Combinatorics KNU Math 254 Classnotes

We have already proved the 'if' part of the 'if and only if', we need to prove the 'only if':. that there is a lid just ifb veriekenmorn. If there is some universeX and S is a subset of the universe, then whereS=X\S is the complement of S inX, we have.

Figure 1.1: A white-black coloured 8 × 8 chessboard and some dominos

PERMUTATIONS AND COMBINATIONS 10

PERMUTATIONS AND COMBINATIONS 11 The last question was asking for the number of circular permutations of an n-element set

Combinations of Sets

Binomial Theorem For any integer n ≥ 0

PERMUTATIONS AND COMBINATIONS 13

Permutations of Multisets

PERMUTATIONS AND COMBINATIONS 14 Nice! That generalises to the following

The multiset

Combinations of Multisets

PERMUTATIONS AND COMBINATIONS 15 between them (there are gaps before the first fruit and after the last) we choose three gaps that we

Finite Probability

PERMUTATIONS AND COMBINATIONS 16 Lets see a simple example of an experiment

The pigeonhole principle

Simple form
The pigeonhole principle
THE PIGEONHOLE PRINCIPLE 19 Fact 3.1.2: Still the pigeonhole principle
THE PIGEONHOLE PRINCIPLE 20

Pigeonhole Principle: Strong Form
Ramsey Theory

THE PIGEONHOLE PRINCIPLE 21

The next more general version of the pigeonhole principle can be used to say that within a set of values, a value must be at least the mean value. Find a coloring of the edges of K5 with the colors red and blue, with no triangle (the vertices of which are the vertices of the graph), and each edge of which has the same color.

Generating Permutations and Combinations

Generating Permutations
GENERATING PERMUTATIONS AND COMBINATIONS 24 An algorithm to order the permutations of [n]
GENERATING PERMUTATIONS AND COMBINATIONS 25 listing of [3]!
GENERATING PERMUTATIONS AND COMBINATIONS 26

Inversions of Permutations

GENERATING PERMUTATIONS AND COMBINATIONS 27 Theorem 4.2.1

Generating Combinations

GENERATING PERMUTATIONS AND COMBINATIONS 28 We let a combination C ⊂ [n] of 2 [n] correspond to its characteristic vector

Gray codes

GENERATING PERMUTATIONS AND COMBINATIONS 30 With this notation, we are ready to prove the theorem

More generally for the list of[n]!, the number n will move every step except when it reaches an end. In the first case, we started Gn(i) and Gn(i+ 1) both with 0, so have the same parity as Gn−1(i) and Gn−1(i+ 1), respectively, and so the result is trivial by induction (since adding a 0 doesn't change what the rightmost1 is).

The Binomial Coeﬀicient

Pascal’s Triangle
THE BINOMIAL COEFFICIENT 32 This magical little triangle yields lots of cool identies. Here is a new proof of one that we have

The Binomial Theorem

THE BINOMIAL COEFFICIENT 33 Assuming now that the identity holds for (x + y) n − 1 we have
THE BINOMIAL COEFFICIENT 34 You can also get this last identity with calculus: take the derivative of

Unimodality of Binomial Coeﬀicients

THE BINOMIAL COEFFICIENT 35 Proof. Consider the ratio
THE BINOMIAL COEFFICIENT 36

Multinomial Coeﬀicients

THE BINOMIAL COEFFICIENT 37

The observation that the sum of the entries in a row is twice the sum of the entries in the previous row show that. This is the answer to the problem you will see in the exercises of finding the number of shortest walks along the grid from one point to another. Using a combinatorial argument about the number of ways to choose k different x in the extension (x+y)n, we proved the binomial theorem.

BINOMIAL COEFFICIENT 33 Assuming that the identity now holds for (x+y)n−1, we have. Assuming that the identity now holds for (x+y)n−1, we have We conclude by noting that we can replace the outer binomial coefficients with those in the desired identity. We conclude this section simply with a more general definition of binomial coefficients.

What is the largest family A⊂2[n] of subsets of [n] such that no subset in the family is in another. definitions of partial orders or poses are given in more detail in section 4.5 of the text. Since no two elements in A can be in the same longest chain in 2[n] and every element in at least ⌊n/2⌋!⌈n/2⌉! we have this. Note that every monomial in the expansion of this polynomial is of the form xn11xn22.

The Inclusion Exclusion Principle and its Applications

The Inclusion Exclusion Principle
THE INCLUSION EXCLUSION PRINCIPLE AND ITS APPLICATIONS 39 S
THE INCLUSION EXCLUSION PRINCIPLE AND ITS APPLICATIONS 40

Combinations with Repetition

THE INCLUSION EXCLUSION PRINCIPLE AND ITS APPLICATIONS 41 We do this now using PIE

Derangements

For n ≥ 1
THE INCLUSION EXCLUSION PRINCIPLE AND ITS APPLICATIONS 42

Now we have an idea and are ready to state the Principle of Inclusion and Exclusion (PIE). Elements in S that have none of the Pi properties are counted in a0 and contribute nothing else to the sum, so it suffices to check that for elements that have some of the Pi properties, the element contributes 0 to the sum. The following complementary version is also often useful when we want to count elements that satisfy one of a set of properties instead of none of the set of properties.

