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```MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Labelled combinatorial classes
Labelled combinatorial classes
Week
Date
Sections
from FS2009
Part/ References
Topic/Sections
1
Sept 7 I.1, I.2, I.3
Contents
1
Combinatorial
Structures
Part A.1, A.2
Labelling Atoms FS:
Comtet74
3
21
II.1, II.2, II.3
1.1 Well-labellingHandout
. . .#1. .
(self study)
4
28
II.4, II.5, II.6
Symbolic methods
6
Multivariable GFs
2
14
I.4, I.5, I.6
Notes/Speaker
Unlabelled structures
. . . .Labelled
. . . structures
. . . .I.
Labelled structures
II
1.2 The Exponential Generating Function
(egf)
1.3
Examples
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Combinatorial
Combinatorial
5
Oct 5
III.1, III.2
parameters
Parameters
1.4 Aside: comparing
to
unlabelled
families
.
.
FS A.III
12
IV.1, IV.2
(self-study)
constructions
19
IV.3, IV.4
Analytic Methods
2.1 Labelled products
. . . . .
FS: Part B: IV, V, VI
8
26
Appendix
2.2 Labelled
product
.
.B4. . . .
IV.5 V.1
Stanley 99: Ch. 6
9 2.3NovSequences
2
. Handout
. . . .#1. . . .
2.49 SetsVI.1. . . . . (self-study)
. . . . . . . .
10
2.512 Cycles
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. . . . . . . .
A.3/ C
11
18
IX.1
3 Surjections
20
IX.2
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.
.
Random Structures
and Limit Laws
FS: Part C
(rotating
presentations)
Complex Analysis
. . . . . . . . .
.Singularity
. . . .Analysis
. . . .
.Asymptotic
. . . . methods
. . . .
. . . . . . . . .
.Introduction
. . . . to. Prob.
. . .
. . . . . . . . .
Limit Laws and Comb
. . . . . .
. . . . . .
. Asst
. .#1. Due
. .
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1
2
2
2
3
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. Asst
. .#2. Due
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. Sophie
. . . . .
. Mariolys
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3
4
5
5
5
5
6
Marni
Discrete Limit Laws
6
Sophie
Combinatorial
IX.3
4 12Set23Partitions
instances of discrete
4.125 Definition
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. . . . . . . .
IX.4
Continuous Limit Laws
4.2 Set partitions as a labelled class . . . . . .
Quasi-Powers and
134.330 Set IX.5
partitions as an unlabelled
classlimit. laws
. . .
Gaussian
4.4 Wrapping up surjections and set partitions
. Marni
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. Sophie
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7
7
7
8
9
5 Permutations
5.1 Sub-classes of permutations . . . . . . . .
5.1.1 Involutions . . . . . . . . . . . . . .
5.1.2 Derangements . . . . . . . . . . . .
5.1.3 Other variants . . . . . . . . . . . .
5.2 The number of cycles in a permutation . .
5.3 Application: 100 Prisoners . . . . . . . . .
5.3.1 The game . . . . . . . . . . . . . . .
Dr. Marni MISHNA,
of Mathematics,
5.3.2Department
The strategy
. SIMON
. . . FRASER
. . . .UNIVERSITY
. . . . .
Version of: 11-Dec-09
5.3.3 The analysis . . . . . . . . . . . . .
5.4 Computing Stirling cycle numbers . . . .
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9
10
10
10
11
11
11
11
12
13
13
14
Dec 10
Presentations
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Mariolys
Asst #3 Due
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6 Trees maps and graphs
1
14
Labelling Atoms
A great many objects in combinatorics like to be labelled — indeed in many cases counting labelled
objects is significantly easier than counting their unlabelled siblings — graphs and trees particularly.
So what do we mean by labelled? Since the objects we consider are constructed from atomic pieces,
simply attatch a label (wlog a natural number) to each atomic piece.
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 1/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
Date
Sections
from FS2009
1
Sept 7
I.1, I.2, I.3
2
14
I.4, I.5, I.6
3
21
II.1, II.2, II.3
4
28
II.4, II.5, II.6
5
Oct 5
III.1, III.2
6
12
IV.1, IV.2
7
19
IV.3, IV.4
8
26
IV.5 V.1
9
Nov 2
The
first
unlabelled
9
VI.1
3.
10
1.1
11
12
Part/ References
1
Topic/Sections
2
Labelled combinatorial classes
Notes/Speaker
2
1
3
2
4
2
4
1
4
3
1
1
2
4
3
Symbolic methods
Combinatorial
Structures
FS: Part A.1, A.2
Comtet74 3
Handout #1
(self study)
4
Combinatorial
parameters
1
FS A.III
(self-study)
2Parameters 1
Unlabelled structures
3
Labelled structures I
Labelled structures II
Combinatorial
Asst #1 Due
4
Multivariable GFs
Complex Analysis
Analytic Methods
2
3 V, VI 4
3
FS: Part B: IV,
Singularity Analysis
Appendix B4
Stanley 99: Ch. 6
Asst #2 Due
Asymptotic
methods
Handoutcorresponds
#1
graph
to 4 different
labelled
(self-study)
Sophie
A.3/ C
IX.1
Well-labelling
18
Introduction to Prob.
Mariolys
Limit Laws and Comb
Marni
graphs, while the second corresponds to
Random Structures
Discrete Limit Laws
Sophie
Labelling is a very intuitive
notion, but we have a formal defintion anyhow:
and Limit Laws
FS: Part C
Combinatorial
23
IX.3
Mariolys
(rotating
instances of discrete
Definition.
A labelled
class
of
combinatorial
objects is a combinatorial class such that every atomic
12
presentations)
component
(atom)
is labelled by a distinct
integer,
andMarni
the set of labels associated to an object of size n
25
IX.4
Continuous
Limit Laws
20
IX.2
is the set {1, 2, . . . , n}. In this case weQuasi-Powers
say thatand
the objects are well-labelled. If the set of labels is not
13
30
IX.5
Sophie
limit laws
{1, 2, . . . , n}, but still distinct, then weGaussian
say theobject
is weakly labelled.
14
Dec 10
Presentations
Asst #3 Due
4
Going back to the above graph example — there are 2(2) = 64 labelled graphs of 4 vertices (every
edge is there or not), but only 11 unlabelled graphs.
1.2
The Exponential Generating Function (egf)
We do not use ogfs to enumerate labelled classes (though one can), instead we use exponential generating functions.
Definition. The exponential generating function (egf) of a sequence (An ) is the formal power series
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
n
X
A(z) =
An
n≥0
z
n!
=
X z |α|
|α|!
α∈A
Again we need the neutral class that contains which has size zero and no label. Similarly the
atomic class contains a single well-labelled object of size 1. So we have Z = {}.
1 The egfs are
E(z) = 1
Z(z) = z,
which coincides with their ogf, incidentally.
