Discrete mathematics

Slides ⋅ Week 01 ⋅ 2nd and 3rd August, 2023

Grading Policy

Each of the three exams account for 20% of the grade.
Each of the eleven tutorials will have a quiz worth 5 points. The total number of points you can earn through quizzes is capped at 40, and accounts for 40% of the grade.
There will be weekly assignments that are not graded but are recommended for practice.

TAs

Harshil Mittal (mittal_harshil@iitgn.ac.in)
Saraswati Nanoti (nanoti_saraswati@iitgn.ac.in)
Eshwar Dhande (eshwar.dhande@iitgn.ac.in)

Office hours

ADH details will be announced shortly
OH with instructors by appointment

Google classroom

For announcements, we will use Google Classroom. Please sign up here or use the code nzl3hkr.

Course website

Mathmatize

For in-class quizzes, we will use Mathmatize. All quizzes are open-notes, closed-internet, and solo efforts unless otherwise mentioned.

Materials for the course are available from this slide deck
and the course website, which you can access from here.

Reach 100 Game

Player 1 starts with a number between 1 and 10 (inclusive).

Players take turns to add X to the current number, where X is between 1 and 10 (inclusive).

The first person to reach 100 wins.

Let's play!

The Game of SET

Link to play

The Game of SET

How many sets are there?

How many completions are there for two cards?

How many completions are there for one card?

Can you create a collection of 12 cards with no set in it?

Can you create a collection of 21 cards with no set in it?

How many cards are there?

The Mutilated Chessboard Puzzle

another tiling puzzle

The Game of Sim

Two players take turns coloring any uncolored lines.

One player colors in one color, and the other colors in another color.

Each player trying to avoid
the creation of a triangle made solely of their color.

(Only triangles with the dots as all corners count;
intersections of lines are not relevant.)

The player who completes such a triangle loses immediately.

The pigeon-hole principle

The pigeonhole principle states that if n items are put into m containers, with n > m, then at least one container must contain more than one item.

Deceptively simple - ~~crazy~~ wide-ranging applications! :)

A chess tournament has N participants, and any two players play one game against each other.

Then it is true that in any given point of time, there are two players who have finished the same number of games?

Discrete mathematics

Slides ⋅ Week 02 ⋅ 8th and 9th August, 2023

SETS

“A set is a gathering together into a whole of definite, distinct objects of our perception or of our thought—which are called elements of the set.”

– Georg Cantor

SETS

A set is an arbitrary collection of objects, absolutely no restriction being placed on the nature and number of these objects, the elements of the set in question.

The elements constitute and determine the set as such, without any ordering or relationship of any kind between them.

von Neumann and Morgenstern

before 1901:

Most sets commonly encountered are not members of themselves.

For example, consider the set of all students in ES214.
This set is not a student, much less one enrolled in ES214, thus it is not a member of itself.

Let us call a set normal if it is not a member of itself,
& abnormal if it is a member of itself.

Clearly every set must be either normal or abnormal.

The set of all students enrolled in ES214 is a normal set.

In contrast, the complementary set that contains everything that is not a student enrolled in ES214 is not a student enrolled in ES214,
& so it is one of its own members and is therefore abnormal.

russel's paradox - informal

Now we consider the set of all normal sets, R, and try to determine whether R is normal or abnormal.

russel's paradox - informal

If R were normal, it would be contained in the set of all normal sets (itself), and therefore be abnormal.

If R were abnormal, it would not be contained in the set of all normal sets (itself), and therefore be normal.

This leads to the conclusion that R is neither normal nor abnormal: Russell's paradox.

(formulation borrowed from Wikipedia)

naive set theory

A set is an unordered collection of distinct objects, called elements or members of the set.

A set is said to contain its elements.

We write a ∈ A to denote that a is an element of the set A.

The notation a ∉ A denotes that a is not an element of the set A.

A set is an unordered collection of distinct objects, called elements or members of the set.

A set is said to contain its elements.

We write a ∈ A to denote that a is an element of the set A.

The notation a ∉ A denotes that a is not an element of the set A.

Ops: union, intersection, difference, symmetric difference

naive set theory

How do we represent sets in a computer?

What do we care about?

Check for membership

Updates (addition, deletion)

SETS in memory

Find out more in DSA :)

Want: Efficiency in time & space

PIE

A. There are 72 students enrolled in ES214.

B. There are 145 students enrolled in ES242.

The intersection is 59. What is the union?

PIE

Source

\(S(A \cup B \cup C)=\)

\(S(A)+S(B)+S(C)\)

\(-S(A \cap B)-S(A \cap C)-S(B \cap C)\)

\(+S(A \cap B \cap C)\)

To compute the size of a union of multiple sets, it is necessary to

sum the sizes of these sets separately,

and then subtract the sizes of all pairwise intersections of the sets,

then add back the size of the intersections of triples of the sets,

subtract the size of quadruples of the sets,

and so on, up to the intersection of all sets.

