New answers tagged complexity-classes
1
$C_2 \subseteq NL$: Let $L \in C_2$. Let the log-space verifier for $L$ be $T$. We can construct a log-space NTM $T'$ which on any input simulates $T$ on the same input. Whenever $T$ tries to read the witness-tape, $T'$ just non-deterministically guesses an alphabet.
This will have the same effect as having a read-once witness tape.
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Let $N$ be the total number of configurations of the machine other than the witness tape, namely state, location of the input tape head, contents of the work tape, and location of the work tape head. Note that $N$ is polynomial in $n$.
We can assume without of generality that at each step, the machine reads a bit from the witness tape, and it affects its ...
1
What I recommend first is to notice that if you're looking at the complexity of the function $3x^2+2x+1$, really all you should care about is the function $x^2$.
because if you will prove that $x^2 = \omega(xlogx)$
then adding the $2x + 1$ won't ruin that proof since $x^2$ is polynomially bigger than $2x + 1$ and so we can just look at the $x^2$.
(I will ...
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The easiest way is to check that $\lim_{x \to \infty} \frac{3x^3 + 2x +1}{ x \log x} = +\infty$, which is a sufficient condition for $3x^3 + 2x +1 \in \omega(x \log x)$.
$$
\lim_{x \to \infty} \frac{3x^3 + 2x +1}{ x \log x} =
\lim_{x \to \infty} \frac{3x^3}{ x \log x} =
\lim_{x \to \infty} \frac{3x^2}{ \log x} =
\lim_{x \to \infty} 6x^2
= +\infty.
$$
3
Consider the "singleton" languages: $L=\{w\}$ for some fixed word $w$. (e.g., the language $L=\{00010100\}$ would be a singleton language) Most "nice" complexity classes include the singleton languages.
Every language can be written as an infinite union of singleton languages. So, if all singleton languages are in the class, and if the class is closed ...
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