1, 3) of (12345)

P2(1, 3) = (3, 1) ◦•◦•• (

1234534152

)I take the Union of the resulting permutations



P1 =

(1 2 32 3 1

)P2 =

(1 22 1

)Compute P1?P2

I for each “subset” S = (s1, s2, s3) of size 3 of (1, 2, 3, 4, 5)I apply the permutation P1 to SI (respect each pairwise order in P1(S) with respect to P1)

12345◦•◦••

S = 2 45

∣∣∣∣∣∣(

123231

)≡(

245452

)I apply P2 to the remaining “subset” (1, 3) of (12345)

P2(1, 3) = (3, 1) ◦•◦•• (

1234534152

)I take the Union of the resulting permutations



Compute P =

(1 2 33 2 1

)?

(1 22 1

)

•••◦◦ 32154••◦•◦ 42513••◦◦• 52431•◦••◦ 45322•◦•◦• 54321•◦◦•• 53241◦•••◦ 54321◦••◦• 45312◦•◦•• 35142◦◦••• 21543

P =

{(32154), (42513), (52143), (43252), (53142),(54132), (43251), (53241), (54231), (54321)

}|P | = 5!

3!2!



Compute P =

(1 2 33 2 1

)?

(1 22 1

)•••◦◦ 32154••◦•◦ 42513••◦◦• 52431•◦••◦ 45322•◦•◦• 54321•◦◦•• 53241◦•••◦ 54321◦••◦• 45312◦•◦•• 35142◦◦••• 21543

P =

{(32154), (42513), (52143), (43252), (53142),(54132), (43251), (53241), (54231), (54321)

}|P | = 5!

3!2!



Compute P =

(1 2 33 2 1

)?

(1 22 1

)•••◦◦ 32154••◦•◦ 42513••◦◦• 52431•◦••◦ 45322•◦•◦• 54321•◦◦•• 53241◦•••◦ 54321◦••◦• 45312◦•◦•• 35142◦◦••• 21543

P =

{(32154), (42513), (52143), (43252), (53142),(54132), (43251), (53241), (54231), (54321)

}|P | = 5!

3!2!


Labelled Products of sets - Generating Functions

A,B: sets of labelled objects

C = A ?B :=⋃α∈Aβ∈B

α?β

Ar = number of objects of A of size rBs = “ “ B “ “ sCn = “ “ C “ “ n

Cn =∑r+s=n

(r + s)!

r!s!ArBs =⇒ Cn

n!=∑r+s=n

Arr!

Bss!

=⇒ C(z) =∑n≥0

Cnzn

n!=∑r≥0

Arzr

r!×∑s≥0

Bszs

s!= A(z)×B(z)

Associativity: C = B1? . . . ?Bk C(z) = B1(z)× . . .×Bk(z)




C = A ?B :=⋃α∈Aβ∈B

α?β


Cn =∑r+s=n

(r + s)!

r!s!ArBs =⇒ Cn

n!=∑r+s=n

Arr!

Bss!

=⇒ C(z) =∑n≥0

Cnzn

n!=∑r≥0

Arzr

r!×∑s≥0

Bszs

s!= A(z)×B(z)





C = A ?B :=⋃α∈Aβ∈B

α?β


Cn =∑r+s=n

(r + s)!

r!s!ArBs =⇒ Cn

n!=∑r+s=n

Arr!

Bss!

=⇒ C(z) =∑n≥0

Cnzn

n!=∑r≥0

Arzr

r!×∑s≥0

Bszs

s!= A(z)×B(z)



k-Sequences, k-Sets and Sets

Setk(B) = Seqk(B)/S

S: Equivalence RelationI identifying two sequences S1 and S2I if the components of the first are a permutation of thecomponents of the second

B(z) =∑α∈B

z|α|

|α|!BSeq,k(z) =

∑γ∈Seqk(B)

z|γ|

|γ|!= B(z)k

BSet,k(z) =∑

κ∈Setk(B)

z|κ|

|κ|!BSet,k(z) =

BSeq,k(z)

k!=B(z)k

k!

