Eric Chapdelaine
Student at Northeastern University Studying Computer Science.


CS1800 Discrete Structures

Module 2 - Statement Representation

Circuits and Logic

Half Adder Circuit

Half Adder Circuit

S Is the sum and c is the carry. Uses xnor for the sum because 0 xnor 0 = 0, 1 xnor 0 = 1, and 1 xnor 1 = 0 (where the carry is).

Ripple Adder

What if you want a third input? You have to expand the Half Adder


Boolean Algebra

Boolean Equivalence

$(A \wedge \neg B) \vee (A \vee B) \equiv A$

Distributive Law: $((A \wedge \neg B) \vee A) \wedge ((A \wedge \neg B) \vee B)$

Commutative Law: $(A \vee (A \wedge \neg B)) \wedge ((A \vee B) \wedge (B \vee \neg B))$

Absorption Law and Known Always True: $A \wedge (A \vee B)$

Absorption: $A$

Implications, Inverses and Contrapositives



p: Ben left the house after 8am

q: Ben arrived at work after 9am

If Ben left after 8am, then Ben arrived after 9pm

$p \to q$ (we use double arrow notation $\implies$)

“If p then q”

What if we flipped the truth table?

This proves that

\[\neg (p \to q) \equiv p \wedge \neg q\] \[p \implies q \equiv \neg(p \land \neg q) \equiv \neg p \lor q\]


$p \to q \equiv \neg q \to \neg p$

If Ben arrived before 9, he must’ve left before 8

Converse of $p \to q$ is $q \to p$

If Ben arrives after 9, then Ben left after 8

Inverse $\neg p \to \neg q$

(Contrapositive of the Converse)

Existential Qualifiers


It rained every first day of the month last string (March, April, May)

On every first day of the month, it rained

On every x, it rained

$p(x)$= it rained on $x$ →on every $x$, $p(x)$

for all first days of the month, it rained

Denotated as:

\[\forall x, p(x)\]

$x$ is the first day of the month

$x \in$ first days of the month

$\forall$ - Universial Quantifiers


$Q(x)$ : $x$ has green leaves, $x$ is a tree

$\forall x, Q(x)$ : Every tree has green leaves

Some trees have red leaves → There is a tree with red leaves (only have to prove one)

$R(x)$ : $x$ has red leaves

\[\exists x, R(x)\]

There exists a tree with red leaves


$\forall x, R(x)$ - All trees have red leaves

Negate: Exhibit a tree with leaves that aren’t real. Some trees don’t have red leaves


$\forall x, R(x)$

Negate: $\exists x, \neg R(x)$

Example: There are some trees that have purple leaves.

Every tree doesn’t have purple leaves

$\exists x, purple(x)$

Negate: $\forall x, \neg purple(x)$


$\neg \forall x, p(x) \equiv \exists x, \neg p(x)$

$\neg \exists x, R(x) \to \forall x, \neg R(x)$

None: $\forall x \neg p(x)$

All : $\forall$

Some: $\exists$

Module 3 - Collection Representation

Lesson 1 - Set Representation

Set of locations that Northeastern has a campus in (each one of them is called an element):

NOTE: Sets cannot have duplicates


NU_CAMPUS = {London, Portland, …, Boston}

$\in$ - an element of

Ex. London $\in$ NU_CAMPUS

Infinite Sets

In order to represent infinite sets (ex. multiples of 5), we need to use set builder notation

Multiple of 5:

{ x x = 5 * n, n ≤ 1}

This example isn’t too clear because the ncan be any number like 1.3, not just natrual numbers.


Therefore the multiple of 5 example should look like this:

{ x x = 5 * n, n $\in \N$} or { 5n n $\in \N$}

The first one is perferred because it represents computer programming languages

General Notation

{ x the rules governing membership}

Lesson 2: Subsets, The Empty Set and The Power Set


NU_CAMPUS = {London, Portland, …, Boston}

NU_US_CAMPUSES = {Boston, Seattle, …}

Subsets - $\subseteq$


Another example: Fives = { x | x = 5 * n, n $\in \N$}

Tens = { x x = 10 * n, n $\in \N$}

Tens $\subseteq$ Fives

General Notation

If and only if - $\iff$


A $\subseteq$ B $\iff$ $\forall$ x $\in$ A, x $\in$ B

If not in a subset:

A $\nsubseteq$ B $\iff$$\exists$ x $\in$ A, x $\notin$ B

Proper Subset - $\subset$

C $\subset$ D $\iff$$\forall$ x $\in$ C, x $\in$ D and $\exists$ y $\in$ D, y $\in$ C

E $\subseteq$ F and F $\subseteq$ E $\iff$E $\equiv$ F (very helpful for proofs)

The Empty Set

The Power Set


IC = {chocolate, strawberry, vanilla}

$\wp (IC)$ = { {}, {chocolate, strawberry, vanilla}, {chocolate}, {strawberry}, {vanilla}, {chocolate, strawberry}, …}

Every set is a subset of itself

How many elements are in $\wp (IC)$?

