[英]Is the language {⟨A⟩∣A is an NFA and L(A)={0,1}∗} decidable? decidable?
How would one go about proving/disproving the language {⟨A⟩∣A is an NFA and L(A)={0,1}∗} is/isn't decidable?如何证明/反驳语言 {⟨A⟩∣A 是 NFA 并且 L(A)={0,1}∗} 是/不可判定的?
I assumed at first since it was an NFA involved it would be decidable, but since there is no input string to simulate does this change things?我一开始认为因为它涉及 NFA,所以它是可判定的,但是由于没有输入字符串来模拟,这会改变事情吗? If so, how?如果是这样,如何? I can't think of a turing machine that would decide this.我想不出能决定这个的图灵机。 Since {0, 1}* is theoretically infinite does mean a turing machine may never halt thus the language is undecidable?由于 {0, 1}* 理论上是无限的,这是否意味着图灵机可能永远不会停止,因此语言是不可判定的? If so how do I go about proving this?如果是这样,我该如何证明这一点?
Speaking informally, you can show this by constructing a Turing Machine constructs a DFA D_A equivalent to NFA A. It then constructs the DFA D_0 that accepts the language {0,1}*, we can then simulate the decider for EQ_DFA on .非正式地说,您可以通过构造图灵机来证明这一点,它构造了一个等效于 NFA A 的 DFA D_A。然后构造接受语言 {0,1}* 的 DFA D_0,然后我们可以模拟 EQ_DFA 上的决策器。
Formally speaking, construct TM S: S = "On input :正式地说,构造 TM S: S = "On input :
Less formally:不那么正式:
Because we can describe an algorithm to do this, and because we don't suppose we have more computational power than Turing machines (at least the above computations do not require such), the problem must be decidable.因为我们可以描述一个算法来做到这一点,并且因为我们不认为我们比图灵机拥有更多的计算能力(至少上面的计算不需要这样),所以问题必须是可判定的。
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