1.3
Examples
We have already seen permutations. We now think of them as a labelled line of n vertices. Since Pn = n!
the egf is
P (z) =
X
n≥0
n!
zn
1
=
n!
1−z
which is nice and simple.
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 2/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
Date
Sections
Part/ References
Topic/Sections
Labelled combinatorial classes
Notes/Speaker
from FS2009
Example. The
class U of urns consists of completely disconnected labelled graph. Since there are no
edges,
there
can
only be a single labelling
and
thus Un = 1. Hence
1
Sept 7 I.1, I.2, I.3
Symbolic
methods
2
14
I.4, I.5, I.6
3
21
II.1, II.2, II.3
4
28
II.4, II.5, II.6
Combinatorial
Structures
FS: Part A.1, A.2
Comtet74
Handout #1
(self study)
X
Unlabelled structures
U (z) =
Labelled structures I
n≥0
zn
= ez .
n!
Labelled structures II
Example. Circular graphs,
C, are oriented
cycles (so 1 → 2 → 3 → 1 6≡ 1 → 3 → 2 → 1).
Combinatorial
Combinatorial
5
Oct 5
III.1, III.2
6
12
IV.1, IV.2
7
19
IV.3, IV.4
8
26
9
Nov 2
IV.5 V.1
Asst #1 Due
parameters
FS A.III
(self-study)
Parameters
Analytic Methods
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
Complex Analysis
1
2
1
3
1
4
Multivariable GFs
4
3
2
4
3
2
1
2
1
3
1
4
2
3
Singularity Analysis
4
3
4
Asymptotic methods
2
Asst #2 Due
9
VI.1
Sophieon either side of the One can
simply
decompose such cycles by cutting them
1 and unrolling them.
10
This gives
aA.3/
permutation
of 2, . . . , n and
theretoare
1)! such objects. This is a bijective construction
12
C
Introduction
Prob.(n −
Mariolys
and thus
CnIX.1
= (n − 1)!.
18
Limit Laws and Comb
Marni
11
12
20
23
IX.2
IX.3
Random Structures
and Limit Laws
C(z)
FS: Part C
(rotating
presentations)
=
X zn
X
Discrete
Laws
Sophie
1
(nLimit
− 1)!
zn =
= log
.
Combinatorial
n!
n
1
−
z
Mariolys
n≥1
instances of discrete
n≥1
25
Limit Laws Marni
Remark
thatIX.4labelled cycles are much Continuous
easier than
unlabelled ones.
13
1.4
14
30
IX.5
Quasi-Powers and
Gaussian limit laws
Sophie
Aside: comparing
to unlabelled families
Presentations
Asst #3 Due
Dec 10
More generally one can find examples of enumerations of labelled objects A and their unlabelled counterparts Â such that
1≤
An
≤ n!
Aˆn
To show both sides of this — consider the class of Urns. Here Un = Uˆn = 1. And similarly, consider
permutations: Pn = n!, but one can consider them as labellings of a linear graph on n-vertices, so
P̂
= 1. Note that one can also consider permutations to be sets of cycles, then the unlabelled version
Dr.nMarni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version
11-Dec-09is a partition.
of theof:object
2
So the combinatorial sum still works just fine, but clearly we are going to have difficulties with cartesian
product — the labels don’t work quite right. In particular, if we take β ∈ B and γ ∈ C and glue them
together to get (β, γ) the label 1 appears twice, so the object is not well-labelled. Clearly we need to
relabel things carefully to make this work.
Proposition (Binomial convolution formula). Let A(z), B(z), C(z) be egfs of sequences An , Bn , Cn such
that A(z) = B(z) · C(z) then
n X
n
An =
Bk Cn−k
k
k=0
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 3/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
Date
Sections
1
Sept 7
I.1, I.2, I.3
2
14
I.4, I.5, I.6
3
21
II.1, II.2, II.3
4
28
II.4, II.5, II.6
5
Oct 5
III.1, III.2
from FS2009
Proof. Expand:
6
12
IV.1, IV.2
7
19
IV.3, IV.4
8
26
9
Nov 2
10
11
12
2.1
IV.5 V.1
9
VI.1
12
A.3/ C
18
IX.1
20
IX.2
23
IX.3
Part/ References
Combinatorial
Structures
A(z) =
FS: Part A.1, A.2
Comtet74
Handout #1
(self study)
Notes/Speaker
Symbolic methods
X
zn
zn X
·
Cn
B(z)Unlabelled
· C(z) structures
=
Bn
n! n
n!
n
Labelled structures I
n
X X
Bk Cn−k
Labelled structures
II
=
k! (n − k)!
n
Combinatorial
parameters
FS A.III
(self-study)
Combinatorial
Parameters
!
zn
k=0
Asst #1 Due
X
Multivariable=
GFs
n
n
X
Bk Cn−k
n!
k! (n − k)!
k=0
Analytic Methods
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
Complex Analysis
Random Structures
and Limit Laws
FS: Part C
(rotating
presentations)
Discrete Limit Laws
Sophie
Combinatorial
instances of discrete
Mariolys
Continuous Limit Laws
Marni
!
zn
n!
n
XX
zn
n!
Bk Cn−k
k!(n − k)!
n!
n k=0Asst
| #2 {z
}
Due
Asymptotic methods
n
Sophie ( k )
n
X XMariolys
n
zn
Introduction to Prob.
=
Bk Cn−k .
n!
k
Limit Laws and Comb
Marni
n
Singularity Analysis
=
k=0
Labelled
products
IX.4
25
Topic/Sections
Labelled combinatorial classes
Quasi-Powers
and
labelled
products.
Gaussian limit laws
Next
how to generate
InSophie
the unlabelled case, a pair (β, γ) with β ∈ B and
13 we
30 describe
IX.5
γ ∈ C gave rise to a single object of size |β| + |γ|. In the labelled case we will generate a set of objects.
14
Dec 10
#3 Due
We
describe
how we get thisPresentations
set. Essentially, we keepAsst
the
structure of the pair (β, γ) and we generate
new labels that preserve the order relation among labels. We can do this formally using the following
two operations:
• reduction: For a weakly labelled structure size n, this reduces its labels to the standard interval
[1, . . . , n] while keeping the relative order fixed. Eg h4, 6, 2, 9i becomes h2, 3, 1, 4i. Denote this
operation by ρ(α).
• expansion: roughly — the inverse of reduction. Let e : N → Z be any strictly increasing function.
Then the expansion of a well-labelled object α by e is denoted e(α) and results in a weakly-labelled
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
object
in which the label j is replaced by e(j). So h2, 3, 1, 4i may expand to h7, 10, 2, 88i or h3, 9, 1, 15i
Version of:
11-Dec-09
etc etc.
Now, given any two objects β ∈ B and γ ∈ C their labelled product (or product) is β ? γ. This is a set of
well-labelled pairs (β 0 , γ 0 ) that reduce back to β, γ.