A = Set(B) =⋃k≥0

Setk(B) A(z) =∑k≥0

B(z)k

k!= exp(B(z))


k-Cycles and Cycles

Setk(B) = Seqk(B)/C

C: Equivalence RelationI identifying two sequences S1 and S2I if the components of the first are a cyclic permutationof the components of the second

B(z) =∑α∈B

z|α|

|α|!BSeq,k(z) =

∑γ∈Seqk(B)

z|γ|

|γ|!= B(z)k

BCyc,k(z) =∑

κ∈Cyck(B)

z|κ|

|κ|!BCyc,k(z) =

BSeq,k(z)

k=B(z)k

k

A = Cyc(B) =⋃k≥1

Cyck(B) A(z) =∑k≥1

B(z)k

k= log

(1

1−B(z)

)Pierre Nicodeme - 46 November 21, 2013

Permutations as set of Cycles

I Cn= Cycles of size n of a permutation= Cycles of a n-Sequence of a singleton (1)

Cn(z) =(z/1!)n

n=zn

n

(=n!

n

zn

n!

)I C= all Cycles of a singleton

C(z) =∑n≥1

zn

n= log

(1

1− z

)I Permutation P = Set of Cycles

P (z) = exp

(log

(1

1− z

))=

1

1− z=∑n≥0

n!z

n!


The cloakroom and hats problem

We need avoiding cycles of length 1 that give back a hat to itsowner: no fixed point in the permutation

The generating function of derangements or set of cycles all ofthem of size at least 2 is

D(z) = exp

(z2

2+z3

3+ . . .

)= exp

(log

(1

1− z

)− z)

=e−z

1− z

Dn number of permutations of size n without fixed point

D(z) =∑n≥0

Dnzn

n!=⇒ Dn = n!× [zn]D(z)

pn := P(nobody finds its own hat in a group of n) =Dn

n!= [zn]

e−z

1− z


Exact and asymptotic computation of pn

pn = [zn]

(1− z

1!+z2

2!+ · · ·+ (−1)j z

j

j!+ . . .

)× (1 + z + z2 + . . . )

= 1− 1

1!+

1

2!+ · · ·+ (−1)n 1

n!

p20 ≈ 0.3678794411714423216142442limn→∞ pn = e−1 ≈ 0.3678794411714423215955238

Asymptotic Computation: Cauchy integral

pn =1

2iπ

∮|z|=A

e−z

zn+1(1− z)dz = e−1 +O((1/A)n) (A > 1)


Exact and asymptotic computation of pn

pn = [zn]

(1− z

1!+z2

2!+ · · ·+ (−1)j z

j

j!+ . . .

)× (1 + z + z2 + . . . )

= 1− 1

1!+

1

2!+ · · ·+ (−1)n 1

n!

p20 ≈ 0.3678794411714423216142442limn→∞ pn = e−1 ≈ 0.3678794411714423215955238

Asymptotic Computation: Cauchy integral

pn =1

2iπ

∮|z|=A

e−z

zn+1(1− z)dz = e−1 +O((1/A)n) (A > 1)


Permutations with only even size cycles

E(z) =∑n≥1

enzn =

∑n≥1

Enzn

n!= exp

(∑n≥1

z2n

2n

)= exp

(1

2

∑n≥1

(z2)n

n

)= exp

(1

2log

1

1− z2

)=

1√1− z2

H(z)=E′(z)(1− z2)− zE(z) = 0

[zn]H(z)=0 =⇒ en+2 = en(n+ 1)/(n+ 2), (e0 = 1, e1 = 0)

E2n+1 = 0 E2n = (2n)!× e2n =(1.3.5 . . . (2n− 1)

)2



E(z) =∑n≥1

enzn =

∑n≥1

Enzn

n!= exp

(∑n≥1

z2n

2n

)= exp

(1

2

∑n≥1

(z2)n

n

)= exp

(1

2log

1

1− z2

)=

1√1− z2

H(z)=E′(z)(1− z2)− zE(z) = 0

[zn]H(z)=0 =⇒ en+2 = en(n+ 1)/(n+ 2), (e0 = 1, e1 = 0)

E2n+1 = 0 E2n = (2n)!× e2n =(1.3.5 . . . (2n− 1)

)2



E(z) =∑n≥1

enzn =

∑n≥1

Enzn

n!= exp

(∑n≥1

z2n

2n

)= exp

(1

2

∑n≥1

(z2)n

n

)= exp

(1

2log

1

1− z2

)=

1√1− z2

H(z)=E′(z)(1− z2)− zE(z) = 0

[zn]H(z)=0 =⇒ en+2 = en(n+ 1)/(n+ 2), (e0 = 1, e1 = 0)

