![
20
State Diagrams and Examples (8)
State Diagrams and Examples (8)
strings over {a, b} that contain an
strings over {a, b} that contain an even
even number of
number of
a
a’s
’s AND
AND an
an odd
odd number of
number of b
b’s
’s
[ea, eb] [ea, ob]
a
[oa, eb] [oa, ob]
b
a a
a
b
b
b](https://image.slidesharecdn.com/finitwautomata2-240811180118-c3e07e96/85/finitw-automata2-Computer-theory-computure-science-20-320.jpg)
![
21
State Diagrams and Examples (9)
State Diagrams and Examples (9)
Let
Let M
M be the DFA previous slide
be the DFA previous slide
A DFA
A DFA M’
M’ that accepts all strings over {a, b} that
that accepts all strings over {a, b} that do not
do not
contain an
contain an even
even number of
number of a
a’s
’s AND
AND an
an odd
odd number of
number of b
b’s
’s
is shown below
is shown below
L(M’) = {a, b}*
- L(M) = *
- L(M)
Any string accepted by
Any string accepted by M
M is rejected by
is rejected by M’
M’ and vice versa
and vice versa
[ea, eb] [ea, ob]
a
[oa, eb] [oa, ob]
b
a a
a
b
b
b](https://image.slidesharecdn.com/finitwautomata2-240811180118-c3e07e96/85/finitw-automata2-Computer-theory-computure-science-21-320.jpg)
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- 1. 1 Chapter 2 Chapter 2 Finite Automata Finite Automata
- 2. 2 A Finite-State Machine (1) A Finite-State Machine (1) A finite state machine is a mathematical model of a A finite state machine is a mathematical model of a system, with discrete inputs and outputs system, with discrete inputs and outputs
- 3. 3 A Finite-State Machine (2) A Finite-State Machine (2) Finite Automata FA: Finite Automata FA: a finite set of states a set of transitions (edges) a start state a set of final states Defining a FA is a kind of programming. Defining a FA is a kind of programming. Problem definition Problem definition Includes defining possible actions & accepting condition. States States structure of program structure of program Includes designating which are initial & final. Transitions Transitions program program
- 4. 4 2.1: Deterministic Finite Automata (1) 2.1: Deterministic Finite Automata (1) DFA M = (Q, DFA M = (Q, , , , , q q0 0, F) , F) Q = a finite set of states = a finite set called the alphabet = transition function total function Q Q q0 = start state q0 Q F = final or accepting states F Q
- 5. 5 Deterministic Finite Automata (2) Deterministic Finite Automata (2) DFA M DFA M Q = {q0, q1} = {a, b} F = {q1} The transition function is given in a tabular form called the transition table (q0, a) = q1 (q0, b) = q0 (q1, a) = q1 (q1, b) = q0
- 6. 6 Deterministic Finite Automata (3) Deterministic Finite Automata (3) A A DFA DFA M can be considered to be a language M can be considered to be a language acceptor. acceptor. The The language language of M, of M, L(M) L(M), is the set of strings , is the set of strings * * accepted by M. accepted by M. A A DFA DFA M reads an input string from left to right. M reads an input string from left to right. The The next state next state depends on the depends on the current state current state and and the the unread (unprocessed) symbol unread (unprocessed) symbol. .