β ? γ = {(β 0 , γ 0 ) s.t. it is well-labelled and ρ(β 0 ) = β, ρ(γ 0 ) = γ}
So how big is this set? If |β| = n1 and |γ| = n2 then
n1 + n2
n
card(β ? γ) =
=
n1 , n2
n1
where n = n1 + n2 . You can see this by considering all the “new” labels and allocating n1 of them to the
part of the object coming from β and the rest go to the part coming from γ. Then assign each block of
labels so as to preserve the original orderings.
Definition (Labelled product). The labelled product B ? C is obtained by forming all ordered pairs from
B × C and computing all possible order-consistent relabellings. That is
B ? C = ∪β∈B,γ∈C (β ? γ)
Now we can look at the admissible constructions.
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 4/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
2.2
Date
Sections
Part/ References
Topic/Sections
from FS2009
Labelled
product
Labelled combinatorial classes
Notes/Speaker
1
Sept 7 I.1, I.2, I.3
Symbolic methods
Combinatorial
When
A = B ? C the corresponding
counting
sequences satisfy
Structures
2
14
I.4, I.5, I.6
Unlabelled structures
FS: Part A.1, A.2
X n Comtet74
ALabelled
Bn1 Cn2
3
21
II.1, II.2, II.3
n = structures I
Handout #1
n1 , n2
n +n =n
4
28
(self study)
II.4, II.5, II.6
1
2
Labelled structures II
The product Bn1 Cn2 takes
care of all Combinatorial
possible pairings of objects and the binomial takes care of the
Combinatorial
5
Oct 5
III.1, III.2
Asst #1 Due
parameters
Parameters convolution of the two sequences and thus
possible relabellings. This
is just the binomial
6
12
IV.1, IV.2
7
19
IV.3, IV.4
8
2.3
9
26
FS A.III
(self-study)
Multivariable GFs
A=B?C
⇒
Analytic Methods
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
Handout #1
power
of B is just
(self-study)
Sequences
IV.5 V.1
Nov 2
The kth (labelled)
9
VI.1
The
10 sequence class is
12
11
18
A.3/ C
A(z) = B(z) · C(z)
Complex Analysis
Singularity Analysis
Asst #2 Due
Asymptotic methods
the
k-fold labelled product of B with itself. It is denoted S EQk (B).
Sophie
Introduction to Prob.
Mariolys
S EQ(B) = E + B + B ? B + · · · = ∪k≥0 S EQk (B)
IX.1
Limit Laws and Comb
Marni
These 20translate
following
operations
on the
Structures
IX.2 to theRandom
Discrete Limit
Laws egf Sophie
and Limit Laws
12
13
2.4
14
23
IX.3 A
FS: Part
C
= S EQ
k (B)
25
IX.4 A
= S EQ(B)
30
(rotating
presentations)
⇒
⇒
Sets
Continuous Limit Laws
A(z)Marni
=
Quasi-Powers and
Gaussian limit laws
IX.5
Dec 10
Combinatorial
A(z)Mariolys
= B(z)k
instances of discrete
Presentations
1
1 − B(z)
B0 = ∅
Sophie
Asst #3 Due
The class of sets of k-components of B is denoted S ETk (B). Formally it is defined as a k-sequence
counted modulo permutation of the components. That is S ETk (B) = S EQk (B)/R where R is an equivalence relation that identifies two sequences if one is simply a permutation of the components of the
other. The labelled set construction is then defined by
S ET(B) = ∪k≥0 S ETk (B)
If we take a labelled k-set, we can order its components in k! different ways to get k! labelled ksequences. Each component must be distinct since the k-set must be well-ordered. In reverse we
Dr.
Marni
MISHNA, Department
of Mathematics,
FRASER
UNIVERSITY of a k-sequnce (they must all be different) and then
can
consider
the smallest
labelSIMON
in each
component
Version of: 11-Dec-09
reduce these labels to get a permutation of k. Hence we must have
A = S ETk (B)
A = S ET(B)
1
B(z)k
k!
A(z) = exp(B(z))
⇒
A(z) =
⇒
Note that this is so much easier than the multiset construction for the unlabelled case since components
must be distinct thanks to their labels.
2.5
Cycles
The class of k-cycles of B is denoted C YCk (B). It is defined as a k-sequence counted modulo cyclic
shifts of it components: C YCk (B) = S EQk (B)/S where S is an equivalence relation that identifies two
sequences if the components of one is a cyclic shift of the components of the other. Since any k-sequence
is well-labelled, its components are necesarrily distinct and thus there is a k-to-one mapping between
k-sequences and k-cycles. Thus
A = C YCk (B)
A = C YC(B)
⇒
⇒
1
B(z)k
k
1
A(z) = log
1 − B(z)
A(z) =
Again — this is much easier because the components are distinct thanks to their labelling.
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 5/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
2.6
Date
Sections
Part/ References
Topic/Sections
from FS2009
Theorem
1
Sept 7 I.1, I.2, I.3
Combinatorial
Thus
we
have now proved
14
I.4, I.5, I.6
5
Oct 5
III.1, III.2
6
12
IV.1, IV.2
7
19
IV.3, IV.4
8
26
9
Nov 2
10
11
12
VI.1
12
A.3/ C
18
IX.1
20
IX.2
23
IX.3
25
IX.4
13
30
14
Dec 10
Combinatorial
parameters
FS A.III
(self-study)
Combinatorial
Parameters
A(z) = B(z) + C(z)
Asst #1 Due
Multivariable GFs
A=B?C
labelled product
9
labelled product, sequence, set and cycle all admis-
A=B+C
sum
IV.5 V.1
Notes/Speaker
Symbolic methods
Structures
Unlabelled structures
FS: Part A.1, A.2
Theorem. The constructions
of combinatorial sum,
Comtet74
3
21
II.1, II.2, II.3
Labelled structures I
Handout #1
sible.
The
operators
are
(self
study)
4
28
II.4, II.5, II.6
Labelled structures II
2
Labelled combinatorial classes
Analytic Methods
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
sequence
Handout #1
(self-study)
A(z) = B(z)C(z)
Complex Analysis
Singularity Analysis
A = S EQ(B)
Asst #2 Due
A(z) =
Asymptotic methods
Sophie
A(z) = B(z)k
A = S EQk (B)
set
Random Structures
and Limit Laws
FS: Part C
(rotating
presentations)
IX.5
Introduction to Prob.
Mariolys
Limit Laws and Comb
Marni
A = S ET(B)
Sophie
A(z) = exp(B(z))
1
A(z) = B(z)k
k!
Discrete Limit Laws
Combinatorial
A = S ETk (B)
Mariolys
instances of discrete
Continuous Limit Laws
Marni
Quasi-Powers and
Sophie
= laws
C YC(B)
GaussianA
limit
cycle
Presentations
1
1 − B(z)
A(z) = log
Asst #3 Due
A = C YCk (B)
A(z) =
1
1 − B(z)
1
B(z)k
k
where for sequence, set and cycle it is assumed that B0 = ∅.