PIE

Source

PIE

Source

The above definition can be expressed mathematically as follows:

\(\left|\bigcup_{i=1}^n A_i\right|=\)

And in a more compact way:

\(\left|\bigcup_{i=1}^n A_i\right|=\sum_{\emptyset \neq J \subseteq\{1,2, \ldots, n\}}(-1)^{|J|-1}\left|\bigcap_{j \in J} A_j\right|\)

\(\sum_{i=1}^n\left|A_i\right|-\sum_{1 \leq i<j \leq n}\left|A_i \cap A_j\right|+\sum_{1 \leq i<j<k \leq n}\left|A_i \cap A_j \cap A_k\right|-\cdots+(-1)^{n-1}\left|A_1 \cap \cdots \cap A_n\right|\)

PIE

Source

\(T=\left(\begin{array}{l}k \\ 1\end{array}\right)-\left(\begin{array}{l}k \\ 2\end{array}\right)+\left(\begin{array}{l}k \\ 3\end{array}\right)-\cdots+(-1)^{i-1} \cdot\left(\begin{array}{l}k \\ i\end{array}\right)+\cdots+(-1)^{k-1} \cdot\left(\begin{array}{l}k \\ k\end{array}\right)\)

\((1-x)^k=\left(\begin{array}{l}k \\ 0\end{array}\right)-\left(\begin{array}{l}k \\ 1\end{array}\right) \cdot x+\left(\begin{array}{l}k \\ 2\end{array}\right) \cdot x^2-\left(\begin{array}{l}k \\ 3\end{array}\right) \cdot x^3+\cdots+(-1)^k \cdot\left(\begin{array}{l}k \\ k\end{array}\right) \cdot x^k\)

\(0 = 1 - T \Longrightarrow T = 1\)

T := # of times an element is counted on the RHS

PIE: Applications

Source

Given two numbers n and r,

count the number of integers in the interval [1 ; r]
that are relatively prime to n

(their greatest common divisor is 1 ).

Suppose \(n = p_1^{b_1} \cdot p_2^{b_2} \cdots p_k^{b_k}\).

Let \(A_i\) be the set of numbers between 1 and r that are a multiple of \(p_i\).

Apply PIE on the \(A_i\)'s to get a count of the number of integers in [1;r] that are not relatively prime to n.

PIE: Applications

Source

Consider n patterns of strings of the same length, consisting only of letters \((a \ldots z\)) or question marks.

You're also given a number k.

A string matches a pattern if it has the same length as the pattern, and at each position, either the corresponding characters are equal or the character in the pattern is a question mark.

The task is to count the number of strings that match
exactly k of the patterns.

PIE: Applications

Source

PIE: Applications

Source

Let Z be a subset of patterns.

Easy: count the number of strings that satisfy at least this set of patterns.

Hard: count the number of strings that satisfy exactly this set of patterns.

PIE: Applications

Source

Fix a specific subset X from the set of patterns consisting of k patterns.

Then we have to count the number of strings that satisfy this set of patterns, and only matches it, that is, they don't match any other pattern.

PIE: Applications

Source

Sum on all supersets Y, and either add to the current answer or subtract it from the number of strings:

\(\operatorname{ans}(X)=\sum_{Y \supseteq X}(-1)^{|Y|-k} \cdot f(Y)\)

Where f(Y) is the number of strings that match at least Y.

PIE: Applications

Source

Note: if we sum up on all X, we will get the final answer:

\(\text { ans }=\sum_{X:|X|=k} \operatorname{ans}(X)\)

Food for thought: what is the running time of this procedure? Can you improve it?

Equality

Do two sets A and B have the same size?

A and B have the same size if there is a bijection between them.

Equality

Do two sets A and B have the same size?

A: the set of all natural numbers

B: the set of all even natural numbers

\(f(x) = 2x\)

Equality

Do two sets A and B have the same size?

A: the set of all integers

B: the set of all natural numbers

\(f(x) = 2x \text{ if } x > 0\)

\(f(x) = -2x+1 \text{ if } x \leqslant 0\)

Equality

Do two sets A and B have the same size?

A: the set of all rational numbers

B: the set of all natural numbers

\(f(p/q) = 2^p 5^q, \text{ for } p \geqslant 0, q > 0 \)

\(f(-p/q) = 3^p 5^q, \text{ for } p,q > 0\)

Equality

Source: Rosen

Hilbert's Grand Hotel has infinitely many rooms, each a luxurious full-floor suite, numbered like the natural numbers: 0, 1, 2, and so on, endlessly.

Each room has a guest, and the hotel is completely full, with infinitely many guests.

But meanwhile, a new guest has just arrived, wanting a room.

What is the manager to do?

Source: Proof and the Art of Mathematics, J. D. Hamkins

Room n → Room n + 1

hilbert's hotel

Hilbert's Grand Hotel has infinitely many rooms, each a luxurious full-floor suite, numbered like the natural numbers: 0, 1, 2, and so on, endlessly.