- 7. 7 Deterministic Finite Automata (5) Deterministic Finite Automata (5) The DFA The DFA M M accepts the set of strings over {a, b} that accepts the set of strings over {a, b} that contain the substring contain the substring bb bb M : Q = {q0, q1, q2}, = {a, b}, F = {q2} The transition function is given in a tabular form called the transition table (q0, a) = q0 (q0, b) = q1 (q1, a) = q0 (q1, b) = q2 (q2, a) = q2 (q2, b) = q2 Is Is abba abba L(M)? Yes, since the computation halts in state L(M)? Yes, since the computation halts in state q q2 2, which is a final state , which is a final state Is Is abab abab L(M)? No, since the computation halts in state L(M)? No, since the computation halts in state q q1 1, which is , which is NOT NOT a final state a final state
- 8. 8 Deterministic Finite Automata (6) Deterministic Finite Automata (6) Extended transition function Extended transition function * * of a DFA with of a DFA with transition function transition function is a function from Q is a function from Q * * Q Q defined recursively on the length of the input string defined recursively on the length of the input string w w Basis: |w| = 0. Then w = and * (qi, ) = qi Recursive step: Let |w| = 1. Then * (qi, av) = * ((qi, a), v) qi Q, a , v *
- 9. 9 Deterministic Finite Automata (7) Deterministic Finite Automata (7) A string A string w w is accepted if is accepted if * * (q (q0 0, , w w) ) F F The language of a DFA The language of a DFA M M is is L(M) = {w * | * (q0, w) F} DFA M = (Q, DFA M = (Q, , , , q , q0 0, F) accepts , F) accepts w w * * * (q0, w) F
- 10. 10 Deterministic Finite Automata (8) Deterministic Finite Automata (8) Two possibilities for DFA M running on Two possibilities for DFA M running on w w M accepts w M accepts w iff the computation of M on w ends up (halts) in an accepting configuration * (q0, w) F M rejects w M rejects w iff the computation of M on w ends up (halts) in a rejecting configuration * (q0, w) F
- 11. 11 The state diagram of a DFA M = The state diagram of a DFA M = (Q, (Q, , , ,q ,q0 0, F) , F) is a is a labeled graph labeled graph G G defined by the following defined by the following conditions: conditions: The nodes of G are the elements of Q The labels on the arcs of G are elements of q0 is the start node, denoted by F is the set of accepting nodes, denoted by There is an arc from node qi to qj labeled a if (qi, a) = qj For every node qi and symbol a , there is exactly one arc labeled a leaving qi State Diagrams and Examples (1) State Diagrams and Examples (1) q0 a
- 12. 12 State Diagrams and Examples (2) State Diagrams and Examples (2) Deterministic Finite Automata DFA Deterministic Finite Automata DFA all outgoing edges are labelled with an input character no state has - transition, transition on input no two edges leaving a given state have the same label for each state s and input symbol a, there is at most one edge label a leaving s Therefore: the next state can be determined uniquely, given the current state and the current input character
- 13. 13 State Diagrams and Examples (1) State Diagrams and Examples (1) strings over { strings over {a a, ,b b} with at least 3 } with at least 3 a a’s ’s 1 a 2 a 3+ a 0 a a a a b b b
- 14. 14 State Diagrams and Examples (2) State Diagrams and Examples (2) strings over { strings over {a a, ,b b} with length mod 3 = 0 } with length mod 3 = 0 1 2 0
- 15. 15 State Diagrams and Examples (3) State Diagrams and Examples (3) strings over { strings over {a a, ,b b} } without without 3 consecutive 3 consecutive a a’s ’s A simple example of “strings not of the form …”. A simple example of “strings not of the form …”. 1 a 2 a Has aaa 0 a a a a b b b
- 16. 16 State Diagrams and Examples (4) State Diagrams and Examples (4) Draw a state diagram for DFA Draw a state diagram for DFA M M that accepts the that accepts the set of strings over {a, b} that contain the substring set of strings over {a, b} that contain the substring bb bb The string The string ababb ababb is accepted since the halting state is accepted since the halting state is the accepting state is the accepting state q q2 2 q1 q2 q0 b b a a a, b
