MINIMIZATION OF DFA WITH EXAMPLE IN AUTOMATA THEORY || DFA MINIMIZATION || TOC
Summary
TLDRIn 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
🤖 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.
🔍 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.
🔄 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.
🏁 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.
Mindmap
Keywords
💡Automata Theory
💡DFA (Deterministic Finite Automaton)
💡Minimization
💡Transition Table
💡Equivalence
💡Zero Equivalence
💡One Equivalence
💡Non-final States
💡Final States
💡Complete DFA
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
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