Note how the cycle and set operations are nearly inverses of each other.
3
Surjections
Consider the class of surjections:
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
R = S EQ(S ET≥1 (Z)).
http://www.nourishingdays.com/2009/02/make-yogurt-in-your-crock-pot/ A mapping from A to B is a
(r)
surjection if every element of B is mapped to be at least one element of A. Now let Rn denote the set
(r)
(r)
of surjections from [1, n] onto [1, r]. Further let R = ∪n Rn .
Consider the following surjection, φ from [1, 9] to [1, 4]
1→3
2→2
3→4
4→2
5→1
6→1
7→2
8→3
9→4
Now consider the preimage of each point
φ−1 (1) = {5, 6}
φ−1 (2) = {2, 4, 7}
φ−1 (3) = {1, 8}
φ−1 (4) = {3, 9}
This is clearly an ordered sequence of non-empty sets. More generally one can decompose a surjection
to [1, r] as an r-sequence of sets. Thus
R(r) = S EQr (V)
V = S ET≥1 (Z)
V (z) = exp(z) − 1
M ARNI M ISHNA , S PRING 2011
R
(r)
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
(z) = (ez − 1)r
PAGE 6/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
Date
Sections
Part/ References
Topic/Sections
Notes/Speaker
∼
fromV
FS2009
Note that here
= U − {} — non-empty urns
(r)
We
can
expand
to get at Rn Symbolic methods
1
Sept 7 I.1, I.2, I.3 thisCombinatorial
2
14
I.4, I.5, I.6
3
21
II.1, II.2, II.3
4
28
II.4, II.5, II.6
5
Oct 5
III.1, III.2
6
12
Structures
FS: Part A.1, A.2
Comtet74
Handout #1
(self study)
Unlabelled structures
r
X
(r)
n
j r
RnLabelled
= n![z
] I (−1)
e(r−j)z
structures
j
j=0
Labelled structures II
r
X
j r
n
Combinatorial
Combinatorial
parameters
FS A.III
(self-study)
IV.1, IV.2
=
Parameters
4
26
9
Nov 2
4.1
9
VI.1
(r#1
−Due
j)
Asst
, which
we
Complex
Analysis
r
Singularity Analysis
Set Partitions
10
j
Multivariable GFs
n
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
IV.5 V.1
(−1)
j=0
which
is related
numbers
7
19
IV.3, IV.4to stirling
Analytic Methods
8
Labelled combinatorial classes
Asymptotic methods
Asst #2 Due
Sophie
Definition
A.3/ C
12
Introduction to Prob.
Mariolys
IX.1 set-partitions. Consider
Limit
Laws
and Comb [1,Marni
Let
us18look at
the
interval
n]. We can partition this into r distinct subsets
11
Random Structures
(called20blocks).
[1, 2, 3, 4] can
beLimit
partitioned
into
IX.2 For example
Discrete
Laws
Sophie
• 123set =IX.3
[1, 2, 3, 4]
12
• 225sets =IX.4[1, 2, 3|4]
13
• 330sets =IX.5[1, 2|3|4]
14
10
• 4Dec
sets
= [1|2|3|4]
and Limit Laws
FS: Part C
(rotating
presentations)
[1, 2, 4|3]
[1, 3|2|4]
Combinatorial
instances of discrete
Limit 3,
Laws
[1, 3,Continuous
4|2] [1|2,
4]
Quasi-Powers and
[1, 4|2|3]
3|4]
Gaussian[1|2,
limit laws
Presentations
Mariolys
Marni
[1, 2|3, 4]
Sophie
[1|2, 4|3]
[1, 3|2, 4]
[1, 4|2, 3]
[1|2|3, 4]
Asst #3 Due
(r)
So let Sn be the number of ways of partitioning an n-set into r-blocks, then we have
(1)
S4
(2)
=1
S4
=7
(3)
S4
=6
(4)
S4
=1
(r)
The numbers Sn are called the
Stirling partition numbers (or Stirling numbers of the second kind)
and are frequently denoted nr .
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
4.2
Set partitions as a labelled class
Now let S (r) denote the class of set partitions in r-blocks. Unlike the surjections we just studied, these
blocks are not ordered. Thus we have a set of non-empty sets
S (r) = S ETr (V)
1
S (r) (z) = (ez − 1)r
r!
V = S ET≥1 (Z)
so we clearly have the relation
Sn(r) =
1 (r)
R
r! n
=
r
1 X
r
(−1)j
(r − j)n
r! j=0
j
Which is quite obvious, since we can take an r-partition and order the r-blocks in r! different ways to
get surjections into [1, r].
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 7/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
Date
Sections
Part/ References
Topic/Sections
Labelled combinatorial classes
Notes/Speaker
FS2009
It is the from
work
of a moment to extend these to get the total number of surjections from [1, n] and
setpartitions
of
[1,
1
Sept 7 I.1, I.2, I.3 n]
Symbolic methods
2
14
I.4, I.5, I.6
3
21
II.1, II.2, II.3
4
28
II.4, II.5, II.6
5
Oct 5
6
12
7
19
R=
R=
Combinatorial
Structures
(r)
Unlabelled structures
Part A.1, A.2
∪FS:
rR
Comtet74
Labelled structures I
SHandout
EQ (S#1
ET ≥1 (Z))
(self study)
Labelled
2 structures
3 II
S = ∪r S (r)
S = S ET(S ET≥1 (Z))
z
1
z
= 1 + z + Combinatorial
3 + 13 + . . .
Combinatorial
z
2
−
e
2!
3!
III.1, III.2
Asst #1 Due
parameters
Parameters
3
2
FSz A.III
z
z
e
−1
IV.1,S(z)
IV.2 = e(self-study)
+ 5 GFs
+ ...
= 1 + z + 2Multivariable
2!
3!
R(z) =
IV.3, IV.4
Complex Analysis
Analytic Methods
FS: Part B: IV, V, VI
8
26
Singularity Analysis
Appendix B4
IV.5 V.1
Stanley 99:
Ch. 6
The
Rn are called
surjection
numbers while the
SDue
n are called Bell numbers.
9 numbers
Nov 2
Asst #2
Asymptotic methods
#1
One can then quiteHandout
easily
compute the egfs of set partitions with any / odd / even number of blocks
(self-study)
9
VI.1
Sophie
of 10any / odd / even sizes, by making suitable restrictions of the set / sequence operators in the construcA.3/quite
C
tions. 12
In fact,
generally we have Introduction to Prob. Mariolys
18
IX.1
23
IX.3
25
IX.4
Limit Laws and Comb
Marni
11
Lemma.
