✎ lfa:2023:lab05 [books]

This page is read only. You can view the source, but not change it. Ask your administrator if you think this is wrong.
====== 5. Minimal DFAs ======

Consider the following DFAs: 

^ **DFA 1** ^ **DFA 2** ^
|{{ :lfa:2022:lfa2022_lab5_ex2_4_cerinta.png?350 |}} | {{ :lfa:screenshot_2021-11-03_at_11.25.00.png?400 |}} |

<note important>The first DFA is the solution to exercise 4.3.4 from Lab4.</note>

===== 5.1. Equivalence between states =====

5.1.1. Identify a pair of states which are **indistinguishable**. (Solve the exercise for the 2 given DFAs.)


<hidden DFA 1>
<note tip>
** DFA 1**

States 2 and 5 are indistinguishable.

Looking at the DFA we can see that states 2 and 5 both go to state 6 on "1" and to 3 on "0".
</note>
</hidden>

<hidden DFA 2>
<note tip>
** DFA 2**

States 1 and 2 are indistinguishable.

Looking at the DFA we can see that states 1 and 2 both go to state 2 on "a" and to 5 on "b".
</note>
</hidden>


5.1.2. Identify a pair of final or non-final states which are **distinguishable**. The pair must be distinguished by a word different from the empty word. (Solve the exercise for the 2 given DFAs.)


<hidden DFA 1>
<note tip>** DFA 1**

States 4 and 5 are distinguishable because we can accept "1" from 4, but not from 5.</note>
</hidden>

<hidden DFA 2>
<note tip>** DFA 2**

States 0 and 5 are distinguishable because from 0 we can accept word "b", but not from 5.</note>
</hidden>



5.1.3. Compute the table of indistinguishable states for the DFA. (Solve the exercise for the 2 given DFAs.)


<hidden DFA 1><note tip>
** DFA 1 **

^      ^1 ^2 ^3 ^4 ^5 ^6
|**1** | -| -| -| -| -| -
|**2** | X| -| -| -| -| -
|**3** |  | X| -| -| -| -
|**4** |  | X|  | -| -| -
|**5** | X|  | X| X| -| -
|**6** | X| X| X| X| X| -

Initially marked pairs of (final, non-final) states:
  * (1, 6)
  * (2, 6)
  * (3, 6)
  * (4, 6)
  * (5, 6)

Predecessors for (6, 1):
  * 0: -
  * 1: -

Predecessors for (6, 2):
  * 0: -
  * 1: (2, 1), (5, 1) => mark them


Predecessors for (6, 3):
  * 0: -
  * 1: -

Predecessors for (6, 4):
  * 0: -
  * 1: -

Predecessors for (6, 5):
  * 0: -
  * 1: (2, 3), (5, 3), (2, 4), (5, 4) => mark them

Predecessors for (2, 1):
  * 0: -
  * 1: -

Predecessors for (5, 1):
  * 0: -
  * 1: -

Predecessors for (3, 2):
  * 0: -
  * 1: -

Predecessors for (4, 2):
  * 0: -
  * 1: -

Predecessors for (5, 3):
  * 0: -
  * 1: -

Predecessors for (5, 4):
  * 0: -
  * 1: -

The partitions of undistinguishable states are given by the missing marks:
  * (5, 2)
  * (3, 1), (4, 3), (4, 1)

=> The partitions are: {2, 5} and {1, 3, 4}

</note>
</hidden>

<hidden DFA 2>
<note tip>
** DFA 2 **

^      ^0 ^1 ^2 ^3 ^4 ^5
|**0** | -| -| -| -| -| -
|**1** | X| -| -| -| -| -
|**2** | X|  | -| -| -| -
|**3** |  | X| X| -| -| -
|**4** | X|  |  | X| -| -
|**5** | X| X| X| X| X| -

Initially marked pairs of (final, non-final) states: 
  * (0, 1)
  * (0, 2)
  * (0, 4)
  * (3, 1)
  * (3, 2)
  * (3, 4)
  * (5, 1)
  * (5, 2)
  * (5, 4)