This seems to be the "direct" form of PIE, but we started with the complement form, since simple uses of the direct form are more contrived. However, we could not deal with the requirement that x1≤3, for which we would have to find the number of rcombinations. PRINCIPLE OF INCLUSION EXCLUSION AND ITS USES 41 We do this now using PIE We do this now using PIE.

We'll see that in the next chapter when we look at how to solve recurrence relations. We can do the same enumeration for disorders withan=i for any of the then−1 values of i∈[n−1]. It is good to know these even if we can calculate the numbers in closed form, because then we can recognize them in combinatorial problems.

Recurrence Relations and Generating Functions

Recurrence Relations
RECURRENCE RELATIONS AND GENERATING FUNCTIONS 44 have to write out a couple terms to solve this recurrence relation. The closed formula was probably
RECURRENCE RELATIONS AND GENERATING FUNCTIONS 45 In Section 7.4 we will solve the the fibonacci recurrence relation, in fact, we will solve it twice

Generating Functions

RECURRENCE RELATIONS AND GENERATING FUNCTIONS 46 so is n+k k − −1 1

Exponential Generating Functions

RECURRENCE RELATIONS AND GENERATING FUNCTIONS 47 Theorem 7.3.1
RECURRENCE RELATIONS AND GENERATING FUNCTIONS 48 Thus we can express the above more compactly as

Linear Homogeneous Recurrence Relations

RECURRENCE RELATIONS AND GENERATING FUNCTIONS 49 The relation is linear because the function is linear in all previous terms h i that occur. It is
RECURRENCE RELATIONS AND GENERATING FUNCTIONS 51 Setting

RECURRENCE RELATIONS AND GENERATING FUNCTIONS 44must write out some terms to solve this recurrence relation. RECURRENCE RELATIONSHIPS AND GENERATING FUNCTIONS 45In Section 7.4 we will solve the fibonacci recurrence relation, in fact, we will solve it twice. Finding a generating function for a sequence will not always make calculating a term of the sequence easier, but sometimes it will.

Using similar combinatorial arguments, one can find the generating function of a sequence, although calculating the co-efficiency may not be easy. In the same way that the generating function was useful for inr combination problems, the exponential generating function is useful for inr permutation problems. REPETITION RELATIONS AND GENERATING FUNCTIONS 48Thus we can express the above more compactly asThus we can express the above more compactly as.

You should be able to find the exponential generating function and then put it into a form where you can read off the coefficients. The ones we consider are linear homogeneous recurrence relations with constant co-efficiencies: recurrence relations of the form RECURENCE RELATIONS AND GENERATING FUNCTIONS 49The relation is linear because the function is linear in all previous expressions hi that occur.

Special Counting Sequences

The Catalan numbers
SPECIAL COUNTING SEQUENCES 53 she likes jokes, but jokes shouldn’t be so hurtful
SPECIAL COUNTING SEQUENCES 54 This yields a recursive formula

Difference Sequences

SPECIAL COUNTING SEQUENCES 57 Theorem 8.2.2
SPECIAL COUNTING SEQUENCES 58 With this, we can do the following, which is a theorem in the text

Partition Numbers

SPECIAL COUNTING SEQUENCES 59

We have skipped this, but in Section 7.6 it is shown, using generating functions, that the number of planar triangulations of a plane cycle on n+ 2 vertices is Cn. Show that Cn counts the number of walks from the origin to (n, n) on an n×n grid that never goes below the diagonal line from (0,0)to(n, n). Show by induction that Cn∗= (4n−6)Cn∗−1 counts the number of multiplication schemes for applying a non-associative, non-abelian operation×numbers.

Not only does it allow us to write the polynomial function in the binomial basis, instead of the standard basis; but from this we get a nice formula for the partial sums of a polynomial sequence. It is convenient to specify the notation for the entries of the 0th diagonal of the difference table h0, h1, h2,. i! is the number of i-permutations of [n]. A reciprocal relation is derived for them in the text, and it is shown that they count the number of ways in which we can split [d]intoknon-empty circular permutations.

We also counted the partitions of indistinguishable items into distinguishable parts: the number of non-negative integer solutions for something like x1+. The conjugate partition λ∗ of a partition λ is the partiton whose diagram we get by interchanging rows and columns of (or rotating in 3 dimensions) the Ferrer's diagram. Wheresn is the number of self-adjoint partitions of nandptn is the number of partitions in distinct odd integers, we havepsn=ptn.

