P Vs NP

1. P Vs NP Turing Machine

2. Definitions - Turing Machine • Turing Machine M has a tape of squares • Each Square is capable of storing a symbol from set Γ (including blank Symbol b) • Σ is an input string such that Σ⊆ Γ and b ∈ Γ − Σ. • Machine Starts with input String Σ, • Initial State is q₀, final state is qaccept or qreject. • At each step M is in a state q, scans symbol s • The action is to output a symbol, enter a new state and move tape head left/right

3. Definitions - Turing Machine • Formally, Turing Machine M is a Tuple: (Σ, Γ,Q, δ) • Γis the finite, non-empty alphabet of M • Q is the finite, non-empty set of States of M • Σ ⊆ Γ and b ∈ Γ − Σ. • δis the transition function, such that: δ : (Q − {qaccept, qreject}) × Γ → Q × Γ × {−1, 1}

4. Definitions - Turing Machine • given δ(q, s) = (q1, s1, h) in Γ, • If M is in q, • Scanning s Then • q1is the new state • s1is the output symbol • The tape moves left if h is –1 or right if h is +1 • M halts if it reaches a state qaccept or qreject

5. Definitions - Turing Machine • Example of simulator: • http://ironphoenix.org/tril/tm/

6. Definitions – L(M) • L(M) is a Language accepted by M, such that • M is a Turing Machine • Σ is finite, non-empty set of alphabet • Σ* is the set of finite strings over Σ • If w ∈ Σ* is Input to M • L(M) = {w ∈ Σ∗ | M accepts w} • M halts at an accepting state.

7. Definitions – tM(w) • Time M takes to accept w - number of steps • tM(w) = ∞ means M never halts.

8. Definitions – TM(n) • Worst case time for M • For n ∈ N, TM(n) is worst case run time for M • TM(n) = max{tM(w) | w ∈ Σn}, where • Σnis set of all string in Σ of length n

9. Definitions – Polynomial Time • M runs in Polynomial time: • There exists k, such that for all n • TM(n) ≤ nk+ k

10. P • P in P vs NP stands for Polynomial • More specifically – Deterministic Polynomial • P is a Set of Languages: • P = {L | L = L(M), M runs in Polynomial Time }

11. Definitions – TM(n) • Worst case time for M • For n ∈ N, TM(n) is worst case run time for M • TM(n) = max{tM(w) | w ∈ Σn}, where • Σnis set of all string in Σ of length n

12. Definitions – Checking Relation • Checking Relation is a binary relation: R ⊆ Σ*× Σ1*, Over some alphabet Σ and Σ1 • Sort of all sets of combinations over Σ and Σ1 • A Language LR overΣ ∪ Σ1 ∪ {#} is defined as: LR= {w#y | R(w, y)} • # is used to separate w and y • R is Polynomial iffLR∈ P.

13. NP • NP in P Vs NP stands for Non-deterministic Polynomial. • A Language L over Σis in NP iff there is k ∈ N , Polynomial Time checking Relation R, and for all w ∈ Σ* w ∈ L ⇐⇒ ∃y(|y| ≤ |w|kand R(w, y)) • Where |w|is length of w, |y|is the length of y

14. Is P = NP • The question is: Suppose that solutions to a problem can be verified quickly. Then, can the solutions themselves also be computed quickly? • Quickly is defined as in Polynomial time • In Travelling Salesman Problem a route can be quickly verified, but the problem can be quickly solved !

15. History – c.e • c.e. – Computably enumerable • L is c.e. (or semi-decidable) iff L = L(M) for some Turing machine M • L is c.e.iff there is checking relation R(x, y), L = {x | ∃yR(x, y)}

16. History – Decidable • Lis decidable iffL = L(M) for some Turing machine M which satisfies the condition that M halts on all input strings w (compare with c.e.)

17. History - c.e. and Decidable • Halting Problem (HP): HP = {M | M is a Turing machine which halts on input M} • HP is c.e. but it is not decidable.

18. History - Reducibility • Symbol for one to many reducibility is ≤m • Definition: Suppose that Liis a language over Σi, i = 1, 2. Then L1 ≤m L2iff there is a (total) computable function f : Σ1*→ Σ2*, such that x ∈ L1⇐⇒ f(x) ∈ L2, for all x ∈ Σ1*

19. History – Reducibility: Consequences • if L1 ≤m L2and L2is decidable, then L1is decidable. • Alternatively: if HP ≤mL then L is undecidable

20. History – Reducibility: Example • Travelling Salesman Problem can be reduced to Hamiltonian Graph Problem

21. History - NP-Complete, c.e. Complete • c.e. complete definition: A language Lis c.e.-complete iff Lis c.e., andL’ ≤mL for every c.e. language L • HP is c.e. Complete • ≤m is transitive.

22. History - NP-Complete, Polynomial time • Polynomial Time computation first introduced in 1960s by Cobham [Cob64] and Edmonds [Edm65]. • Polynomial Time Algorithms – “good Algorithms” • Is Polynomial of n100 a “good Algorithm” ? • NP-Complete is plynomial time equivalent of c.e. Complete.

23. History - NP-Complete, Definition • Suppose that Liis a language over Σi, i = 1, 2. Then L1 ≤p L2(L1is p-reducible to L2) iff there is a polynomial-time computable function f : Σ1*→ Σ2* , such that x ∈ L1⇐⇒ f(x) ∈ L2, for all x ∈ Σ1* • A language L is NP-complete iff L is in NP, and L’ ≤p L for every language L’in NP • Focus on reducibility is polynomial

24. History - NP-Complete, Propositions • If L1 ≤p L2and L2∈ P then L1∈ P. • If L1is NP-complete, L2∈ NP, and L1 ≤p L2then L2is NP-complete. • If L is NP-complete and L ∈ P, then P=NP.

25. Some NP-Complete Problems - SAT • Satisfiability (SAT): given a boolean function F of M variables x1, x2 .. xM. F can involve the operators and, or and not. The problem is to find values for these variables which give the function F the value true

26. Some NP-Complete Problems - 3SAT • An important special case of Satisfiability is 3-SAT Satisfiability (SAT). Instances of 3-SAT are restricted to formulas in conjunctive normal form with three literals per clause: (P ∨ Q ∨ R) ∧ ( ~P ∨ Q ∨ ~R) ∧ (P ∨ ~Q ∨ S) ∧ ( ~P ∨ ~R ∨ ~S)

27. Attempts at proving • Over the years several attempts have been made to Prove P = NP or P != NP • The latest in Aug 2010 by Vinay Deolalikar announcing P != NP. Shown to be no correct • Since 1987, 34 papers try to Prove P = NP • Since 1987, 25 papers try to Prove P != NP Ref : http://www.win.tue.nl/~gwoegi/P-versus-NP.htm