Predecessors for (5, 4):
  * a: -
  * b: (3, 4), (3, 1), (3, 2), (3, 5) => mark them

Predecessors for (5, 1):
  * a: -
  * b: (5, 0) => mark it

Predecessors for (3, 1):
  * a: -
  * b: -

Predecessors for (3, 2): 
  * a: (0, 1), (0, 2), (0, 4) => already marked
  * b: -

Predecessors for (5, 2): 
  * a: (5, 1), (5, 2), (5, 4) => already marked
  * b: -

Predecessors for (3, 4):
  * a: -
  * b: -

Predecessors for (0, 1):
  * a: -
  * b: -

Predecessors for (0, 2):
  * a: (3, 1), (3, 2), (3, 4) => already marked
  * b: -

Predecessors for (0, 4):
  * a: -
  * b: -

Predecessors for (0, 5):
  * a: (3, 5) => mark it
  * b: -

Predecessors for (3, 5):
  * a: (0, 5) => already marked
  * b: -

The partitions of undistinguishable states are given by the missing marks:
  * (0, 3)
  * (4, 1), (4, 2), (2, 1)

=> The partitions are: {0, 3} and {1, 2, 4}
</note>
</hidden>


===== 5.2. Minimisation ======

5.2.1. Minimise the DFA. (Solve the exercise for the 2 given DFAs.)


<hidden DFA 1>
** DFA 1 **

{{ :lfa:2022:lfa2022_lab5_dfa1_min.png?400 |}}
</hidden>

<hidden DFA 2>
** DFA 2 **

{{ :lfa:2022:lfa2022_lab5_dfa2_min.png?400 |}}
</hidden>


5.2.2. How can we algorithmically determine if two minimal DFAs accept the same language?


<hidden><note tip>
  * assume the states are labeled as integers starting with 0, and that 0 is the unique initial state of both DFAs (this can be easily obtained from any previous transformation).
  * give an ordering to the symbols of the alphabet. Write $math[\delta] as a string by the following rules:
    * we sort each transition $math[\delta(q,c)=q'], by $math[q] first and by $math[c], second. We append each transition given this ordering (//smaller// states first, then for each state, output the order of the $math[\delta(q,c)] transitions according to $math[c].
  * two minimal DFAs accept the same language **iff**:
    * they have the same string representation of $math[\delta] and
    * have the same final states.

Another more elegant approach is to use state equivalence. We can treat two (not necessarily minimal) DFAs as one and the same, and compute the (in)distinguishability table over all states, in a single go. Finally, the DFAs accept the same language iff their initial states are indistinguishable.
</note></hidden>




5.2.3. Determine if the following regexes are **equivalent**: $math[(1\cup\epsilon)(00^*1)^*0^*] and $math[(10\cup 0)^*(01 \cup 1)^*(0 \cup \epsilon)]


<hidden>
<note tip>The two languages are not equivalent because of word "110" which is accepted by the second regex, but rejected by the first regex.</note>
</hidden>


5.2.4. Remember exercise 3.3.4 from Lab3. Show that the two regexes are equivalent: $ E1 = ((ab^*a)^+b)^* $ and $ E2 = (a(b\mid aa)^*ab)^* $


<hidden>
<note tip>
There are several methods to do this.

