What happens to the coefficient in O(t(n)b^{t(n)})?

[gnome]

In Sipser, "Introduction to the Theory of Computation", a proof that "every t(n) time nondeterministic single-tape Turing machine (where t(n) ≥ n) has an equivalent 2^O(t(n)) time deterministic single-tape Turing machine" shows that the running time of the equivalent NTM is $O(t(n)b^{t(n)})$ and then concludes with the statement:

$$O(t(n)b^{t(n)}) = 2^{O(t(n))}$$

which is not entirely clear to me.

b in this case is just a constant representing the maximum branching factor of the NTM.

I guess that $2^{O(t(n))}$ is equivalent to $b^{t(n)}$ since b is a constant and therefore $log_2{b}$ is a constant so we get

$$b^{t(n)} = 2^{(log_2 b)* t(n)} = 2^{O(t(n))}$$

Correct?

But, then what happened to the other O(t(n)) (the coefficient of the original expression)? That was not a constant term. It might be n, or nlogn, or n², or whatever, but certainly not a constant. So why did it just disappear? Is it just that the exponential term so dominates that the t(n) coefficient is dropped as insignificant?

 

[Hurkyl]

It didn't just disappear -- you have to break the RHS into two parts, each part asymptotically bigger than one of the factors of the LHS.

 

[gnome]

I don't understand your meaning. Where are the two parts in $2^{O(t(n))}$?

Or to put it another way: for argument's sake, suppose that every branch of the original NTM has a length of at most n², so t(n) = n² in the original NTM. And further, suppose that the maximum branching factor was just 2. Then we would have the total number of nodes $= O(2^{n^2})$ and the time for traveling from the root to a node = $O({n^2})$. Therefore, the running time of the equivalent deterministic machine = $O(n^22^{n^2})$. Can this be simplified to just $2^{O(n^2)}$? Why?

 

[Hurkyl]

You can almost always break things into two parts if you're creative! How about writing it as the square of 2^{(1/2) O(t(n))}?

 

[gnome]

But what is the point of that? (And anyway, ½O(t(n)) is the same as O(t(n)), isn't it?)

 

[Hurkyl]

Because you can prove 2^O(t(n)) is asymptotically larger than t(n), and equal to O(b^t(n)), therefore 2^O(t(n)) * 2^O(t(n)) must be asymptotically at least as large as O(t(n) * b^t(n)).

But what is the point of that? (And anyway, ½O(t(n)) is the same as O(t(n)), isn't it?)

Yep! But writing it that way makes it clear how I'm breaking it into two parts.

 

[gnome]

OK, I see your point.

Thanks, Hurkyl.