Taille de l'intersection de la taille de 2 langues n'est pas décembre

Question

Je pense que la langue suivante n'est pas décembre, mais je ne peux pas penser à la réduction pour le montrer. J'apprécierais un indice ou une intuition

$ eq= $ {< $ $ | $ m1 \, \, \, \, \, \, m2 \, \, \, sont \, \, \, \, tms \, \, \, \, \, \, \, \, \, \, \, \,L (m1) \, ∩ \, l (m2) \, | \, $$= 2020 \, $ }

Answer 1

considérer une instance de problème d'arrêt $ \ utilmm m, x \ rangs $ . Construisez une nouvelle machine Turing qui $ m_x $ qui acceptera les chaînes $ 0 $ , $ 00 $ , $ 000 \ ldots 0 ^ {2020} $ Si la machine M $ M $ s'arrête sur l'entrée $ x $ ; Et cela rejette toutes les autres cordes. Laissez $ m_u $ Soyez le TM qui accepte toutes les chaînes.

Convaincuez-vous que $ | L (M_X) \ CAP L (M_U) | $ sera 2020 si M $ $ s'arrête sur $ x $ , et sinon $ 0 $

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La preuve suit facilement.

(informel) indice: chaque fois que vous avez une telle question, essayez de penser à la manière dont vous pouvez résoudre le problème d'arrêt (ou un autre langage indéfectable) à l'aide de l'hypothétique décider pour la langue donnée.

Answer 2

welcome to the site :)

I assume you already know that the Halting problem $L_H$ is undecidable (as it is usually the first specific undecidable language students learn of). So, let us try and find a reduction $L_H \leq EQ$.

We want to know whether some given TM $M$ halts given the input word $w$ by transforming the pair $\langle M, w \rangle$ to some $EQ$-instance $f(\langle M, w \rangle) = \langle M_1, M_2 \rangle$ such that $\langle M, w \rangle \in L_H$ if and only if $\langle M_1, M_2 \rangle \in EQ$. Let us consider some dummy TM $M_{2020}$ which just accepts 2020 inputs (any set of 2020 words will do the trick) and modify $M_{2020}$ such that it only ever accepts if $M$ halts on $w$ to get another TM $M_{2020}(M, w)$. Since we know $M$ and $w$ when we construct $M_{2020}(M, w)$, we can implement it to do the following:

1. Read the input and if it is one of the 2020 words accepted by $M_{2020}$, make note of that (we can use the states to store this).
2. Clear the tape and write $w$ to it.
3. Run $M$ on $w$.
4. If $M$ halts on $w$, accept the original input.

Such a construction can be performed by another TM, this is a bit tedious to show formally but by noting that we can essentially "embed" $M$ in $M_{2020}(M, w)$ it should be intuitive that this can be done (as $M_{2020}$ is a fixed TM).

Now observe that $M_{2020}(M, w)$ accepts some input $x \in L(M_{2020})$ precisely when $M$ halts on $w$, hence we have that $L(M_{2020}) = L(M_{2020}(M, w))$ if and only if $\langle M, w \rangle \in L_H$, otherwise we have $L(M_{2020}(M, w)) = \emptyset$. It follows that $$ | L(M_{2020}(M, w)) \cap L(M_{2020}) | = \begin{cases} 2020, & \text{if $M$ halts on $w$} \\ 0, & \text{otherwise} \end{cases} $$ and thus (denoting $\langle M_{2020}, M_{2020}(M, w) \rangle$ as $f(\langle M, w \rangle$) we have $\langle M, w \rangle \in L_H \Leftrightarrow f(\langle M, w \rangle)$ and therefore a suitable reduction.

An easier way to prove that the language is not decidable is by using Rice's theorem which (roughly speaking) says that properties of functions computed by TMs (we call these semantic properties) cannot be decided algorithmically unless they are trivial. If we fix some TM $M_1$ and consider an arbitrary TM $M$ then asking whether $L(M_1) \cap L(M)$ has size 2020 is such a semantic property and hence Rice's theorem immediately gives us that $EQ$ is undecidable. The theorem (for me, at least) is basically the formal underpinning of the informal intuition I gave above.