1. The first way will be presented in the next seminar. Using closure properties we will prove both the following properties:
  * L(E1) ∩ L(E2) = ∅
  * L(complement(E1)) ∩ L(complement(E2)) = ∅

2. Check that the initial state of minDfa(E1) is indistinguishable from the initial state of minDfa(E2). 

We will solve 2, but we will assume we have already computed minDfa for the 2 regular expressions and got the following DFAs as result:


{{ :lfa:2022:lfa2022_lab5_ex2_4.png?500 |}}


**Obs**: We can tell they are the same by looking at them but the computer needs an algorithm to check this equivalence.

^      ^0 ^1 ^2 ^3 ^4 ^5
|**0** | -| -| -| -| -| -
|**1** | X| -| -| -| -| -
|**2** | X| X| -| -| -| -
|**3** |  | X| X| -| -| -
|**4** | X|  | X| X| -| -
|**5** | X| X|  | X| X| -

Initially marked pairs of (final, non-final) states: 
  * (0, 1)
  * (0, 2)
  * (0, 4)
  * (0, 5)
  * (3, 1)
  * (3, 2)
  * (3, 4)
  * (3, 5)

Predecessors for (0, 1):
  * a: -
  * b: -

Predecessors for (0, 2):
  * a: -
  * b: (1, 2) => mark it

Predecessors for (0, 4):
  * a: -
  * b: -

Predecessors for (0, 5):
  * a: -
  * b: (1, 5) => mark it

Predecessors for (2, 1):
  * a: (1, 2), (0, 2) => already marked
  * b: -

Predecessors for (3, 1):
  * a: -
  * b: -

Predecessors for (5, 1):
  * a: (3, 2), (4, 2) => mark them
  * b: -

Predecessors for (3, 2):
  * a: -
  * b: (4, 2) => already marked

Predecessors for (4, 2):
  * a: (5, 1), (5, 0) => already marked
  * b: -

Predecessors for (4, 3):
  * a: -
  * b: -

Predecessors for (5, 3):
  * a: -
  * b: (5, 4) => mark it

Predecessors for (5, 4):
  * a: (3, 5), (4, 5) => already marked
  * b: -

So, we get 3 sets of indistinguishable states: {0, 3}, {1, 4} and {2, 5}.

Because states 0 and 3, which are initial states in the 2 DFAs, we can conclude that the 2 DFAs accept the same language.

</note></hidden>




5.2.5. (extra) Apply the **Brzozowski** minimisation algorithm on DFA 1.



<note important>Brzozowski minimisation: **minDfa(A) = toDfa(reverse(toDfa(reverse(A))))**</note>


<hidden>
<note tip>
Step 1: ** Reverse(A) **
  * reverse transitions
  * the initial state becomes final in Reverse(A)
  * the final state becomes initial in Reverse(A), but here we have more final states:
    * normally, to compute Reverse(A) we should add a new initial state in Reverse(A); the final states in A become non-final in Reverse(A); then add $\varepsilon$ - transitions from the new initial state of Reverse(A) to the former final states of A
    * for Brzozowski's minimization algorithm however, we will not add this new initial state and, instead, we will start the subset construction in Step 2 from the set of final states that were in A, that now become "multiple" initial states in Reverse(A)
  * we can also not include the sink state in the reversal of the DFA
  * the set of initial states is: {1, 2, 3, 4, 5}
</note>

{{ :lfa:2022:lfa2022_lab5_bmin_step1.png?400 |}}

<note tip>
Step 2: **toDfa(Reverse(A))**
  * rename the subsets as follows: {1,2,3,4,5} -> 1 and {1,3,4} -> 2
</note>

{{ :lfa:2022:lfa2022_lab5_bmin_step2.png?400 |}}


<note tip>
Step 3: **Reverse(toDfa(Reverse(A)))**
  * we can ignore the sink state
  * as previously, we remember a set of initial states: {1,2}
</note>

{{ :lfa:2022:lfa2022_lab5_bmin_step3.png?400 |}}

<note tip>
Step 4: **toDfa(Reverse(toDfa(Reverse(A))))**
  * this is minDfa(A)
</note>

{{ :lfa:2022:lfa2022_lab5_bmin_step4.png?400 |}}

</hidden>


===== Conclusion =====

<note important>
Now we can do a complete transformation of a regex into a minimal DFA.

  * Regex -> NFA (Thomson's algorithm)
  * NFA -> DFA (Subset Construction)
  * DFA -> minDfa

**To get a better performance, include minDfa conversion in your project.**
</note>