1 / 63

# ATM, Halting Problem, P vs. NP - PowerPoint PPT Presentation

ATM, Halting Problem, P vs. NP. Chapter 4, 5 & 7. Russel’s Paradox. 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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

## PowerPoint Slideshow about 'ATM, Halting Problem, P vs. NP' - reina

An Image/Link below is provided (as is) to download presentation

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 - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### ATM, Halting Problem, P vs. NP

Chapter 4, 5 & 7

Russel’s Paradox

• 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.

ATM reduces to HALTTM

• 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).

ATM reduces to 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.

ATM reduces to 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 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

O(n2)

• 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)

Non-determinism  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

Non-determinism  O(kn)

“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