Finite Automata A simple model of computation Slide 2 Jaruloj
Chongstitvatana 2301379Chapter 2 Finite Automata2 Outline
Deterministic finite automata (DFA) How a DFA works How to
construct a DFA Non-deterministic finite automata (NFA) How an NFA
works How to construct an NFA Equivalence of DFA and NFA Closure
properties of the class of languages accepted by FA Slide 3 Jaruloj
Chongstitvatana 2301379Chapter 2 Finite Automata3 Finite Automata
(FA) CONTROL UNIT INPUT TAPE TAPE HEAD Finite set of states Move to
the right one cell at a time Store input of the DFA Can be of any
length Start state Final state(s) is in exactly one state at a time
Slide 4 Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata4
What does an FA do? Read an input string from tape Determine if the
input string is in a language Determine if the answer for the
problem is YES or NO for the given input on the tape Slide 5
Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata5 How does
an FA work? At the beginning, an FA is in the start state (initial
state) its tape head points at the first cell For each move, FA
reads the symbol under its tape head changes its state (according
to the transition function) to the next state determined by the
symbol read from the tape and its current state move its tape head
to the right one cell Slide 6 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata6 When does an FA stop working?
When it reads all symbols on the tape Then, it gives an answer if
the input is in the specific language: Answer YES if its last state
is a final state Answer NO if its last state is not a final state
Slide 7 Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata7
Example of a DFA qa (q,a)(q,a) s0s s1f f0f f1s s f 01 0 1
Transition diagram Transition function Slide 8 Jaruloj
Chongstitvatana 2301379Chapter 2 Finite Automata8 How to define a
DFA a 5-tuple (Q, , , s, F), where a set of states Q is a finite
set an alphabet is a finite, non-empty set a start state s in Q a
set of final states F contained in Q a transition function is a
function Q Q See formal definition Thats why its called a finite
automaton. Slide 9 Jaruloj Chongstitvatana 2301379Chapter 2 Finite
Automata9 How an FA works sf 01 0 1 Input: 001101 Input: 01001
00110101001 ACCEPT REJECT Slide 10 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata10 Configurations Definition Let M
=( Q, , , s, F ) be a deterministic finite automaton. A
configuration of M is an element of Q configurations of M (s, ),
(s, 00), (s, 01101), (f, ), (f, 0011011001), (f, 1111), (f, 011110)
s f 01 0 1 Transition diagram of M Unread input string Current
state Slide 11 Jaruloj Chongstitvatana 2301379Chapter 2 Finite
Automata11 Yielding next configuration Definition Let M = ( Q, , ,
s, F ) be a deterministic finite automaton, and ( q 0, 0 ) and ( q
1, 1 ) be two configurations of M. We say ( q 0, 0 ) yields ( q 1,
1 ) in one step, denoted by ( q 0, 0 ) M ( q 1, 1 ), if ( q 0, a )
= q 1, and 0 =a 1, for some a . When a configuration yields another
configuration, the first symbol of the string is read and discarded
from the configuration. Once a symbol in an input string is read,
it does not effect how the DFA works in the future. Slide 12
Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata12 Example
: yielding next configuration (s, 001101) M (s, 01101) M (s, 1101)
M (f, 101) M (s, 01) M (s, 1) M (f, ) sf 01 0 1 0 0 1 1 0 1 Slide
13 Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata13 Yield
in zero step or more Definition Let M = (Q, , , s, F) be a DFA, and
(q 0, 0 ) and (q 1, 1 ) be two configurations of M. (q 0, 0 )
yields (q 1, 1 ) in zero step or more, denoted by (q 0, 0 ) * M (q
1, 1 ), if q 0 = q 1 and 0 = 1, or (q 0, 0 ) M (q 2, 2 ) and (q 2,
2 ) * M (q 1, 1 ) for some q 2 and 2. This definition is an
inductive definition, and it is often used to prove properties of
DFAs. Slide 14 Jaruloj Chongstitvatana 2301379Chapter 2 Finite
Automata14 Example: Yield in zero step or more (s, 001101) * M (f,
101) * M (s, 01) * M (f, ) s f 01 0 1 (s, 01001) * M (f, 001) * M
(f, 1) * M (s, ) Slide 15 Jaruloj Chongstitvatana 2301379Chapter 2
Finite Automata15 Accepting a string Definition Let M = (Q, , , s,
F) be a DFA, and *. We say M accepts if (s, ) * M (f, ), when f F.