The class R(A,B) of surjections in which the cardinalities of the premiages lie in A ⊆ N and
Random Structures
20
IX.2
Discrete Limit Laws
Sophie
the cardinality of the range
and Limitlies
Lawsin B ⊆ N is given by
12
13
30
14
Dec 10
FS: Part C
(rotating
R(A,B) (z)
presentations)
Combinatorial
instances of discrete
= β(α(z))
Limit Laws
XContinuous
za
α(z) = Quasi-Powers and
a!
IX.5
a∈A
Gaussian limit laws
Presentations
Mariolys
Marni
β(z) =
Sophie
X
zb
b∈B
Asst #3 Due
The class S (A,B) of set partitions with block sizes in A ⊆ N and number of blocks in in B ⊆ N is given
by
S (A,B) (z) = β(α(z))
X za
α(z) =
a!
β(z) =
a∈A
4.3
X zb
!
b
b∈B
Set partitions as an unlabelled class
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version
11-Dec-09
Now, of:we
can also consider set partitions using an
unlabelled class. This shows that we need to keep
(r)
an open mind when deciding these properties. We encode set partitions as a family of words! Let Sn
n
(r)
denote the set partitions of an n-set into r-blocks, so that r = card(Sn ).
We can encode any such set partition as a word on B = {b1 , . . . , br }. In particular take each block
and sort its elements smallest to largest. Then order the blocks according to the smallest element in
each block. Label everything in the first block by b1 , the second block by b2 etc etc. Eg
{1, 2, . . . , 9} = {6, 4} + {5, 1, 2} + {3, 8, 7}
= {1, 2, 5} + {3, 7, 8} + {4, 6}
| {z } | {z } | {z }
b1
b2
b3
Then simply read off the label of each number in order
≡ [b1 , b1 , b2 , b3 , b1 , b3 , b2 , b2 ]
Now this is not just any word in the letters bi there are some extra conditions
• all r letters must occur
• the first occurrance of b1 is before that of b2 which is before the first occurrance of b3 etc etc.
So this means that our word looks like
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 8/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
Date
Sections
Part/ References
Topic/Sections
from FS2009
• a sequence
of at least 1 b1 , then
1
2
3
Notes/Speaker
I.1, I.2, I.3
Symbolic methods
Combinatorial
• aSept
b2 7then
a sequence
of {b1 , b2 }, then
Structures
14
I.4, I.5, I.6
21
II.1, II.2, II.3
Unlabelled structures
FS: Part A.1, A.2
• a b3 then a sequence
of {b1 , b2 , b3 }, then etc etc
Comtet74
This gives
4
28
5
6
Labelled combinatorial classes
Oct 5
II.4, II.5, II.6
III.1, III.2
IV.1, IV.2
12
A.3/ C
Labelled structures II
b1 S EQ
(b1 )b2 S EQ(b1 +
b2 )b3 S EQ(b1 + b2 + b3 ) . . . br S EQ(b1 + · · · + br )
Combinatorial
Combinatorial
Hence the ogf is
12
Labelled structures I
Handout #1
(self study)
parameters
FS A.III
(self-study)
Asst #1 Due
Parameters
Multivariable GFs
zr
(r)
S
(z)
=
7
19
IV.3, IV.4
Complex (1
Analysis
− z)(1 − 2z) . . . (1 − rz)
Analytic Methods
FS: Part B: IV, V, VI
8
26we can re-writeAppendix
Singularity
Analysis
Which
using
partial
fraction
decomposition,
or other methods as:
B4
IV.5 V.1
Stanley 99: Ch. 6
9
Nov 2
Asst
#2
Due
r
Asymptotic methods
Handout #1
1 X r (−1)r−j
(self-study)
S(z) =
9
VI.1
Sophie
r! j=1 j 1 − jz
10
and so18
IX.1
20
IX.2
23
IX.3
25
IX.4
11
as12before.
4.4
13
Introduction to Prob.
Random Structures
and Limit Laws
FS: Part C
(rotating
presentations)
Mariolys
Comb
Marni LimitLaws andX
r
n
1
r−j r
Discrete=
Limit Laws (−1)
Sophie
jn,
j
r! j=1
r
Combinatorial
instances of discrete
Mariolys
Continuous Limit Laws
Marni
Quasi-Powers and
Gaussian limit laws
Wrapping
up surjections and set partitions
IX.5
Sophie
30
Proposition.
The class R(A,B)
of surjections in which
the
cardinalities of the premiages lie in A ⊆ N
14
Dec 10
Presentations
Asst
#3 Due
and the cardinality of the range lies in B ⊆ N is given by
R(A,B) (z) = β(α(z))
X za
α(z) =
a!
β(z) =
a∈A
Proposition. The class S
B ⊆ N is given by
(A,B)
X
zb
b∈B
of set partitions with block sizes in A ⊆ N and number of blocks in in
(A,B)
S
(z)
= β(α(z))
Dr. Marni MISHNA, Department of Mathematics,
SIMON
FRASER UNIVERSITY
Version of: 11-Dec-09
α(z) =
X za
a!
β(z) =
a∈A
5
X zb
!
b
b∈B
Permutations
We just looked at sequences and sets of sets. We could do the same for cycles. Sequences of cycles are
called alignments, but do not appear that frequently. Let us instead consider sets of cycles, also known
as permutations:
P = S ET(C YC(Z))
As suggested by this formula, every cycle can be naturally decomposed into a set of cycles. Consider
the permutation
σ = (10, 12, 2, 7, 1, 8, 6, 9, 4, 5, 3, 11)
Let us track where 1 moves under the action of this permutation
1 → 10 → 5 → 1
2 → 12 → 11 → 3 → 2
4→7→6→8→9→4
So the above decomposes into 3 cycles of lengths 3,4,5; its cycle number is 3. Schematically this gives
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 9/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
Date
1
Sept 7
I.1, I.2, I.3
2
14
I.4, I.5, I.6
3
21
II.1, II.2, II.3
4
28
II.4, II.5, II.6
5
Oct 5
Using
this
6
12
7
19
8
26
9
Nov 2
Sections
from FS2009
Part/ References
Topic/Sections
1
5
2
9
VI.1
18
IX.1
Analytic Methods
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
4
9
7
Labelled structures I
10
Combinatorial
parameters
FS A.III
decomposition
as a set
IV.1, IV.2
(self-study)
IV.5 V.1
12
Unlabelled structures
3
III.1, III.2
IV.3, IV.4
Notes/Speaker
Symbolic methods
Combinatorial
Structures
FS: Part A.1, A.2
Comtet74
Handout #1
(self study)
Labelled combinatorial classes
Labelled structures II
11
Combinatorial
Parameters
Asst #1 Due
8
6
of cycles
we get
Multivariable
GFs
Complex
∼ Analysis
P = S ET(C YC(Z)) ∼
= S EQ(Z)
1
1
P (z) = exp log Asst #2 Due
=
Asymptotic methods 1 − z
1−z
Singularity Analysis
Sophie
10 fact that exp and log are inverses is reflected in cycle-decomposition of permutations.
