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
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
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
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
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
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
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
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
= 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
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
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
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
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