Otherwise, we say M rejects . (s, 001101) * M (f, ) M accepts
001101 (s, 01001) * M (s, ) M rejects 01001 Slide 16 Jaruloj
Chongstitvatana 2301379Chapter 2 Finite Automata16 Language
accepted by a DFA Let M = (Q, , , s, F ) be a DFA. The language
accepted by M, denoted by L(M ) is the set of strings accepted by
M. That is, L(M) = { *|(s, ) * M (f, ) for some f F } Example: L(M)
= {x {0,1}* | the number of 1s in x is odd}. s f 01 0 1 Slide 17
Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata17 Example
s 0 1 1 1 1 1 1 1 0 0 0 0 0 0, 1 0 Slide 18 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata18 How to construct a DFA Determine
what a DFA need to memorize in order to recognize strings in the
language. Hint: the property of the strings in the language
Determine how many states are required to memorize what we want.
final state(s) memorizes the property of the strings in the
language. Find out how the thing we memorize is changed once the
next input symbol is read. From this change, we get the transition
function. Slide 19 Jaruloj Chongstitvatana 2301379Chapter 2 Finite
Automata19 Constructing a DFA: Example Consider L= { {0,1}*| has
both 00 and 11 as substrings}. Step 1: decide what a DFA need to
memorize Step 2: how many states do we need? Step 3: construct the
transition diagram Slide 20 Jaruloj Chongstitvatana 2301379Chapter
2 Finite Automata20 Example s 0 1 1 1 1 1 1 1 0 0 0 0 0 0, 1 0 Got
nothing Got 0 as substring Got 00 as substring Got 11 and got 0 Got
00 and 11 as substrings Got 1 as substring Got 11 as substring Got
00 and got 1 Got 11 and 00 as substrings Slide 21 Jaruloj
Chongstitvatana 2301379Chapter 2 Finite Automata21 Constructing a
DFA: Example Consider L= { {0,1}*| represents a binary number
divisible by 3}. L = {0, 00, 11, 000, 011, 110, 0000, 0011, 0110,
1001, 00000,...}. Step 1: decide what a DFA need to memorize
remembering that the portion of the string that has been read so
far is divisible by 3 Step 2: how many states do we need? 2 states
remembering that the string that has been read is divisible by 3
the string that has been read is indivisible by 3. 3 states
remembering that the string that has been read is divisible by 3
the string that has been read - 1 is divisible by 3. the string
that has been read - 2 is divisible by 3. Choose this!! Why? Slide
22 Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata22 Using
2 states Reading a string w representing a number divisible by 3.
Next symbol is 0. w 0, which is 2* w, is also divisible by 3. If w
=9 is divisible by 3, so is 2* w =18. Next symbol is 1. w 1, which
is 2* w +1, may or may not be divisible by 3. If 8 is indivisible
by 3, so is 17. If 4 is indivisible by 3, but 9 is divisible. Using
these two states is not sufficient. Slide 23 Jaruloj
Chongstitvatana 2301379Chapter 2 Finite Automata23 Using 3 states
Each state remembers the remainder of the number divided by 3. If
the portion of the string that has been read so far, say w,
represents the number whose remainder is 0 (or, 1, or 2), If the
next symbol is 0, what is the remainder of w 0? If the next symbol
is 1, what is the remainder of w 1? Slide 24 Jaruloj
Chongstitvatana 2301379Chapter 2 Finite Automata24 How remainder
changes Current number Current remainder Next symbol New number New
remainder 3n006n0 3n016n+11 3n+1106n+22 3n+1116n+30 3n+2206n+41
3n+2216n+52 Slide 25 Jaruloj Chongstitvatana 2301379Chapter 2
Finite Automata25 Transition table Current number Current remainder
Next symbol New number New remainder 3n006n0 3n016n+11 3n+1106n+22
3n+1116n+30 3n+2206n+41 3n+2216n+52 Current state Next state Slide
26 Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata26