The
12
A.3/ C
Introduction to Prob.
Mariolys
Now we can go further by playing with restrictions (as we did above) to get
11
Limit Laws and Comb
Marni
Proposition.
P (A,B)
of permutations
with cycle
Random
Structures
20
IX.2 The class
Discrete Limit Laws
Sophie lengths in A ⊆ N and cycle number in B ⊆ N
and
Limit Laws
is given by
FS: Part C
12
23
IX.3
25
IX.4
13
30
IX.5
14
Dec 10
Combinatorial
instances of discrete
(rotating
presentations)
P (A,B) (z)
Mariolys
= β(α(z))
Continuous Limit Laws
X za
α(z) = Quasi-Powers and
Gaussian
a limit laws
Marni
β(z) =
Sophie
a∈A
5.1
5.1.1
X zb
b!
b∈B
Presentations
Asst #3 Due
Sub-classes of permutations
Involutions
The permutation σ is an involution if σ ◦ σ = identity. This means that all cycles are length 1 or
2, but the number of cycles is unconstrained. In the language of the above theorem, A = {1, 2} and
B = {0, 1, 2, . . . }. Hence
I = S ET(C YC1,2 (Z))
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
I(z) = exp z + z 2 /2
Version of: 11-Dec-09
=
X zk
k≥0
=
k!
(1 + z/2) =
n/2
X
j=0
k j+k
XX
k z
k≥0 j=0
n/2 XX
n−j
n≥0 j=0
In =
k
j
j
2j k!
1
zn
2j (n − j)!
n!
2j (n − 2j)!j!
One can of course extend this to consider permutations in which all cycles are of length ≤ r.
5.1.2
Derangements
In the opposite direction, a derangement is a permutation in which no number stays put — hence
all cycles must be 2 or longer. In the language of the above theorem, A = {2, 3, 4, 5, . . . } and B =
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 10/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
Date
Sections
1
Sept 7
I.1, I.2, I.3
2
14
I.4, I.5, I.6
3
21
4
28
5
Oct 5
{0, 1, 2, . . . }. from FS2009
6
Part/ References
Topic/Sections
Labelled combinatorial classes
Notes/Speaker
Symbolic methods
D = SCombinatorial
ET (C YC >1 (Z))
Structures
Unlabelled
structures
A.1, A.2
FS: Part
1
e−z
Comtet74
− Labelled
z = structures I
II.1,D(z)
II.2, II.3 = exp log
Handout #1 1 − z
1−z
(self
study)
II.4, II.5, II.6
Labelled structures II
2 Combinatorial
(−1)n
1
Combinatorial
+
−
·
·
·
+
D
=
n!
1
−
III.1, III.2 n
parameters1!
2! Parameters n!
12
IV.1, IV.2
FS A.III
(self-study)
9
VI.1
Handout #1
(self-study)
12
A.3/ C
18
IX.1
20
IX.2
=
X
(−1)n
n≥0
zn X n
×
z
n!
n≥0
Asst #1 Due
Multivariable GFs
Hence Dn /n! = probability that a permutation leaves nothing in place, is the truncation of the expansion
IV.3, IV.4
Complex Analysis
Analytic Methods
of 7e−119
. Since
this converges
very quickly,
the probability of a permutation being a derangment is
FS: Part B: IV, V, VI
8
26
Singularity
Analysis
asymptotically
1/e ≈ 0.37.
Appendix B4
IV.5 V.1
99: Ch. 6 to be length > r or longer one gets
— 2if you requireStanley
all cycles
9 So Nov
Asst #2 Due
Asymptotic methods
10
11
12
5.1.3
Sophie
D(r) = S ET(C YC>r (Z))


Introduction
to Prob.
Mariolys
z
z2
zr
X zk
e− 1 − 2! −···− r!
1
Limit
Laws and Comb
Marni 
(r)

=
−
D (z) = exp log
1−z
k
1−z
Random Structures
Discrete Limit Laws
Sophie
k≤r
and Limit Laws
FS: Part C
23
IX.3
(rotating
Other variants
presentations)
25
IX.4
Combinatorial
instances of discrete
Mariolys
Continuous Limit Laws
Marni
This approach will work well to consider
other classes of permutations such as permutations with all
Quasi-Powers and
13 cycles,
30
IX.5
Sophie
even
all odd cycles, cycles of even
length,
Gaussian
limit laws all of odd length etc. It is also not hard to get all
d
permutations
so that σ = identity
for a given d. Give Asst
these
a shot.
14
Dec 10
Presentations
#3 Due
5.2
The number of cycles in a permutation
(r)
Let P (r) be the class of permutations that decompose into r cycles. The number Pn
Stirling cycle numbers (or Stirling numbers of the first kind), and their egf is
≡
n
r
are the
P (r) = S ETr (C YC(Z))
r
1
1
(r)
P (z) =
log
r!
1−z
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
1
.
So — the probability that a permutation has exactly r cycles is nr n!
What do these numbers look like? Let pn,k be this probability, and consider n = 100.
k
pn,k
1
.01
2
.05
3
.12
4
.19
5
.21
6
.17
7
.11
8
.06
9
.03
10
.01
How do we interpret this? A random permutation of 100 has on average a few more than 5 cycles.
It rarely has more than 10. What does this imply about the average length of a cycle?
5.3
Application: 100 Prisoners
This example is a reformulation of an applied problem on probing and hashing. The wording is modified
5.3.1
The game
A prison warden wants to make room in his prison and is considering liberating one hundred prisoners,
thereby freeing one hundred cells. He assembles one hundred prisoners and asks them to play the
following game:
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 11/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
Date
Sections
Part/ References
Topic/Sections
Labelled combinatorial classes
Notes/Speaker
from
FS2009
1. he lines
up
one hundred urns in a row, each containing the name of one prisoner, where every
prisoner’s
name
occurs exactly once;
1
Sept 7 I.1, I.2, I.3
Symbolic methods
Combinatorial
Structures
isFS:
allowed
to lookUnlabelled
insidestructures
fifty urns. If he or she does not find his or
Part A.1, A.2
Comtet74
the
fifty
will
immediately
be executed, otherwise the game
II.1, II.2,
II.3 urns, all prisoners
Labelled
structures
I
Handout #1
(self
study)
II.4, II.5, II.6
Labelled structures II
22.
14
I.4, I.5, I.6
every
prisoner
3
one
21 of
her name in
continues.
