MINIMIZATION OF DFA WITH EXAMPLE IN AUTOMATA THEORY || DFA MINIMIZATION || TOC

Summary

In this educational video, the presenter delves into the concept of DFA minimization in automata theory. Starting with a five-state DFA, the tutorial guides viewers through constructing a transition table and then systematically reduces the number of states by identifying and merging equivalent states. The process involves partitioning states into non-final and final sets, followed by iteratively refining these sets based on state equivalence. The presenter clearly explains how to determine state equivalence using input symbols and the resultant states, ultimately demonstrating how to minimize the DFA to a more efficient four-state model. The video is designed to help viewers understand the practical steps of DFA minimization and is complemented by a clear visual example.

Takeaways

• 😀 The video discusses the minimization of Deterministic Finite Automata (DFA), a topic in automata theory.
• 🔍 It covers the construction of DFA and the conversion of Non-deterministic Finite Automata (NFA) to DFA, which is necessary for implementation.
• 📋 The presenter uses a transition table to illustrate the process, with input symbols 0 and 1, and states q0, q1, q2, q3, and q4.
• 🤔 The minimization process begins by dividing states into non-final and final state sets, identifying initial equivalences.
• 🔄 The video explains how to find state equivalences by comparing transitions for each input symbol and grouping equivalent states.
• 📉 The process of finding equivalences continues iteratively, refining sets until consecutive iterations yield the same results.
• 🚫 The script emphasizes that states with different transitions for the same input symbol are not equivalent and must be separated.
• 🔢 The example given reduces the number of states in a DFA from five to four by identifying non-equivalent states.
• 🔧 The video demonstrates how to create a new, minimized DFA using the equivalence classes found during the minimization process.
• 🔗 The presenter suggests renaming states for clarity in the minimized DFA, proposing labels a, b, c, and d instead of q0, q1, q2, q3, and q4.

Q & A

What is the main topic discussed in the video?

The main topic discussed in the video is the minimization of Deterministic Finite Automata (DFA).

Why is it necessary to convert NFA to DFA before implementation?

It is necessary to convert NFA to DFA before implementation because NFA can have multiple transitions for the same input symbol from the same state, which is not feasible for implementation, whereas DFA has a single transition for each input symbol from each state.

How many states are initially present in the DFA example discussed in the video?

Initially, there are five states in the DFA example discussed in the video.

What are the input symbols used in the DFA example?

The input symbols used in the DFA example are 0 and 1.

What is meant by zero equivalence in the context of DFA minimization?

Zero equivalence in DFA minimization refers to the initial division of states into two sets: one containing non-final states and the other containing final states.

How are states determined to be equivalent during DFA minimization?

States are determined to be equivalent during DFA minimization by comparing their transitions for each input symbol and checking if they lead to states in the same set in the previous equivalence.

What is the process of finding equivalences among states called?

The process of finding equivalences among states during DFA minimization is called the partition refinement method or the table-filling algorithm.

What is the final number of states after minimization in the DFA example?

After minimization, the DFA example has four states instead of the initial five.

How does the video ensure that the minimized DFA is still complete?

The video ensures that the minimized DFA is complete by checking that there is a transition for every state over every input symbol.

Can the minimization process be applied to DFAs with multiple final states?

Yes, the minimization process can be applied to DFAs with multiple final states by following the same procedure of finding equivalences among states.

What is the purpose of renaming states in the minimized DFA?

The purpose of renaming states in the minimized DFA is to simplify the representation and make it easier to understand, often using letters like a, b, c, and d instead of q0, q1, q2, and q3.

Outlines

00:00

🤖 Introduction to DFA Minimization

The video begins by welcoming viewers to the channel and introducing the topic of DFA (Deterministic Finite Automaton) minimization within the field of automata theory. The presenter discusses previous topics covered, including the construction of DFAs and NFAs (Nondeterministic Finite Automata), and the conversion of NFAs to DFAs for practical implementation. The focus then shifts to the minimization of DFAs, aiming to reduce the number of states. An example DFA with five states is presented, and the process of creating a transition table for this DFA with input symbols 0 and 1 is outlined. The initial step in minimization is dividing the states into non-final and final state sets, with the aim of finding equivalences among the states to reduce their number.

