Lecture 005

Ada's Lecture

Undecidable Languages



Turing reduction: \leq_{T} In order to define it formally, we need to define Turing machines that have access to an oracle. (definition omitted)

// Exercise (Practice with reduction definition) // Exercise (Practice with reduction proofs)

Semidecidability and Recursive Enumerability

Semidecidable languages: there is TM (not necessarily a decider) s.t. L(M) = L \subseteq \Sigma^*.

Co-semidecidable languages: if \bar{L} is semidecidable

// Exercise (HALTS is semidecidable)

Enumerator: a TM that when started with a blank input tape, outputs a list of strings, possibly with repetition, on a second output tape. An enumerator may run forever if it outputs an unbounded number of strings.

Recursively enumerable: exists enumerator E only outputs strings in L and eventually outputs every string in L.

// TODO: Check Your Understanding // TODO: High Order Bits

Sutner's Lecture

Decidable and Undecidable

Universal TM (\mu): simulate turning machines

Decidable Problems

Friedman's Self-Avoiding Words

Subsequence: A is subsequence of B if you can erase some of A to get B. [abc] is a subsequence of [cbabbabcac] but [ccb] is not.

x[i] = x_ix_{i+1}...x_{2i}

Self-avoiding: if for all a \leq i < j \leq |x|/2, block x[i] is not a subsequence of block x[j]

Undecidable Problems

Hilbert's 10th problem (Solutions of Diophantine Equations)

Diophantine Equations: P(x_1, x_2, ... x_n) = 0

Theorem: wether integer solutions of Diophantine equation exists is undecidable

Tiling Problem

Tailing Problem

Tailing Problem

Tailing Solution 1

Tailing Solution 1

Tailing Solution 2

Tailing Solution 2

Tailing Solution 3

Tailing Solution 3

Tailing Solution 4

Tailing Solution 4

Theorem: Tailing Problem is undecidable

Riemann Hypothesis

\zeta(s) = \sum_{n\geq 1}n^{-s} = 1/1^s + 1/2^s + 1/3^s +... where Re(s)>1

Zeta Function

Zeta Function

Riemann Hypothesis: all non-real roots s of this function have Re(s) = 1/2

Simplifying Zeta to Computation Problem

Simplifying Zeta to Computation Problem

Claim: M halts iff RH is false.

Computability Hierarchy

Semidecidable Diagram

Semidecidable Diagram
Halting: Does M_e halt on input e? (where e is TM index) Full Halting: Does M_e halt on input x\in \mathbb{N}? (where e is TM index) Pure Halting: Does M_e halt on empty tape? (where e is TM index)

Oracle Machine

Oracle Machine: can solve any math problem and decide whether element x is in A \subseteq \mathbb{N} for all A.

Oracle Process: a normal TM write a number on tape, and enter q_{query} give it to oracle machine. Oracle machine look at the tape and force normal TM to enter either q_N or q_Y.

Oracle Turing Machine (o-machine): normal TM who can access to oracle machine.

Turing Reducibility: B \leq_{t} A (B can be reduced to A)

Hardness and Completeness:

Many-one reducible: A \leq_m B \iff (\exists \text{ computable } f:\mathbb{N} \rightarrow \mathbb{N})(x\in A \iff f(x) \in B) (A is many-one reducible to B if when asking A if x belongs to a set, A manipulate x using f and return B(f(x)))

Models of Computation

Many Equivalent Models of Computation

Many Equivalent Models of Computation

Similarity in Models:

\mu-Recursive Functions

\mu-Recursive Functions: primitive recursive with unbounded search. Given P.R. g : \mathbb{N}^{n+1} \rightarrow \mathbb{N}, f : \mathbb{N}^n \nrightarrow \mathbb{N} with f(x) = \text{min}(\{z \in \mathbb{N} | g(z, x) = 0\}) is \mu-recursive

Prove TM decidable implies \mu-recursive

  1. C\xrightarrow[M]{t}C' is primitive recursive for fixed t
  2. find t such that C = C^{init}_x and C' is final configuration. (unbounded search for t)
  3. \text{TM decidable } \implies \mu\text{-recursive} \land \lnot \text{TM decidable } \implies \lnot \mu\text{-recursive}

Prove \mu-recursive implies TM decidable: simulate \mu-recursion on TM

  1. All P.R. is TM decidable
  2. to simulate f(x) = \text{min}(\{z \in \mathbb{N} | g(z, x) = 0\}), we keep two tapes. One tape keep counter of z from 0 while the other does the P.R. calculation.

Unbounded Search: Let L \subseteq \Sigma^* be a language. Then K = \{y \in \Sigma^* | (\exists x \in \Sigma^*)(\langle{x, y}\rangle \in L)\} is unbounded search language of L




DIV, PDE, UTM Problems

DIV, PDE, UTM Problems

\Delta_1: all decidable sets \Sigma_1 = \{K \subseteq \Sigma^* | K \text{ is unbounded search of decidable language}\}: all semidecidable sets \Pi_1 = \{K \subseteq \Sigma^* | (\Sigma^* - K) \in \Sigma_1\}: all co-semidecidable sets \Delta_k = \Sigma_{k} \cap \Pi_{k} \Sigma_{k+1}: all sets obtained from \Pi_k by unbounded search \Pi_{k+1}: all complements of sets in \Sigma_k \emptyset^{(n)} = K^{(n-1)} = \Sigma_n\text{-complete} (with many-one reduction) \emptyset^{(\omega)} = \{\langle{e, n}\rangle | e \in \emptyset^{(n)}\} (any set in arithmetical hierarchy is reducible to \emptyset^{(\omega)}, which captures all arithmetic truth), but you can also apply jump to \emptyset^{(\omega)}...

Note: all higher level language includes lower level languages


Hierarchy Diagram

Hierarchy Diagram

Halting Set of OTM: a set in which all OTM halt.

jump: the jump A' of A \subseteq \mathbb{N} is the Halting Set for OTMs with oracle A: A' = \{e \in \mathbb{N} | \{e\}^A(e) \downarrow\}

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