objective is to help satan out by sending Cantor to the depths of hell for practicing mathematics for some fucking reason????????
Seriously though, might wanna get this clarified by someone…
Kolmogrov Complexity  The number of lines that are required to “prove”/code up a program to solve/prove the problem/statement
Turing Machines
The class of languages that can be recognized by Turing machines are called Recursively Enumerable Languages.
Church Turing Thesis
Turing machines and recursive function theory can represent any effectively computable language. (“Thesis” and not a theorem because “effectively” is loosely defined)
Not all languages are enumerable, even if $\Sigma$ is finite. This can be easily proven using diagonalization.
A class of languages is said to be decidable if given a word $w$, an algorithm exists to check if $w\in L$ for any $L$ that belongs to that language class. Recursively Enumerable Languages are decidable, but $L^C$ are NOT !
Halting Theorem
Given by Godel, if an algorithm can tell if any program terminates, then we have solved the halting problem.
Turing Machines consist of:

Input Tape: The “cells” in the tape are enumerated as $w_1,\ldots w_n$ from the left to right after which all cells are the blank character $B$. The set of tape alphabets is represented by $\tau$. Note that $Σ \sube τ  {B}$

Control: A pointer which points to a cell of the tape. The control has states and transitions amongst these states tell the pointer to move to the left/right.
$w_i/0,R$ transition on the state $q_j$ …. the pointer points to $w_i$ on the tape, replace $w_i$ with 0, and move 1 cell to the right.
For example, consider the language $L = { 0^n 1^n\vert n\geq0 }$. The Turing machine for this language would be given by:
draw and put
More formally, a Turing machine would be given by the 7tuple: \(M = (Q, Σ, \tau, δ, q₀,B, F)\)
Symbol  Meaning 

$Q$  Set of states 
$\Sigma$  Set of alphabet 
$\tau$  Set of tape alphabet, $Σ\sub\tau$ 
$\delta$  Transition function, $δ:Q\times\tau\to Q\timesτ×{L,R}$ 
$q₀$  Start state 
$B$  Blank tape symbol 
$F$  Final state 
Instantaneous Descriptions (ID)
Similar to PDA, we have a notion of Instantaneous Descriptions (IDs), where the symbol $\vdash$ refers to a single move of the M. IDs are of the form $(\alpha q\beta)$, where $q$ refers to the current state of $M$ and the tape head points to the first character of $\beta$. For example, Consider the tape to be $\ldots BB00100BB\ldots$ with the head at $1$ and the state being $q$. The ID would be given by $(00q100)$.
Variations of the Turing Machine
Note that all these variations have the same expressive power. The first three variations are called “Syntactic Sugar” because they don’t add ant computational capability.
 2way infinite tape
 Multiple tape heads
 nondeterministic Turing Machine
Also note that the halting and accepting variants of Turing Machines are equivalent. Moreover, all these Turing machines are equivalent to $(Q,\Sigma={0,1}, \Gamma = {0,1,B}, \delta, q, B, F = {q_f})$ which has a single final state.
The above machine can be “encoded” using only a string of 0s and 1s. Let the automaton move from $x_i$ to $x_j$, with the tape’s transition being $x_k\vert x_l,D_m$. \(111 \ldots 11 0^i10^j10^k10^l10^m11\ldots111\) ==Now that we have a binary encoding, we can clearly see that Turing machines can be enumerated.==
Definitions
Name  Definition 

L is Recursively Enumerable  A recognizer TM exists that accepts L 
L is Recursive  A recognizer TM that accepts L must halt on all inputs 
==Note that RE languages are not closed under Complementation and Difference, but recursive languages are!==
Define $L_d$ to be the diagonal formed by all words and all Turing Machines. Define $L_u$ as follows: \(L_u = \{ \text{Enc}(M)\vert\vert \text{Enc}(w) : M\text{ is a TM which accepts }w \}\) $L_d$ is not recursively enumerable, whereas $L_u$ is Recursively Enumerable but it isn’t Recursive in nature. This can be proven using Reduction.
We can use diagonalization to give an example of a language that is not Recursive Enumerable in nature $(L_d)$. Moreover, the language of binary encoded tuples $<M,w>$ can be used to give language that is Recursive Enumerable, but not recursive. The language is: \(L_u = \{ <M,w>\vert \text{ Turing Machine }M\text{ accepts the word }w \}\) The TM for $L_u$ would contain 3 Tapes;
 Holding the input $<M,w>$
 Holding $M$‘s tape
 Holding $M$‘s state
The following steps would be performed by the UTM:
 Check that the encoding for $M$ is valid for a turing machine, reject and halt if not valid
 Examine $M$ and figure out how many blocks are needed to represent all alphabets used by $M$
 Initialize Tape2 to contain $w$ as the input for $M$
 Simulate $M$ by using Tape1 and Tape3, and writing onto Tape2
This is obviously not recursive as $M$ is not guaranteed to halt.
Language Properties
A set of languages would be a property of languages. This can be represented as a TM problem: Given a property $P$ we would like the set of all Turing Machines $L_p$ such that every $L(M), M\in L_p$ has the property $P$.
Rice’s Theorem
$L_p$ is undecidable for every property other than the trivial alwaysfalse(empty set) and alwaystrue (all languages accepted).
We shall use Reduction to prove this theorem.
Reduction
A reduction from $L$ to $L’$ is given by a TM that always halts (algorithm) which takes $w$ and outputs $x$ such that $w\in L \iff x\in L’$.
Therefore, if we reduce $L$ to $L’$ and we know that $L’$ is decidable, it follows that $L$ is decidable as well. Conversely, if $L$ is not decidable then $L’$ would not be decidable as well.
Proof of Rice’s Theorem
…didn’t understand that well…
PostCorrespondence Problem
This is a general problem that $L_u$ can be reduced to, and is usually used to show undecidability of other problems.
Given two sets $W = { w_1, w_2\ldots w_k }$ and $X = { x_1, x_2\ldots x_k }$; the solution set is defined as the sequence $(i_1, \ldots, i_l)$ where $w_{i_1}\vert\vert w_{i_2}\ldots\vert\vert w_{i_l} = x_{i_1}\vert\vert x_{i_2}\ldots\vert\vert x_{i_l}$. This problem is Recursively Enumerable, but is not recursive in nature. (undecidable!)
This problem can be used to prove that the Ambiguity Problem is undecidable. The following grammar is ambiguous IFF PCP has a solution set. \(\begin{align*} S&\to S_1\vert S_2\\ S_1&\to w_iS_1a_i\vert w_ia_i \\ S_2&\to x_iS_2a_i\vert x_ia_i \\ \end{align*}\) Similarly, ==CFL Equivalence, and Regularity of a CFL== are also undecidable.