05:02

🔍 Exploring Zero Equivalence in DFA Minimization

This segment delves into the process of finding zero equivalence in DFA minimization. The presenter explains how to compare states using input symbols to determine if they result in the same set of states, which would indicate equivalence. The comparison starts with the initial state 'q naught' against other states. Through examples, it's shown that 'q naught' is not equivalent to 'q one', 'q two', or 'q three' as their resultant states after input symbols do not fall into the same set. Consequently, new sets are created for non-equivalent states. The process is repeated for each state comparison, aiming to refine the sets based on equivalence.

10:02

🔄 Continuing Equivalence Testing for DFA Minimization

The video continues with the process of DFA minimization, focusing on finding two equivalences. The presenter compares states 'q one' and 'q two', and then 'q one' with 'q three', using the previously established equivalences to guide the comparison. It's determined that 'q one' is equivalent to 'q three' as their resultant states match in the current equivalence sets. However, 'q one' and 'q two' are not equivalent. The presenter emphasizes the importance of checking each state's transitions for all input symbols to ensure the sets are correctly divided based on equivalence. The process is iterative, aiming to refine the sets until no further changes occur.

15:05

🏁 Completing DFA Minimization and Final Thoughts

In the final part of the video, the presenter concludes the DFA minimization process by identifying the three equivalences. It's shown that 'q naught', 'q one', 'q three', 'q two', and 'q four' form distinct sets after the equivalences are determined. The DFA is then redrawn with the minimized number of states, reducing from five to four. The presenter explains how to update the transition table according to the new state sets and ensures that the minimized DFA is still complete, with transitions for every state and input symbol. The video wraps up with advice on renaming states for clarity and offers assistance for any doubts in the comments. The presenter also encourages viewers to like, share, and subscribe for more content.

Keywords

💡Automata Theory

Automata Theory is a branch of theoretical computer science and mathematics that deals with abstract machines and automata, as well as the computational processes that they can perform. In the context of the video, Automata Theory is the overarching theme, focusing on the minimization of Deterministic Finite Automata (DFA). The script discusses how to construct and minimize DFAs, which are essential concepts in this theory.

💡DFA (Deterministic Finite Automaton)

A DFA is a theoretical model of computation used to design and understand finite state machines. It is a type of automaton where each state has exactly one transition for each possible input symbol. The video script describes the process of minimizing a DFA, which involves reducing the number of states in the automaton without changing its functionality.

💡Minimization

Minimization in the context of DFA refers to the process of reducing the number of states in a DFA to the minimum required to represent the same language. This is important for optimizing the efficiency of state machines used in various applications. The script provides a step-by-step guide on how to minimize a DFA by finding equivalent states.

💡Transition Table

A transition table is a representation of a DFA that lists all possible transitions between states for each input symbol. In the script, the creation of a transition table is discussed as a preliminary step before beginning the minimization process. It is used to systematically compare states and determine their equivalence.

💡Equivalence

In the context of DFA minimization, equivalence refers to the property of two states being indistinguishable based on their behavior for all possible input sequences. The script explains how to determine equivalence by comparing the resulting states after applying input symbols, which is crucial for merging states during minimization.

💡Zero Equivalence

Zero equivalence is the initial partitioning of states into two sets: non-final states and final states, which serves as the starting point for the minimization process. The script uses zero equivalence to set up the comparison of states and begins the iterative process of finding equivalent states.

💡One Equivalence

One equivalence is the result of the first iteration of comparing states in the minimization process, where states are divided into sets based on whether they lead to the same resultant state for all input symbols. The script describes how one equivalence is calculated and how it leads to the creation of new sets of equivalent states.

💡Non-final States

Non-final states are states in a DFA that are not accepting or final states. In the script, the distinction between non-final and final states is made during the zero equivalence step, which is essential for understanding the structure of the DFA and its minimization.

💡Final States

Final states, also known as accepting states, are the states in a DFA where the machine halts and accepts the input string. The script mentions that the DFA in the example has only one final state, which simplifies the minimization process as all other states are non-final.

💡Complete DFA

A complete DFA is a DFA where every state has a transition for every input symbol. The script ensures that the minimized DFA remains complete, meaning every state has transitions for all input symbols, which is a requirement for the DFA to function correctly.

Highlights

Introduction to the topic of DFA minimization in automata theory.

Explanation of constructing DFA and the necessity of converting NFA to DFA for implementation feasibility.