Notice that $8 = 2^{3}$

We can make this into a general form:

Lesson 3: Operations on Sets

Favorite Green Vegetables (FGV) - there is an implied context. It is Ben’s favorite things in the universe green vegetables

Universe - $\mho$

FGV = {cucumber, peas, lettuce, broccoli}

Venn Diagram Representations of Sets


Complement of FGV

The complement of a set A, is written $\bar{A}$

$\bar{A}$ is the set of all elements in a universe that aren’t in A

Formation Definition: $\bar{A} = {x x \notin A, x \in \mho}$

NOTE: The $x \in \mho$ is optional and most often left out

FGV = {cucumber, peas, lettuce, broccoli}

MGV = {cucumber, lettuce, asparagus}


Addition of sets

FGV $\cup$ MGV = {cucumber, lettuce, asparagus}

$A \cup B = B \cup A$

The union of a set A with a set B written, $A \cup B$, is th set of all elements in A and all elements in B

Formal Definition:

$A \cup B = {x x\in A \vee x \in B}$


All of the elements in common

$FGV \cap MGV = {cucumber, lettuce}$

$A \cap B = B \cap A$

The intersection of a set A with a set B is the set of all elements that are in both A and B

Formal Definition:

$A \cap B = {x x \in A \wedge x \in B }$

Unions and intersections of more than two sets:

$A \cup B \cup C = {x x \in A \vee x \in B \vee x \in C }$
$A \cap B \cap C = {x x \in A \wedge x \in B \wedge x \in C }$

Formal Explanation:

$\bigcup\limits_{i=1}^{n} A_{i} = A_{1} \cup A_{2} \cup A_{3} … \cup A_{n}$

$\bigcap\limits_{i=1}^{n} A_{i} = A_{1} \cap A {2} \cap A{3} … \cap A_{n}$


What’s in one set and not the other


FGV = {cucumber, peas, lettuce, broccoli}

MGV = {cucumber, lettuce, asparagus}

$FGV - MGV = { peas,brocolli}$

Sometimes it’s written $FGV / MGV$

The difference between set A and B, written $A-B$, is the set of all elements in A that not in B

Formal Definition:

$A - B = {x x\in A \wedge x \notin B }$

Cartesian Product

All possible pairs

Recall the Cartesian plane which represents all possible pairs of x and y positions

$GV = {cucumber, peas}$

$berry = {blue, straw, black }$

$GV \times berry = {(cucumber, blue), (cucumber, straw), (cucumber, black), (peas, blue), (peas, straw), (peas, black)}$

Parenthesis mean pair


The Cartesian product of set A with set B, written $A \times B$, is the set of all possible pairs from A with the elements in B.

$A \times B = {(x,y) x \in A \wedge y \in B}$

This is useful for pairing things in databases

Symmetric Difference

Everything but the intersection

$A \vartriangle B = (A - B) \cup (B -A)$

Lesson 5 - Binary Strings

Consider the universe:

$u = [vegtables, fruits, grains, protein, dairy]$

Length 5

If we have the set:

$s = {vegtables, protein}$

This set corresponds to the 1st and 4th elements in the given universe so we can denote it as:


It’s 5 bits because the universe is length 5. Notice how only the 1st and 4th digit (from the left) is 1.

Lesson 6 - Writing Proofs

Proof - A mathematical reason that something is true

Prove that $A - B \equiv A \cap \bar{B}$

Belief Phase

Explain the statements to their formal definitions:

$A - B = {x x \in A \wedge x \notin B }$
$A \cap \bar{B} = {y y \in A \wedge y \in \bar{B}}$

They already look very similar

Prove Idea

A rough skeleton of the formal proof. This is called a direct proof.

Set Equivalence Laws

In order to prove this, we need to first prove this:

$A - B \subseteq A \cap \bar{B}$ and

$A \cap \bar{B} \subseteq A - B$

$x \in A - B \implies x \in A \wedge x \notin B \implies x \in A \wedge x \in \bar{B}$

$x \in A \cap \bar{B} \implies x \in A \wedge x \in \bar{B} \implies x \in A \wedge x \notin B$

