In their book Fundamentals of Parameterized Complexity, Downey and Fellows claim (in chapter 27.1) that $\mathrm{FPT}\subsetneq \mathrm{XP}$, and that this is a "basic result" that follows by "standard diagonalization", without any further reference.

I have neither been able to find a proof of this "basic result" in the literature (although many state it as a basic fact without reference), nor am I familiar enough with the "standard diagonalization" technique to easily produce a proof on my own.

Is there a full proof or more detailed sketch available in the literature? Alternatively, can you provide such a proof or sketch here? I suppose this diagonalization technique would be a basic tool in complexity theory, so looking near that field might be useful.


1 Answer 1


By the time hierarchy theorem, there is a family of languages $(L_k)_{k \in \mathbb{N}}$ such that $L_k$ is decidable in time $O(n^{2^{k+1}})$, but not in time $O(n^{2^k})$. We define $\mathcal{L} = \{\langle k, w\rangle \mid w \in L_k\}$, and use $k$ as the parameter.

Then $(\mathcal{L}, k)$ is in $\mathrm{XP}$, as every fixed slice of $\mathcal{L}$ is just some language $L_k$, which by our choice belongs to $\mathrm{P}$. However, if $(\mathcal{L},k)$ were fixed-parameter tractable, it would have some $O(f(k)\cdot n^\ell)$-decision process. But by fixing the parameter $k$ to be $\ell$, we'd get a $O(n^k)$ algorithm for $L_k$, in contradiction to our choice of $L_k$.

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    $\begingroup$ This is a very clean proof, thanks. I suppose Downey and Fellows had a proof in mind that would directly use diagonalization, rather than hiding it in the time hierarchy theorem. However, I think I would prefer this approach. It seems more elegant. $\endgroup$
    – Discrete lizard
    Commented Aug 22, 2021 at 12:03
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    $\begingroup$ @Discretelizard Actually, I'd consider this a diagonalization argument in a broad sense; specifically the fact that we stitch together $\mathcal{L}$ from more and more complex languages inside $\mathrm{P}$. My guess would that this is more or less what Downey and Fellows had in mind. $\endgroup$
    – Arno
    Commented Aug 22, 2021 at 12:10
  • $\begingroup$ Hmm, yes, I suppose that's true as well. I think your guess would be right. $\endgroup$
    – Discrete lizard
    Commented Aug 22, 2021 at 12:11

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