Lecture 1

1/22/25 - Cantor’s Diagonalization Argument §

• Theoretical foundations of CS
• Follow up to CS 205 and CS 344
• Contemplation along with mathematical proofs
• Convey ideas

There are an infinite number of primes

• Suppose there are a finite number of primes
• So there is a largest prime $p_n$
• Then $p_1\cdot p_2\cdot \dots \cdot p_n+1$ is prime, but not in the list

Understanding the limitations of computations

• To get a realistic view of CS
• Truth behind what is done

Grades:

HW	Due	Exam
1	02/24	02/26
2	03/31	04/02
3	05/05	?

HW is a tool for learning something, and testing you

Most of the exam is on HW 1 - checking the bare minimum

• Writing rigorous proofs

• No extension on the deadline max(HW1, Ex1) + max(HW2, Ex2) + max(HW3, Ex3) Out of 1000, above 900 to get an A

• Scaling policy

• Bonus points from exam

• 2 parts of the exam

• Hard questions on the HW

• Lecture - tell you what questions will be on the exam

• All HW should be done in $\text{LATEX}$

• OH on Monday - 2-3 in CoRE

• First part - simple mathematical things

• Second part - formal but outdated

• Third part - more modern

Natural Numbers - countably infinite

• Through Peano’s axioms
  • There exists a 1
  • There is a thing called adding 1 to a number
• Natural numbers - extend to negatives - extend to fractions - extend to irrationals
• Start with something very simple, then add to it

$f:\mathbb{N}\to \mathbb{R}$ $f$ is a bijection We want to show that $f$ is not surjective, so there exists an $x\in \mathbb{R}$ such that there does not exist an $a\in \mathbb{N}$ where $f(a)=x$.

For each $i\in \mathbb{N},x_i=9-a_{ii},x={\sum }_{i=1}^{\infty }10^{-i}{x}_i$

Claim: $\forall y\in \mathbb{N},x\ne f(y)$, so $f$ is not surjective.

$x_y\ne f(y)$

$$\begin{array}{r}f(1)={\mathbf{a}}_{11}{a}_{12}{a}_{13}\dots \\ f(2)=a_{21}{\mathbf{a}}_{22}{a}_{23}\dots \\ \dots \\ f(n)=a_{n1}{a}_{n2}{a}_{n3}\dots \\ \dots \end{array}$$

$$\begin{array}{r}.a_{11}{a}_{22}{a}_{33}\dots \\ \pm 1\pm 1\pm 1\dots \end{array}$$

this specific number does not have a preimage

• Set up notations early on

• Be as generous with English words

• Cantor’s diagonalization

  • Non-explicitly show that something exists

• $x$ exists because we wrongly assumed $f$ exists

Suppose for the sake of contradiction Then we would like to show that… so that $f(a)$ is not surjective. Let $f(1)=\dots ,f(2)=\dots$ Then $x_i=a-a_{ii}$ Then for all $y\in \mathbb{N},x_y\ne f(y{)}_{yy}$

• Following Sipser’s book from lectures 3-15

What is computation

• The study of space and time (also like physics)

• All of philosophy is just a misunderstanding of language

• Space - on hard drive

• Time - how long it takes before it spits out an answer

• Things that you want to compute, but can’t compute

• Can’t prove your solutions

  • Finding the weather
  • Diophantine equations

• Modeling

  • Input: weighted graph
  • Languages - strings
  • If you see 01, that is a delimiter
  • Give all pairs as integers
  • Sequence of bits

$L_{MST}\subseteq \{0,1{\}}^{\ast }={\bigcup }_{i\in \mathbb{N}}\{0,1{\}}^i$

• All possible bitstrings of length $i$

• $x\in L_{MST}⟺$ if it is interpretable as $(G_x,\alpha )$ (In the graph $G_x$, is there a graph with weight at most $\alpha$) and the answer is Yes

• Who is making the interpretation

• DFA

• Can a DFA compute it

• Context-free grammar - PDA

• Turing Machine - considered to be universal - answer is yes

• If you want the solution, we talk about a relation

$$R_{MST}\subseteq \{0,1{\}}^{\ast }\text{(input)}\times \{0,1{\}}^{\ast }\text{(output)}$$

$$\begin{array}{rl}(x,y)\in R_{MST}⟺& x\to (G_x,\alpha )\\ & y\to \text{Tree T in Gx whose weight is }\le \alpha \end{array}$$

• Strong if you talk about languages
• Canonical way of encoding