The Actual Proof

THM: A - B \equiv A \cap \bar{B}
To show that the set A - B is equivalent to the set A \cap \bar{B} we will
show that A - B \subseteq A \cap \bar{B} and A \cap \bar{B} \subseteq A - B.
First we will show A - B \subseteq A \cap \bar{B}. Consider the definition
of A - B, A - B = \{x|x \in A \wedge x \notin B \} since x \in A \wedge x \notin
B, we know htat x \in A \wedge x \in \bar{B} by the definition of set complement
so x \in A \cap \bar{B}. Now to show A \cap \bar{B} \subseteq A - B we again
consider the definition A \cap \bar{B} which is the set of all x, x \in A
\wedge x \in \cap{B} but since x \in \bar{B} is the same as x \notin B by
definition, we know x \in A \wege x \notin B which is the same as x \in A - B.
So when x \in A \cap \bar{B} \implies x \in A - B. So we have A - B \subseteq A
\wedge \bar{B} and A \wedge \bar{B} \subseteq A - B so A - B \equiv A \wedge
\bar{B}. \boxtimes

You end formal proofs with either $\boxtimes$ or $Q.E.D$




Sets are collections of unique objects, meaning there are no duplicates. We can write them either in list notation or set builder notation.

$S = {0,1,2,3}$

$x \in A$: x in an element of A

$A \equiv B$: A is equivalent to B if both sets have the exact same elements

$A \subseteq B$: A is a subset of B if all elements in A are also in B.

$A \subset B$: A is a proper subset of B if A is a subset of B but A is not equivalent to B.

Special Sets

We learned of a number of special sets.

$\empty$ is the empty set, written as ${}$in list notation. Note that for all sets A, $\empty \subseteq A$.

$\mathcal{P}(S)$ is the power set of S, or all subsets of S.

$\mathbb{Z}$ is the set of all integers

$\mathbb{Q}$ is the set of rationals, or all fractions.

$\mathbb{R}$ is the set of reals, which are rationals and all irrational numbers like $\pi$.

$U$ is the universe and contains all objects.

Set Operations

Just like boolean statements, we can create new sets using set operations. We can also visualize these sets with Venn Diagrams.

Proving Equivalence of Sets

We can prove two sets A, B are equivalent using the set equivalence laws or by proving A, B are subsets of each other.

Recall $A \equiv B \iff A \subseteq B \land B \subseteq A$

By definition of AB, we can show ABby showing for all objects x, if xA, then xB. So our proofs look as follows:

  1. Announce we will prove $A \equiv B$ by showing $A \subseteq B$ and $B \subseteq A$.
  2. Show $A \subseteq B$ by showing if $x \in A$, then $x \in B$
  3. Then show $B \subseteq A$ by showing if $x \in B$ then $x \in A$.
  4. Conclude we have indeed shown $A \equiv B$.

    Module 4 - Set Cardinality

Lesson 1 - Counting Unions

Counting Sets

$nucampus = {London, Portland, Boston}$

Set cardinality - $ nucampus $

GV = {cucumber, kale, asparagus, peas}

Berries = {Black, straw, raspberry, blue berry}

$ GV \cup Berries $= {cucumber, kale, asparagus, peas, black, straw, rasp, blue} = 8
$ GV \cup B = GV + B $

Remove duplicates:

Inclusion/Exclusion Principle

2 Sets General Form:

$ A \cup B = A + B - A \cap B $

What about for 3 sets?

3 Sets General Form:

$ A \cup B \cup C = A + B + C - A \cap B - A \cap C - B \cap C + A \cap B \cap C $

4 Sets General Form:

$ A \cup B \cup C \cup D = A + B + C + D - A \cap B - A \cap C - B \cap D - C \cap D + A \cap B \cap C + A \cap B \cap D + A \cap C \cap D + B \cap C \cap D - A \cap B \cap C \cap D $

Lesson 2: Counting Cartesian Products

WestCoast = {Vancoover, Seattle, San Fran, Fan Jose}

EastCoast = {Portland, Charlotte, Boston}

LT = {London, Toronto}

Need one person from each region

FwC = {Bethany, Ian, Jodi}

FeC = {Lany, Maria}

FLT = {Nate, Paul, Samantha}

1 Person to join from each region

How Many Different Combinations?

FwC x FeC x FLT = {(Bethany, Laney, Nate), (Bethany, Maria, Nate)…}

Let’s visualize this in a tree:

How Many Possible 4 Digit Pin Numbers Are There?

Because we multiply how choices we have by the amount of the times we have to make a choice, we get

Digits = {0,1,2,3,4,5,6,7,8,9} $\therefore$ Digits = 10

10 * 10 * 10 * 10

Lesson 3: Pigeonhole Principle

Example 1:

7 Teams that are going to have students assigned to

8 Students in the class

What can I say happens when I assign students to a team?

No matter what, one team has at least 2 students

Example 2:

7 Teams

22 Students

No matter the placement/assignment, if all students are on some team then there is 1 team that has at least 4 students

Example 3:

25 Professors

6 TAs


What can I say about the number of professors per TA?

