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MORE RESOURCES, MORE POWER using .net toadd code 3 of 9 for web,windows application NW-7 (e.g., indeed, fo visual .

net bar code 39 r starters, assume that is the identity mapping). Thus, by construction, f DTIME(t1 ). The issue is presenting a relatively ef cient algorithm for computing f , that is, showing that f DTIME(t2 ).

The algorithm for computing f as well as the de nition of f (sketched in the rst paragraph) are straightforward: On input x, the algorithm computes t = t1 (. x. ), determines the Code 39 Full ASCII for .NET machine M = (x) that corresponds to x (outputting a default value if no such machine exists), emulates M(x) for t steps, and returns the value 1 M (x). Recall that M (x) denotes the time-truncated emulation of M(x) (i.

e., the emulation of M(x) suspended after t steps); that is, if M(x) halts within t steps then M (x) = M(x), and otherwise M (x) may be de ned arbitrarily. Thus, f (x) = 1 M (x) if M = (x) and (say) f (x) = 0 otherwise.

In order to show that f DTIME(t1 ), we show that each machine of time complexity t1 fails to compute f . Fixing any such machine, M, we consider an input x M such that M = (x M ), where such an input exists because is onto. Now, on the one hand, M (x M ) = M(x M ) (because M has time complexity t1 ), while on the other hand, f (x M ) = 1 M (x M ) (by the de nition of f ).

It follows that M(x) = f (x). We now turn to upper-bounding the time complexity of f by analyzing the time complexity of the foregoing algorithm that computes f . Using the time constructibility of t1 and ignoring the easy computation of , we focus on the question of how much time is required for emulating t steps of machine M (on input x).

We should bear in mind that the time complexity of our algorithm needs to be analyzed in the two-tape Turing machine model, whereas M itself is a two-tape Turing machine. We start by implementing our algorithm on a three-tape Turing machine, and next emulate this machine on a two-tape Turing machine. The obvious implementation of our algorithm on a three-tape Turing machine uses two tapes for the emulation itself and designates the third tape for the actions of the emulation procedure (e.

g., storing the code of the emulated machine and maintaining a step-counter). Thus, each step of the two-tape machine M is emulated using O(.

M . ) steps on the th 39 barcode for .NET ree-tape machine.1 This also includes the amortized complexity of maintaining a step counter for the emulation (see Exercise 4.

7). Next, we need to emulate the foregoing three-tape machine on a two-tape machine. This is done by using the fact (cf.

, e.g., [123, Thm.

12.6]) that t steps of a three-tape machine can be emulated on a two-tape machine in O(t log t ) steps. Thus, the complexity of computing f on input x is upper-bounded by O(T (x) (.

x. ) log T (x) (. x. )), where TM (n) = O(. M . t1 (n)) repres USS Code 39 for .NET ents the cost of emulating t1 (n) steps of the two-tape machine M on a three-tape machine (as in the foregoing discussion). It turns out that the quality of the separation result that we obtain depends on the choice of the mapping (of inputs to machines).

Using the naive (identity) mapping (i.e., (x) = x) we can only establish the theorem for t2 (n) = O(n t1 (n)) rather than t2 (n) = O(t1 (n)), because in this case T (x) (.

x. ) = O(. x. t1 (. x. )). (Note that, i .NET bar code 39 n this case, x M = M is a description of (x M ) = M.

) The theorem follows by associating the machine M with the input x M = M 01m , where m = 2. M . ; that M . is, we may use t barcode code39 for .NET he mapping such that (x) = M if x = M 012 and (x) equals some xed machine otherwise. In this case .

(x). < log2 x. < log t1 (. x. ) and so T (x) (. x. ) = O((log t1 (. x. )) t1 (. x. )). The theorem follows. 1 This overhead a ccounts both for searching the code of M for the adequate action and for the effecting of this action (which may refer to a larger alphabet than the one used by the emulator)..
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