- 17. 17 State Diagrams and Examples (5) State Diagrams and Examples (5) The DFA The DFA accepts ( accepts (b b| |ab ab) )* * ( (a a| | ) ) the set of strings over {a, b} that the set of strings over {a, b} that do not do not contain contain the substring the substring aa aa 1 a 2 a 0 a a a b b
- 18. 18 State Diagrams and Examples (6) State Diagrams and Examples (6) The language { The language {a an n b bn n , , n n 0} is not regular, so we can not 0} is not regular, so we can not build a DFA that accept this language build a DFA that accept this language It needs an infinite number of states It needs an infinite number of states But { But {a an n b bn n , 1 , 1 n n 3} is regular and its DFA is 3} is regular and its DFA is q0 a q1 a q2 a q3 q4 b q5 b b q6 b b This DFA is NOT Complete
- 19. 19 State Diagrams and Examples (7) State Diagrams and Examples (7) strings over {a, b} that contain the substring strings over {a, b} that contain the substring bb bb OR OR do not do not contain the substring contain the substring aa aa This language is the union of the languages of the This language is the union of the languages of the previous examples previous examples 1 a 2 a 0 a a a b a 1 b 2 b b b a 2 a b a b
- 20. 20 State Diagrams and Examples (8) State Diagrams and Examples (8) strings over {a, b} that contain an strings over {a, b} that contain an even even number of number of a a’s ’s AND AND an an odd odd number of number of b b’s ’s [ea, eb] [ea, ob] a [oa, eb] [oa, ob] b a a a b b b
- 21. 21 State Diagrams and Examples (9) State Diagrams and Examples (9) Let Let M M be the DFA previous slide be the DFA previous slide A DFA A DFA M’ M’ that accepts all strings over {a, b} that that accepts all strings over {a, b} that do not do not contain an contain an even even number of number of a a’s ’s AND AND an an odd odd number of number of b b’s ’s is shown below is shown below L(M’) = {a, b}* - L(M) = * - L(M) Any string accepted by Any string accepted by M M is rejected by is rejected by M’ M’ and vice versa and vice versa [ea, eb] [ea, ob] a [oa, eb] [oa, ob] b a a a b b b
- 22. 22 State Diagrams and Examples (10) State Diagrams and Examples (10) Let Let = {0, 1, 2, 3}. A string in = {0, 1, 2, 3}. A string in * * is a sequence of integers is a sequence of integers from from The DFA The DFA M M determines whether the determines whether the sum of elements sum of elements of a of a string is string is divisible by 4 divisible by 4 The string The string 12302 12302 and and 0130 0130 should be accepted and should be accepted and 0111 0111 rejected by M rejected by M w mod 4 = 0 0 2 2 1 3 w mod 4 = 1 0 w mod 4 = 2 0 1 3 w mod 4 = 3 0 2 2 1 3 1 3
- 23. 23 State Diagrams and Examples (11) State Diagrams and Examples (11) DFA DFA M1 M1 accepts accepts (ab) (ab)* * c c M1 M1 is is incomplete determinism incomplete determinism The string The string abcc abcc is is rejected rejected since since M1 M1 is unable to process is unable to process the final the final c c from state from state q q2 2 q0 q1 c q2 a b M1
- 24. 24 State Diagrams and Examples (12) State Diagrams and Examples (12) M2 M2 accepts the same language as accepts the same language as M1 M1 in previous example in previous example (ab) (ab)* * c c The state The state q qe e is the error state (dead end) is the error state (dead end) q0 q1 c q2 a b qe b a, c a, b, c M2
- 25. 25 State Diagrams and Examples (13) State Diagrams and Examples (13) strings over { strings over {a a, ,b b} with next-to-last symbol = } with next-to-last symbol = a a …aa …ab a …ba …bb a b b a b b b a a a a a b b b
- 26. 26 State Diagrams and Examples (14) State Diagrams and Examples (14) start 0 3 b 2 1 b a b a b a a What Language is Accepted?
- 27. 27 2.2: Nondeterministic Finite Automata (1) 2.2: Nondeterministic Finite Automata (1) Finite Automata : A recognizer that takes an input string & determines whether it’s a valid sentence of the language Non-Deterministic : Has more than one (or no) alternative action for the same input symbol. Deterministic : Has at most one action for every given input symbol.
