This comment by @DerekElkins suggests a general method of constructing decision problems for problems with bit-strings as output, of which a slightly formalised version is the following:
Given a problem $X$, where we expect output for input $x$ to be $X(x)$, construct the following decision problem $B_X$: Given $(n,x)$ for some $n\in \mathbb{N}$, does the $n$-th bit of $X(x)$ exist and is it equal to $1$?
Clearly, this produces a decision problem for any problem $X$. However, the resulting 'bit-checking' problem may be easier than the original problem $X$. A simple case is when $X(x)$ is exponential in $x$: Suppose $X(m)$ is listing all permutations of $1,\ldots,m$ in lexicographical order. Then clearly $X$ can only be solved in $\Omega(m!)$, but detecting a single bit means we only have to check a single permutation, which can be found in polynomial time. So, in this case, the 'bit-checking' problem is easier than the original.
Therefore, I wondered if there is a general procedure that creates a decision problem that isn't 'easier', for any of the following definitions of 'isn't easier':
- $X$ can be solved in $\Theta(f(x))$ if and only if $B_X$ can be solved in $\Theta(f(x))$
- $X$ is in $P$ if and only if $B_X$ is in $P$
I'm ok with additional assumptions if those are convenient, such as assuming that the output $X(x)$ is polynomial in $x$ for case 1 or any other (general) subset of problems.
