Loading in 5 sec....

ATM, Halting Problem, P vs. NPPowerPoint Presentation

ATM, Halting Problem, P vs. NP

- By
**reina** - Follow User

- 103 Views
- Uploaded on

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.

Presentation Transcript

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

### Chapter 7

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).

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

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.

Deeply Understanding ATM

- 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).

Time Complexity

O(n2)

O(n log2n)

Polynomial Verification

- 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.

Hamiltonian Path: Decider vs. Verifier

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

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

Understand NP in English

- 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

- Given a graph with n vertices, 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

Download Presentation

Connecting to Server..