- 28. 28 Nondeterministic Finite Automata (2) Nondeterministic Finite Automata (2) NFA: Formal Definition NFA: Formal Definition NFA M = (Q, NFA M = (Q, , , , q , q0 0, F) , F) Q = a finite set of states = a finite set alphabet = transition function total function Q (Q) = 2Q - power set of Q q0 = initial/starting state q0 Q F = final or accepting states F Q
- 29. 29 Nondeterministic Finite Automata (3) Nondeterministic Finite Automata (3) Q = { 0, 1, 2, 3 } q0 = 0 F = { 3 } = { a, b } start 0 3 b 2 1 b a a b What Language is defined ? What is the Transition Table ? s t a t e i n p u t 0 1 2 a b { 0, 1 } { 2 } { 3 } { 0 }
- 30. 30 Nondeterministic Finite Automata (4) Nondeterministic Finite Automata (4) Change in Change in For an DFA M, (q, a) results in one and only one state for all states q and alphabet a For an NFA M, (q, a) can result in a set of states, zero, one, or more states: qn a qi qn a qi qj qk a a qn (q (qn n, a) = {q , a) = {qi i} } (q (qn n, a) = {q , a) = {qi i, q , qj , j , q qk k } } (q (qn n, a) = {} = , a) = {} =
- 31. 31 Nondeterministic Finite Automata (5) Nondeterministic Finite Automata (5) 1 2 3 4 a a a b a,b a,b •Why is this only an NFA and not an DFA? 5
- 32. 32 Nondeterministic Finite Automata (6) Nondeterministic Finite Automata (6) Computing with NFA’s Computing with NFA’s Computations are different Computations are different We always start from start state. Call it the root of We always start from start state. Call it the root of the computation. the computation. Then we might go to different states on one symbol. Then we might go to different states on one symbol. Then from those states to new sets of states, Then from those states to new sets of states, creating a creating a tree-like tree-like computation. computation. If one path ends up in a final state, then ACCEPT, If one path ends up in a final state, then ACCEPT, else REJECT else REJECT
- 33. 33 Nondeterministic Finite Automata (7) Nondeterministic Finite Automata (7) start 0 3 b 2 1 b a a b • Given an input string, we trace moves • If no more input & in final state, ACCEPT EXAMPLE: Input: ababb Path 1: 0 -> 0 -> 0 -> 0 -> 0 -> 0 (REJECT) Path 2: 0 -> 0 -> 0 -> 1 -> 2 -> 3 (ACCEPT)
- 34. 34 Nondeterministic Finite Automata (8) Nondeterministic Finite Automata (8) Extended transition function Extended transition function * * of a NFA with of a NFA with transition function transition function is a function from Q is a function from Q * * 2 2Q Q (power set) defined recursively on the length of (power set) defined recursively on the length of the input string the input string w w Basis: |w| = 0. Then w = and * (qi, ) = {qi} Recursive step: Let |w| = 1. Then * (qi, av) = U * (qj, v), qj (qi, a) qi Q, qj Q , a , v * The language of a NFA The language of a NFA M M is is L(M) = {w * | * (q0, w) F ≠ }
- 35. 35 NFA (9) NFA (9) The state diagram The state diagram DFA DFA M1 and M1 and NFA NFA M2 accepts M2 accepts (a|b) (a|b)* * bb(a|b) bb(a|b)* * a a, b start 0 2 1 b b a M1 start 0 2 1 b b a, b a, b M2
