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Let $X$ be some finite alphabet. Given $(x,y) \in X^{\ast}\times X^{\ast}$, how to show that $K(x) \le K(y)$ is not decidable?

I know that $K(x) \le k$ for some fixed $k$ is not decidable, so I tried to reduce the problem. I also know that $K(1^n) \le O(\log n)$, so but reducing it by simply plugging the word $y = 1^{2^k}$ in $K(y)$ would not give exactly $k$, just $K(y) \le O(k)$. So this reduction (i.e. decide $K(x) \le k$ if we know that $K(x) \le K(y)$ is possible) will not work.

StefanH
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1 Answers1

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This answer is only a hunch - I am not familiar with the details.

Some strings are incompressible. If $x$ is incompressible, then $K(x)$ is the largest possible number for a string of this length. The optimal encoding of an incompressible string is writing it verbatim.

Suppose $K(x) \leq K(y)$ is decidable. I believe that it's then possible to decide incompressibility. The algorithm, given $y$, checks whether $K(x) \leq K(y)$ is true for every $x$ whose length is the same as $y$.

  • If $y$ is incompressible, then $K(y)$ will be at least as large as $K(x)$ for every $x$.
  • Suppose $y$ is not incompressible. Let's pick an incompressible $x$ of the same length; it should exist because most strings are incompressible. Then $K(x) > K(y)$.

However, it's undecidable if a string is incompressible.

sdcvvc
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