Discussion on the possibility of reducing the number of states in a DFA.

Illustration of the DFA with five states and the goal to minimize it.

Introduction to the concept of zero equivalence in the minimization process.

Division of states into non-final and final states for the zero equivalence set.

Methodology for comparing states to determine their equivalence using input symbols.

Procedure for creating new sets when states are found to be non-equivalent.

Explanation of the one equivalence process and its role in state minimization.

Demonstration of the process to compare states using the input symbols 0 and 1.

Identification of states that are not equivalent and the creation of new sets for them.

Continuation to the two equivalence process after establishing one equivalence.

Finding and comparing two equivalences among the remaining states.

Procedure to stop when consecutive equivalences yield the same result.

Final step of minimizing the DFA to a new DFA with a reduced number of states.

Renaming of states for clarity in the minimized DFA.

Conclusion on the successful minimization of the DFA from five to four states.

Encouragement for viewers to ask questions and engage with the content.

Transcripts

[Music]

hello friends welcome back to our

channel so in today's session we'll

discuss about one more topic in automata

theory that is the minimization of dfa

so so far we have discussed about how to

construct the dfa and how to construct

the nfa and also we have seen the

conversion of nfa to bfa because the

implementation is not feasible for nfu

so that we have to convert it to a dfa

and the implementation can be done

further dfu and now in this topic so we

will see the minimization of dfa so this

is a

complete dfa but

so is it possible to reduce the number

of states so here you can see in this

example we are having around five states

and is it possible to reduce the states

that means minimization of dft

right so first of all uh we'll take this

example and we'll

draw the transition table for this one

so the input symbols are 2 0 and 1 so

for 0 and for 1 and the states are q

naught

q 1

q 2

q 3

and

q 4 so total 5 states are there

right so these are the 5 states now we

have to fill the

transition table so q naught on zero

q naught on zero q one it moves to q one

state and q naught on one it moves to q

three

and q one on zero

it moves to q two

q one on 1 it moves to q4

q2 on 0

it moves to q1

q2 on 1 it moves to q4

q3 on 1

it moves sorry q3 on 0 it moves to q2

q3 on 1 it moves to q4

q4 on 0 and 1 the self transition it

remains in the same state so this is the

transition table for the given diagram

now the next step is

we need to divide the complete states

into two sets that is one with the

non-final states and another with the

final states so we know that the dfa can

have a multiple final states but here in

this example we are holding we are

having only one final states right so

but the first step is we need to find

the two different sets one set with the

non-final states and another set within

final state so that we call it as a zero

equivalence so in this minimization

problem so we need to find the

equivalence right so

first one is zero equivalence

zero equivalence

so here we are having two sets one is

the non-final sense and another one with

the final states so here non-final

states are

q naught

q one

q two

and a q three so in from this diagram

you can see all these four are the

non-final states and there is one more

state called

q4 which is a final state non-final

states

okay non-final states

and

final state

the first step is this one okay and from

this one we need to find the one

equivalence that means we have to find

whether these two states are equal or

not so that means equivalence of states

you should be fined and if they are

equal it remains in the same set

otherwise it fall in the new set okay so

let us

check with the one equivalence

see whenever you are finding the one

equivalence we have to compare

everything with the zero equivalence now

we have to compare q naught with all the

remaining states whether it is equal or

not

right how how can we

compare this one so

over the input symbol 0 and 1 compare

the two states the resultant states if

the resultant states fall in the same

