We say that two languages $L_1,L_2$ are decreasing reducible if there exists a polynomial time reduction $f:\Sigma^*\to\Sigma^* $ and there exists $n\in\mathbb{N}$ such that for every $x\in\Sigma^*$ satisfying $|x|\ge n \implies |f(x)|\lt |x|$.

Assuming $P\ne NP$

Prove\Disprove: Every two NP-complete languages $L_1,L_2$ are decreasing reducible.

I'd appreciate a hint or direction

  • $\begingroup$ What have you tried? Can you explain the problem in your own words? $\endgroup$
    – Pål GD
    Jun 9, 2019 at 15:57
  • $\begingroup$ @PålGD I know every two NP-complete languages are polynomial reducible. Choosing some two NP-complete languages, I'd like to show they can not be decreasing reducible, i.e. for large enough $x$ the length of the image of $x$ decreases for every $x$. $\endgroup$
    – Um Shmum
    Jun 9, 2019 at 16:14

1 Answer 1


Let $L$ be any NP-complete language, and consider what happens when $L_1 = L_2 = L$. Given an instance $x$, by applying $f$ a linear number of times we would get that $x \in L$ iff $y \in L$, where $|y| < n$. Since there are only finitely many strings of length smaller than $n$, we can hardcode the correct answer for these strings, thus obtaining a polynomial time algorithm for $L$.


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