Example DFA Current state Next symbol Next state q0q0 0q0q0 q0q0
1q1q1 q1q1 0q2q2 q1q1 1q0q0 q2q2 0q1q1 q2q2 1q2q2 M = ({q 0, q 1, q
2 }, {0, 1}, , q 0, {q 0 }) q0q0 q1q1 q2q2 0 1 0 0 1 1 Slide 27
Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata27
Nondeterministic Finite Automata Similar to DFA Nondeterministic
move On reading an input symbol, the automaton can choose to make a
transition to one of selected states. Without reading any symbol,
the automaton can choose to make a transition to one of selected
states or not. 0 0 Slide 28 Jaruloj Chongstitvatana 2301379Chapter
2 Finite Automata28 How to define an NFA a 5-tuple (Q, , , s, F),
where a set of states Q is a finite set an alphabet is a finite,
non-empty set a start state s in Q a set of final states F
contained in Q a transition function is a function Q ( { }) 2 Q See
formal definition Slide 29 Jaruloj Chongstitvatana 2301379Chapter 2
Finite Automata29 Example of NFA An NFA accepting { w {0,1} |w ends
with 11 } 0,1 11 0 00 1 1 1 0 1 1 0 10 1 0 DFA Slide 30 Jaruloj
Chongstitvatana 2301379Chapter 2 Finite Automata30 Yielding next
configuration Definition Let M = (Q, , , s, F) be a
non-deterministic finite automaton, and (q 0, 0 ) and (q 1, 1 ) be
two configurations of M. We say (q 0, 0 ) yields (q 1, 1 ) in one
step, denoted by (q 0, 0 ) M (q 1, 1 ), if q 1 (q 0, a,), and 0 =a
1, for some a { }. Slide 31 Jaruloj Chongstitvatana 2301379Chapter
2 Finite Automata31 Yield in zero step or more Definition Let M =
(Q, , , s, F) be an NFA, and (q 0, 0 ) and (q 1, 1 ) be two
configurations of M. (q 0, 0 ) yields (q 1, 1 ) in zero step or
more, denoted by (q 0, 0 ) * M (q 1, 1 ), if q 0 = q 1 and 0 = 1,
or (q 0, 0 ) M (q 2, 2 ) and (q 2, 2 ) * M (q 1, 1 ) for some q 2
and 2. Same as that for DFA Slide 32 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata32 Accepting a string Definition
Let M = (Q, , , s, F) be an NFA, and *. We say M accepts if (s, ) *
M (f, ), when f F. Otherwise, we say M rejects . Same as the
definition for a DFA Slide 33 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata33 Language accepted by an NFA Let
M = (Q, , , s, F) be an NFA. The language accepted by M, denoted by
L(M) is the set of strings accepted by M. That is, L(M) = { *| (s,
) * M (f, ) for some f F } Same as the definition for a DFA Slide
34 Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata34 How
NFAs work (s,01111) - (s,1111) - (s,111) - (s,11) - (s,1) - (s, )
(s,01111) - (s,1111) - (s, 111) - (s,11) - (s,1) - (q, ) (s,01111)
- (s, 1111) - (s, 111) - (s,11) - (q,1) - (f, ) (s,01111) - (s,
1111) - (s, 111) - (q,11) - (f,1) (s,01111) - (s, 1111) - (q, 111)
- (f,11) 0,1 11 s, 01111 sqf s, 1111 s, 111q, 111 s, 11 f, q, 1 q,
11 s, s, 1 q, f, 1 f, 11 Slide 35 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata35 Other examples of NFA { w {0,1}*
| w has either 00 or 11 as substring} {w {0,1}* | w has 00 and 11
as substrings and 00 comes before 11 } 0,1 0 11 0 0 11 0 Slide 36
Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata36 DFA and
NFA are equivalent M d and M n are equivalent L(M d ) = L(M n ).
DFA and NFA are equivalent For any DFA M d, there exists an NFA M n
such that M d and M n are equivalent. (part 1) For any NFA M n,
there exists a DFA M d such that M d and M n are equivalent. (part
2) Slide 37 Jaruloj Chongstitvatana 2301379Chapter 2 Finite
Automata37 Part 1 of the equivalence proof For any DFA M d, there
exists an NFA M n such that M d and M n are equivalent. Proof: Let
M d be any DFA. We want to construct an NFA M n such that L(M n ) =
L(M d ). From the definitions of DFA and NFA, if M is a DFA then it
is also an NFA. Then, we let M n = M d. Thus, L(M d ) = L(M n ).