set then automatically it shows that two

states are equal right so let me explain

you with an example now we'll compare q

naught and q one okay q naught and q one

so i i will explain here itself okay for

the one equivalence and see

here there is a one set and

automatically uh here we are having a

one more set called q4

okay q4 because it's having only one uh

state so we we did not compare with any

other so it will remain in the same

speed okay now we have to compare q

naught with q one q two and q three now

we will compare q naught with

q one

should not with q one so compare

so we know that q naught on zero

and

q one on zero similarly

q naught on

sorry q naught on one and

q naught sorry q one on one

q one on one okay so we are comparing

the q naught and q one over the input

symbol zero and q naught and q one over

the input symbol one

now we can observe q naught on zero q

naught on zero it moves to q one

q one on zero

it moves to q

two just compare the resultant states

whether these states fall in the same

set of previous equivalence here we are

finding the one equals why we have to

check in the zero equivalence you can

see q1 and q2 so far in the same set

okay q one and q two fall in the same

set right now compare this one the

states with input symbol one q naught on

one it is a q three

and q one on one it is a q4 and here you

can observe if these two states are in

the same

set or not so q3 is in one state one set

and q4 is in another set so this implies

so this is not possible okay this is not

possible so that implies q naught

is not equal to q one q naught is not

equal to q one so that's why q naught

will be here and there will be one more

set created and q naught will be remains

here itself okay so we have to divide

the unequivalent say state two and use

it so q naught and q one if you compare

q naught and q one both are not

equivalent so we have to divide this q

naught and q one so open the new set and

just move the state which is not equal

to q naught two in the other

and now the next step compare q naught

with q two

same procedure now we need to compare q

naught with q two

q two right say q naught on zero

okay q naught on zero and q two on zero

similarly q naught on

one q naught on one and

q two on one see q naught on zero q

naught on zero it is a q one

and q two one zero

is a q one

yes so you can compare so q one both are

falling in the same set q1 right next q

not on one it's a q3

and q2 on one it's a q4

so again you can observe here both are

different

both are different q3 in one set and q4

is in another set so

q naught is

not equal to q two q naught is not equal

to q two so we have to divide the q two

also so just

uh

insert the

q2 in the new cell okay and now we need

to compare q naught with a q three

because we have compared q naught with q

one q naught with q now we have to

compare q naught with q three now

see

cube

and here also we need to find a q three

on zero and

q three on

one now

so let us repeat the same process

so q naught on zero is a q one

and q naught on zero is a q two both are

in the same set q naught q three sorry q

one and q right q naught on one it is a

q three q3 q3 on one it's a q4

it's a q4 again this is not equal so

again we need to divide q naught with

the q3 so

q3 falls in the new set and q2 q9 will

be remaining in the same set so

this is the

one equivalence so after one equivalence

the q naught is in one set q one q two q

three are in another set because q

naught is not equivalent to q1 q2 and q3

so that's why we are dividing the set so

we are reducing the states of a set okay

now

so this is

the procedure to form to find out the

one equivalence after that we need to

find the two equivalents and then three

equivalents so we need to repeat the

same process until we get the same

result in the consecutive equivalences

now let us check with the two

equivalence okay right here itself i'll

write here itself so two

equivalence right so here you can

observe q naught is a single

state so you need not change it and we

need to compare the q one with q2 and q3

and we have to

divide that with the equivalence and

non-equivalence

now

just

compare

q1 with

q2

so

one on zero

q two on zero similarly q one on one

[Music]