4
28
The
prisoners
have a few moments to decide on a strategy, knowing that once the game has begun, they
Combinatorial
will
not
be
able
to communicate
withCombinatorial
each other, mark
5
Oct 5
III.1, III.2
Asst the
#1 Dueurns in any way or move the urns or the
parameters
Parameters
names inside them. Choosing
urns at random, their chances of survival are almost zero, but there is a
FS A.III
6
12 giving
IV.1, IV.2
Multivariable GFs
strategy
them a(self-study)
30% chance of survival,
assuming that the names are assigned to urns randomly
what
is
7
19 it? IV.3, IV.4
Complex
Analysis
Analytic Methods
99 100
FS: Part B: IV, V, VI
(49)
1
8 First
26 of all, the survival
Singularity
Analysis
= 2100
so this is definitely
probability
using
random
choices
is
Appendix B4
IV.5 V.1
(100
50 )
Stanley 99: Ch. 6
9
Nov 2
Asst
#2
Due
Asymptotic methods
Handout #1
not a practical strategy.
10
5.3.2
9
(self-study)
VI.1
Sophie
12The A.3/
C
strategy
Introduction to Prob.
Mariolys
18
IX.1
Limit Laws and Comb
Marni
1
BOX
The
11 30% survival strategy is to consider the contents of the urns to be a permutation of the prisoners,
Random Structures
20
IX.2
Discrete Limit Laws
Sophie
and traverse
cycles. To
the notation
simple, assign
a number to each prisoner, for example by
and keep
Limit Laws
FS: Part C
Combinatorial
sorting23 theirIX.3names alphabetically.
The
urns may thereafter
be considered to contain numbers rather
Mariolys
(rotating
instances of discrete
12 names. Now clearly the contents
than
of
the
urns
define
a
permutation.
The first prisoner opens the
presentations)
25
Continuous
Limit Laws
first urn.
IfIX.4
he finds his name, he has
finished
and Marni
survives. Otherwise he opens the urn with the
number
he found
in the first urn. The
processand
repeats:
the prisoner opens an urn and survives if he
Quasi-Powers
13
30
IX.5
Sophie
Gaussian limit laws
finds his name, otherwise he opens the
urn with the number just retrieved, up to a limit of fifty urns.
14 second
Dec 10 prisoner starts Presentations
#3 Duewith urn number three, and so on. This
The
with urn number two, theAsst
third
strategy is precisely equivalent to a traversal of the cycles of the permutation represented by the urns.
Every prisoner starts with the urn bearing his number and keeps on traversing his cycle up to a limit
of fifty urns. The number of the urn that contains his number is the pre-image of that number under
the permutation. Hence the prisoners survive if all cycles of the permutation contain at most fifty
elements. We have to show that this probability is at least 30%.
Let us see how this works on a smaller example. Consider the permutation from the beginning of
these notes:
BOX
2
BOX
3
BOX
4
BOX
5
BOX
6
Dr. Marni
Department of
UNIVERSITY
10 MISHNA,12
2 Mathematics,
7 SIMON FRASER
1
8
Version of: 11-Dec-09
1
2
BOX
7
BOX
6
12
8
BOX
9
4
9
BOX
5
10
BOX
3
11
12
11
BOX
4
9
7
5
10
3
11
8
6
This simulates the situation where the warden has put a 10 in box 1, a 12 in box 2, a 2 in box 3 and in
general σ(i) in box i. If the prisoners are following the strategy, they will each have 6 boxes to look in
to, and since the maximum cycle size is 5, they will win.
How will it go down? Prisoner one will open box 1. It will say 10. He then opens box 10, and it will
say 5. He opens box 5, it says 1, and he breathes a sigh of relief. Prisoner two’s turn. He opens box 2: it
says 12. He continues along the cycle and the fourth box he opens will have a 2 inside. Later, prisoner
4 is up. He follows this and he starts to sweat because it takes him 5 turns, but this is still less than 6,
so he is okay. We know what this permutation looks like, and that every cycle is of length less than 6,
so, as we already noted, if this were the scheme, the prisoners would win.
Note that this assumes that the warden chooses the permutation randomly; if the warden anticipates this strategy, he can simply choose a permutation with a cycle of length 51. To overcome this, the
prisoners may agree in advance on a random permutation of their names.
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 12/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
Date
5.3.3
Sections
Part/ References
from FS2009
The analysis
Topic/Sections
Labelled combinatorial classes
Notes/Speaker
1
Sept 7 I.1, I.2, I.3
Symbolic methods
Combinatorial
First,
let us consider the
general problem
with 2n prisoners choosing n urns. Let us calculate the egf
Structures
2
14
I.4,
I.5,
I.6
Unlabelled
that a random permutation
(ofA.2any size)
has astructures
cycle of length n + 1, and then deduce the probability for
FS: Part A.1,
Comtet74
the3 case
of length n, and
these
are
21 of permutations
II.1, II.2, II.3
Labelled
structures
I the cases in which case this strategy will fail. Fix
Handout #1
n and
consider
the
class
of
all
permutations
which
have a cycle of length at least n + 1. We take a sum
(self
study)
4
28
II.4, II.5, II.6
Labelled structures II
over all permutations σ of the term z |σ| if it has no such cycle and z |σ| u if it does
5
Oct 5
III.1, III.2
Combinatorial
Combinatorial
Analytic Methods
Complex Analysis
Asst #1 Due
parameters
Parameters
1 (if σ has a cycle of length n);0
otherwise
X z |σ| u
2z 2
zn
z n+1
FS A.III
=1+z+
+ · · · + n!
+ (n!u + nn!)
+ ...
g
(z,
u)
=
6
12
IV.1, IV.2
Multivariable GFs
(self-study) |σ|!
2!
n!
(n + 1)!
(n)
7
19
σ∈P
IV.3, IV.4
We first expand theFS:decomposition
P = S ET(C YC(Z)) = S ET(C YC≤n (Z)+ C YC>n (Z)) which we then
Part B: IV, V, VI
8
26
Singularity Analysis
Appendix B4
expand as follows:
IV.5 V.1
Stanley 99: Ch. 6
9
Nov 2

 Asymptotic methods Asst #2 Due
Handout #1
10
11
12
9
(self-study)
VI.1
12
A.3/ C
18
IX.1
20
IX.2
23
IX.3
25
IX.4
g
Sophie


z n+1
z2
z3
zn
z n+2

 Introduction
+ to Prob.
+ · · · +Mariolys+ u
+u
+ . . .
(z, u) = exp z +
2
3{z
n}
n+1
n+2
|
|
{z
}
Limit Laws and Comb
Marni
C YC≤n (Z) Sophie
C YC>n (Z)
Random Structures
Discrete Limit Laws
n+1
and Limit Laws
n+2
1
z
z
FS: Part C=
Combinatorial
exp (u − 1)
+
+ ...
.
Mariolys
(rotating
1 − z instances of discreten + 1 n + 2
(n)
presentations)
Continuous Limit Laws
Marni
The desired probability is
Quasi-Powers and n+1
13
30
IX.5
Sophie
Gaussian limit
1 laws
z
z n+2
2n
(n)
[z ] [u]g (z, u) = [z 2n ][u]
1 + (u − 1)
+
+ ...
