- By
**skyla** - 1069 SlideShows
- Follow User

- 63 Views
- Presentation posted in: General

NP-complete and NP-hard problems

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 - - - - - - - - - - - - - - - - - - - - - - - - - -

- Transitivity of polynomial-time many-one reductions
- Definition of complexity class NP
- Nondeterministic computation
- Problems that can be verified

- The P = NP Question
- Concept of NP-hard and NP-complete problems

- Let R1 be the reduction used to prove P1 ≤pP2
- Let R2 be the reduction used to prove P2 ≤pP3
- Let x be an input to P1
- Define R3(x) to be R2(R1(x))

- Because R1 is a reduction between P1 and P2, we know that R1(x) is a yes input instance of P2 iff x is a yes input instance of P1
- Because R2 is a reduction between P2 and P3, we know that R2(R1(x)) is a yes input instance of P3 iff R1(x) is a yes input instance of P2
- Applying transitivity of iff, we get that R3(x) is a yes input of P3 iff x is a yes input instance of P1

- Let R1 take time nc1
- Let R2 take time nc2
- Let n be the size of x
- Then the R1 call of R3 takes time at most nc1
- Furthermore, R1(x) has size at most max(n,nc1)
- Therefore, the R2 call of R3 takes time at most max(nc2, (nc1)c2) = max (nc2, nc1 c2)
- In either case, the total time taken by R3 is polynomial in n

- Transitivity of polynomial-time many-one reductions
- Definition of complexity class NP
- Nondeterministic computation
- Problems that can be verified

- The P = NP Question
- Concept of NP-hard and NP-complete problems

- Turing machine model of computation
- Simple model where data is on an infinite capacity tape
- Only operations are reading char stored in current tape cell, writing a char to current tape cell, moving tape head left or right one square

- Deterministic versus nondeterministic computation
- Deterministic: At any point in time, next move is determined
- Nondeterministic: At any point in time, several next moves are possible

- NP: Class of problems that can be solved by a nondeterminstic turing machine in polynomial time

A Turing machine has a finite-state-control (its program), a two way infinite tape (its memory) and a read-write head (its program counter)

Finite State

Control

Head

….

….

1

0

1

0

0

0

1

1

0

1

1

Tape

Deterministic Computation

Nondeterministic Computation

- We measure running time by looking at height of computation tree, NOT number of nodes explored
- Both computation have same height 4 and thus same running time

Yes Result

No Result

- If any leaf node returns yes, we consider the input to be a yes input.
- If all leaf nodes return no, then we consider the input to be a no input.

- Hamiltonian Path
- Input: Undirected graph G = (V,E)
- Y/N Question: Does G contain a HP?

- Nondeterministic polynomial-time solution
- Guess a hamiltonian path P (ordering of vertices)
- V! possible orderings
- For binary tree, V log V height to generate all guesses

- Verify guessed ordering is correct
- Return yes/no if ordering is actually a HP

- Guess a hamiltonian path P (ordering of vertices)

2

Guess Phase

Nondeterministic

---------------

Verify Phase

Deterministic

3

1

Yes input graph

123

132

231

321

213

312

2

Guess Phase

Nondeterministic

---------------

Verify Phase

Deterministic

3

1

No input graph

123

132

231

321

213

312

- Preliminary Definitions
- Let P be a decision problem
- Let I be an input instance of P
- Let Y(P) be the set of yes input instances of P
- Let N(P) be the set of no input instances of P

- P belongs to NP iff
- For any I e Y(P), there exists a “certificate” [solution] C(I) such that a deterministic algorithm can verify I e Y(P) in polynomial time with the help of C(I)
- For any I e N(P), no “certificate” [solution] C(I) will convince the algorithm that I e Y(P).

2

Guess Phase

Nondeterministic

---------------

Verify Phase

Deterministic

3

1

Yes input graph

123

132

231

321

213

312

2

Guess Phase

Nondeterministic

---------------

Verify Phase

Deterministic

3

1

No input graph

123

132

231

321

213

312

- Certificate [Solution]
- A Hamiltonian Path
- C(I1): 123 or 321
- C(I2): none
- Verification Alg:
- Verify certificate is a possible HP
- Check for edge between all adjacent nodes in path

- Clique Problem
- Input: Undirected graph G = (V,E), integer k
- Y/N Question: Does G contain a clique of size ≥k?