Then 1 TA is assigned to 5 professors

Pigeon Hole Principle - General Definition

If you have k+1 objects in k bins, at least one bin has more than 1 object

General Formula

If you have n objects to pack into k bins, then at least one bin has at least $\lceil \frac{n}{k} \rceil$ objects


25 professors

6 TAs

Every professor has exactly 1 TA

One TA with $\frac{25}{6} = 4 \frac{1}{6}$

$\therefore$ one TA has at least 5 professors if every professor has 1 TA

Example 2:

25 professors

6 TA’s


At least 3 TA’s are assigned to 4 or more professors


If we have 25 professors that need to be assigned to 6 TAs and every professor must be assigned to TA AND no TA is assigned to more than 5 professors then at least 3 TA’s are assigned to 4 or more professors.

A - 25 professors are assigned 6 TA’s and each professor has 1 TA and no TA has more than 5 professors

B - At least 3 TA’s are assigned to 4 or more professors

$A \implies B$

Negation : $A \land \neg B$

Proof By Contradiction

Proof by contradiction. Assume not, we have 25 professors assigned to 6 TAs such that every professor is assigned to exactly one TA and no TA has more than 5 professors. We also have by our assumption that fewer than 3 TA’s have 4 or more professors. Then we have at most 2 TAs with 4 or more professors. Consider the case when 2 are assigned 4 and all others are smaller. Then we have 17 professors to assign to 4 TA’s. By the pigeonhole principle we know at least 1 has $\lceil \frac{17}{4}\rceil = \lceil4.25 \rceil = 5$ leaving at least 3 with more than 4. Causing a contradiction $\Rightarrow \Leftarrow$

Module 5 - Combinatorics and Permutations


CS1802 - Counting Handout

Recitation Notes

Lesson 1: Permutations (order matters and no repetitions)

How many ways can we order n objects?

A = {a,b,c}


6 Ways!

General Formula

3 elements - 321

4 elements - 432*1

$\therefore n!$

Sidenote: $\sum$ for multiplication is $\prod$ therefore $n! =\prod_{i=n}^{1} i$


How can we choose k elements from a set of n when order matters?

D = {0,1,2,3,4,5,6,7,8,9}

D = 10

Pick 4 elements

1st choice : 10 possibilities

2nd choice : 9 possibilities

3rd choice : 8 possibilities

4 choice : 7 possibilities


Remember: We stopped at k, so it’s no entire factorial. We need to cut off the rest of it

$\frac{n!}{(n-k)!}$ or $p(n,k)$

Lesson 2: Combinations (Order doesn’t matter and no repetition)

No repetitions

ex. Choose 2 elements from a set of 4

A = {a,b,c,d}

Recall: order matters would be $\frac{4!}{(4-2)!} = 12$

However $ab=ba$. In this example, that would mean that we cut the options in half

Written as ${4}\choose{2}$ or $c(4,2)$

General Definition:

${n}\choose{k}$$= \frac{n!}{(n-k)!k!}$

k! is added because we overshot by how many times we can order k elements


Let’s consider we have 9 people, we want to know the number of possible teams we can make of size 5.

${9}\choose{5}$$= \frac{9!}{(9-5)!5!}$


9 people, teams of size 5. But 2 people, Logan and Andrew must together if picked.

Two cases

${7 \choose 3} + {7 \choose 5} =$


9 people, need team of 5, but Ben can’t work with Nate.

3 Cases

${7 \choose 4} + {7 \choose 4} + {7 \choose 5} =$

Logical Evaluation:

$\neg Ben \lor \neg Nate$

Negate: $Ben \land Nate$

Things that can’t happen:

Difference Method

${9 \choose 5} - {7 \choose 3} = 91$

Teams with Types

12 People

Teams of size 5

How many teams have 3 faculty members and 2 students?

${5 \choose 3} \cdot {7 \choose 2} = 210$

How many teams of size 5 contain at most 1 faculty?

2 Cases

${7 \choose 5} \cdot {5 \choose 0} + {7 \choose 4} \cdot {5 \choose 1}$

How many teams have at least 1 faculty members?

5 cases

${5 \choose 1} \cdot {7 \choose 4} + {5 \choose 2} \cdot {7 \choose 3} + {5 \choose 3} \cdot {7 \choose 2} + {5 \choose 4} \cdot {7 \choose 1} + {5 \choose 5} \cdot {7 \choose 0}$

Lesson 3: Counting with Repetitions when Order Matters

ex. 4 Digit pin

Digits = {0,1,2,3,4,5,6,7,8,9}

1st choice - 10 options

2st choice - 10 options

3st choice - 10 options

4st choice - 10 options



Lesson 4: Balls and Bins: Counting With Repetitions When Order Doesn’t Matter

ex. Ordering pizza

4 pizza choices

How many ways from this set can I get 3 pizzas?

