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by LPisGood 300 days ago
There is no relationship between n and N, of course. N is some fixed constant.

Log(n) is unbounded, yes, but if a Turing machine halts in sub-linear time then there is some length input for which that TM halts before it reads every cell of the input tape. Therefore, it doesn’t matter how the size of the input grows, that TM must halt in constant time. Make sense?
1 comments
Droobfest 300 days ago
I’m sorry but no. Maybe it’s my ignorance of TM’s, but O(log n) doesn’t read all input by definition. It doesn’t follow that it is therefore _independent_ of the input size.

What makes a/your TM special that this is the case?

I don’t mean to take too much of your time though, maybe I’m too dumb for this.

Edit: I can sort of see how O(log n) is impossible or at least O(n) in a TM, but to reduce it to O(1) makes no sense to me

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umanwizard 300 days ago
Let me try to explain the proof in more detail. Assume the TM is sublinear. Then there exists some fixed N such that for every input of size n <= N, the number of steps the TM halts in is less than N.

Now consider any input I, and let I_N be the subset of I that fits in N characters (I.e. I_N is the same as I except with all characters past the Nth replaced by blanks).

Then by our earlier statement defining N, the TM must halt on I_N after fewer than N steps. But the TM’s behavior on I for the first N steps must be the same as its behavior on I_N, because it can’t possibly have read any different characters (since in N steps it can only have read at most the first N characters, and those two inputs are the same on that interval). Thus the machine must halt in less than N steps on I. But I was chosen arbitrarily, so we’ve shown that the machine halts in less than N steps on any input, and recall that N was a fixed constant.

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Droobfest 299 days ago
This looks logically sound to me, thanks!
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LPisGood 300 days ago
Remember that TM’s read their input left to right. Therefore if there is some fixed number N where the TM halts on all length N inputs, then it would halt in the same number of steps if input where length N+1 because TM halts before it can read the additional input.
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