# Final Exam Syllabus – Discrete Math (V63.0120)

- Relations: 4.1-4.5
- Combinatorics: 5.1-5.4 (5.5 skipped)
- Probability: 6.1-6.3
- Graph theory: 7.1, 7.3 (7.2 skipped)
- Any previous material will only be covered implicitly through the above material.

## Sections 4.1-4.5 – Relations

A **relation** {$R$} with domain {$A$} and codomain {$B$} is a subset of {$A \cross B$}: {$R \subseteq A \cross B$}.

Can be defined using one of the following:

- a formula for {$R$} in set-builder notation
- an exhaustive enumeration of pairs in {$R$}
- a table
- an arrow diagram

### Possible properties of a relation

You should be able to prove or disprove any of the properties below.

**function:**For every element {$a \in A$}, there is exactly one element {$b \in B$} such that {$(a,b) \in R$}.**image:**An image of a function is the set of all possible outputs the function. This is a subset of the co-domain.

**inverse:**{$(b,a) \in R^{-1} \subseteq B\cross A$} exactly when {$(a,b) \in R \subseteq A\cross B$}**invertible function:**{$R$} is an invertible function of both {$R$} and {$R^{-1}$} are functions. In this case, {$R \o R^{-1} = \id_{B}$} and {$R^{-1} \o R = \id_{A}$} (the identity functions).

For relations with the same domain and co-domain ({$A=B$}), the following additional properties may hold:

**reflexive:**{$\forall a \in A$}, {$(a,a)\in R$}.**irreflexive:**{$\forall a \in A$}, {$(a,a)\notin R$}.**symmetric:**{$\forall a,b \in A$}, if {$(a,b)\in R$} then {$(b,a)\in R$}.**antisymmetric:**{$\forall a,b \in A$}, if {$a \ne b$} and {$(a,b)\in R$} then {$(b,a)\notin R$}.- Alternatively: {$\forall a,b \in A$}, if {$(a,b)\in R$} and {$(b,a)\in R$}, then {$a = b$}.

**transitivity:**{$\forall a,b,c \in A$}, if {$(a,b)\in R$} and {$(b,c)\in R$}, then {$(a,c)\in R$}.

Special combinations

**Partial order:**reflexive, antisymmetric, transitive**Strict partial order:**irreflexive, antisymmetric, transitive**Equivalence relation:**reflexive, symmetric, transitive- results in a partitioning of the domain/codomain {$A$}

### Function {$f: A \rightarrow B$}

**one-to-one**{$\equiv$}**injective**: Every element in the domain is mapped to a**unique**element in the codomain. i.e. {$\forall x_1, x_2 \in A, f(x_1) = f(x_2) \rightarrow x_1 = x_2$}- {$f$} is one-to-one → {$n(A) \leq n(B)$}
- Proof strategy: "Let {$x_1,x_2\in A$} such that {$f(x_1)=f(x_2)$}." Then show that {$x_1=x_2$}.

**onto**{$\equiv$}**surjective**: Every element in the codomain is produced by some element in the domain. i.e. {$\forall x_1, x_2 \in A, f(x_1) = f(x_2) \rightarrow x_1 = x_2$}- {$f$} is onto → {$n(A) \geq n(B)$}
- Proof strategy: "Let {$y\in B$}." Then produce an {$x$} satisfying {$f(x)=y$}.

**one-to-one correspondence**{$\equiv$}**invertible**{$\equiv$}**bijective**: Both one-to-one and onto.- {$f$} is a one-to-one correspondence → {$n(A) = n(B)$}

All three properties above are conserved via composition. E.g. the composition of two invertible functions is invertible.

### Set cardinality

- Two sets have the same
**cardinality**if there is a one-to-one correspondence between them. - A set {$A$} is
**infinite**if it has the same**cardinality**as a strict subset of itself. i.e. There is a map from {$A$} to itself that is one-to-one, but not onto. **Cantor-Bernstein Theorem:**If you have two functions {$f:A\rightarrow B$} and {$g:B\rightarrow A$} that are both one-to-one, then the sets {$A$} and {$B$} have the same cardinality.**Cantor's Theorem**: For every set {$A$}, {$A$} and {$\P(A)$} do not have the same size.- All sets that have the same cardinality at {$\ZZ$} are said to be
**countable**. E.g. {$\NN, \QQ, \ZZ^2$}, but not {$\P(\ZZ)$} nor {$\RR$}.

