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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
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
Pierre Nicodeme - 4 November 21, 2013
Course II - Discrete Random Walks
Cyril Banderier - U. Paris13
I Keywords: Discrete Lattices, Functional Equations, Fastenumeration schemes
Pierre Nicodeme - 5 November 21, 2013
Course III: Random Generation of Combinatorial Structures
Frédérique Bassino - U. Paris13
I Keywords: Large Data, Models, Numerical Validation
Pierre Nicodeme - 6 November 21, 2013
Course IV - Analysis of Random Trees
Hosam Mahmoud - George Washington U.
I Keywordsdata structure, algorithms, recursive decompositions
Pierre Nicodeme - 7 November 21, 2013
Course V -Statistics of Motifs
Pierre Nicodème - U. Paris13
I KeywordsRandom texts, counting occurrences, languages, automatas
Pierre Nicodeme - 8 November 21, 2013
Course VI - Eigenvalues of Graphs
Subhi Ruzieh - An Najah U.
I KeywordsAdjacency and Distance Matrix, Bounds on Eigenvalues
Pierre Nicodeme - 9 November 21, 2013
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
Pierre Nicodeme - 10 November 21, 2013
Course VIII - Urn models
Nicolas Pouyanne - U. Versailles
I KeywordsPólya urn, replacement rules, enumerative and probabilisticmethods
Pierre Nicodeme - 11 November 21, 2013
Two introductory examples to Analytic Combinatorics
1. Word Statistics
2. Derangements in Permutations
Pierre Nicodeme - 12 November 21, 2013
Part I
The Waiting Time
of First Occurrence of Words
Pierre Nicodeme - 13 November 21, 2013
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?
Pierre Nicodeme - 14 November 21, 2013
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?
Pierre Nicodeme - 14 November 21, 2013
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?
Pierre Nicodeme - 14 November 21, 2013
00100
11110010111111100011000100111001110001010001101000010010111111010111111100000101000101100
1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111
Pierre Nicodeme - 15 November 21, 2013
00100111100
10111111100011000100111001110001010001101000010010111111010111111100000101000101100
1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111
Pierre Nicodeme - 15 November 21, 2013
0010011110010111111100
011000100111001110001010001101000010010111111010111111100000101000101100
1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111
Pierre Nicodeme - 15 November 21, 2013
001001111001011111110001100
0100111001110001010001101000010010111111010111111100000101000101100
1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111
Pierre Nicodeme - 15 November 21, 2013
0010011110010111111100011000100
111001110001010001101000010010111111010111111100000101000101100
1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111
Pierre Nicodeme - 15 November 21, 2013
0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100
1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111
Pierre Nicodeme - 15 November 21, 2013
0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100
100010100111
1100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111
Pierre Nicodeme - 15 November 21, 2013
0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100
1000101001111100010100111
100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111
Pierre Nicodeme - 15 November 21, 2013
0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100
10001010011111000101001111000010111
00110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111
Pierre Nicodeme - 15 November 21, 2013
0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100
100010100111110001010011110000101110011000011000000111
0101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111
Pierre Nicodeme - 15 November 21, 2013
0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100
100010100111110001010011110000101110011000011000000111010101010101000111
1101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111
Pierre Nicodeme - 15 November 21, 2013
0010011110010111111100011000100111001110001010001101000010010111111010111111100000101000101100
1000101001111100010100111100001011100110000110000001110101010101010001111101101100110110101001011001100000110101110111111001100010111010001001001010000010011100111000101100101010000111101010101000101011001010100000010000110111
Pierre Nicodeme - 15 November 21, 2013
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. . .
Pierre Nicodeme - 16 November 21, 2013
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. . .
Pierre Nicodeme - 16 November 21, 2013
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. . .
Pierre Nicodeme - 16 November 21, 2013
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. . .
Pierre Nicodeme - 16 November 21, 2013
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. . .
