Read e-book online Introduction to complexity theory, lecture notes PDF

By Goldreich O.

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That is, we let i'th machine to make t steps during i2 Pt rounds of the simulating machine. Also the number of steps made in these r = i2 t rounds is j 1 j jr j < r. Such a construction would fail if some of these machines would provide wrong answers. To solve this di culty we "force" the machines to verify the validity of their outputs. So, without loss of generality, we assume that each machine is augmented by a verifying component that checks the validity of machine's output. Since the problem is in NP , verifying the output takes polynomial amount of time.

Yet the computation we care about refers to one speci c input. Similarly the initial state should be q0 and the head should be located at time 0 in the rst location. This will be done by setting all triplets (0 j ) as following: 4 j(hM i x 1t )j the length of the input. Let x = x1 x2 x3 :::xm and n = ( 1 j m constants set by input x a0 j = xy j j ;m m < j t these are the inputs to the circuit h0 j = ( ( 1 j=1 0 j 6= 1 = 1 where q0 is the initial state of M q0 j = 0q0 jj 6= 1 The y elements are the variables of the circuit.

2 Gates at level i of the circuit represent the exact con guration of M at time i on input (x y). Proof: By induction on time i. i = 0, stage 1 of the construction ensures correctness. Assume C 's gates on level i correctly represent M 's con guration at time i and prove for i + 1: Set j as the position of the head at time i (hi j = 1). { The letter contents of all cells (i + 1 k) k 6= j does not change. Same happens in the circuit since (ai k V(:hi k )) = ai k . { Likewise the head can not reach cells (i + 1 k) where k < (j ; 1) or k > (j + 1).

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Introduction to complexity theory, lecture notes by Goldreich O.


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