Each room has a guest, and the hotel is completely full, with infinitely many guests.

On the weekend, a crowd shows up all at once 1000 guests.

What is the manager to do?

Source: Proof and the Art of Mathematics, J. D. Hamkins

Room n → Room n + 1000

hilbert's hotel

Hilbert's Grand Hotel has infinitely many rooms, each a luxurious full-floor suite, numbered like the natural numbers: 0, 1, 2, and so on, endlessly.

The following week, Hilbert's bus pulls up. The bus has infinitely many seats, and in every seat there is a new guest wanting to check into the hotel.

What is the manager to do?

Source: Proof and the Art of Mathematics, J. D. Hamkins

Room n → Room 2n; Seat s → Room 2s + 1

hilbert's hotel

Hilbert's Grand Hotel has infinitely many rooms, each a luxurious full-floor suite, numbered like the natural numbers: 0, 1, 2, and so on, endlessly.

The following week, Hilbert's train comes. The train has infinitely many train cars, each with infinitely many seats, and every seat is occupied.

What is the manager to do?

Source: Proof and the Art of Mathematics, J. D. Hamkins

Room n → Room 2n; Car c, Seat s → Room \(3^c5^s\)

hilbert's hotel

Hilbert's Grand Hotel has infinitely many rooms, each a luxurious full-floor suite, numbered like the natural numbers: 0, 1, 2, and so on, endlessly.

Source: Proof and the Art of Mathematics, J. D. Hamkins

Next, Cantor's cruise ship pulls into the harbor, with a passenger for every real number. Every ship passenger has a ticket, with a distinct ticket number printed on it, a real number, and every real number arises on some ticket.

What is the manager to do?

Resign.

hilbert's hotel

A diagonalization argument to show that the reals cannot be placed in one-to-one correspondence with the naturals:

Source: Proof and the Art of Mathematics, J. D. Hamkins

A set is countable if
it can be placed in one-to-one correspondence with a set of natural numbers.

The union of two countable sets is countable.
In other words, if A and B are countable sets, so is A U B.

Definition

Theorem

...by the Odd/Even trick;

or interleaving.

We say that sets A and B are equinumerous, written \(A \simeq B\), if they can be placed into one-to-one correspondence with each other.

In other words, \(A \simeq B\) if there is a bijection \(f: A \rightarrow B\), a one-to-one onto function from A to B.

No set X is equinumerous with its power set P(X),
the set of all subsets of X.

Definition

Theorem

Let f be a proposed bijection that maps X to P(X).

Consider: \(A=\{x \in X \mid x \notin f(x)\}\).

There is a natural partial order* underlying equinumerosity, namely, \(A \lesssim B\) if there is a one-to-one function \(f: A \rightarrow B\), not necessarily onto.

Definition

There is a natural partial order* underlying equinumerosity, namely, \(A \lesssim B\) if there is a one-to-one function \(f: A \rightarrow B\), not necessarily onto.

Definition

It is easy to see that \(X \lesssim P(X)\),

since we may map any element \(a \in X\) to its singleton \(a \mapsto\{a\}\), which is a subset of X, and this is an injective function.

Definition

By simply iterating this process, we thereby produce larger and larger sizes of infinity:

\(\mathbb{N}<P(\mathbb{N})<P(P(\mathbb{N}))<P(P(P(\mathbb{N})))<\cdots\)

(Shröder-Cantor-Bernstein). If sets A and B are at least as large as each other, then they are equinumerous.

Definition

Theorem

Specifically, if \(A \lesssim B\) and \(B \lesssim A\), then \(A \simeq B\).

In other words, if there is an injective function \(f: A \rightarrow B\) and another injective function \(g: B \rightarrow A\), then there is a bijective function \(h: A \rightarrow B\).

Source: Proof and the Art of Mathematics, J. D. Hamkins

Function composition

A SET variation: let's play!

Discrete mathematics

Slides ⋅ Week 03 ⋅ 16th August, 2023

Induction

Source: Folklore

If there is a natural number with certain property, then there is a smallest number with that property.

Least-number principle

Every nonempty subset S of the positive integers
has a least element.

The set of positive integers does not contain any
infinite strictly decreasing sequences.

(AKA the well-ordering principle)

Let \(A\) be a set of integers with some \(b \in A\), and furthermore, that whenever \(n \in A\), then also \(n+1 \in A\).

Then \(A\) contains all integers above \(b\).

Principle of induction

Suppose \(\mathcal{P}\) is a property of an integer such that:

(1) \(\mathcal{P}(b)\) is true, and
(2) \(\mathcal{P}(n)\) being true implies that \(\mathcal{P}(n+1)\) is true.

Then \(\mathcal{P}\) holds for all integers above \(k\).

Assume that A is a set of natural numbers with the property that:
for every natural number n,
if every number smaller than n is in A, then n itself is in A.

Then every natural number is in A.