How many binary numbers have 3 0’s and 3 1’s (6 bit)

ex 2. 10 pizzas, 4 types

${13 \choose 3}$

Or we could also ask where the 10 0’s are which would be

${13 \choose 10}$ (it’s the same answer)

General Formula

n - categories (n-1 1’s)

k - things to pick (k 0’s)

$(k+(n-1)) \choose k$

Module 6 - Probability


CS1802 - Probability

Lesson 1: Overview of Probability

Represents uncertainty

Lesson 2 Part 1: Events, Experiments, Sample Space and Probabilities

An event is something that might happen

An experiment or observation is a situation with one or more possible outcomes

A probability is a real number in the range [0,1] (inclusive)

$Pr(A)$ - Probability of A

$Pr(heads) = 0.5$ or $Pr(C=H) =0.5$

Dice Example

S = {1,2,3,4,5,6}

$Pr(event) = \frac{ E }{ S }$

Either 5 or 6:

E = {5,6}

$Pr(\text{roll 5 or 6}) = \frac{1}{3}$

Coin Example

S = {H,T}

E = {E}


Two Dice Example

S = {(1,1),(1,2)….,(6,6)}

$ S = 6\times6= 36$

What is $pr(\text{doubles})$?

E = {(1,1), (2,2) …}

$Pr(\text{doubles}) = \frac{ E }{ S } = \frac{1}{6}$

Sum to 6:

E = {(1,5), (2,4) …}

$Pr(\text{roll a 6}) = \frac{5}{36}$

Flipping 4 Coins

$ S = 2^{4} = 16$

Example: Exactly 2 heads

$ E = {4 \choose 2} = 6$

$Pr(\text{2 heads}) = \frac{3}{8}$

Lesson 2 Part 2: Fallacies to Avoid

Inclusion-Exclusion Principle

$Pr(E_{1} \cup E_{2}) = Pr[E_1] + Pr[E_2] - Pr[E_1 \cap E_2]$

Lesson 2 Part 3: Sampling with and Without Replacement

Example: Urns and balls

20 balls

With Replacement

S = {red1, red2, green1…}

Valid Draw: {r1,r1,g1}

Draw 2 Balls:

$ S \times S \times S = 20\times 20\times 20 = 8,000$

$Pr(\text{0 green balls}) = \frac{18^3}{8,000} = 0.729$

$Pr(\text{3 green balls}) = \frac{2^3}{8,000} = 0.001$

Without Replacement

Valid Draw: {r1,r2,g1}

Sample Size Space: $C(20,3) = 1140$ ←Outcomes

$Pr(\text{0 green}) = \frac{C(18,3)}{1,140} = 0.716$

Axons of Probability

Lesson 3: Independence and Conditional Probability

Events are independent if and only if $Pr(A\land B) = Pr(A) \times Pr(B)$

Example: At least one double in 3 rolls → Independence

$1-Pr(\text{no doubles on 3 rolls})$


$Pr(\text{sum of die1 and die2 is 7 die1 is 6})$

When die1 is a 6: {(6,1),(6,2)…}

When sum is 7 and die1 is a 6: {(6,1)}

$Pr(\text{die1 + die2 = 7 die is a 6}) = \frac{1}{6}$

General Formulas - Dependence

$Pr(A B) = \frac{Pr(A\land B)}{Pr(B)}$
$Pr(A B) = \frac{ (A\cap B) }{ B }$ (holds by definition)

General Formulas - Independence

If and only if A and B are dependent: $Pr(A\land B) = Pr(A) \times P(B)$

Independence holds if and only if $P(A B) = Pr(A)$

Example: Critical Bugs in Code

$Pr(\text{Bugs}) = Pr (\text{Bugs}\land\text{SB}) + Pr(\text{Bugs}\land\neg\text{SB})$

$= Pr(\text{Bugs} \text{SB})\times Pr(\text{SB}) + Pr(\text{Bugs} \neg\text{SB}) \times Pr(\neg\text{SB})$

= 0.12

Lesson 4 Part 1: Mathematical Expectation

Expected Value

The expectation of a random variable (r.v.) X is $E[X] = \sum_x Pr(X=x) \times x$


D = outcome of a 6-sided die

$E[D] = \frac{1}{6}\times 1 + \frac{1}{6} \times 2 + \frac{1}{6} + 3 … \frac{1}{6} \times 6 = 3.5$


L = outcome of the lottery with 1 in 10 million chance of winning, $2 million payout

$E[L] = 2,000,000 (\frac{1}{10,000,000}) + 0\times\frac{9,999,999}{10,000,000} = $0.20$

Linearity of Expectation

Given rv (random variables) x, y and z=x+y then

$E[z] = E[x]+E[y]$

Example: Let z = sum of 2 six-sided dice. Let x and y be the first and 2nd rolls

$E[z] = E[x] + E[y] = 3.5 + 3.5 = 7$

Look out for this in exams!

