[英]Let T = {<M> | M is a TM that accepts $w^R$ whenever it accepts w}. Show that T is undecidable
Let T = {<M> |
设T = {<M> | M is a TM that accepts w r whenever it accepts w}.
M是一个接受W R每当它接受瓦特}一个TM。
Show that T is undecidable.表明T是不可判定的。
I have two answers to this question - San Diego : 我对这个问题有两个答案 - 圣地亚哥 :
5.9
5.9
Let T = { <M> |设T = {<M> | M is a TM that accepts w r whenever it accepts w }.
M是一个接受W R每当它接受瓦特}一个TM。
Assume T is decidable and let decider R decide T. Reduce from A TM by constructing a TM S as follows:
假设T是可判定的并且让决策者R决定T.通过如下构造TM S从A TM减少:
- S: on input <M,w>
S:输入<M,w>
- create a TM Q as follows:
按如下方式创建TM Q:
On input x:输入x:
- if x does not have the form 01 or 10 reject.
如果x没有01或10表格拒绝。
- if x has the form 01, then accept.
如果x的形式为01,则接受。
- else (x has the form 10), Run M on w and accept if M accepts w.
else(x的格式为10),在w上运行M并接受M是否接受w。
- Run R on
运行R开启
- Accept if R accepts, reject if R rejects.
如果R接受则接受,如果R拒绝则拒绝。
Because S decides A TM , which is known to be undecidable, we then know that T is not decidable
因为S决定已知不可判定的A TM ,我们就知道T不可判定
Undisclosed source: 未披露的消息来源:
5.12 We show that A TM ≤ m S by mapping ‹ M , w › to ‹ M' › where M' is the following TM:
5.12我们表明,通过映射A TM≤m S 后 <M,W>至<M”>其中M'是以下TM:
- M' = “On input x :
M' =“输入x :
- If x = 01 then accept .
如果x = 01则接受 。
- If x ≠ 10 then reject .
如果x ≠10则拒绝 。
- If x = 10 simulate M on w .
如果x = 10则模拟W上的 M.
If M accepts w then accept ;如果M接受w则接受 ; if M halts and rejects then reject .”
如果M停止并拒绝,那么拒绝 。“
If ‹ M , w › ∈ A TM then M accepts w and L ( M' ) = {01,10}, so ‹ M' › ∈ S .
如果<M,W>∈A TM然后M接受w和L(M ')= {01,10},因此<M'>∈S上 。
Conversely, if ‹ M , w › ∉ A TM then L ( M' ) = {01}, so ‹ M' › ∉ S .相反,如果<M,W>∉A TM则L(M ')= {01},因此<M'>∉S上 。 Therefore,
因此,
‹ M , w › ∈ A TM ⇔ ‹ M' › ∈ S .<M,W>∈A TM⇔<M”>∈S上 。
But I do not understand the following: 但我不明白以下几点:
1- what is the relation between x and w? 1- x和w之间的关系是什么?
2- why we consider the 2 cases ‹ M , w › ∈ A TM and ‹ M , w › ∉ A TM ? 2-为什么我们考虑2案件<M,W>∈A TM和<M,W>∉A TM?
3- why if A is mapping reducible to S this makes S undecidable? 3-为什么如果A映射可简化为S这会使S不可判定?
could anyone clarify these points for me? 谁能为我澄清这些要点?
I think it is not suitable for asking in SO because it is not a educational website, but I answered it. 我认为它不适合在SO中询问,因为它不是一个教育网站,但我回答了它。
1- what is the relation between x and w? 1- x和w之间的关系是什么?
Answer 1: x is a symbol that used for using a symbol for operate.
答案1: x是用于使用符号进行操作的符号。 This symbol should not be in alphabet of language, just it.
这个符号不应该是语言的字母,只是它。 It hasn't any relation to w.
它与w没有任何关系。
2- why we consider the 2 cases ‹M, w› ∈ ATM and ‹M, w› ∉ ATM? 2-为什么我们考虑2个案例<M,w>∈ATM和<M,w>∉ATM?
Answer 2: For proofing a language like L is decidable or not, we need to determine a string like w is member of language or not.
答案2:为了证明像L这样的语言是可判定的,我们需要确定一个像w是语言成员的字符串。 So we have to consider two type of string w∉L and w∈L.
所以我们必须考虑两种类型的字符串w∉L和w∈L。
3- why if A is mapping reducible to S this makes S undecidable? 3-为什么如果A映射可简化为S这会使S不可判定?
Answer 3: It means the process of checking a string is in language in A and S is similar and if we can't find a algorithm for checking this for A, we can't find any algorithm for S.
答案3:这意味着检查字符串的过程是用A语言和S语言是相似的,如果我们找不到用于检查A的算法,我们找不到任何S的算法。
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