构建一个仅接受单词 baa、ab 和 abb 且不接受其他更长或更短字符串的 FA

Question

I have been trying solve this problem for a while now for a university assignment. I'm required to build a DFA and an NFA for the above question. So far I have been able to solve the DFA, but can not find a solution for a proper NFA after multiple tries.

The solution of my DFA for the language provided above:

My attempts for the NFA are down below. I apologize for my messy handwriting but these were just rough works that I was drawing out on the go.

My first attempt:

My second attempt:

My third attempt:

Answer 1

There's only three words, so just make three parallel paths for your NFA, using a transition function like the following:

Input State	Input Symbol	Output States
[start]	a	[ab], [a.bb]
[start]	b	[b.aa]
[ab]	b	[ab#]
[a.bb]	b	[ab.b]
[ab.b]	b	[abb#]
[b.aa]	a	[ba.a]
[ba.a]	a	[baa#]

Here, state names are in brackets ([..]) and state names that end with "#" are terminal.

Generally it's considered easier to make an NFA than a DFA, so the usual method is to first make an NFA and then make the DFA by modifying the NFA by changing multiple output states to a single intermediate state.

If you followed this method for the above NFA, then the resulting DFA would look something like this (I have appended an "*" to the intermediate state names):

Input State	Input Symbol	Output State
[start]	a	[ab*]
[start]	b	[b.aa]
[ab*]	b	[ab.*]
[ab.*]	(e)	[ab#]
[ab.*]	b	[abb.*]
[abb.*]	(e)	[abb#]
[b.aa]	a	[ba.a]
[ba.a]	a	[baa#]

I've been a bit loose about all of the empty symbol/end-of-input transitons to terminal states. If you need me to fill al of them in, them I can do that.