Lesson 5: Bayes’ Rule

How to combine:

$Pr(\text{hypothesis evidence}) \propto Pr(\text{hypothesis}) \times Pr(\text{evidence hypothesis})$

More precise:

$Pr(\text{hypothesis evidence}) = \frac{Pr(\text{hypothesis}) \times Pr(\text{evidence hypothesis})}{Pr(\text{evidence})}$

General Formula

$Pr(\text{A B}) = \frac{Pr(A) \times Pr(B A)}{Pr(B)}$

Bayes’ Rule follows from the definition conditional probability


$Pr(A\land B) = Pr(A B) \times Pr(B) = Pr(B A) \times Pr(A)$
$Pr(A B) = \frac{Pr(B A) \times Pr(A)}{Pr(B)}$

Example: If I have a bag with 6-sided die and 20-sided die and I pull one out at random and roll a 6. Is it more likely to be a 6 or 20 sided die?

$Pr(\text{6 6-sided-die}) = \frac{1}{6}$
$Pr(\text{6 20-sided-die}) = \frac{1}{20}$
$Pr(\text{6-sided-die 6}) \propto Pr(\text{6 6-sided}) \times Pr(\text{6-sided}) = \frac{1}{12}$
$Pr(\text{20-sided-die 6}) \propto Pr(\text{6 20-sided}) \times Pr(\text{20-sided}) = \frac{1}{40}$

It is more likely for it to be a 6-sided die

Modify the prior:

$Pr(\text{6-sided}) = \frac{1}{3}$

$Pr(\text{20-sided}) = \frac{2}{3}$

$Pr(\text{6-sided-die 6}) \propto Pr(\text{6 6-sided}) \times Pr(\text{6-sided}) = \frac{1}{18}$
$Pr(\text{20-sided-die 6}) \propto Pr(\text{6 20-sided}) \times Pr(\text{20-sided}) = \frac{1}{30}$

Still more likely for it to be a 6-sided die

Lesson 6: Variance

Variance: How spread out a distribution is. How far are the samples to the mean on average?

$E[x-E[x]] = 0$ (because the negatives and positives negate each other by definition)

Definition: For a r.v. X, $var(x) = E[(x-E[x]^2)]$ or $var(x) = E[x^2] - E[x]^2$

The standard deviation of x is $\sqrt{var(x)}$

Module 7 - Sequences, Series, and Induction

CS1802 - Sequences and Series

Lesson 1: Sequences

Order matters

ex. 3, 10, 17, 24, 31, 38



Arithmetic Sequence

Getting an Arbitrary Term:

ex. 3, 10, 17, 24, 31, 38, …

General Form:

$t(n) = a_1 + (n-1)d$

Geometric Sequence

1, 5, 25, 125, 625, 3125, …

$t(n) = 1 \times5^{n-1}$

General Form:

$t(n) = a_1\times r^{n-1}$

Quadratic Sequence

ex. 13, 23, 37, 55, 77

But what if we look at the differences as a sequence?

10, 14, 18, 22, …

In this case, 4 would be the second difference

Quadratic Sequence: **when the 2nd difference is a constant

General Form:

$t(n) = an^{2} + bn + c$

ex. 13, 23, 37, 55, 77

What is the nth term for this?

$t(1) = 13 = a+b+c$

$t(2) = 23 = a*2^{2} + 2b + c = 4a+2b+c$

$t(3) = 37 = a3^2 + b3 +c = 9a+3b+c$

System of Linear Equations

Plug in values and you get:

$t(n) = 2n^2 + 4n +7$

Lesson 2: Series

Summing up terms:

$\sum_{i=1}^{5} a_i$ ←Sum up the first 5 terms

General Form:

$\sum_{i=1}^{n} a_i$


1+2+3+4+5 = s

— ———————

 6+6+6+6+6 = 2s

General Form: Arithmetic Series

$\sum_{i=1}^{n} a_i = \frac{(n+1)n}{2}$

$\sum_{i=1}^{n} a_i = \frac{(a+(a+(n-1)d))n}{2}$

General Form: Geometric Series

Example: $3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}$

$\frac{s}{2} = \frac{3}{2} + \frac{3}{4} + \frac{3}{8} + \frac{3}{16}$

$s-r*s = 1-r^n$

$s(1-r) = 1-r^n$

$\therefore s = \frac{a-r^n}{ar}$

Lesson 3: Proof By Induction


We proves that we can fit the L shaped piece into any 2x2 grid

What about 4x4?

What about 8x8?