Pierre Nicodeme - 16 November 21, 2013
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 )
Pierre Nicodeme - 18 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 )
Pierre Nicodeme - 18 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 )
Pierre Nicodeme - 18 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 )
Pierre Nicodeme - 18 November 21, 2013
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
Pierre Nicodeme - 20 November 21, 2013
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)
Pierre Nicodeme - 22 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
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)
Pierre Nicodeme - 22 November 21, 2013
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+. . .
Pierre Nicodeme - 23 November 21, 2013
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}
Pierre Nicodeme - 24 November 21, 2013
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}
Pierre Nicodeme - 24 November 21, 2013
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}
Pierre Nicodeme - 24 November 21, 2013
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
1− z2I Set of words of odd length
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
Pierre Nicodeme - 25 November 21, 2013
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
1− z2I Set of words of odd length
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
Pierre Nicodeme - 25 November 21, 2013
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
Pierre Nicodeme - 26 November 21, 2013
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
Pierre Nicodeme - 26 November 21, 2013
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
Pierre Nicodeme - 26 November 21, 2013
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
=
Pierre Nicodeme - 27 November 21, 2013
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
=
Pierre Nicodeme - 27 November 21, 2013
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
=
Pierre Nicodeme - 27 November 21, 2013
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
Pierre Nicodeme - 28 November 21, 2013
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
Pierre Nicodeme - 29 November 21, 2013
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
Pierre Nicodeme - 29 November 21, 2013
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
Pierre Nicodeme - 29 November 21, 2013
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•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]
Pierre Nicodeme - 30 November 21, 2013
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•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
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•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
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•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
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•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|
Pierre Nicodeme - 31 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|
Pierre Nicodeme - 31 November 21, 2013
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
Pierre Nicodeme - 32 November 21, 2013
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?
Pierre Nicodeme - 34 November 21, 2013
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
Pierre Nicodeme - 36 November 21, 2013
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?
Pierre Nicodeme - 37 November 21, 2013
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)
Pierre Nicodeme - 38 November 21, 2013
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)
Pierre Nicodeme - 38 November 21, 2013
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
Pierre Nicodeme - 39 November 21, 2013
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)
Pierre Nicodeme - 40 November 21, 2013
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?
Pierre Nicodeme - 41 November 21, 2013
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?
Pierre Nicodeme - 41 November 21, 2013
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?
Pierre Nicodeme - 41 November 21, 2013
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
Pierre Nicodeme - 42 November 21, 2013
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
Pierre Nicodeme - 42 November 21, 2013
Labelled Product - Example
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!
Pierre Nicodeme - 43 November 21, 2013
Labelled Product - Example
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!
Pierre Nicodeme - 43 November 21, 2013
Labelled Product - Example
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!
Pierre Nicodeme - 43 November 21, 2013
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)
Pierre Nicodeme - 44 November 21, 2013
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)
Pierre Nicodeme - 44 November 21, 2013
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)
Pierre Nicodeme - 44 November 21, 2013
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))
Pierre Nicodeme - 45 November 21, 2013
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!
Pierre Nicodeme - 47 November 21, 2013
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
Pierre Nicodeme - 48 November 21, 2013
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)
Pierre Nicodeme - 49 November 21, 2013
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)
Pierre Nicodeme - 49 November 21, 2013
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
Pierre Nicodeme - 50 November 21, 2013
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
Pierre Nicodeme - 50 November 21, 2013
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
Pierre Nicodeme - 50 November 21, 2013
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
Pierre Nicodeme - 50 November 21, 2013
Automatic AsymptoticsI Permutations with only even size cycles
E2n =(1.3.5 . . . (2n− 1)
)2= n![zn]
1√1− z2
Pierre Nicodeme - 51 November 21, 2013
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
Pierre Nicodeme - 52 November 21, 2013
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→∞)
Pierre Nicodeme - 53 November 21, 2013
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→∞)
Pierre Nicodeme - 53 November 21, 2013
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
Pierre Nicodeme - 54 November 21, 2013
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
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.
Pierre Nicodeme - 55 November 21, 2013
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.
Pierre Nicodeme - 56 November 21, 2013