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I'm trying to understand the proof of the time hierarchy theorem appearing in sipser's book. The proof requires a TM M to simulate an arbitrary TM N without too much slowdown. In particular, it is assumed that the encoding of N's tape alphabet using M's alphabet causes only a constant factor slowdown. This seems plausible since if N's alphabet is size k then M can use $\log k$ cells to represent each symbol that N writes to the tape.

But my question is this: If this is how the simulation works then before the simulation starts M will have to change the input so that each bit is repeated $\log k$ times and I don't know how to do this without adding a quadratic term to the time. I should say its assumed that N's computation is no faster than $O(n\log(n))$.

Raphael
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Nick
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1 Answers1

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This has been clarified in the book's online errata:

Page 369, Proof of the Time Hierarchy Theorem 9.10.

The proof requires some additional technical discussion at one place. In Stage 4 of D, it simulates M on w. This simulation may require representing each cell of M's tape with several cells on D's tape because M's tape alphabet is arbitrary and D's tape alphabet is fixed. However, initializating the simulation by converting D's input w to this representation involves rewriting w so that its symbols are spread apart by several cells. If we use the obvious copying procedure for spreading w, this conversion would involve O(n^2) time and that would exceed the O(t(n)) time bound for small t. Instead, we observe that D operates on inputs w of the form x10^k where x = ‹M›, and we only need to carry out the simulation when k is large. We consider only k > |x|^2. We can spread w by first using the obvious copying procedure for x and then counting trailing 0s and rewriting these by using that count. The time for spreading x is O(|x|^2) which is O(n). The time for counting the 0s is O(n log n) which is O(t(n)) because t is time constructible.

Reported 11/15/12 by Kaveh Ghasemloo of the University of Toronto.

Juho
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