- Certificate
- A clique C of size at least k

- Verification algorithm
- Verify this is a potential clique of size k
- Verify that all nodes in C are connected in E

- You need to describe what the certificate C(I) will be for any input instance I
- You need to describe the verification algorithm
- usually trivial

- You need to argue that all yes input instances and only yes input instances have an appropriate certificate C(I)
- also usually trivial (typically do not require)

- Vertex Cover
- Input: Undirected graph G = (V,E), integer k
- Y/N Question: Does G contain a vertex cover of size ≤k?
- Vertex cover: A set of vertices C such that for every edge (u,v) in E, either u is in C or v is in C (or both are in C)

- Certificate
- A vertex cover C of size at most k

- Verification algorithm
- Verify C is a potential vertex cover of size at most k
- Verify that all edges in E contain a node in C

- Satisfiability
- Input: Set of variables X and set of clauses C over X
- Y/N Question: Is there a satisfying truth assignment T for the variables in X such that all clauses in C are true?

- Certificate?
- Verification algorithm?

- Unsatisfiability
- Input: Set of variables X and set of clauses C over X
- Y/N Question: Is there no satisfying truth assignment T for the variables in X such that all clauses in C are true?

- Certificate and Verification algorithm?
- Negative certificate and Negative verification algorithm?

- Exact Vertex Cover
- Input: Undirected graph G = (V,E), integer k
- Y/N Question: Does the smallest vertex cover in G have size exactlyk?
- Vertex cover: A set of vertices C such that for every edge (u,v) in E, either u is in C or v is in C (or both are in C)

- Certificate and Verification algorithm?
- Negative certificate and Negative verification algorithm?

- Transitivity of polynomial-time many-one reductions
- Definition of complexity class NP
- Nondeterministic computation
- Problems that can be verified

- The P = NP Question
- Concept of NP-hard and NP-complete problems

- A problem П is NP-hard if
- for all П’ e NP П’ ≤p П holds.

- Intuitively, an NP-hard problem П is at least as hard (defined by membership in P) as any problem in NP

NP

P

NP-complete

- A problem П is NP-complete if
- П is NP-hard and
- П is in NP

- Intuitively, an NP-complete problem П is the hardest problem in NP
- That is, if П is in P, then P=NP
- If P ≠ NP, then П is not in P

P=NP=NP-complete

OR

- Practitioners view
- There exist a large number of interesting and seemingly different problems which have been proven to be NP-complete
- The P=NP question represents the question of whether or not all of these interesting and different problems belong to P
- As the set of NP-complete problems grows, the question becomes more and more interesting

Graph Theory

Network Design

Sets and Partitions

Storage and Retrieval

Sequencing and Scheduling

Mathematical Programming

Algebra and Number Theory

Games and Puzzles

Logic

Automata and Languages

Program Optimization

Miscellaneous

- Theoretician’s view
- NP is exactly the set of problems that can be “verified” in polynomial time
- Thus “Is P=NP?” can be rephrased as follows:
- Is it true that any problem that can be “verified” in polynomial time can also be “solved” in polynomial time?

- Hardness Implications
- It seems unlikely that all problems that can be verified in polynomial time also can be solved in polynomial time
- If so, then P≠NP
- Thus, proving a problem to be NP-complete is a hardness result as such a problem will not be in P if P≠NP.

- Proving a problem П is NP-complete
- Show П is in NP (usually easy step)
- Prove for all П’ e NP П’ ≤p П holds.
- Show that П’ ≤p П for some NP-hard problemП’
- This only works if we have a known NP-hard problem П’ to reduce from
- Also depends on transitivity property proven earlier

- We need to prove the existence of a first NP-hard problem
- Cook-Levin Thm

- Show that П’ ≤p П for some NP-hard problemП’
- Developing new reductions is a skill or art form
- Over time, it gets easier

3-SAT: The old reliable. When none of the other problems seem to work, this is the one to come back to.

Integer Partition: A good choice for number problems.

3-Partition: A good choice for proving “strong” NP-completeness for number problems.

Vertex Cover: A good choice for selection problems.

Hamiltonian Path: A good choice for ordering problems.

- Cook: “The complexity of theorem-proving procedures” STOC 1971, pp. 151-158
- Polynomial-time reductions
- NP complexity class
- SAT is NP-complete

- Levin: “Universal sorting problems”, Problemi Peredachi Informatsii 9:3 (1973), pp. 265-266
- Independent discovery of many of the same ideas

- Karp: “Reducibility among combinatorial problems”, in Complexity of Computer Computations, 1972, pp. 85-103
- Showed 21 problems from a wide variety of areas are NP-complete