- 36. 36 NFA (10) NFA (10) An NFA that accepts string over {a, b} with An NFA that accepts string over {a, b} with substring substring aa aa or or bb bb There are 2 distinct acceptance paths for the string There are 2 distinct acceptance paths for the string abaaabb abaaabb start 0 2 1 a a a, b a, b 4 3 b b a, b
- 37. 37 NFA (11) NFA (11) The state diagram The state diagram DFA DFA M1 and M1 and NFA NFA M2 accepts M2 accepts (a|b) (a|b)* * abba(a|b) abba(a|b)* * start 0 4 1 b a b a, b a M1 2 b 3 a a b start 0 4 1 b a a, b a, b M2 2 b 3 a
- 38. 38 NFA with NFA with -Transitions (1) -Transitions (1) NFA with NFA with -transitions, denoted by NFA- -transitions, denoted by NFA- Formal Definition Formal Definition NFA- NFA- M = (Q, M = (Q, , , , q , q0 0, F) , F) Q = a finite set of states = a finite set alphabet = transition function total function Q ( (Q) q0 = initial/starting state q0 Q F = final or accepting states F Q transitions j i Switch state but do not use any input symbol
- 39. 39 -Transitions (2) -Transitions (2) The input string The input string w w is accepted using is accepted using NFA- NFA- if there if there is a computation that processes the is a computation that processes the entire string entire string and halts in an and halts in an accepting state accepting state A computation may continue using A computation may continue using -transition -transition after the input string has been completely after the input string has been completely processed processed
- 40. 40 -Transitions (3) -Transitions (3) L(M L(M1 1) = (a|b) ) = (a|b)* * bb(a|b) bb(a|b)* * , L(M , L(M2 2) = (b|ab) ) = (b|ab)* * (a| (a| ) ) L(M) = L(M L(M) = L(M1 1) ) L(M L(M2 2) ) q1,1 q1,2 a, b b b q1,0 b a q2,0 0 a, b q2,1 b q1,1 q1,2 a, b b b q1,0 a, b b a q2,0 q2,1 b M1 M M2
- 41. 41 -Transitions (4) -Transitions (4) L(M L(M1 1) = (a|b) ) = (a|b)* * bb(a|b) bb(a|b)* * , L(M , L(M2 2) = (b|ab) ) = (b|ab)* * (a| (a| ) ) L(M) = L(M L(M) = L(M1 1)L(M )L(M2 2) ) q1,1 q1,2 a, b b b q1,0 b a q2,0 a, b q2,1 b q1,1 q1,2 a, b b b q1,0 a, b b a q2,0 q2,1 b M1 M M2
- 42. 42 -Transitions (5) -Transitions (5) M M accepts all strings over {a, b} of accepts all strings over {a, b} of even even length length q1 q2 a, b q0 a, b
- 43. 43 -Transitions (6) -Transitions (6) Let M Let M1 1 and M and M2 2 be 2 NFA- be 2 NFA- ’s as follows ’s as follows q1,0 q1,f M1 q2,0 q2,f M2 M1 = (Q1, , , q1,0, F1) M2 = (Q2, , , q2,0, F2)
- 44. 44 -Transitions (7) -Transitions (7) L(M) = L(M L(M) = L(M1 1) ) L(M L(M2 2) ) q1,0 q1,f M1 q2,0 q2,f M2 q0 M = (Q1 Q2 {q0}, , , q0, F1 F2) is the same as in 1 and 2, but add: (q0, ) = {q1,0, q2,0}
- 45. 45 -Transitions (8) -Transitions (8) L(M) = L(M L(M) = L(M1 1)L(M )L(M2 2) ) q1,0 q1,f M1 q2,0 q2,f M2 M = (Q1 Q2, , , q1,0, F2) is the same as in 1 and 2, but add: (q, ) = {q2,0}, where q F1
- 46. 46 -Transitions (9) -Transitions (9) L(M) = L(M L(M) = L(M1 1) )* * q1,0 q1,f M1 qf q0 M = (Q1 {q0, qf}, , , q0, {qf}) is the same as in 1, but add: (q0, ) = {q1,0 ,qf} (q1,f, ) = {q1,0 ,qf}, where q1,f F1
- 47. 47 -Transitions (10) -Transitions (10) Given the regular expression : a(b* c) | a( b | c+ | ) Find a transition diagram NFA- that recognizes it.
