Compression algorithms for low-complexity strings? Let $s$ denote a string over a finite alphabet, $n_s = |s|$ be the length of $s$, and $n_s^{*}$ denote the minimum description size of $s$ under a given computational model (TM, CFG, etc.). Are there compression algorithms that, when restricted to input $s$ where $n_s / n_s^*$ is large, achieve provably tighter approximations than algorithms with unrestricted input?

In other words, can a compression algorithm specifically tailored for low complexity inputs outperform the best general input algorithm on inputs of low complexity?

Admittedly, the question isn't completely precise, but I'm hoping to find results of roughly the same flavor. 

Edit: See domotorp's answer below for a proof of the impossibility of a large class of low-complexity compression algorithms under the TM model. Note that this result does not bar the possibility of compression algorithms where the function $A$ (see the answer for context) is non-computable. For example, setting $A(|x|) = 100*BB^{-1}(|x|)$ where $BB^{-1}$ is the inverse busy beaver function, could make for a useful compression algorithm and is not covered by the proof.
No, at least not in the TM model, i.e., for Kolmogorov complexity. Given $g(x) \leq f(x) \leq |x|$ for all $x$ where $g$ is monotonic and unbounded and $f$ is recursive, it is not possible to enumerate an infinite subset of $\{ x : K(x) &gt; f(x)\}$. 

Suppose for the sake of contradiction there exists a recursive function $h$, the low-complexity compression algorithm, which on input $x$ outputs a TM that reproduces $x$ and has the following additional property: Let $A:\mathbb{N} \rightarrow \mathbb{N}$ and $B:\mathbb{N} \rightarrow \mathbb{N}$ be given such that both are recursive, $A(|x|) \leq B(|x|) \leq |x|$ for all $x$, and $A$ is unbounded. Then $\forall x, K(x) \leq A(|x|) \implies |h(x)| \leq B(|x|)$.

Now choose $f(x) = A(|x|)$. We wish to enumerate an infinite subset of $\{x: K(x) &gt; A(|x|)\}$. Note that there are infinite $x$ such that $K(x) &gt; B(|x|)$ because $B(|x|) \leq |x|$. For any such $x$, $|h(x)| \geq K(x)$ by the definition of $K$ so $|h(x)| &gt; B(|x|)$. Then for a generic string $x$, $|h(x)| &gt; B(|x|) \implies K(x) &gt; A(|x|)$ and we can enumerate an infinite subset of $\{x: K(x) &gt; A(|x|)\}$, a contradiction.