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无法在输入上写入的固定大小磁带图灵机等价于 DFA

[英]Fixed size tape Turing machine which cannot write on input is equivalent to DFA

原文 2019-01-03 16:39:54 2 1 computer-science/ automata/ formal-languages

I have to prove that Turing machine with fixed size tape which cannot write on input is equivalent to finite automaton (DFA or NFA).我必须证明带有无法在输入上写入的固定大小磁带的图灵机等效于有限自动机（DFA 或 NFA）。

It's important to add that the size of the tape is the size of the tape which excludes the input.重要的是要补充一点，磁带的大小是不包括输入的磁带大小。 For example, if the input's size is n so the tape's size would be k+n, where k is the length of the tape that excludes the input.例如，如果输入的大小为 n，则磁带的大小将为 k+n，其中 k 是不包括输入的磁带的长度。

I understand the main idea but it is super hard to prove it.我理解主要思想，但要证明它是非常困难的。 Thanks in advance.提前致谢。

1 个解决方案

We can see that you can emulate a DFA on such a Turing Machine - the Turing Machine just has read-only states and consumes one character of input on each step - essentially implementing a DFA on a Turing Machine.我们可以看到，您可以在这样的图灵机上模拟 DFA——图灵机只有只读状态并在每一步消耗一个字符的输入——本质上是在图灵机上实现 DFA。 This is the easy direction.这是容易的方向。

Showing that you can emulate the TM on the DFA is a bit harder, but comes down to the fact that there are only k possible places to write m characters, where m is the size of the writing alphabet of the machine.证明你可以在 DFA 上模拟 TM 有点困难，但归结为只有k可能的地方可以写m字符，其中m是机器的书写字母表的大小。 Therefore your TM has only k^m possible tape states in addition to however many states the machine has, which we'll call n .因此，除了机器具有的许多状态之外，您的 TM 只有k^m可能的磁带状态，我们将其称为n 。 So a DFA with n*k^m states can cover the TM's states.因此，具有n*k^m个状态的 DFA 可以覆盖 TM 的状态。