Principle of strong induction

Suppose \(\mathcal{P}\) is a property of an integer such that:

(1) \(\mathcal{P}(1)\) is true, and
(2) \(\mathcal{P}(k)\) being true for all \(k \leqslant n\) implies that \(\mathcal{P}(n+1)\) is true.

Then \(\mathcal{P}\) holds for all integers.

Least-number principle

\(\equiv\)

Principle of induction

\(\equiv\)

Principle of strong induction

Examples!

🎉

Nostalgia trip

\( \forall n \in \mathbb{N}, 1+3+5+\cdots+(2 n-1)=n^2\)

sum of the first \(n\) odd numbers...

\( \forall n \in \mathbb{N}, 1+2+4+\cdots+2^n=2^{n+1}-1.\)

sum of the first \(n\) powers of two...

\( \forall n \in \mathbb{N}, n^2 < 3^n.\)

\( \forall n \in \mathbb{N}, (n^3 - n)\) is a multiple of \(6\).

\( \forall n \in \mathbb{N}, (6^n - 1)\) is a multiple of \(5\).

Every natural number can be expressed as the sum of a unique set of distinct powers of 2.

A \(2^n \times 2^n\) square grid with one random tile removed can be completely tiled with L-shaped triminos.

Every positive integer can be represented uniquely as
the sum of one or more distinct Fibonacci numbers
in such a way that the sum
does not include any two consecutive Fibonacci numbers.

Zeckendorf's theorem

Every positive integer can be represented uniquely as
the sum of one or more distinct Fibonacci numbers
in such a way that the sum
does not include any two consecutive Fibonacci numbers.

Zeckendorf's theorem

\(F_j\)

\(F_{j+1}\)

\(n\)

A country has \(n\) cities and any two cities are connected by a one-way road.

There is a route that passes through every city.

\(\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\)

Induction Hypothesis 🎁

A country has \(n\) cities and any two cities are connected by a one-way road.

There is a route that passes through every city.

\(\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\)

Induction Hypothesis 🎁

If this road goes this way, we are done.

A country has \(n\) cities and any two cities are connected by a one-way road.

There is a route that passes through every city.

\(\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\)

Induction Hypothesis 🎁

So suppose not.

A country has \(n\) cities and any two cities are connected by a one-way road.

There is a route that passes through every city.

\(\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\)

Induction Hypothesis 🎁

If this road goes this way, we are done.

A country has \(n\) cities and any two cities are connected by a one-way road.

There is a route that passes through every city.

\(\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\)

Induction Hypothesis 🎁

So suppose not.

A country has \(n\) cities and any two cities are connected by a one-way road.

There is a route that passes through every city.

\(\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\)

Induction Hypothesis 🎁

An interim change of state is forced between consecutive cities...

A country has \(n\) cities and any two cities are connected by a one-way road.

There is a route that passes through every city.

Buckets Of Fish

This game is played with two players on the beach.

The game begins with finitely many buckets arranged in a row,
each bucket containing some finite number of fish,
with a large supply of additional fish available nearby, as many as needed, fresh off the boats.

Each player, in turn, takes a fish from one bucket and adds to each of the buckets to the left, if any, as many fish from the extra supply as he or she likes.

The winner is whoever takes the very last fish from the buckets,
leaving them empty.

Every play of the game Buckets of Fish ends in finitely many moves.

All the fish in the buckets, including all the new fish that may have been added during play, will eventually run out by some finite stage during play.

Examples?

🤔

All integers of the form \(2n+1\) are divisible by 2.

Base Case. Easy as usual, left as an exercise.

Induction Step.

Source: Miklós Bóna, A Walk Through Combinatorics

WTS. \(2(n+1)+1\) is divisible by two.

Equivalently: \(2(n+1)+1\) \(-\) \((2n+1)\) is divisible by two.

\(2\) is divisible by two. ✔️

Every positive integer \(n\) is a whole lot less than 1,000,000.

Base Case. 1 is certainly a whole lot less than 1,000,000.

Source: Brilliant Wiki

If the positive integer \(k\) is a whole lot less than 1,000,000,
then \(k+1\), which is just slightly bigger than k,
is still a whole lot less than 1,000,000.

Hence, by mathematical induction, the statement is true for all positive integers. In particular, a billion is a whole lot less than a million.

Induction Step.

For every non-negative number \(n, 2 \times n=0\).

Base Case. \(n=0\). Clearly \(2 \times n=0\).

Induction Step. Say it holds for \(k\), and consider \(k+1\).

Write \(k+1=i+j\), where \(i\) and \(j\) are non-negative numbers. Then,

\(2(k+1)=2(i+j)=2 i+2 j=0+0=0.\)

Hence, by mathematical induction, the statement is true for all non-negative integers. In particular, \(2 \times 1=0\).

Source: Brilliant Wiki

Claim. All positive integers are equal to 1.