- 48. 48 -Transitions (11) -Transitions (11) 3 2 b c a 1 6 5 7 c a c 4 b a (b* c) a( b | c+ | ) Now that you have the individual diagrams, “or” them as follows:
- 49. 49 -Transitions (12) -Transitions (12) 3 2 b c a 1 6 5 7 c a c 4 b 0
- 50. 50 -Transitions (12) -Transitions (12) See examples 2.7, 2.8, 2.9, and 2.10 See examples 2.7, 2.8, 2.9, and 2.10
- 51. 51 2.3: Equivalence of DFA & NFA (1) 2.3: Equivalence of DFA & NFA (1) Is NFA more powerful than DFA? NO! Is NFA more powerful than DFA? NO! NFA inefficient to implement directly, so convert to a DFA that recognizes the same strings Is there a language accepted by an NFA that is not Is there a language accepted by an NFA that is not accepted by any DFA? No accepted by any DFA? No NFA is equivalent to DFA. NFA is equivalent to DFA. Each state in DFA corresponds to a SET of states of the NFA
- 52. 52 Removing Nondeterminism (2) Removing Nondeterminism (2) -closure -closure In an NFA- In an NFA- , the , the -closure( -closure(q q) ) of a state of a state q q is the set is the set of all states that can be reached from of all states that can be reached from q q by by following a path whose edges are all labeled by following a path whose edges are all labeled by . . q0 q1 q2 q3 b a b Start -closure(q0) = {q0, q1, q2, q3} - -closure(q1) = {q1,q3} - -closure(q2) = {q1,q2,q3}
- 53. 53 Removing Nondeterminism (3) Removing Nondeterminism (3) NFA- N = (Q, , , q0, F) -closure(q) : q Q : set of NFA- states that are reachable from q on -transitions only -closure(T) : T Q : NFA- states reachable from some state t T on -transitions only move(T, a) : T Q, a : set of states to which there is a transition on input a from some state t T move({q1, q2, …, qi}, a) = (q1, a) (q2, a) … (qi, a) These 3 operations are utilized by algorithms / techniques to facilitate the conversion process No input is consumed
- 54. 54 Removing Nondeterminism (4) Removing Nondeterminism (4) -closure(0) = {0, 1, 2, 4, 7} (all states reachable from 0 on -moves) -closure(0, 6) = -closure(0) -closure(6) = {0, 1, 2, 4, 7} {6, 1, 2, 4,7} = {0, 1, 2, 4, 6, 7} move({0, 1, 2, 4, 7}, a) = {3, 8} (since move(2, a)=3 and move(7, a)=8) 0 1 2 3 5 4 6 7 8 9 10 a a b b b start
- 55. 55 NFA- NFA- DFA DFA with Subset Construction with Subset Construction NFA- N DFA M construction: Given N = (Q, , , q0, F), define M = (Q’, , ’, q’0, F’) Q’ P(Q), all the possible NFA- states M could be in. q’0 = -closure(q0) F’ = {q’ Q’ | q’ F } ’(q’, a) = -closure(move(q’, a)) DFA M has a state ERR DFA M has a state ERR Q’ = P(Q) Q’ = P(Q) serves as the “error” state, when needed ’(ERR, a) = ERR, a By definition of ’
- 56. 56 Subset Construction Subset Construction Algorithm Concepts Algorithm Concepts First we calculate: -closure(0) (i.e., state 0) -closure(0) = {0, 1, 2, 4, 7} (all states reachable from 0 on -moves) Let A = {0, 1, 2, 4, 7} be a state of new DFA M Start with NFA-: (a | b)*abb 0 1 2 3 5 4 6 7 8 9 10 a a b b b start
