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For any language $A$, define a language $L_{A}$ as:

$L_{A} = \{0^{n}: \text{Number of strings in $A$ of length $n$ is more than $2^{n-1}$} \}$

I am trying to construct an $A$ such that $L_{A} \notin P^{A}$. It should intuitively be true, as a polynomial time machine cannot query an exponential number of strings and check whether a majority of them belongs to a language.

The proof boils down to a diagonalization argument. But I am unsure on how to construct the sets.

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  • $\begingroup$ These arguments typically have two parts: the query lower bound (which you mentioned) and the construction of the oracle. The first part follows by an adversary argument. If any path of the tree queried less than n/2 variables then it couldn't distinguish between a specific (based on the path) "Yes"-instance and a specific "No"-instance. The second part just applies the first part: each poly-time M_j on input 1^n induces a small decision tree (on the oracle) and so there's some oracle A_j for which M_j(1^n) fails to solve L(A_j) on inputs of length n. Then you basically "glue" these together. $\endgroup$ – Sam McGuire Jul 16 at 4:32
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Let $T_1,T_2,\ldots$ be an enumeration of all timed polytime machines, machine $T_i$ running in time $p_i(n)$. We will construct $A$ in stages. At each stage, the truth value of $A$ is defined on only finitely many inputs.

Stage $i$ handles $T_i$. Choose an input length $n$ such that (i) $A$ is completely undefined and (ii) $p_i(n) < 2^{n-1}$. Run $T_i$ on $0^n$. Each time $T_i$ queries some yet undefined string in $A$, answer arbitrarily and fix the value of this string in $A$. Since $p_i(n) < 2^{n-1}$, in particular $T_i$ queries fewer than $2^{n-1}$ strings of length $n$. If $T_i$ answered "Yes", then define all remaining strings of length $n$ to be outside $A$. If $T_i$ answered "No", then define all remaining strings of length $n$ to be inside $A$. Either way, the machine $T_i$ is wrong on input $0^n$.

After infinitely many steps, we have defined $A$ at infinitely many strings, but probably not all strings. Define $A$ arbitrarily on all remaining strings.

We note that the same proof works for every (time constructible) complexity class in which a machine cannot make at least $2^{n-1}$ queries on input of length $n$.

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