Source: Notes

To prove the claim, we will prove by induction that, for all \(n \in \mathbb{N}\), the following statement holds:

Here \(\max (x, y)\) denotes the larger of the two numbers \(x\) and \(y\),
or the common value if both are equal.

If \(P(n)\) is true for all \(n \in \mathbb{N}\), then \(\max (1, n)=n\) for any positive integer \(n\).
Therefore it will follow that \(1=n\) for any positive integer \(n\).
Thus, all positive integers must be equal to 1.

\((P(n)). \quad\) For any \(x, y \in \mathbb{N}\), if \(\max (x, y)=n\), then \(x=y\).

Claim. All positive integers are equal to 1.

\((P(n)). \quad\) For any \(x, y \in \mathbb{N}\), if \(\max (x, y)=n\), then \(x=y\).

Base Case: When \(n=1\), the condition in \(P(1)\) becomes \(\max (x, y)=1\). But this forces \(x=1\) and \(y=1\), and hence \(x=y\).

Induction step: Let \(k \in \mathbb{N}\) be given and suppose \(P(k)\) is true. We seek to show that \(P(k+1)\) is true as well.

Let \(x, y \in \mathbb{N}\) be such that \(\max (x, y)=k+1\).
Then \(\max (x-1, y-1)=\max (x, y)-1=\) \((k+1)-1=k\).

By the induction hypothesis, it follows that \(x-1=y-1\),
and therefore \(x=y\). This proves \(P(k+1)\).

Claim. All cats have the same color.

Claim. For any positive integer \(n\),
any \(n\) cats always have the same color.

Base Case. For \(n = 1\), the statement is obviously true:
any one cat has the same color as itself.

Induction Step. Take \(n + 1\) cats, and line them up.

Then the first \(n\) cats must have the same color, say black, by the IH.

The last \(n\) cats also must have the same color, by the same IH, so they too must be black as we already have seen that all the first \(n\) cats were black, and that included the second, third, fourth,..., nth cats, which are also included among the last \(n\) cats.

Therefore, all \(n + 1\) cats are black.

Discrete mathematics

Slides ⋅ Week 04 ⋅ 22nd and 23rd August, 2023

Black Vienna

27 suspects, whose names are A,B,C,...,Z,Ö.

Ingredients

3 criminals, sometimes referred to as the Black Vienna ring. This is an unknown three-sized subset of A,B,C,...,Z,Ö.

3 investigators, who separately have alibis for 8 suspects each. These eight-sized subsets are disjoint from each other.

Black Vienna

The game progresses in a sequence of queries.
You are dealt three sets containing three suspects each.

Ingredients

You can choose any of these available sets, and pick an investigator (red, blue, or green) and ask them to tell you how many (not which!) of the three suspects they have alibis for.

So they can certify how many people from the given three
are NOT in the gang: 0, 1, 2, or 3.

Black Vienna

Link to Play (us v/s the machine)

Link to Play (teams of threre)

Link to Tracker

and

then:

Proposition. A proposition is a declarative sentence (that is, a sentence that declares a fact) that is either true or false, but not both.

We use letters to denote propositional variables.
The conventional letters used for propositional variables are p, q, r, s, ...

U belongs to the BV ring.

The truth value of a proposition is true (T) if it is a true proposition.
The truth value of a proposition is false (F), if it is a false proposition.

Propositions that cannot be expressed in terms of simpler propositions are called atomic propositions.

p 🤝 q

Make Your Own Truth Tables!

Always False

p 🤝 q

Make Your Own Truth Tables!

Negation of OR (NOR)

p 🤝 q

Make Your Own Truth Tables!

q is right and p is wrong
[CONVERSE NONIMPLICATION]

p 🤝 q

Make Your Own Truth Tables!

p is false.

p 🤝 q

Make Your Own Truth Tables!

p is right and q is wrong
MATERIAL NONIMPLICATION

p 🤝 q

Make Your Own Truth Tables!

q is false.

p 🤝 q

Make Your Own Truth Tables!

Exclusive OR (XOR) | Parity

p 🤝 q

Make Your Own Truth Tables!

Negation of AND (NAND)

p 🤝 q

Make Your Own Truth Tables!

AND

p 🤝 q

Make Your Own Truth Tables!

Negation of Exclusive OR (NOR) | p IFF q

p 🤝 q

Make Your Own Truth Tables!

q is true.

p 🤝 q

Make Your Own Truth Tables!

IMPLICATION
It is not the case that p is true and q is not.

p 🤝 q

Make Your Own Truth Tables!

p is true.

p 🤝 q

Make Your Own Truth Tables!

Converse [i.e, q implies p]

p 🤝 q

Make Your Own Truth Tables!

OR

p 🤝 q

Make Your Own Truth Tables!

Always True

Demonstration of Equivalence between

\(p \lor (q \land r)\) and \((p \lor q) \land (p \lor r)\)

Source: Rosen, 8th Edition, p. 28.

Source: Rosen, 8th Edition, p. 29.