- 57. 57 Conversion Example (1) Conversion Example (1) b : -closure(move(A,b)) = -closure(move({0,1,2,4,7}, b)) adds {5} (since move(4, b) = 5) From this we have : -closure({5}) = {1,2,4,5,6,7} (since 5614, 67, and 12 all by -moves) Let C = {1,2,4,5,6,7} be a new state in D. Define M(A, b) = C 2nd , we calculate : a : -closure(move(A, a)) and b : -closure(move(A, b)) a : -closure(move(A, a)) = -closure(move({0,1,2,4,7}, a))} adds {3, 8} (since move(2, a) = 3 and move(7, a) = 8) From this we have : -closure({3, 8}) = {1,2,3,4,6,7,8} (since 3614, 67, and 12 all by -moves) Let B = {1,2,3,4,6,7,8} be a new state in D. Define M(A, a) = B
- 58. 58 Conversion Example (2) Conversion Example (2) 3rd , we calculate for state B on {a, b} a : -closure(move(B, a)) = -closure(move({1,2,3,4,6,7,8}, a))} = {1,2,3,4,6,7,8} = B Define M(B, a) = B b : -closure(move(B, b)) = -closure(move({1,2,3,4,6,7,8}, b))} = {1,2,4,5,6,7,9} = D Define M(B, b) = D 4th , we calculate for state C on {a, b} a : -closure(move(C, a)) = -closure(move({1,2,4,5,6,7}, a))} = {1,2,3,4,6,7,8} = B Define M(C, a) = B b : -closure(move(C, b)) = -closure(move({1,2,4,5,6,7}, b))} = {1,2,4,5,6,7} = C Define M(C, b) = C
- 59. 59 Conversion Example (3) Conversion Example (3) 5th , we calculate for state D on {a, b} a : -closure(move(D, a)) = -closure(move({1,2,4,5,6,7,9}, a))} = {1,2,3,4,6,7,8} = B Define M(D, a) = B b : -closure(move(D, b)) = -closure(move({1,2,4,5,6,7,9}, b))} = {1,2,4,5,6,7,10} = E Define M(D, b) = E Finally, we calculate for state E on {a, b} a : -closure(move(E, a)) = -closure(move({1,2,4,5,6,7,10}, a))} = {1,2,3,4,6,7,8} = B Define M(E, a) = B b : -closure(move(E, b)) = -closure(move({1,2,4,5,6,7,10}, b))} = {1,2,4,5,6,7} = C Define M(E, b) = C
- 60. 60 Conversion Example (4) Conversion Example (4) State Input Symbol a b A B C B B D C B C E B C D B E … A …b C …a B … abD … abbE start b b b b b a a a a This gives the transition table for the DFA M of: a
- 61. 61 Converting NFA- Converting NFA- to DFA – 2 to DFA – 2nd nd Example Example 0 8 5 4 7 3 6 2 1 b a c RE: c*(ab)*c* 1st we calculate: -closure(0) -closure(0) = {0, 1, 2, 6, 8} (all states reachable from 0 on -moves) Let A={0, 1, 2, 6, 8} be a state of new DFA M
- 62. 62 Conversion Example (1) Conversion Example (1) b : -closure(move(A,b)) = -closure(move({0,1,2,6,8}, b)) There is NO transition on b for all states 0,1,2,6 and 8 Define M(A, b) = Reject 2nd , we calculate : a : -closure(move(A, a)) and b : -closure(move(A, b)) c : -closure(move(A, c)) a : -closure(move(A, a)) = -closure(move({0,1,2,6,8}, a))} adds {3} (since move(2, a) = 3) From this we have : -closure({3}) = {3} (since 33 by -moves) Let B = {3} be a new state. Define M(A, a) = B c : -closure(move(A, c)) = -closure(move({0,1,2,6,8}, c)) adds {7} (since move(6, c) = 7) From this we have : -closure({7}) = {1,2,5,6,7,8} (since 758, 7512, and 7516 by -moves) Let C = {1,2,5,6,7,8} be a new state. Define M(A, c) = C