We are essentially building up a foundation

Example 2:

3 cent coin

5 cent coin

Assume we have figured out how to make k cents worth of change using 3 cent and 5 cent coins. Can we make k+1 cents?

Case 1:

Case 2:

Proof by induction


$P$ is some property defined over the integers and $a$ is some fixed integer and

  1. $P(a)$ is true (base case)
  2. $\forall k\in \mathbb{Z}, n \leq a, P(N)$ (inductive step)

This applies $\forall n \in \mathbb{Z} , n \geq a, P(n)$


Proof by Induction

Consider the base case when n=8, we know that 8 can be made with 1 3 cent coin and 1 5 cent coin. Now consider the inductive hypothesis for an arbitrary k, k≥8. We can make k cents using 3 cent and 5 cent coins. Consider two cases. Case 1 in our k cents. The number of 5 cent coins is zero. Then remove 3 3 cent coins to make k-9 cents. Now add 2 5 cent coins to make k+1 coins. Case 2. We have at least 1 5 cent coin. Remove it to make k-5 cents. Now add two 3 cent coins to make k+1 cents. Since we know P(8) to be true and P(k) implies P(K+1) then P(n) is true for all $n\in \mathbb{Z}, n\geq 8$. Q.E.D.


$\sum_{i=1}^{n} i = 1+2+3…+n = \frac{n(n+1)}{n}$

Prove this by induction

Proof by Induction

Consider the base case when n=1 then we know $\frac{1(1+1)}{1} = 1$ which is the series with one term. Let’s assume for some $k$ $1+2+3…k = \frac{k(k+1)}{2}$ by our inductive hypothesis. Consider $1+2+3 …+k+(k+1)$which is $\frac{k(k+1)}{2} + (k+1)$ by our inductive hypothesis.

$\frac{k(k+1)}{2} = \frac{(k+2)(k+1)}{2}$

So we have $1+2+3…+k+(k+1) = \frac{(k+2)(k+1)}{2}$ as desired. So it is true for all n. Q.E.D.

Lesson 4: Strong Induction

Strong Mathematical Induction

$P(n)$ property defined over the integers with a, b fixed integers, a≤b. Base P(a), P(a+1), P(a+2), … P(b) are all true. Inductive $\forall k \geq b$ if P(i) is true $\forall i, a\leq i\leq k$ then P(k+1) is true. This implies $\forall n, n\geq a, P(n)$ is true.

Consider the Sequence:

$a_0 = 0,\ a_1=4,\ a_k = 6 \times a_{k-1} - 5 \times a_{k-2}$

$\therefore a_{2} = 6\times 4 - 5 \times 0 = 24$

$a_3 = 6\times 24 - 5 \times 4 = 124$

Claim $a_n = 5^n - 1$

$P(n) = a_k = 6\times a_{k-1} - 5 \times a_{k-2}$

$a_n = 5^n -1$

Proof by Strong Induction

Consider the base case $P(0),\ P(1),\ P(2)$

$P(0) = a_0=0=5^0 -1 = 1-1 =0$

$P(1) = 4 = 5^1-1$

$P(2) = 24 = 5^2 - 1$

Consider the case when n is 0, we know that $5^0-1 = 0$, which is $a_0$ as defined. We know $a_{k+1}$is $6\times a_k - 5\times a_{k+1}$ by definition. Inductive hypothesis is that P(i) is true for all 0≤i≤k, $a_i = 5^1 -1.$ Consider $a_{k+1} = 6\times(a_k)-5 \times a_{k-1}$ which is $6(5^k -1) - 5(5^{k-1}-1)$ (string induction). Therefore $a_{k+1} = 5^{k+1} -1$. $\forall n, a_n = 5^n -1$. Q.E.D.

Lesson 5: Recurrences

Looking at recursion.

Substitution Method:


T(0) = 5 (base case)

T(n) = T(n-1) + n (next step)

How can we define T(n) without T?

T(0) = 5

T(n) = T(n-1) + n

T(1) = T(0) + 1

T(3) = T(1) + 2 = T(0) 1 + 2

T(4) = T(2) + 3 = T(0) 1 + 2 + 3

T(5) = T(3) + 4 = T(0) 1 + 2 + 3 + 4

T(n) = T(0) + 1 + 2 + 3 …

$T(n) = T(0) + \frac{n(n+1)}{2} = 5 + \frac{n(n+1)}{2}$

Example 2:

T(1) = 1

T(n) = n + 2 * T(n/2)

$T(\frac{n}{2}) = \frac{n}{2} + 2\times T(\frac{n}{4})$

$T(n) = n+ 2( \frac{n}{2} + 2\times T(\frac{n}{4}))$

$T(\frac{n}{4}) = \frac{n}{4}+ 2\times T(\frac{n}{8})$

$T(\frac{n}{8}) = \frac{n}{8} + 2\times T(\frac{n}{16})$

$T(n) = n+2(\frac{n}{2} + 2(\frac{n}{4} + 2(\frac{n}{8} + 2\times T(\frac{n}{16}))))$

