2种语言尺寸的交叉量的大小不可解除

Question

我认为下一个语言不是可判定的，但我无法想到减少才能显示它。 我希望一些提示或直觉

$ eq= $ { $ $ | $ M1 \，\，\和\，\，\，m2 \，\，\，\，\，\，tms \，\，\，\，\，\，\，\，\，\，\，| \，l（m1）\，∩\，l（m2）\，| \，$$= 2020 \，$ }

Answer 1

考虑一个停止问题的实例 $ \ langle m，x \ rangle $ 。构建一个 $ m_x $ 的新图形机器，它将接受字符串 $ 0 $ ， $ 00 $ ， $ 000 \ ldots 0 ^ {2020} $ 如果机器 $ m $ 停止输入 $ x $ ;它将拒绝所有其他字符串。让 $ m_u $ 是接受所有字符串的tm。

说服自己 $ | l（m_x）\ cap l（m_u）| $ 将是2020如果 $ m $ 停止 $ x $ ，否则 $ 0 $ 。

证明随之而来。

（非正式）提示：每当您有这样的问题时，请尝试考虑使用给定语言的假设求解器如何解决停止问题（或其他一些未定语言）。

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.