[英]Nondeterministic finite automata (NFA) correction
I am trying to solve a question about NFA. 我正在尝试解决有关NFA的问题。 The instructions are as following: Alphabet {a, b, c}. 说明如下:字母{a,b,c}。 • L1 is all strings that their last character is the same as their fifth-last character. •L1是所有最后一个字符与第五个最后一个字符相同的字符串。 For instance, the string aaacbacbca should be accepted, because the fifth-last character and the last character are both a. 例如,字符串aaacbacbca应该被接受,因为倒数第二个字符和最后一个字符都是a。 The string ccaab should be rejected because the fifth-last character is c and the last character is b. 字符串ccaab应该被拒绝,因为倒数第二个字符是c,最后一个字符是b。 Here is with what I came up, but I am really beginner in this topic and I am not sure whether is correct or not: 这是我的想法,但我确实是该主题的初学者,并且不确定是否正确:
The automata you have at the moment accepts only the strings that end with acbca
. 您目前拥有的自动机仅接受以acbca
结尾的字符串。 Here are steps to arrive to the solution: 以下是到达解决方案的步骤:
a
更改您拥有的自动机,使其可以接受所有具有倒数和倒数第五个符号a
字符串 b
, c
对符号b
, c
You were almost right, but the automata you drew was only to accept strings ending in acbca
. 您几乎是正确的,但是您绘制的自动机只接受以acbca
结尾的acbca
。 This one will accept the strings you want 这将接受您想要的字符串
a,b,c a a,b,c a,b,c a,b,c a a,b,c
,--->[q0]--->[q1]--->[q2]--->[q3]--->[q4]--->{q5}----+>[q16]-----.
| /| b a,b,c a,b,c a,b,c b a,b,c| ^ a,b,c |
`---´ +----->[q6]--->[q7]--->[q8]--->[q9]--->{q10}---+ `-------´
| c a,b,c a,b,c a,b,c c a,b,c|
`----->[q11]-->[q12]-->[q13]-->[q14]-->{q15}---´
States like {q5}
are accepting states, while states like [q0]
are not accepting. 像{q5}
这样的状态正在接受状态,而像[q0]
这样的状态则不在接受状态。 The meaning of q16
is to make sure that strings that have two equal letters distant 4 characters but not ending there, sink to a non accepting state. q16
的含义是确保两个相等的字母的字符串相距4个字符,但不以该字符结尾,该字符串陷入非接受状态。 Same might be for letters b,c
in state [q4]
, for letters a,c
in state [q9]
, and for letters a,b
in state [q14]
, but for clarity of the drawing, I have omitted them. 同样可能是字母b,c
状态[q4]
对于字母a,c
状态[q9]
并为字母a,b
状态[q14]
但为了绘图清楚起见,我省略了它们。
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