Ladner's Theorem

Transcript

1. Ladner’s Theorem Sai Krishna Vipul Harsh September 3, 2014 Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 1 / 16
2. Beliefs A large number of NP problems were proven to be NP complete suggesting a possibility that all problems were in P or NP complete Ladner’s Theorem tells us that if one were to believe that P = NP, then infact, this is not true. Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 2 / 16
3. Ladner’s Theorem Definition Suppose P = NP. Then there exists a language L ∈ NP \ P that is not NP-complete. Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 3 / 16
4. Basic Idea : Proof 1 We take a NP complete language say SAT. We pad each word in the language with unnecessary characters at the end, blowing up the size. Intuitively, this makes the language ”easier”. However, we pad just enough that the language is not in P. Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 4 / 16
5. Basic Idea : Proof 2 We construct a language in NP, say A such that it is not accepted by any turing machine in the enumeration of poly time deterministic turing machines. That is, for each poly time running Turing machine, Mi there is a word in A that is not accepted by Mi or vice-versa. We then prove that if we enumerate all possible poly time functions fi : N → N(Why are they countable?), and apply this function on each of the words in SAT, the resulting language will differ from A at infinite number of words. This would mean that there is no possible polynomial time reduction from A to SAT and hence, A is not in NP-complete. Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 5 / 16
6. Proof 1 Definition For a function H : N → N, we define the language SATH to contain all length n satisfiable formulae that are padded with nH(n) 1’s ; i.e, SATH = { ψ01nH(n) : ψ ∈ SAT and n = |ψ| } Definition We define the function H : N → N as: H(n) is the smallest number i < loglog n such that for every x ∈ {0,1}∗ with |x| ≤ log n, Mi outputs SATH(x) within i|x|i steps. If there is no such i then H(n) = loglog n. Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 6 / 16
7. Proof 1 : Continued CLAIM 1 : SATH ∈ P iff H(n) = O(1) (i.e., H has an upper bound) PROOF : (SATH ∈ P) =⇒ (H(n) = O(1)) Suppose there is a machine M solving SATH in at most cnc steps. Then there exists a number i>c such that M=Mi . The definition of H(n) implies that ∀ n>22i ,H(n)≤i. Thus, H(n) = O(1). Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 7 / 16
8. Proof 1 : Continued CLAIM 1 : SATH ∈ P iff H(n) = O(1) (i.e., H has an upper bound) Proof Continued : (H(n) = O(1)) =⇒ (SATH ∈ P) If H(n)=O(1), then H can take only finitely many values. So, there exists an i such that H(n)=i for infinitely many values of n. This implies the TM Mi solves SATH in ini time. Otherwise, if there was an input x on which Mi fails to output SATH(x) within this time, then ∀ n> 2|x| we would have H(n) = i . Hence, TM Mi solves SATH in O(ni )(i.e, SATH∈ P). Moreover, if SATH / ∈ P then H(n) tends to infinity with n. Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 8 / 16
9. Proof 1 : Continued Theorem : If P = NP then SATH is neither in P nor NP-complete PROOF : Suppose that SATH ∈ P. By CLAIM1, H(n)≤C for some constant C. This implies SATH is simply SAT padded with atmost polyomial (nC ) number of 1’s. So, we can solve SAT in polynomial time using the polynomial time algorithm of SATH , implying that P=NP ! Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 9 / 16
10. Proof 1 : Continued Theorem : If P = NP then SATH is neither in P nor NP-complete Proof Continued : Suppose that SATH is NP-complete. Now, we have a reduction f from SAT to SATH that runs in time O(ni ) for some constant i and the instance of SATH(say ’ψ’) should be of length O(ni ). Since SATH is not in P, CLAIM 1 impiles that H(n) tends to infinity. So, |ψ| + |ψ|H(|ψ|) = O(ni ). So,|ψ| = o(n). This implies a poly-time reduction from a SAT instance of size O(n) to SAT instance of size o(n), which in turn implies SAT can be solved in polynomial time.This contradicts P = NP. Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 10 / 16
11. Proof 2 Definition Define language A as follows : A = {x | x ∈ SAT f (|x|) is even } Where f is yet to be defined. Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 11 / 16
12. Proof 2 : Continued Definition Define f recursively. Let f(0) = f(1) = 2. To compute f(n), assume f(0) through f(n-1). We have two cases, f(n-1) = 2i Note that, we have f(0) through f(n-1). Thus, we can check membership of x for A for 0 to n-1. If the membership of any of these x mismatch with the ith Turing Machine Mi , then f(n) = f(n-1) + 1, else f(n) = f(n-1). Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 12 / 16
13. Proof 2 : Continued Definition Continued f(n-1) = 2i + 1 If we apply Fi to 1 to n-1 (problem instances of SAT), to obtain Fi (1), Fi (2).....Fi (n − 1). If the membership of any of these Fi (k) mismatch with that of k in SAT, then f(n) = f(n-1) + 1, else f(n) = f(n-1). Note that, it will not be possible to check the membership for all elements in Fi (1), Fi (2).....Fi (n − 1), since we have calculated f only for 0 through n-1. We simply skip those elements. Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 13 / 16
14. Proof 2 : Continued Since, f is computable in polynomial time and SAT is in NP, it can be inferred that A is in NP. CLAIM: A / ∈ P PROOF: For the sake of contradiction, suppose there is a polynomial time deterministic turing machine Mi for A. Thus, the function f would become constant at 2i after some point(Refer to the first case for f, when f(n-1) = 2i). But, then SAT differs from f in only a finite number of words. By this logic, SAT is also in P which contradicts our hypothesis. Hence, A / ∈ P Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 14 / 16
15. Proof 2 : Continued CLAIM: A / ∈ NP-complete PROOF: For the sake of contradiction, suppose there is a polynomial time reduction function Fi which reduces instances of SAT to A. Thus, the function f would become constant at 2i+1 after some point(Refer to the second case for f, when f(n-1) = 2i+1). But, then A has only a finite number of words and hence A ∈ P. Since, we assumed A to be NP-complete, P=NP which again contradicts our Hypothesis. Hence, A / ∈ NP-complete Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 15 / 16
16. Addendum Assuming P = NP, no naturally arising language is known to be in NP and not in P or NP-complete. However, people believe Graph Isomorphism, Integer Factorization to be potential candidates. Sai Krishna, Vipul Harsh Ladner’s Theorem September 3, 2014 16 / 16