,
1−z
n+1 n+2
14
Dec 10
Presentations
Asst #3 Due
since exp(•) = 1 + • + •2 + . . . , and so once you square the argument to the exp, you will only create
terms which are a power of z greater than 2n. This is equal to
n+1
z n+2
1
z
2n
(n)
2n
+
+ ...
[z ] [u]g (z, u) = [z ]
1−z n+1 n+2
!
2n
`
X
X
X
1
1
2n
= [z ]
z` =
.
k
k
`
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Theseof:are
related to Harmonic numbers and have
Version
11-Dec-09
we compute that
[z 2n ] [u]g (n) (z, u) < log 2 =⇒
k=n+1
k=n+1
good estimates. Using one of these good estimates
1 − [z 2n ] [u]g (n) (z, u) > 1 − log 2 = 0.3068528 . . .
That is, at least 30 %.
5.4
Computing Stirling cycle numbers
We can follow a similar analysis to get formulas for Stirling cycle numbers nr which we recall are the
number of permutations of n that decompose into r cycles. Start with the bivariate g.f.
P (z, u) =
∞
X
P (r) (z)ur
r=0
∞ X
r
1
1
=
u log
1−z
r!
r=0
1
= exp u log
= (1 − z)−u
1−z
X
n −u
=
(−1)
zn
n
n≥0
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 13/15
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
f ac ulty of sc ience
depar tment of mathem atics
Week
Date
Sections
1
Sept 7
I.1, I.2, I.3
2
14
I.4, I.5, I.6
3
21
II.1, II.2, II.3
Part/ References
Topic/Sections
from
so the coeff of
z nFS2009
is an expansion in u
Combinatorial n Symbolic methods
Structures X n
r
structures
uUnlabelled
= u(u
+ 1)(u
FS: Part A.1, A.2
r
Comtet74
r=0
Labelled structures I
Handout #1
(self study)
Labelled structures II
Labelled combinatorial classes
Notes/Speaker
+ 2) . . . (u + n − 1)
4
28
II.4, II.5,
II.6 and setting u = 1 gives
Differentiating
this
Combinatorial
Combinatorial
5
Oct 5
III.1, III.2
Asst #1 Due
n
1 X
parameters
Parameters
r
= 1 + 1/2 + 1/3 + . . . 1/n
FS A.III
n! Multivariable
r
6
12
IV.1, IV.2
GFs
(self-study)
7 is
19the expected
IV.3, IV.4
Complex
Analysis
Analytic Methods
This
number
of cycles in
a permutation
of length n and is approximately log n.
6
8
26
9
Nov 2
IV.5 V.1
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
Singularity Analysis
Asymptotic methods
Trees maps and graphs
10
9
VI.1
Asst #2 Due
Sophie
12
A.3/ C
to Prob.
For trees,
the
labelled case is not soIntroduction
different
to theMariolys
unlabelled case. We can consider planar and
non-planar
rooted
trees.
18
IX.1
Limit Laws and Comb
Marni
11
Random
Structures
20
IX.2 A be the
Discrete
Limit Laws
Sophie
Example.
Let
class
of rooted
labelled
planar
trees whose vertex outdegrees must lie in the
and Limit Laws
set Ω. Then we have FS: Part C
Combinatorial
12
23
IX.3
25
IX.4
(rotating
presentations)
A = Z S EQΩContinuous
(A)
Limit Laws
13 so30usingIX.5
and
similar reasoning as the
14
Dec 10
instances of discrete
Mariolys
Marni
Quasi-Powers and
unlabelled
case Sophie
Gaussian
limit laws
Presentations
Asst #3 Due
A(z) = zφ(A(z))
φ(u) =
X
uω
ω∈Ω
and indeed this is identical to the ogf for the unlabelled version. Hence An = n!Aˆn — this is easy
to prove by reading the vertices of a labelled tree in a canonical order (eg breadth first) to obtain a
permutation and an unlabelled tree of the same shape. Using this idea or lagrange inversion we get
that all labelled rooted planar trees are
1 2n − 2
= 2n−1 · 1 · 3 · · · (2n − 3)
An = n!
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER
n−1
n UNIVERSITY
Version of: 11-Dec-09
So — to the non-planar case
Example. Let T be the class of all non-planar rooted labelled trees. By deleting the root vertex one
obtains a set of labelled trees. Thus
T = Z ? S ET(T )
T (z) = zeT (z)
Now φ(u) = eu and Lagrange inversion gives
Tn = n![z n ]T (z)
1 n−1 u n
[u
] (e )
= n!
n
nn
= (n − 1)!
= nn−1
n!
Which is a famous result due to Cayley. These are usually called Cayley trees. A k-forest of Cayley
trees gives a very similar result
k
n − 1 n−k
(k)
n T (z)
Fn = n![z ]
=
n
k!
k−1
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 14/15
f ac ulty of sc ience
depar tment of mathem atics
Week
Date
Sections
1
Sept 7
I.1, I.2, I.3
2
14
I.4, I.5, I.6
3
21
II.1, II.2, II.3
4
28
II.4, II.5, II.6
5
Oct 5
III.1, III.2
6
12
IV.1, IV.2
7
19
IV.3, IV.4
8
26
9
Nov 2
Part/ References
MATH 895-4 Fall 2010
Course Schedule
A DDITIONAL N OTES
Topic/Sections
Labelled combinatorial classes
Notes/Speaker
from FS2009
Similar games
with restricted vertex degree lead to
10
11
12
IV.5 V.1
9
VI.1
12
A.3/ C
18
IX.1
20
IX.2
23
IX.3
25
IX.4
13
30
14
Dec 10
Symbolic methods
Combinatorial
T (Ω) = Z ? S ETΩ (T (Ω) )
Structures
Unlabelled structures
FS: Part A.1, A.2
(Ω)
(Ω)
Comtet74
T (z) = z φ̄(T Labelled
(z)) structures I
Handout #1
(self study)
Labelled structures II
Combinatorial
parameters
FS A.III
(self-study)
Combinatorial
Parameters
Analytic Methods
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
Complex Analysis
Random Structures
and Limit Laws
FS: Part C
(rotating
presentations)
IX.5
φ̄(u) =
X uω
|ω|!
ω∈Ω
Asst #1 Due
Multivariable GFs
Singularity Analysis
Asymptotic methods
Asst #2 Due
Sophie
Introduction to Prob.
Mariolys
Limit Laws and Comb
Marni
Discrete Limit Laws
Sophie
Combinatorial
instances of discrete
Mariolys
Continuous Limit Laws
Marni
Quasi-Powers and
Gaussian limit laws
Sophie
Presentations
Asst #3 Due
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
M ARNI M ISHNA , S PRING 2011
MATH 343: A PPLIED D ISCRETE M ATHEMATICS
PAGE 15/15
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