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# lect02 - 6.841 Advanced Complexity Theory February 9 2009...

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6.841 Advanced Complexity Theory February 9, 2009 Lecture 2 Lecturer: Madhu Sudan Scribe: Sam McVeety Today, we will study diagonalization and Ladner’s Theorem, which states roughly that: P 6 = NP ⇒ ∃ NP-intermediate Problem Later, we will look at relativization, in the context of why diagonalization can’t prove P 6 = NP. 1 Diagonalization All subsequent applications of diagonalization are based on the method discovered by Cantor (1880s), which he used to prove the uncountability of rational numbers. We outline that method below: We can enumerate the rational numbers, r i , and use them as rows in a matrix Examine the binary expansion of each number, one digit in each column Construct a new number n such that n does not equal r i at the i th place, negating along the “diagonal” Recall that we first applied this to languages to show that they are not computable, and then moved to prove further results. The proofs are very similar, as we can simply change the rows from rational numbers to languages, and the column headings to be an element in the set of all finite strings. For entries in the table, 1 implies membership in the language, 0 implies otherwise. Proving the uncomputability of the Halting Problem represents a significant step forward (1940s). The jump from “a language is not computable” to “this language is not computable” is nontrivial, particularly given that we have a concise description of the language. Taking the technique still further, we can use diagonalization to prove results for time and space com- plexity. Rabin and Blum contributed significantly to these results. Theorem 1 (Hartmanis, Stearns, ’65) time ( f ( n )) ( time ( ω ( f ( n )) log f ( n )) space ( f ( n )) ( space ( ω ( f ( n ))) Sketch of Proof Describe language L (or Turing Machine M ) such that M time ( ω ( f ( n )) log f ( n )) For every L i time ( f ( n )), L i 6 = L We ensure this inequality by using different “phases” for different i On input x , simulate L i ( x ) and negate its answer. This is the language L . Note that negation is easy for deterministic machines, but it is more subtle for nondeterministic machines. Now, we will look at Ladner’s Theorem. First, a bit of context for this particular result. In his paper on NP-completeness, Levin managed to prove many things to be NP-complete, with the major exceptions of LP, GRAPHISO, and Factoring. Because so many problems were NP-complete, theorists began to wonder whether problems were either NP-complete or in P. (Concisely, is P NP-complete = NP?) Ladner’s Theorem answers this question in the negative (unless P = NP).

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