## Sections 5.1-5.4 (5.5 skipped) – Combinatorics

### Counting rules

**Rule of sums**for disjoint sets- {$n(A \union B)=n(A)+n(B)$}
- {$n(A \union B \union C)=n(A)+n(B)+n(C)$}

**Rule of sums with overlap:**(more general) {$n(A \union B)=n(A)+n(B)-n(A\inter B)$}**Rule of complements:**{$n(A')=n(U)-n(A)$}**Rule of products:**- {$n(A\cross A\cross … \cross A)$} (r times) {$ = n(A) \cdot n(A) \cdot … \cdot n(A) = n(A)^r$}
- {$n(A\cross B) = n(A)\cdot n(B)$}
- {$n(A\cross B\cross C) = n(A)\cdot n(B)\cdot n(C)$}

### Useful operators

**Factorial:**{$n! = n \times (n-1) \times (n-2) \times … \times 2\times 1$}**Permutations:**{$P(n,r) = {}_nP_r = P^n_r = \frac{n \times (n-1) \times (n-2) \times … \times (n-r+1) \times (n-r) \times … \times 2 \times 1}{(n-r) \times … \times 2 \times 1} = \frac{n!}{(n-r)!}$}**Combinations:**{$C(n,r) = {}_nC_r = C^n_r = {n \choose r} = \frac{P(n,r)}{r!} = \frac{n!}{(n-r)!\;r!} = C(n,n-r)$}- Forms the entries of the Arithmetic Triangle, also called Pascal's Triangle.
**Binomial theorem:**The coefficient of {$x^r$} in {$(1+x)^n$} is {$C(n,r)$}. i.e. {$(1+x)^n = \sum_{i=0}^n C(n,r) x^r$}- Together, these facts imply that the sum of the {$n^{th}$} row in Pascal's Triangle is {$\sum_{i=0}^n C(n,r) = (1+1)^n = 2^n$}.

### Ways to pick {$r$} items from {$n$}

Ways to pick {$r$} items from {$n$} |
order matters? | ||
---|---|---|---|

E.g. Pick 2 from {$\{a,b,c\}$} | Y | N | |

repetitions allowed? | Y | {$n^r$} ordered lists(a,a), (a,b), (a,c) (b,a), (b,b), (b,c) (c,a), (c,b), (c,c) |
{$C(r+n-1,r)=C(r+n-1,n-1)$}unordered lists or bags (a,a), {a,b}, {a,c} (b,b), {b,c} (c,c) |

N | {$P(n,r)$} permutations(a,b), (a,c) (b,a), (b,c) (c,a), (c,b), |
{$C(n,r)=C(n,n-r)$} sets{a,b}, {a,c} {b,c} |

### Intuition and examples

**Ordered lists**(order matters, repetitions allowed)- Determine the number of {$r$}-letter words that can be made from an alphabet of size {$n$}.
- Determine the number of non-negative integers less than 1000 with no 9's.
- Number of possible birthday combinations of 23 people.

**Permutations**(order matters, no repetitions)- Anagrams
- Selecting a committee where the order matters. E.g. number of ways to select 3 people to be President, VP, and secretary.
- Number of ways for 23 people to have unique birthdays.

**Sets**(order doesn't matter, no repetitions)- Selecting a sub-group of people with irrelevant order. E.g. number of ways to select 3 people to be on the board of directors.
- Number of ways of picking 23 distinct days from a year.

**Unordered lists, bags**(order doesn't matter, repetitions allowed)- Pick 7 fruits from an unlimited supply of apples, bananas, and oranges.
- Equivalent to the number of non-negative integers satisfying {$a+b+c=7$}.
- Number of ways to fill a shopping bag with exactly {$r$} fruits from an unlimited supply of {$n$} distinct objects.
- Equivalent to the number of binary strings of length {$r+n-1$} with exactly {$r$} zeros (or exactly {$n-1$} ones).
- Number of ways of picking 23 days from a year, where you can pick the same date multiple times.

## Sections 6.1-6.3 – Probability

### Properties

- {$Prob(E) = \frac{n(E)}{n(E)+n(\overline{E})} = \frac{n(E)}{n(S)}$}
- {$Prob(E) + Prob(\overline{E})=1 \;\tor\; Prob(E) = 1 - Prob(\overline{E})$}
- {$Prob(E \tor F) = Prob(E) + Prob(F) - Prob(E \tand F)$}
- {$Prob(E \tand F) = Prob(E) Prob(F | E) = Prob(F) Prob(E | F)$} (Bayes' Law)
- Splitting an event into possible cases, or contexts.
- {$Prob(E) = Prob(E \tand F) + Prob(E \tand \overline{F})$}
- If {$F, G, H$} disjoint and encompass the set of all possibilities – i.e. {$Prob(F \tor G \tor H) = 1$}, then

{$Prob(E) = Prob(E \tand F) + Prob(E \tand G) + Prob(E \tand H)$}

**Disjoint events** = **Mutually-exclusive** events

- {$Prob(E \tand F) = Prob(E | F) = Prob(F | E) = 0$}
- {$Prob(E \tor F) = Prob(E) + Prob(F)$}