A compound proposition that is always true,
no matter what the truth values of the propositional variables that occur in it,
is called a tautology.

A compound proposition that is always false
is called a contradiction.

A compound proposition that is
neither a tautology nor a contradiction
is called a contingency.

Always T/F

A compound proposition is satisfiable if there is an assignment of truth values to its variables that makes it true
(that is, when it is a tautology or a contingency).

When no such assignments exists, that is, when the compound proposition is false for all assignments of truth values to its variables, the compound proposition is unsatisfiable.

Satisfiability

Note that a compound proposition is unsatisfiable if and only if its negation is true for all assignments of truth values to the variables, that is, if and only if its negation is a tautology.

\(p_A := \) A belongs to the blue team.

\(p_B := \) B belongs to the blue team.

\(p_C := \) C belongs to the blue team.

\(\vdots\)

\(p_Z := \) Z belongs to the blue team.

\(q_A := \) A belongs to the red team.

\(q_B := \) B belongs to the red team.

\(q_C := \) C belongs to the red team.

\(\vdots\)

\(q_Z := \) Z belongs to the red team.

\(r_A := \) A belongs to the green team.

\(r_B := \) B belongs to the green team.

\(r_C := \) C belongs to the green team.

\(\vdots\)

\(r_Z := \) Z belongs to the green team.

\(s_A := \) A belongs to the Black Vienna gang.

\(s_B := \) B belongs to the Black Vienna gang.

\(s_C := \) C belongs to the Black Vienna gang.

\(\vdots\)

\(s_Z := \) Z belongs to the Black Vienna gang.

27 * 4 = 108 atomic propositions 🎉

\(p_A \implies (\neg q_A \land \neg r_A \land \neg s_A)\)

\(q_A \implies (\neg p_A \land \neg r_A \land \neg s_A)\)

\(r_A \implies (\neg p_A \land \neg q_A \land \neg s_A)\)

\(s_A \implies (\neg p_A \land \neg q_A \land \neg r_A)\)

Everyone belongs to at most one team.

\(p_Z \implies (\neg q_Z \land \neg r_Z \land \neg s_Z)\)

\(q_Z \implies (\neg p_Z \land \neg r_Z \land \neg s_Z)\)

\(r_Z \implies (\neg p_Z \land \neg q_Z \land \neg s_Z)\)

\(s_Z \implies (\neg p_Z \land \neg q_Z \land \neg r_Z)\)

\(\ldots\)

\(p_A \lor q_A \lor r_A \lor s_A\)

Everyone belongs to at least one team.

\(p_B \lor q_B \lor r_B \lor s_B\)

\(p_Z \lor q_Z \lor r_Z \lor s_Z\)

\(\vdots\)

\(\cdots \large{\color{indianred}\lor} (p_A \land p_B \land p_C \land p_D \land p_W \land p_X \land p_Y \land p_Z) {\color{indianred}\lor} \cdots \)

\(\cdots \large{\color{indianred}\lor} \underbrace{(q_A \land q_B \land q_C \land q_D \land q_W \land q_X \land q_Y \land q_Z)}_{\text{a typical 8-sized subset of} \{ \text{ABCDEFGHIJKLMNOÖPQRSTUVQXYZ} \}} {\color{indianred}\lor} \cdots\)

\(\cdots\large{\color{indianred}\lor} (r_A \land r_B \land r_C \land r_D \land r_W \land r_X \land r_Y \land r_Z){\color{indianred}\lor} \cdots \)

\(\huge{\land}\)

\(\cdots \large{\color{indianred}\lor} (s_D \land s_M \land s_E){\color{indianred}\lor}\cdots\)

\(\huge{\land}\)

Any assignment of truth values to the 108 atomic variables will correspond to a valid partition of the population into R, B, G and BV.

We are looking for a specific partition among these.

\(\Huge{\oplus(q_E, q_G, q_Ö)}\)

\(\Huge{q_E \land q_H \land q_N}\)

Source: Rosen, 8th Edition, p. 76.

Sometimes, propositional logic is not expressive enough.

Suppose you want to assert that there are at least 5 primes among the numbers \(\{1,2,\ldots,100\}\).

Let \(p_k\) denote the proposition that \(k\) is a prime.

\(\cdots \lor \underbrace{( p_3 \land p_7 \land p_{42} \land p_2 \land p_9 )}_{\text{A typical 5-sized subset of} \{1,2,\ldots,100\}.} \lor \cdots\)

Sometimes, propositional logic is not expressive enough.

Suppose you want to assert that
there are infinitely many primes...?

Predicate Logic

A statement involving the \(n\) variables

\((x_1, \ldots x_n)\)

can be denoted by
\(P(x_1, \ldots, x_n)\).

\(P(x_1,x_2) := x_1 > x_2 + 1024\)

Predicate Logic

The universal quantification of \(P(x)\) is the statement:

\(P(x)\) for all values of \(x\) in the domain.