- 63. 63 Conversion Example (2) Conversion Example (2) b : -closure(move(B, b)) = -closure(move({3}, b)) adds {4} (since move(3, b) = 4) From this we have : -closure({4}) = {1,2,4,5,6,8} (since 458, 4512, and 4516 by -moves) Let D = {1,2,4,5,6,8} be a new state. Define M(B, b) = D 3rd , we calculate : a : -closure(move(B, a)) and b : -closure(move(B, b)) c : -closure(move(B, c)) a : -closure(move(B, a)) = -cclosure(move({3}, a))} There is NO transition on a for state 3 Define M(B, a) = Reject c : -closure(move(B, c)) = -closure(move({3}, c)) There is NO transition on c for state 3 Define M(B, c) = Reject
- 64. 64 Conversion Example (3) Conversion Example (3) b : -closure(move(C, b)) = -closure(move({1,2,5,6,7,8},b)) There is NO transition on b for all states 1,2,5,6, and 7 Define M(C, b) = Reject 4th , we calculate : a : -closure(move(C, a)) and b : -closure(move(C, b)) c : -closure(move(C, c)) a : -cclosure(move(C, a)) = -closure(move({1,2,5,6,7,8},a))} adds {3} (since move(2, a) = 3) -closure({3}) = {3} Define M(C, a) = B c : -closure(move(C, c)) = -closure(move({1,2,5,6,7,8},c)) adds {7} (since move(6, c) = 7) From this we have : -closure({7}) = {1,2,5,6,7,8} (since 758, 7512, and 7516 by -moves) Define M(C, c) = C
- 65. 65 Conversion Example (4) Conversion Example (4) b : -closure(move(D, b)) = -closure(move({1,2,4,5,6,8}, b)) There is NO transition on b for all states 1,2,4,5,6, and 8 Define M(D, b) = Reject 5th , we calculate : a : -closure(move(D, a)) and b : -closure(move(D, b)) c : -closure(move(D, c)) a : -closure(move(D, a)) = -closure(move({1,2,4,5,6,8},a))} adds {3} (since move(2, a) = 3) -closure({3}) = {3} Define M(D, a) = B c : -closure(move(D, c)) = -closure(move({1,2,4,5,6,8},c)) adds {7} (since move(6, c) = 7) From this we have : -closure({7}) = {1,2,5,6,7,8} (since 758, 7512, and 7516 by -moves) Define M(D, c) = C
- 66. 66 The Resulting DFA M The Resulting DFA M Which States are FINAL States ? 1, 2, 5, 6, 7, 8 1, 2, 4, 5, 6, 8 0, 1, 2, 6, 8 3 c b a a a c c D C A B c b a a a c c
- 67. 67 NFA- NFA- DFA: Example DFA: Example strings over {a,b} with next-to-last symbol = a a a/b a/b q2 q1 q0 a b q0 {q0, q1} {q0} q1 {q2} {q2} q2 M = ({q0, q1, q2}, {a,b}, , q0, {q2}) q0 q2 q0 q0 q1 q0 q1 q2 a a a a b b b b Equivalent DFA Equivalent DFA
- 68. 68 An Example of NFA- An Example of NFA- DFA DFA Consider a simple NFA: Consider a simple NFA: Construct a corresponding DFA: Construct a corresponding DFA: 0 1 1 Start 0 1 q0 q1 {q0} {q0, q1} {q1} ERR ERR Start 1 0 1 0 1,0 1,0
- 69. 69 Examples Examples See examples 2.11, 2.12, and 2.13 See examples 2.11, 2.12, and 2.13
- 70. 70 Class Discussion Class Discussion Construct a DFA equivalent to this NFA: Construct a DFA equivalent to this NFA: Start 0,1 q0 q1 0 q2 1 {q0, q1} Start 1 {q0} 0 1 {q0,q2} 0 1 0 0
- 71. 71 b a,b,c An NFA a 1 2 a 0 b,c c start Notice that the state with label {0, 1, 2} is from the set of states given by the nondeterministic transition (0, a) = {0, 1, 2}. Also notice that any state whose label contains an accepting state is defined as an accepting state in the deterministic machine. {0,1,2} {2} a a {1,2} b c b c c Converted DFA 0 start b,c Class Discussion Class Discussion
- 72. 72 2.4 Reduction of the Number of States in FA 2.4 Reduction of the Number of States in FA Self study Self study but not included in the exams material