Slide 38 Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata38
Part 2 of the equivalence proof For any NFA M n, there exists a DFA
M d such that M d and M n are equivalent. Proof: Let M n = (Q, , ,
s, F) be any NFA. We want to construct a DFA M d such that L(M d )
= L(M n ). First define the closure of q, denoted by E(q). Second,
construct a DFA M d =(2 Q, , ', E(s), F') Finally, prove f F (s, )
|- * Mn (f, ) f ' F ' (E(s), ) |- * Md (f ', ). Slide 39 Jaruloj
Chongstitvatana 2301379Chapter 2 Finite Automata39 Eliminating
transitions to multiple states sq f 0,1 11 s s,q 1 s,q,f 1 0 0 1 0
Slide 40 Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata40
Eliminating empty-string transitions spq r f 0 0 0 ,1 1 11 sp,q, r
rp,q,r,f 0 0,1 1 1 Slide 41 Jaruloj Chongstitvatana 2301379Chapter
2 Finite Automata41 Closure of state q Let M = (Q, , , s, F) be an
NFA, and q Q. The closure of q, denoted by E(q), is the set of
states which can be reached from q without reading any symbol. {p
Q| (q, ) |- M * (p, )} If an NFA is in a state q, it can also be in
any state in the closure of q without reading any input symbol.
Slide 42 Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata42
Example of closure sqf abc M=({s, q, f}, {a, b, c}, , s, {f}) where
is defined below. L(M)={a i b j c k | i, j, k 0} stateclosure sqfs,
q, f qfq, f ff Slide 43 Jaruloj Chongstitvatana 2301379Chapter 2
Finite Automata43 Another example of closure q0q0 q1q1 q2q2 q3q3
q4q4 1q1q1 q 2, q 3 q3q3 2 3 q(q,a,r)(q,b,r) (q, ,r) q0q0 q2q2 q1q1
q1q1 q 0,q 4 q2, q3q2, q3 q2q2 q4q4 q3q3 q4q4 q4q4 q3q3 q0q0 q1q1
q2q2 q3q3 q4q4 b a b a a Slide 44 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata44 Constructing the equivalent DFA
Let M n = (Q, , , s, F) be any NFA. We construct a DFA M d =(2 Q, ,
', E(s), F'), where : '(q',a) = {r E(p)| p (q,a) } and F' = {f Q |
f F }) q q' q pr a a DFA Slide 45 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata45 Example of DFA construction E(q
0 ) E(q 1 ) E(q 2 )E(q 3 ) E(q 4 ) q 0, q 1, q 2, q 3 q 1, q 2, q 3
q2q2 q3q3 q 3,q 4 q0q0 q1q1 q2q2 q3q3 q4q4 b a b a a Slide 46
Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata46 Example
of DFA construction q 0,q 1,q 2,q 3 q 0,q 1,q 2,q 3,q 4 q 2,q 3,q 4
q 3,q 4 a b b a a b b a a,b E(q 0 ) E(q 1 ) E(q 2 )E(q 3 ) E(q 4 )
q 0, q 1, q 2, q 3 q 1, q 2, q 3 q2q2 q3q3 q 3,q 4 (q,a,r)(q,b,r)
(q, ,r) q0q0 q2q2 q1q1 q1q1 q 0 q 4 q 2 q 3 q2q2 q4q4 q3q3 q4q4
q4q4 q3q3 Slide 47 Jaruloj Chongstitvatana 2301379Chapter 2 Finite
Automata47 Prove property of and ' Let M n = (Q, , , s, F) be any
NFA, and M d = (2 Q, , ', E(s), F') be a DFA, where '(q', a) = {r
E(p)| p (q,a)} and F' = {f Q | f F } Prove , f F (s, ) |- * Mn (f,
) f ' F ' (E(s), ) |- * Md (f', ) and f f' by induction. Prove a
more general statement , p, q Q (p, ) |- * Mn (q, ) (E(p), ) |- *
Md (q', ) and q q'. q q' Slide 48 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata48 Proof Part I: For any string in
*, and states q and r in Q, there exists R Q such that (q, ) * Mn
(r, ) (E(q), ) * Md (R, ) and r R. Slide 49 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata49 Proof Basis: Let be a string in
*, q and r be states in Q, and (q, ) * Mn (r, ) in 0 step. Because
(q, ) * Mn (r, ) in 0 step, we know (1) q=r, and (2) = . Then,
(E(q), ) = (E(r), ). Thus, (E(q), ) * Md (E(r), ). That is, there
exists R=E(r) such that r R and (E(q), ) * Md (R, ). Slide 50
Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata50 Proof
Induction hypothesis: For any non-negative integer k, string in *,
and states q and r in Q, there exists R Q : ( q, ) * Mn (r, ) in k
steps -> (E(q), ) * Md (R, ) and r R. Induction step: Prove, for