The notation \(\forall xP(x)\) denotes the universal quantification of \(P(x)\).

Can be falsified with a single counterexample.

\(P(x) := x \text{ can be written as a sum of two odd numbers}.\)

\(\forall x P(x)\) is false because it is not true for \(x = 3\).

Predicate Logic

The existential quantification of \(P(x)\) is the statement:

There exists an element x in the domain such that \(P(x)\).

The notation \(\exists xP(x)\) denotes the existential quantification of \(P(x)\).

Can be justified with a single counterexample.

\(P(x) := x \text{ can be written as a sum of two odd numbers}.\)

\(\exists x P(x)\) is true because it is true for \(x = 2\).

De Morgan's Laws for Quantifiers.

Source: Rosen, 8th Edition, p. 51.

Lewis CaRroll Examples

All lions are fierce.
Some lions do not drink coffee.
Some fierce creatures do not drink coffee.

Source: Rosen, 8th Edition, p. 80.

Common Fallacy

\(q\)

\(p \rightarrow q\)

\(p\)

\(\neg p\)

\(p \rightarrow q\)

\(\neg q\)

\(\forall x ~ P(x) \rightarrow Q(x)\)

\(\neg P(y)\)

\(\neg Q(y)\)

People who play chess are smart.

If I am bad at chess then I am stupid.

\(P(x)\) := \(x\) plays chess.

\(Q(x)\) := \(x\) is smart.

\(P(y) \rightarrow Q(y)\)

\(\neg P(y)\)

\(\neg Q(y)\)

People who play chess are smart.

If I am bad at chess then I am stupid.

\(P(x)\) := \(x\) plays chess.

\(Q(x)\) := \(x\) is smart.

If you get 100% on the final, then you will get an A.

- Not Me 👀

You don't get 100% on the final.

What can you say about your grade?

Source: Folklore; image from brilliant.org logic puzzles

1. If Jyothi is watching television, so is Tanvi.
2. Either Gaurav or Manish, or both of them, are watching television.
3. Either Tanvi or Sharda, but not both, is watching television.
4. Gaurav and Sharda are either both watching or both not watching television.
5. If Manish is watching television, then Jyothi and Gaurav are also watching.
Who is watching television and who is not?

1. \(J \rightarrow T\)
2. Either Gaurav or Manish, or both of them, are watching television.
3. Either Tanvi or Sharda, but not both, is watching television.
4. Gaurav and Sharda are either both watching or both not watching television.
5. If Manish is watching television, then Jyothi and Gaurav are also watching.
Who is watching television and who is not?

1. \(J \rightarrow T\)
2. \(G \lor M\)
3. Either Tanvi or Sharda, but not both, is watching television.
4. Gaurav and Sharda are either both watching or both not watching television.
5. If Manish is watching television, then Jyothi and Gaurav are also watching.
Who is watching television and who is not?

1. \(J \rightarrow T\)
2. \(G \lor M\)
3. \(T \oplus S\)
4. Gaurav and Sharda are either both watching or both not watching television.
5. If Manish is watching television, then Jyothi and Gaurav are also watching.
Who is watching television and who is not?

1. \(J \rightarrow T\)
2. \(G \lor M\)
3. \(T \oplus S\)
4. \(G {\color{indianred}\oplus} S\)
5. If Manish is watching television, then Jyothi and Gaurav are also watching.
Who is watching television and who is not?

1. \(J \rightarrow T\)
2. \(G \lor M\)
3. \(T \oplus S\)
4. \(G {\color{indianred}\oplus} S\)
5. \(M \rightarrow (J \land G) \)
Who is watching television and who is not?

extras

A logician's wife is having a baby.
The doctor immediately hands the newborn to the dad.

The wife says, 'Is it a boy or a girl?'

The logician says, 'Yes.'

Three logicians walk into a bar

The bartender asks: "Do you all want a drink?"

The first logician says: "I don't know."

The second logician says: "I don't know."

The third logician says: "Yes."

Blue Eyes

a puzzle

Source. Terry Tao's blog

Argument 1

The foreigner has no effect, because his comments do not tell the tribe anything that they do not already know (everyone in the tribe can already see that there are several blue-eyed people in their tribe).

Argument 2

100 days after the address, all the blue eyed people commit suicide.
This is proven as a special case of:

Suppose that the tribe had n blue-eyed people for some positive integer n.
Then n days after the traveller’s address, all n blue-eyed people commit suicide.

Proof: We induct on n.

When \(n=1\), the single blue-eyed person realizes that the traveler is referring to him or her, and thus commits suicide on the next day.

Now suppose inductively that n is larger than 1.

Each blue-eyed person will reason as follows: “If I am not blue-eyed, then there will only be n-1 blue-eyed people on this island, and so they will all commit suicide n-1 days after the traveler’s address”.

But when n-1 days pass, none of the blue-eyed people do so
(because at that stage they have no evidence that they themselves are blue-eyed).

