结合确定性有限自动机

Question

I'm really new to this stuff so I apologize for the noobishness here.

construct a Deterministic Finite Automaton DFA recognizing the following language:

L= { w : w has at least two a's and an odd number of b's}. 

The automate for each part of this (at least 2 a's, odd # of b's) are easy to make separately... Can anyone please explain a systematic way to combine them into one? Thanks.

Answer 1

You can use following simple steps to construct combined DFA.

Let Σ = {a ₁ , a ₂ , ...,a _k } .
1st step: Design DFA for both languages and name their state Q ₀ , Q ₁ , ...

2nd step : Rename every state in both DFA uniquely ie rename all states in DFA as Q ₀ , Q ₁ , Q ₂ , Q ₃ , ... assuming you have started with subscript 0; that means none of the state would have same name.

3rd Step: Construct transition table(δ) by using following steps

3a. Start state of the combined DFA:
Take start state of both DFAs(DFA1 and DFA2) and name them as Q _{[ i , j ]} where i and j are the subscript of start state of DFA1 and DFA2 respectively; ie Q _i is start state of 1st DFA and Q _j is start state of 2nd DFA and mark Q _{[i , j]} as start state of combined DFA.

3b. Map state of both DFAs as
if δ(Q _i ,a _k ) = Q _p1 and δ(Q _j ,a _k ) = Q _p2 , where Q _p1 belongs to DFA1 and Q _p2 belongs to DFA2 then δ(Q _{[ i , j ]} , a _k ) = Q _[p1,p2]

3c . fill entire table while there is any Q _[i,j] remaining in transition table.

3d . Final state of the combined DFA:
For AND case final state would be all Q _{[i , j]} where Q _i and Q _j are final state of DFA1 and DFA2 respectively.
For OR case final state would be all Q _{[i , j]} where either Q _i or Q _j is the final state of DFA1 and DFA2.

4th step: Rename all Q _{[i, j]} (uniquely) and draw DFA this will be your result.

Example:

L= {w: w has at least two a's and an odd number of b's}.

Step1:
DFA for odd number of b's .

DFA for at least 2 a's.

Step2:
Rename the stae of DFA1

Step3(a,b,c):
Constructed transition table will be as.

Step3d:
Since we have to take AND of both DFA so final state would be Q _[2,4] , since it contains final state of both DFA .
If we have to take OR of both DFA the final state would be Q _[0,4] ,Q _[2,3] ,Q _[1,4] ,Q _[2,4] .
Transition table would like this after adding final state .

Step4:
Rename all states Q _[i,j]
Q _[0,3] to Q ₀
Q _[1,3] to Q ₂
Q _[0,4] to Q ₁
Q _[2,3] to Q ₄
Q _[1,4] to Q ₃
Q _[2,4] to Q ₅
So final DFA would will look like as below .

Answer 2

它是使用两个自动机的产品完成的。

Answer 3

The Language L where a are at-least two and b are odd is an regular language. Its DFA is as below:

In this DFA I have combined two DFS s conceptually!

DFA-1 = for odd number of `b`'s (placed vertically three times in diagram)
DFA-2 = for >=  two a           (placed Horizontally two times in diagram)

_{The DFA is too symptomatic and simple so I believe no need in word that how to combine both DFAs}

To draw this DFA you are always keep track how many b s has been come either even or odd. States 0, 2 and 4 means even number of b has been come. So you can divide this DFA in two parts vertically where bottom states at even b s and upper states at odd.

Also String is accepted if odd b hence final state should be in one of state in upper part.

not only number of b s is condition but a should be atleast 2 . So you can divide this DFA horizontally in three parts where number of a s are 0 at state-0 and 1 , a s are one at state-2 and 3 and a s are 2 at state-4 and 5 . After first two a s any number of a s are allow in string so there is self loop on state q4 and q5 .

_{number of state required is six because, 2 state for odd even b and as hould be atleast 2 so 3 states a=0, a=1, a=2, hence 2*3 = 6}