**Mutually-independent**, or simply **independent**

- {$Prob(E | F) = Prob(E)$} and {$Prob(F | E) = Prob(F)$}
- {$Prob(E \tand F) = Prob(E) Prob(F)$}
- Using decision trees/probability trees to compute probability
**Bernoulli trial:**{$Prob($}E exactly {$k$} times in {$n$} indep. trials{$)$} = {$C(n,k) p^k (1-p)^{n-k}$} where {$Prob(E) = p$} for each trial.- Don't always have to do this using the formula explicitly.
**E.g.**Probability of getting 6 exactly twice in five rolls of a fair die can be computed either of the following ways.**With formula above:**{$C(5,2) \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^3$}**Or by counting:**{$\frac{n(6 \text{ exactly two times})}{n(\text{total possibilities})} = \frac{C(5,2)5^3}{6^5}$}

Numerator: (pick two positions out of five for the 6's) × (the number of ways to pick the three non-6's)

**E.g.**Jim has a probability of {$4/5$} of winning a poker game. What is the probability that in 5 games, he would win two and lose three?**Use formula explicitly:**{$C(5,2) \left(\frac{4}{5}\right)^2 \left(\frac{1}{5}\right)^3$}**Use formula implicitly:**The possibilities are WWLLL, WLWLL, LWWLL, etc. In other words, you have 5 spots "_ _ _ _ _". Two of these spots are W ("win") and three are L ("lose").- There are {$C(5,2)$} ways to pick the two winning spots out of 5. (And {$C(3,3)=1$} way to pick the three losing spots from the remaining ones.)
- The probability of particular case with 2 wins and 3 losses (e.g. WWLLL) is {$\left(\frac{4}{5}\right)^2 \left(\frac{1}{5}\right)^3$}
**Putting it all together:**The probability of him winning two out of 5 is: {$C(5,2) \left(\frac{4}{5}\right)^2 \left(\frac{1}{5}\right)^3$}.

**E.g.**Probability of getting exactly three 4's and exactly one 3 when rolling a die 5 times.- In other words, you have five spots "_ _ _ _ _". Two of these are 4 and one is 3. The last one is neither 3 nor 4.
- Pick three spots for 4: {$C(5,3)$}
- Pick 1 spot for 3 from the remaining spots: {$C(2,1)$}
- (Optional) Pick 1 spot for the non-3/4: {$C(1,1)=1$}
- The probability of each case (e.g. 434_4) where you have exactly three 4's and exactly one 3 is {$\left(\frac{1}{6}\right)^3 \left(\frac{1}{6}\right)^1 \left(\frac{4}{6}\right)^1$}
**Putting it all together:**{$C(5,3)C(2,1)\left(\frac{1}{6}\right)^3 \left(\frac{1}{6}\right)^1 \left(\frac{4}{6}\right)^1$}**Or by counting:**So the number of ways to pick exactly three 4's and exactly one 3 is {$C(5,3) C(2,1) 4$} (the last 4 is because we have 4 ways to pick the non-3/4). This means that the total probability is {$\frac{C(5,3) C(2,1) 4}{6^5}$}.

- In other words, you have five spots "_ _ _ _ _". Two of these are 4 and one is 3. The last one is neither 3 nor 4.

## Sections 7.1, 7.3 (7.2 skipped) – Graphs:

### Terminology and definitions

- Graph: G = (V,E)
- V = set of
**vertices/nodes****degree**of node v = deg(v) = number of edges incident to v, except that loops are counted twice

- E = set of
**edges**- has two**endpoints**(vertices)- node/vertex v is an
**endpoint**of edge e ⇆ edge e is**incident with**vertex v **loop:**if the two endpoints are the same**parallel edges:**edges with the same endpoints- {$\sum_{v\in V}\;deg(V) = 2\;n(E)$}

- node/vertex v is an
**walk:**sequence of alternating vertices and edges where every edge lies between its endpoints**length**= number of edges in the walk**trivial walk:**length=0 (E.g. A)**closed walk:**the first and the last vertex are the same (E.g. A 4 D 1 C 3 A)

**trail:**walk with no repeated edges**circuit:**closed trail**cycle:**circuit with first/last vertex being the only repeated node**path:**walk with no repeated vertices (always a trail)

- Terms not included on the final: simple graph, directed graph, connected, subgraph, connected component

### Eulerian trails/circuits

**Eulerian trail/circuit:**traverses every edge exactly once- Provide an Eulerian trail/circuit of a graph
**Eulerian graph:**one that contains an Eulerian*circuit (i.e. a closed Eulerian trail)*- A connected graph is Eulerian if and only if every vertex has even degree.
- Make a graph Eulerian by adding the minimum number of edges.

### Isomorphisms

- Tell whether or not graphs are
**isomorphic**. Prove using a simple argument (e.g. counting vertex degree) that two graphs are not isomorphic. - Identifying
**planar graphs**- A graph is planar if and only if it does not contain {$K_5$} and {$K_{3,3}$} as a subgraph
- Draw
**planar embeddings**of provided graphs. **Euler's formula**for planar graphs: {$V+F=E+2$} (Note that the number {$F$} of faces includes the unbounded face defined by the outer-most cycle of edges.)