2 개 언어 크기의 교차로 크기가 아닙니다.

Question

다음 언어가 알지 못할 필요가 없다고 생각하지만 그것을 보여주기 위해 감소를 생각할 수 없습니다. 어떤 힌트 나 직관을 감상하겠습니다

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

Answer 1

문제 $ \ langle m, x \ rangle $ 의 인스턴스를 고려하십시오. $ m_x $ 을 $ 0 $ , $ 00 $ , $ 000 \ ldots 0 ^ {2020} $ $ m $ $ x $ 입력에서 $ 중지; 그리고 그것은 다른 모든 문자열을 거부 할 것입니다. $ m_u $ 모든 문자열을 받아들이는 TM이되도록하십시오.

$ | l (m_x) \ cap l (m_u) | $ 은 $ m이면 2020이됩니다. $ $ x $ , 그렇지 않으면 $ 0 $ .

증거는 쉽게 뒤집습니다.

(비공식적 인) 힌트 : 그러한 질문이있을 때마다 주어진 언어의 가상의 Decider를 사용하여 문제를 해결할 수있는 방법 (또는 다른 미확인 언어)을 어떻게 해결할 수 있습니까?

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.