- 91 Views
- Uploaded on
- Presentation posted in: General

ATM, Halting Problem, P vs. NP

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

ATM, Halting Problem, P vs. NP

Chapter 4, 5 & 7

- http://www.jimloy.com/logic/russell.htm
- An Index is a book that lists other books in the library
- Index of all biology text books in the library.

- Consider the index of all indices, i.e., book that lists other indices.
- It would contain itself.
- Called a self-referential index (form of recursion).

- Consider the index of all non self-referential indices.
- The book that lists all indices that are non self-referential.
- Call it the non-recursive index

- If the non-recursive index lists itself then its recursive and shouldn’t be in the list.
- But if it doesn’t list itself, then it is non-recursive and it should be listed as one of the non self-referential books.

- Proof by Contradiction
- Assume we have a Turing Machine R that decides HALTTM.
- In other words, we can design an algorithm that determines of another algorithm (M) will halt on a given input (w).

- Assume we have a Turing Machine R that decides HALTTM.

- Proof by Contradiction
- Assume we have a Turing Machine R that decides HALTTM.
- Then, use R to construct a new machine S that can decide ATM.
- We have already proven that is NOT decidable.
- So, if we can construct a machine S that can decide ATM based on HALTTM then we must conclude that cannot be decidable.

- Proof by Contradiction
- Assume we have a Turing Machine R that decides HALTTM.
- Then, use R to construct a new machine S that can decide ATM.
- We have already proven that ATM is NOT decidable.
- So, if we can construct a machine S that can decide ATM based on HALTTM then we must conclude that cannot be decidable.

- ATM is just a set of strings where part of the string is the encoding of a Turing Machine (M) and the other part is an input string (w).
- The string in ATM are the ones were the input w is accepted by the Machine M.
- The string in the complement of are the ones the input w is rejected by the Machine M.

- A decider for ATM cannot exist because there are absurd Turing Machines infinite loop or output nothing.
- Detecting such absurdity is impossible because you are limited by the power of the same machine that allows for such absurdity.
- Simulating an absurd Turing Machine with another Turing Machine cannot “undo” it absurdity and allow it to always produce an answer (accept or reject).

Chapter 7

Time Complexity

- Some problems (Hamiltonian Path) cannot be solved in Polynomial Time (P), but…
- If we could somehow obtain a solution (ask an oracle), the solution could be verified in P-time.

- There is very subtle difference:
- Hamiltonian Decider
- Input: Graph
- Output: Hamiltonian Path

- Hamiltonian Verifier
- Input: Graph, Hamiltonian Path
- Output: Accept/Reject

- Problems where the solutions can be verified in P-Time, but cannot be “discovered” in P-time
- Problems where the solutions can be “discovered” P-Time using a Non-deterministic Turing Machine
- Note: Every non-deterministic Turing machine can be made deterministic by adding O(kn)

- k is usually 2
- Given a graph with n vertices, there are 2n possible subsets.
- 2 options for each vertex
- Either you are in the subset (1) on not (0)
- Binary state condition

- Given a list with n value, there are 2n possible subsets

- Given a graph with n vertices, there are 2n possible subsets.

“Non deterministically select a subset implies 2n operations.”

- This is a loop through a Turing machine state that has a branching non-deterministic transition
- For each n vertex regardless of the tape symbol
- Add (1) the vertex to the subset and continue
- Don’t add (2) the vertex and continue