Systems of Distinct Representatives

General Problem Formulation
SYSTEMS OF DISTINCT REPRESENTATIVES 63

Existence of SDRs

Hall’s Marraige Theorem (1935)
SYSTEMS OF DISTINCT REPRESENTATIVES 64 n = 1 the theorem is that there is an SDR if and only if A 1 contains an element. This is clearly

Stable Marriages

SYSTEMS OF DISTINCT REPRESENTATIVES 66 Betty is less happy, she gets her last choice, but everyone else prefers who they have to Betty, so

The Deferred Acceptence Algorithm for the stable marriage prob- lem

SYSTEMS OF DISTINCT REPRESENTATIVES 67 ii. When a man is proposed to he accepts if the woman who proposed is the highest ranked
SYSTEMS OF DISTINCT REPRESENTATIVES 68

If not all jobs can be filled, how many can be? A} of the subgroups of a set M without SDR, we can still ask what the size of the largest subgroup is. If this is greater than 0, then there can be no SDR, and clearly if the absence is, then at least d of the families in AI must be removed from each subfamily with an SDR.

Thus, the size of the largest subfamily Awith SDR is at most n-d, where the number of defects is the largest in all subfamilies. With a proof similar to that of Marraige's theorem, we can show that this is tight. A perfect marriage is any match between women and men, but some are better than others.

A stable marriage is optimal for women if each woman gets her highest ranked (favorite) man over all the men they get in stable marriages. The delayed acceptance algorithm in which women propose produces a stable marriage that is optimal for women. Call a man viable for a woman if he is paired with that woman in a stable marriage.

Combinatorial Designs

Modular Arithmetic
COMBINATORIAL DESIGNS 70

When Z n is a field

Block Designs

SBIBD

COMBINATORIAL DESIGNS 73 The example for v = 7 and k = 3 that we saw above above can be constructed from the first

Steiner Triple system

COMBINATORIAL DESIGNS 74 Lemma 10.3.3

Latin Squares

COMBINATORIAL DESIGNS 75 Here are a couple of latin squarea of order 5
COMBINATORIAL DESIGNS 76 Theorem 10.4.1
COMBINATORIAL DESIGNS 77 Our twelve 3- blocks are of four types

It is easy to see that the additive inverse of an element is unique, and we denote this by −a. Show that if there is a BIBD with parameters b, k, v, r and λ, then there is also one with parameters cb, k, v, cr and cλ for every integer c≥1. A Latin square of order is an n×n matrixA with entries inZn such that every integer occurs exactly once in every row and in every column.

Two different Latin squares A= [aij] andB = [bij] of order are orthogonal if for every pair(a, b)∈Z2n there is aniandj such that dataij =aandbij =b. In fact, a set of Latin squares of order is a set of MOLS, for 'Mutually Orthogonal Latins Squares' if every pair of Latin squares in the set is a POLS. This follows by finding POLS of orders9en2k for all oddk≥5, (which we will not show) and the following statement.

If there is a pair of POLS of order m and a POLS of order mn, then there is a POLS of order mn. For a Latin square A = [aij] of order m and a Latin square B = [bij] of order mlet A⊕B be the matrix of order mn whose ijth entry is defined as follows. An array of n−1MOLS of order n can be used to make aBIBD with parameters b=n2+n, v= n2, k =n and λ= 1. This is done in some detail in the text, but we show only one example of prime with = 3, which gives us aSTS(9).

Intro to Graph Theory

Basic Definitions
INTRO TO GRAPH THEORY 79
INTRO TO GRAPH THEORY 80 If two graphs are isomorphic, we usually consider them the same, and so the vertices can be
INTRO TO GRAPH THEORY 81 The degree sequence is a graph invariant– isomorphic graphs have the same degree sequences,

Eulerian Trails
Hamilton Paths and Cycles

Bipartite Graphs

Trees

INTRO TO GRAPH THEORY 86 Problems from the text

The degree d(v) of a vertex v in a graph is the number of edges it is on, (plus the number of loops). A list of degrees of the vertices of a graph in non-increasing order is a degree sequence. Each edge of the graph contributes two to degrees to the sequence, so a sequence of degrees must sum to an even number.

It is often said that the first problem of graph theory was the Königsberg Bridges problem. A graph is Eulerian if and only if it is connected and all its vertices have equal degree. Indeed, if it were not, then the degree of vminG is odd by our proof above, so there is another edge associated with T(vm).

When G is connected, there is an edge. If T is not an Euler circuit, then G′=G\T is not empty. A graph G= (V, E) is bipartite if there is a partition V =R∪B of vertices such that all edges have one endpoint in R and one in B. The equivalence of these definitions is quite clear, although it is a bit tedious to prove.

More on Graph Theory

Graph Colouring