$T(n) = n+n+n+… 2^k\times T(\frac{n}{k})$

$T(n) = kn +2^k\times T(\frac{n}{k})$


$T(1) = 1$ so what value of k does $T(\frac{n}{2^k}) = T(1)$

$\frac{n}{2^k} = 1$

$n = 2^k$

$\therefore k = log_2(n)$

$T(n) = n\times log_2 (n) + n$

Module 8 - Growth of Functions with Sorting

Lesson 1: Sorting a List of Number: Insertion Sort

We want to ask: How good are these sorting algorithms?

Input: 7,3,2,4 Output: 2,3,4,7

A list with one element is always sorted

This is an invariant which means that it is always true (never varies).

Insertion Sort:

  1. Swap until it is in the correct location
  2. Advance Bar

    for i from 1 to n-1:
        current := A[i] // Current is one to insert
        j := i-1 // Counter to end
        while j>= 0 and A[j] > current:
            swap(A[j], current)

How expensive is this?

Loop through every element -1

What doe sit do? Possible operations?

(These are called atomic operations)

A swap is 3 assignments

Atomic operations cost 1

We need to count the loop

for loop: n-1 iterations

while loop

$\therefore$ Worst case: $\frac{(n-1)(n)}{2}$

Lesson 2: Sorting a List of Numbers: Merge Sort

Recall: a list of size 1 is already sorted

Merge Sort

Does long does this take?

2n comparisons to merge sorte lists

What we know:

Now we can merge them quickly

Number of ways to have a number n is lg(n)

$\therefore \text{cost} = n\cdot lg(n)$


mergeSort(A, left, right)
    if |A| > 1
        middle = left + (right-left)/2
        B = mergeSort(A, left, middle)
        C = mergeSort(A, middle+1, right)
        merge(B, C)

merge(A, B)
    c = []
    while (|A| > 0 and |B| > 0)
        if (A[0] < B[0])
            remove A[0] add to back of c
            remove B[0] add to back of c

Is merge sort better than insertion sort? Yes!!

$\text{Merge Sort}: \approx n\cdot lg(n)$

$\text{Insertion Sort}: \frac{n(n-1)}{2} \approx n^2$


![[Module 8 - Growth of Functions with Sorting/Screenshot_from_2020-12-01_21-04-43.png]]

Lesson 3: Asymptotic and Big O

Why can we say “about”?

We want to make algorithms efficent

We’re going ot focus on steps

Consider this code:

someFunc(A) //where |A| = n
    n = |N| // Cost 2
    while (n>0) //Cost 1
        temp = x // Cost 2
        x = y // Cost 1
        y = temp // Cost 1
        n = n - 1 // Cost 2

The total cost of this code is 7n+3

How different is 7n+3 and n when comparing function order?

Big O Notiation

Classify function (group them) so we can compare them.

Let f(n) and g(n) be function mapping non-negative integers to real numbers.

f(n) is O(g(n)) if there exists a real constant c>0 and an integer $n_o >= 1$ such that $f(n) <= c\cdot g(n) \forall n >= n_o$

Claim: 7n+3 is O(n)


Let c=8, then we konw that $7n+3<=8n$ for all $n>=3$. By the definition of Big O, 7n+3 is o(n).

Lesson 4: Ordering Functions

Big Omega

F(n) is $\Omega$(g(n)) if there exists a real constant c>0 and integer $n_o >=1$ such that $c\cdot g(n) <= f(n) \forall n >= n_o$

Big Theta

f(n) is $\Theta$(g(n)) if there exists constants $c_1>0,c_2>0\text{and} n_o >=1$ such that $c_1 \cdot g(n) <= f(n) <= c_2 \cdot g(n)$

$f(n) \text{is} O(g(n)) => f(n) <= g(n)$

$f(n) \text{is} \Omega(g(n)) => f(n) >= g(n)$

$f(n) \text{is} \Theta(g(n)) => f(n) = g(n)$


f(n) is o(g(n)) if for all constants c>0, $\exists n_o > 0, f(n) < c\cdot g(n)\forall n >= n_o$

f(n) is $\omega(g(n))$ (little omega) if for all constants c>0, $\exists n_o >0, f(n) < c\cdot g(n) \forall n>= n_o$

f(n) is $\theta(g(n))$ (little theta) if for all constants c>0, $\exists n_o >0, c\cdot g(n) < f(n) \forall n>= n_o$

Size of Functions

In order of size:

constants, logs, polylogorithms, polynomials, exponential

Note: Exponential algorithms are not possible