After nobody commits suicide on the \((n-1)^{st}\) day, each of the blue eyed people then realizes that they themselves must have blue eyes, and will then commit suicide on the \(n^{th}\) day.

1. \(A \rightarrow B\)
2. Either Dale or Ellsworth, or both of them, are watching television.
3. Either Beryl or Cleo, but not both, is watching television.
4. Dale and Cleo are either both watching or both not watching television.
5. If Ellsworth is watching television, then Abner and Dale are also watching.
Who is watching television and who is not?

1. \(A \rightarrow B\)
2. \(D \lor E\)
3. Either Beryl or Cleo, but not both, is watching television.
4. Dale and Cleo are either both watching or both not watching television.
5. If Ellsworth is watching television, then Abner and Dale are also watching.
Who is watching television and who is not?

1. \(A \rightarrow B\)
2. \(D \lor E\)
3. \(B \oplus C\)
4. Dale and Cleo are either both watching or both not watching television.
5. If Ellsworth is watching television, then Abner and Dale are also watching.
Who is watching television and who is not?

1. \(A \rightarrow B\)
2. \(D \lor E\)
3. \(B \oplus C\)
4. \(D {\color{indianred}\oplus} C\)
5. If Ellsworth is watching television, then Abner and Dale are also watching.
Who is watching television and who is not?

1. \(A \rightarrow B\)
2. \(D \lor E\)
3. \(B \oplus C\)
4. \(D {\color{indianred}\oplus} C\)
5. \(E \rightarrow A \land D\)
Who is watching television and who is not?

Discrete mathematics

Slides ⋅ Week 05 ⋅ 29th and 30th August, 2023

A TILING GAME

A Tiling Game

Source: Mathologer

Play

Source: Mathologer

Corners determine the bottom cell for n = 4.

Using this, we can show that the corners determine the bottom cell for n = 10 as well.

PASCAL's TRIANGLE

126

120

210

252

Claim. The \(i\)-th non-zero entry

in the \(n\)-th row of the Pascal's Triangle is
\({n-1 \choose i-1}\).

Proof. For \(i = 1\), the statement is true for any \(n\).

The statement is also true for \(n = 1\).
For \(n > 1, i > 1\), the \(i\)-th non-zero entry

in the \(n\)-th row of the Pascal's Triangle is:

[\(i\)-th non-zero entry in the \((n-1)\)-th row]

[\((i-1)\)-th non-zero entry in the \((n-1)\)-th row]

Claim. The \(i\)-th non-zero entry

in the \(n\)-th row of the Pascal's Triangle is
\({n-1 \choose i-1}\).

Proof. For \(i = 1\), the statement is true for any \(n\).

The statement is also true for \(n = 1\).
For \(n > 1, i > 1\), the \(i\)-th non-zero entry

in the \(n\)-th row of the Pascal's Triangle is:

\({n-2 \choose i-1}\)

\({n-2 \choose i-2}\)

\({n \choose k}\) = \({n-1 \choose k}\) + \({n-1 \choose k-1}\)

\(\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}_{k}\)

\(\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}_{k-1}\)

\(\begin{aligned} ~ & \left(\begin{array}{c}n \\ r\end{array}\right) =\frac{n!}{(n-r)!(r)!} \\~\\ & =\frac{(n{\color{IndianRed}-r+r}) \cdot (n-1)!}{(n-r)!(r)!} \\~\\ & =\frac{(r)(n-1)!}{(n-r)!(r)!}+ \frac{(n-r)(n-1)!}{(n-r)!(r)!} \\~\\ & =\frac{(n-1)!}{(n-r)!(r-1)!}+\frac{(n-1)!}{(n-r-1)!r!} \\~\\ & = \left(\begin{array}{c}n-1 \\ r-1\end{array}\right)+\left(\begin{array}{c}n-1 \\ r\end{array}\right) \cdot \end{aligned}\)

Claim. The \(i\)-th non-zero entry

in the \(n\)-th row of the Pascal's Triangle is
\({n-1 \choose i-1}\).

Proof. For \(i = 1\), the statement is true for any \(n\).

The statement is also true for \(n = 1\).
For \(n > 1, i > 1\), the \(i\)-th non-zero entry

in the \(n\)-th row of the Pascal's Triangle is:

\({n-2 \choose i-1}\)

\({n-2 \choose i-2}\)

...and we are done 🎉

126

120

210

252

\({2 \choose 1}\)

\({3 \choose 1}\)

\({3 \choose 2}\)

\({4 \choose 1}\)

\({4 \choose 3}\)

\({5 \choose 1}\)

\({5 \choose 4}\)

\({6 \choose 1}\)

\({6 \choose 5}\)

\({7 \choose 6}\)

\({7 \choose 1}\)

\({8 \choose 7}\)

\({8 \choose 1}\)

\({9 \choose 8}\)

\({9 \choose 1}\)

126

\({10 \choose 9}\)

\({10 \choose 1}\)

120

210

252