# P the process for constructing an equivalent dtm from

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P . The process for constructing an equivalent DTM from a NTM does not preserve polynomial-time complexity. (See Theorem 15.1.2: tc M ( n ) = f ( n ) tc M’ ( n ) O ( f ( n ) c f ( n ) ).) Due to the additional time complexity of currently known non-deterministic solutions over deterministic solutions across a wide range of important problems, it is generally believe that P NP . The P = ( ) NP problem is a precisely formulated mathematical problem and will be resolved only when either (i) Defn. 15.7.1 A language Q is called NP-hard if for every L NP , L is reducible to Q in polynomial time. 15.7 P = NP ? the equality of the two classes, or (ii) P NP is proved. An NP-hard language that is also in NP is called NP-complete .

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21 Some problems L are so hard that although we can prove they are NP-hard , we cannot prove they are NP-complete , i.e., L NP . P = NP , if there exists a polynomial-time TM, which accepts an NP-complete language, can be used to construct TMs to accept every language in NP in deterministic polynomial time. Theorem 15.7.2 If there is an NP-complete language that is also in P , then P = NP . Proof . Assume that Q is an NP-complete language that is accepted in polynomial time by a DTM, i.e., Q P . 15.7 P = NP ? Let L NP . Since (by Defn. 15.7.1) Q is NP-hard , there is a polynomial time reduction of L to Q . By Theorem 15.6.2 (i.e., if L is reducible to Q in polynomial time and Q P , then L P .), L P .
22 The class consisting of all NP-complete problems, which is non-empty, is denoted NPC . If P NP , then P and NPC are nonempty, disjoint subsets of NP , which is the scenario believed to be true by most mathematicians and computer scientists. If P = NP , then the two sets collapse to a single class. 15.9 Complexity Class Relations P = NP P = NP NPC P P NP
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• Winter '12
• DennisNg
• Computational complexity theory, polynomial time, Ntm

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