any non-negative integer k, string in *, and states q and r in Q,
there exists R Q : (q, ) * Mn (r, ) in k +1 steps -> (E(q), ) *
Md (R, ) and r R. Slide 51 Jaruloj Chongstitvatana 2301379Chapter 2
Finite Automata51 Proof Let be a string in *, q and r be states in
Q, and (q, ) * Mn (r, ) in k +1 steps. Because (q, ) * Mn (r, ) in
k +1 steps and k 0, there exists a state p in Q and a string * such
that (q, ) * Mn (p, a) in k steps and (p, a) Mn (r, ) for some a
{}. Slide 52 Jaruloj Chongstitvatana 2301379Chapter 2 Finite
Automata52 Proof From the induction hypothesis and (q, ) * Mn (p,
a) in k steps, we know that there exists P Q such that (E(q), ) *
Md (P, a) and p P. Since (p, a) Mn (r, ), r (p, a). From the
definition of of M d, E( (p, a)) (P, a) because p P. Because r (p,
a) and E( (p, a)) (P, a), r (P, a). Then, for R= (P, a), (P, a) *
Md (R, ) and r R. Thus, (E(q), ) * Md (P, a) * Md (R, ) and r R.
Slide 53 Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata53
Proof Part II: For any string in *, and states q and r in Q, there
exists R Q such that r R and (E(q), ) * Md (R, ) -> (q, ) * Mn
(r, ). E(q) R q r rRrR Slide 54 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata54 Proof Basis: Let be a string in
*, q and r be states in Q, R be a subset of Q such that r R and
(E(q), ) * Md (R, ) in 0 step. Because (E(q), ) * Md (R, ) in 0
step, E(q)=R and =. From the definition of E, (q, )=R because
E(q)=R. Then, for any r R, (q, ) * Mn (r, ). That is, there exists
R=E(q) such that r R and (q, ) * Mn (r, ). Slide 55 Jaruloj
Chongstitvatana 2301379Chapter 2 Finite Automata55 Proof Induction
hypothesis: For any non-negative integer k, string in *, and states
q and r in Q, there exists R Q such that r R and: (E(q), ) * Md (R,
) in k steps ->(q, ) * Mn (r, ). Induction step: Prove, for any
non-negative integer k, string in *, and states q and r in Q, there
exists R Q such that r R : ( E(q), ) * Md (R, ) in k +1 steps
->(q, ) * Mn (r, ). Slide 56 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata56 Proof Let be a string in *, q
and r be states in Q, and (E(q), ) * Md (R, ) in k +1 steps.
Because (E(q), ) * Md (R, ) in k +1 steps and k 0, there exists P 2
Q (i.e. P Q ) and a string * such that = a, (E(q), ) * Md (P,) in k
steps and (P, a) Md (R, ) for some a . E(q) R q r rRrR P k+1 steps
a Slide 57 Jaruloj Chongstitvatana 2301379Chapter 2 Finite
Automata57 Proof From the induction hypothesis and (E(q), ) * Md
(P, ) in k steps, we know that there exists p P such that (q, ) *
Mn (p,) (i.e. (q, a) * Mn (p, a) ). Since (P, a) Md (R, ), there
exists r R such that r= (p, a). Then, for some r R, (p, a) * Mn (r,
). Thus, (q, ) * Mn (p, a) * Mn (r, ) for some r R. E(q)E(q) R q r
rRrR P p pPpP k +1 steps a a Slide 58 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata58 Closure Properties The class of
languages accepted by FAs is closed under the operations Union
Concatenation Complementation Kleenes star Intersection Slide 59
Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata59 Closure
Property under Union The class of languages accepted by FA is
closed under union. Proof: Let M A = (Q A, , A, s A, F A ) and M B
= (Q B, , B, s B, F B ) be any FA. We construct an NFA M = (Q, , ,
s, F) such that Q = Q A Q B {s} = A A {(s, , {s A, s B })} F = F A
F B sAsA sBsB We prove that L(M) = L(M A ) L(M B ). s Slide 60
Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata60 Proof To
prove L(M) = L(M A ) L(M B ), we prove: I.For any string * L(M A )
or L(M B ) L(M) & II.For any string * L(M A ) and L(M B ). L(M)
For I, consider (a) L(M A ) or (b) L(M B ). For (a), let L(M A ).