E2n+1 = 0 E2n = (2n)!× e2n =(1.3.5 . . . (2n− 1)

)2



E(z) =∑n≥1

enzn =

∑n≥1

Enzn

n!= exp

(∑n≥1

z2n

2n

)= exp

(1

2

∑n≥1

(z2)n

n

)= exp

(1

2log

1

1− z2

)=

1√1− z2

H(z)=E′(z)(1− z2)− zE(z) = 0

[zn]H(z)=0 =⇒ en+2 = en(n+ 1)/(n+ 2), (e0 = 1, e1 = 0)

E2n+1 = 0 E2n = (2n)!× e2n =(1.3.5 . . . (2n− 1)

)2


Automatic AsymptoticsI Permutations with only even size cycles

E2n =(1.3.5 . . . (2n− 1)

)2= n![zn]

1√1− z2


Some more examples

I Permutations with an even number of cyclesEven number of Sets of Cycles:

E∗(z) = 1 +C(z)2

2!+ · · ·+ . . .

C(z)2n

2n!+ . . .

= cosh

(log

1

1− z

)=

1

2

1

1− z+

1− z2

I Permutations with an odd number of cycles

O∗(z) = sinh

(log

1

1− z

)=

1

2

1

1− z− 1− z

2


Number of cycles in a random permutationsP(r): permutations with r cycles

P (r)(z) =1

r!

(log

1

1− z

)rWe compute the multivariate generating function

P (z, u) =

∞∑r=0

ur

r!

(log

1

1− z

)r

By differentiation, with µn the expected number of cycles in arandom permutation of size n

M(z) =∑n≥0

µnzn =

∂P (z, u)

∂u

∣∣∣∣u=1

=1

1− zlog

1

1− z

µn = [zn]M(z) = Hn ≡ 1 +1

2+ · · ·+ 1

n

µ100 = 5.18738, µn ∼ log n (n→∞)


Number of cycles in a random permutationsP(r): permutations with r cycles

P (r)(z) =1

r!

(log

1

1− z

)rWe compute the multivariate generating function

P (z, u) =

∞∑r=0

ur

r!

(log

1

1− z

)rBy differentiation, with µn the expected number of cycles in arandom permutation of size n

M(z) =∑n≥0

µnzn =

∂P (z, u)

∂u

∣∣∣∣u=1

=1

1− zlog

1

1− z

µn = [zn]M(z) = Hn ≡ 1 +1

2+ · · ·+ 1

n

µ100 = 5.18738, µn ∼ log n (n→∞)


Tools for asymptoticsI Cauchy integrals (with variants such as Hankel contour)I Transfer theoremsI Saddle-points integralsI Large powers and semi-large powers theoremsI Analytic Poissonization and Depoissonization

Some books for Analytic Combinatorics

I On-line-PDF


http://algo.inria.fr/flajolet/Publications/books.html

Tools for asymptoticsI Cauchy integrals (with variants such as Hankel contour)I Transfer theoremsI Saddle-points integralsI Large powers and semi-large powers theoremsI Analytic Poissonization and Depoissonization

Some books for Analytic Combinatorics

I On-line-PDFPierre Nicodeme - 54 November 21, 2013

http://algo.inria.fr/flajolet/Publications/books.html

Free on-line MOOC Course on Analytic Combinatorics

I https://www.coursera.org/course/ac

This 6-weeks course will be given from February to May 2014 on aregular basis.


https://www.coursera.org/course/ac

Practical InformationI Location of the School: An Najah University, NablusI Dates: August 18-28I Registration is freeI Financial support and number of students:

I full financial support for 15 students not living in Nablus(hotel costs in 2- or 3-persons rooms, lunch at the Universityand dinner at the hotel)

I full financial support for 15 students living in Nablus (lunchat the University, dinner at the hotel)

I travel expenses are not supportedI Professor Assistants: if we have enough funding, we could

partially support a few Professor-AssistantsI Application is mandatory before May 1st, 2014. (See the

Web Site of the School)I Social event: there will be a one day excursion organized

during the school; participation will be free.


Documents

ManyThanksto Andrea forthisbeautiful posternicodeme/talks/intro_AC.pdf · -EigenvaluesofGraphs-S.Ruzieh Probabilistic Approaches -RandomTrees-B.Chauvin-UrnModels-N.Pouyanne Pierre