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Theorem 7.11 in Introduction to the theory of computation 3rth edition says

Let $t(n)$ be a function where $t(n)>n$. Then every $t(n)$ time nondeterministic single-tape Turing machine has an equivalent $2^{O(t(n))}$ time deterministic single tape Turing machine.

I can understand that in this proof when we have a nondeterministic TM which every node in it can at most go to $b$ other nodes in its computation tree, then the time complexity of a deterministic Turing machine simulating it would be $O(t(n)b^{t(n)})$. But I'm not sure if I understand the $2^{O(t(n))}$ correctly.

Does it means that if $f(n) \in 2^{O(t(n))}$ then $\exists b\in \mathbb{N} \; f(n)\in O(b^{t(n)})?$

On the other hand, the claim 1.5 in the Computational complexity book says

For every $f : \{0, 1\}^∗ \rightarrow \{0, 1\}$ and time-constructible $T : N\rightarrow N$, if $f$ is computable in time $T(n)$ by a TM $M$ using alphabet $\Gamma$, then it is computable in time $4 log |\Gamma|T(n)$ by a TM $M$ using the alphabet $\{0,1,\square,\rhd\}$.

So every nondeterministic TM M running in $t(n)$ time can have an equivalent nondeterministic TM using $\{0,1\}$ alphabet running upper hand in $|\Gamma|^2t(n)$. Thus this equivalent TM's computation tree has at most 2 different branches in each step. Then we can have a deterministic simulator of that TM running in time $O(2^{|\Gamma|^2t(n)})$.

The intersection of 2 above statement suggests that carelessly we can say $O(2^{t(n)})=2^{O(t(n))}$, but we know $\forall b>2,\; b^{t(n)}\notin O(2^{t(n)})$.

Unfortunately, I didn't find any definition of $2^{O(t(n)}$. I wish to know is there any formal definition for $2^{O(t(n))}$?

- Edit: There is a similar question here. But I'm doubting its answer is the same since the big O notation in the power is not just $O(1)$.

Suppose we have nondeterministic TM that each node of it goes to 3 other nodes. Then there is a deterministic Turing machine that simulates it in time $t(n)3^{t(n)}$. So we can't say that there is a constant $k$ that $t(n)3^{t(n)} < 2^{kt(n)}.$

Meanwhile, when I think more, this definition holds if we ignore the Sipser proof. I mean if we begin the proof by the statement that every nondeterministic TM has a version $\{0,1\}$ alphabet running in $c.t(n)$ then the answer's definition is correct.

I wonder if there is bad use of big O notation in the Sipser's book or $2^{O(t(n))}$ has a different definition.

Doralisa
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    Possible duplicate of What does $\log^{O(1)}n$ mean? – David Richerby Nov 16 '18 at 18:14
  • It's not exactly the same question but the same principles apply. – David Richerby Nov 16 '18 at 18:15
  • @DavidRicherby, Thank you for the link. By its best answer If I have $f(n) \in 2^{O(t(n))}$, I should think it as $\exists n_0,c; \forall n> n_0 ; |f(n)| < 2^{ct(n)} $. But if for example I have a nondet. TM that each node of it goes 4 other nodes, then det. TM simulating it runs in $t(n)3^{t(n)}$ and I think it doesn't satisfy that definition. Or I'm misunderstanding something? – Doralisa Nov 16 '18 at 18:26
  • correcting my above comment, " ... nondet. TM that each nodes of it goes 3 other nodes ..." – Doralisa Nov 16 '18 at 18:34
  • I have not read beyond the second paragraph. Here is what I am thinking. $f(n)=2^{O(t(n))}$ means, by applying $log_2$ to both sides of the equality, $\log_2f(n)=O(t(n))$. Please note that everything here is well-defined! (assuming $f(n)>0$ for all $n$, or $f(n)>0$ eventually.) Let me see what happens in the question, comments and the answer then. – John L. Nov 19 '18 at 02:09
  • @Apass.jack Thanks for your comment. The definition you've provided is very clear and well -defined. But In the proof of theorem 7.11 in the book it says $t(n)b^{t(n)}\in 2^{O(t(n))}$. Let $t(n)=n^5$and $b=3$, then $5log3.log,n.,n^5 \notin O(n^5)$. So I thought there should be something wrong. – Doralisa Nov 19 '18 at 07:26
  • I am afraid you made a typo. "$t(n)b^{t(n)}\in 2^{O(t(n))}\ \ $. Let $t(n)=n^5$ and $b=3$, then" $5\log n + n^5\log3\in O(n^5)$. I have failed to see anything wrong. – John L. Nov 19 '18 at 08:43
  • @Apass.Jack ah, It was my mistake by being too hasty in taking the log of $t(n)b^{t(n)}$. Thank you again for providing this clear definition. – Doralisa Nov 19 '18 at 10:58

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This is perfectly standard use of big O notation, despite some purists disliking it. Here is what the passage you quoted means:

Let $t(n)$ be a function where $t(n) > n$. Then every $t(n)$ time nondeterministic single-tape Turing machine has an equivalent $2^{f(n)}$ time deterministic single tape Turing machine, for some function $f(n)$ satisfying $f(n) = O(t(n))$.

This is still somewhat ambiguous, since it’s perhaps not clear what variables the big O depends on. In this case, big O hides a universal constant:

There exists a constant $C>0$ such that the following holds.

Let $t(n)$ be a function where $t(n) > n$. Then every $t(n)$ time nondeterministic single-tape Turing machine has an equivalent $2^{f(n)}$ time deterministic single tape Turing machine, for some function $f(n)$ satisfying $f(n) \leq Ct(n)$ for all $n$.

This allows us to rephrase the statement in yet another way:

There exists a constant $C>0$ such that the following holds.

Let $t(n)$ be a function where $t(n) > n$. Then every $t(n)$ time nondeterministic single-tape Turing machine has an equivalent $2^{Ct(n)}$ time deterministic single tape Turing machine.

(This assumes that by “$f(n)$ time machine” we mean a machine running in time at most $f(n)$.)

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