q two on one so q one one zero

show one one zero it's a q two

q two on 0 it's a q1

now you have to check this one

with the

previous equivalent so we are finding

the two equivalence we have to compare

it with the one equals so before we have

we are finding the one equivalent so we

have to compare the zero equivalence now

you can see q2 and q1 so q2 is in in one

set and q1 is in another set so you did

not find out this one so q1 is not equal

to q2 so just divide the

q1

q2 okay now

hope you understood q1 and q2 now you

just find the compare the q1 with q3

just compare the q1 with should be so i

will compare with the q one with q three

so here also i'll go with the q one and

q three and q one and q three and see

now whether the both the states are

equal or not so q one zero

is a q2

q3 on 0 is a q2 so you can find you can

find these two are the same

state which falls in the same set so

this is equal and q1 on 1 it's a q4

q3 on one it's a q4 so both are in the

same

set okay both are in the same set q4

right so that implies so q1 is equal to

q3 so both are equal

so just

place the place it here q1 and q3 and q2

and again it's a q4

now you can observe here

in the two equivalents q naught q and q

two are same q two and q four

okay now

now we need to find the three

equivalence we need to find the three

equations because

equivalence one and equals two both are

not equal right so we need to find the

equivalence 3

so now just find the

three equals

three equals so in the three equivalence

we need to compare q1 and q3 because q

naught is a single state q 2 is a single

state and q 4 is single state so we are

having the two states here now we need

to compare

q 1

with

q 3

so q 1 on zero

q three

on zero so q one

on one

q three on one so you can see q one on

zero it is a q two and q three on zero

is again q two so both are equal right

coming to the q one on one q one on one

execute four and q three on one it's

again q four both are equal so simply

the third equivalence for this one

is

q naught the single state q two single

state q four in single state so we need

not uh compare them right so q one q

three remains same so

here you can observe that three

equivalence

the three equivalence

is

q naught

q one

q three

q two

q four because q naught q three are

equal here q 1 and q 3 are equal so you

can see the resultant of third

equivalence resultant of second

equivalence both are equal so just stop

finding the equivalences right so we

have to repeat the process until we get

the

last two equivalences result as a same

so here one equivalence and two

equivalence are both are not equal so we

are finding the third equivalence so

third equivalence and second equivalence

are equal so we have to find out i mean

we have to stop the procedure now by

using this one you have to draw the dfa

so here you can observe q naught is one

state q one q three is one state q t is

one state and q four is one state so

total four states so here you can say

the total states are four but before

minimization how many states are there

there are five states so we are

minimizing the five states to four

states so just write down here so q

naught

q1 comma q3 becomes a one state

q2 is one state

q4 is another state okay four states now

just

follow the transition table and find out

the i mean uh give the transitions q

naught on zero q one so where is q one

here this is q one

q one on zero

it goes to q one q naught on one it goes

to q three

so here itself so zero comma one okay

and q one on zero q two

q one on zero

q two so q1 on 0

it moves to

q2

next q1 on 1 it moves to q4 q1 on 1 it

moves to

q4

q2 on 0 q1 q2 on 0 q1

so again

0

and q2 1 q4

q2 on 1 is

q4

q3 on 0 q2 q3 on 0

q2 so here you can see q3 on 0 q2

already there is a transition q3 on 1 q4

q3 on 1 q4 again already there is a

transition and q4 on 0 and 1 it remains

in the same state so

this is a

and

1 so this will be the initial state so

here also q naught is initial state and

q 4 is the final state and here also the

q 4 will remains the same as a final

state so we have we are minimizing the

given dfa

to the

another dfa with a limited number of

states with a minimum number of states

right and here you can observe that

whether it is a complete af or not so

for q naught there is a transition for 0

and 1 yes q 2 for q 2 there is a

transition for 0 and 1 and q 1 q 3 is

having a transition for 0 and 1 q 4 is

having a transition for 0 and 1 yes so

for every state there is a transition

over every input symbol so this is a

complete dfa and we are minimizing this

dfva to

this df that means we are reducing the

five states to four states so simply by

finding the equivalence between the

states so once again i am repeating so

first we need to find out the zero

equivalence which is having a two sets

one set with the non-final states and

our another set with the final states

and from that we need to divide the

complete states by finding the

equivalences right so for every state

with the input symbols we need to find

the resultant state whether they fall in

the same set or not if they fall in the

same set automatically the states are

equal and if they fall in different sets

automatically the states are not equal

so we have def we have to divide in this

one equivalence q naught is completely

different with q one q two q three so we

have divided q naught with the q one q

two q three and q four so here after

zero equivalence we are getting the one

equivalence with the three sets and from

these three states again you can find

the three states with one say we need to

find the equivalence among these three

states q one with q two q three so that

we have we got that q1 and q3 are equal

and they are both are not equal to q2

you can also contact q3 with q2 because

here we are having some q1 and q3 are

equal so q1 and q2 are different

automatically q3 and q2 q2 will also be

different right so like this we have to

repeat the same process by finding the

equivalences until you get the last two

equivalents as the same states

okay so here you can observe q naught q

one q three q two q four four states are

there four sets are there and third

equivalence also the four sets with the

same

states so that's why we have to stop at

third equivalence and just draw the dfa

with the given following states so you

can also rename this as a

b c and d and you can rename instead of

using q naught you can use a a instead

of using q one q three you can use a b

instead of q two using it you can use a

c and instead of q four you can use d

right so this is the procedure to follow

the

i mean uh to find out the minimization

of dfu right and if in some

dfa we are having a multiple final

states we have will be having a multiple

final states so here also then in such

cases we will get this set within

multiple final states and here also we

need to compare each and every state of

this

final states we have to follow the same

procedure

right so hope you understood this one

if you are having any

doubts regarding this procedure feel

free to post your doubts in the comment

section uh definitely i will try to

clarify all your doubts and if you

really enjoyed my session like my

session share my session with your

friends and don't forget to subscribe to

our channel thanks for watching thank

you very much