From the definition of strings accepted by an FA, there is a state
f A in F A such that ( s A, ) |-* MA ( f A, ). Because A , (s A, )
|-* M (f A, ) also. Because s A (s,), (s, ) |- M (s A, ). Thus, (s,
) |- M (s A, ) |-* M (f A, ). Because f A F, L(M). Similarly for
(b). s sAsA sBsB Slide 61 Jaruloj Chongstitvatana 2301379Chapter 2
Finite Automata61 Proof For (II), let L(M A ) L(M B ). Because (s,
, {s A, s B }) , either (s, ) |- M (s A, ) or (s, ) |- M (s B, )
only. Because L(M A ), there exists no f A in F A such that (s A, )
|-* MA (f A,). Because L(M B ), there exists no f B in F B such
that (s B, ) |-* MB (f B, ). Since there is no transition between
states in Q A and Q B in M, there exists no state f in F=F A F B
such that (s, ) |- M (s A, ) |-* M (f A, ) or (s, ) |- M (s B, )
|-* M (f B, ). That is, L(M). Thus, L(M) = L(M A ) L(M B ). s sAsA
sBsB Slide 62 Jaruloj Chongstitvatana 2301379Chapter 2 Finite
Automata62 Closure under concatenation sAsA MAMA sBsB MBMB Slide 63
Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata63 Closure
under complementation sAsA MAMA DFA only Slide 64 Jaruloj
Chongstitvatana 2301379Chapter 2 Finite Automata64 Closure under
Kleenes star sAsA fAfA s f Slide 65 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata65 Closure under intersection The
class of languages accepted by FA is closed under intersection.
Proof: Let L 1 and L 2 be languages accepted by FA. L 1 L 2 = ( L 1
L 2 ) By the closure property under complementation, there are FA
accepting L 1 and L 2. By the closure property under union, there
is an FA accepting L 1 L 2. By the closure property under
complementation, there is an FA accepting( L 1 L 2 ). Thus, the
class of languages accepted by FA is closed under intersection.
Slide 66 Jaruloj Chongstitvatana 2301379Chapter 2 Finite Automata66
DFA accepting the intersection of two languages Let M A = (Q A, ,
A, s A, F A ) and M B = (Q B, , B, s B, F B ) be any FA. We
construct an NFA M = (Q, , , s, F) such that Q = Q A Q B = A A
(i.e. ((q A,q B ),a) = A (q A,a) B (q B,a)) s = (s A, s B ) F = F A
F B Slide 67 Jaruloj Chongstitvatana 2301379Chapter 2 Finite
Automata67 Example q0q0 q1q1 q2q2 0 1 0 0 1 1 0,1 11 p0p0 p1p1 p2p2
p 0,q 1 p 0,q 0 p 0,q 2 0 1 0 0 1 1 p 1,q 1 1 p 1,q 0 1 p 1,q 2 1 p
2,q 0 1 p 2,q 1 1 p 2,q 2 1 Slide 68 Jaruloj Chongstitvatana
2301379Chapter 2 Finite Automata68 Check list Basic Explain how
DFA/NFA work (configuration, yield next configuration) Find the
language accepted by DFA/NFA Construct DFA/NFA accepting a given
language Find closure of a state Convert an NFA into a DFA Prove a
language accepted by FA Construct FA from other FAs Advanced Prove
DFA/NFA accepting a language Prove properties of DFA/NFA
Configuration change Under some modification etc. Prove some
properties of languages accepted by DFA/NFA Under some modification
Surprise!
