Running an NFA & the subset algorithm (NFA->DFA) CS 350 Fall 2018 gilray.org/classes/fall2018/cs350/

Transcription

1 Running n NFA & the suset lgorithm (NFA->DFA) CS 350 Fll 2018 gilry.org/lsses/fll2018/s350/ 1

2 NFAs operte y simultneously exploring ll pths nd epting if ny pth termintes t n ept stte.!2

3 Try n exmple: L is the lnguge L(( )*). q0 q1,!3

4 Let s run this nondeterministilly on the string nd trk wht pths we n tke: q0 q1,!4

5 Let s run this nondeterministilly on the string nd trk wht pths we n tke: q0 q1,!5

6 Let s run this nondeterministilly on the string / nd trk wht pths we n tke: q0 q1,!6

7 Let s run this nondeterministilly on the string // nd trk wht pths we n tke: q0 q1,!7

8 Let s run this nondeterministilly on the string /// nd trk wht pths we n tke: q0 q1,!8

9 Let s run this nondeterministilly on the string //// nd trk wht pths we n tke: q0 q1,!9

10 Let s run this nondeterministilly on the string ///// nd trk wht pths we n tke: If ny possile pth would ept, then the NFA epts. q0 q1,!10

11 Consider tht wht n e deterministilly known out n NFA, fter onsuming string, is not single stte the mhine must oupy, ut the set of NFA sttes the mhine ould oupy.!11

12 Try n exmple: L(( )( )*) L is the lnguge.!12

13 Try n exmple: L(( )( )*) L is the lnguge. q0 q1 q4 q5!13

14 Try n exmple: L(( )( )*) L is the lnguge. q0 q1 q6 q4 q5!14

15 Try n exmple: L(( )( )*) L is the lnguge. q0 q1 q7 q8 q6 q9 q4 q5 q10!15

16 Try n exmple: L(( )( )*) L is the lnguge. q0 q1 q11 q7 q8 q6 q9 q4 q5 q10!16

17 Try n exmple: L(( )( )*) L is the lnguge. q0 q1 q11 q7 q8 q6 q9 q4 q5 q10!17

18 Now onsider the sttes visited t eh step, running the string: {q 9 } q0 q1 q11 q7 q8 q6 q9 L(( )( )*) q4!18 q5 q10

19 Now onsider the sttes visited t eh step, running the string: {q 9 } {q 9, q 4, q 6, q 10, q 11, q 7, q 2, q 0 } q0 q1 q11 q7 q8 q6 q9 L(( )( )*) q4!19 q5 q10

20 The ε-losure of set of sttes is the set of sttes whih re rehle without onsuming ny hrters/strings. (Another fixpoint itertion.) losure δ : P(Q) P(Q) losure δ (qs) = ext N δ (qs) where ext N 1 (qs) = ext δ N δ (qs) ext δ (qs) = qs {q q qs (q,, q ) δ}!20

21 Now onsider the sttes visited t eh step, running the string: {q 9 } {q 9, q 4, q 6, q 10, q 11, q 7, q 2, q 0 } q0 q1 q11 q7 q8 q6 q9 L(( )( )*) q4!21 q5 q10

22 Now onsider the sttes visited t eh step, running the string: {q 9, q 4, q 6, q 10, q 11, q 7, q 2, q 0 } {q 5 } / q0 q1 q11 q7 q8 q6 q9 L(( )( )*) q4!22 q5 q10

23 The move of set of sttes nd hrter s, is the set of sttes whih re rehle from ny input stte, vi single trnsition on the hrter s. move δ : P(Q) Σ P(Q) move δ (qs, s) = {q q qs (q, s, q ) δ}!23

24 Now onsider the sttes visited t eh step, running the string: {q 9, q 4, q 6, q 10, q 11, q 7, q 2, q 0 } {q 5 } / q0 q1 q11 q7 q8 q6 q9 L(( )( )*) q4!24 q5 q10

25 Now onsider the sttes visited t eh step, running the string: / {q 9, q 4, q 6, q 10, q 11, q 7, q 2, q 0 } {q 5, q 10, q 11, q 7, q 2, q 0 } q0 q1 q11 q7 q8 q6 q9 L(( )( )*) q4!25 q5 q10

26 Now onsider the sttes visited t eh step, running the string: {q 5, q 10, q 11, q 7, q 2, q 0 } {q 3 } // q0 q1 q11 q7 q8 q6 q9 L(( )( )*) q4!26 q5 q10

27 Now onsider the sttes visited t eh step, running the string: {q 5, q 10, q 11, q 7, q 2, q 0 } // {q 3, q 8, q 11, q 7, q 2, q 0 } q0 q1 q11 q7 q8 q6 q9 L(( )( )*) q4!27 q5 q10

28 Now onsider the sttes visited t eh step, running the string: {q 3, q 8, q 11, q 7, q 2, q 0 } {q 1 } /// q0 q1 q11 q7 q8 q6 q9 L(( )( )*) q4!28 q5 q10

29 Now onsider the sttes visited t eh step, running the string: {q 3, q 8, q 11, q 7, q 2, q 0 } /// {q 1, q 8, q 11, q 7, q 2, q 0 } q0 q1 q11 q7 q8 q6 q9 L(( )( )*) q4!29 q5 q10

30 Now onsider the sttes visited t eh step, running the string: {q 1, q 8, q 11, q 7, q 2, q 0 } {q 3 } //// q0 q1 q11 q7 q8 q6 q9 L(( )( )*) q4!30 q5 q10

31 Now onsider the sttes visited t eh step, running the string: {q 1, q 8, q 11, q 7, q 2, q 0 } {q 3, q 8, q 11, q 7, q 2, q 0 } //// q0 q1 q11 q7 q8 q6 q9 L(( )( )*) q4!31 q5 q10

32 Now onsider the sttes visited t eh step, running the string: //// {q 3, q 8, q 11, q 7, q 2, q 0 } {q 11 } Ø q0 q1 q11 q7 q8 q6 q9 q10 L(( )( )*) q4!32 q5

33 So, if we n deterministilly step through n NFA using sets of NFA sttes insted of single NFA sttes, n we produe DFA from n NFA tht enodes these sets of sttes? {q 9, q 4, q 6, q 10, q 11, q 7, q 2, q 0 } YES {q 5, q 10, q 11, q 7, q 2, q 0 } {q 3, q 8, q 11, q 7, q 2, q 0 } {q 1, q 8, q 11, q 7, q 2, q 0 }!33

34 So, if we n deterministilly step through n NFA using sets of NFA sttes insted of single NFA sttes, n we produe DFA from n NFA tht enodes these sets of sttes? {q 9, q 4, q 6, q 10, q 11, q 7, q 2, q 0 } YES {q 5, q 10, q 11, q 7, q 2, q 0 } {q 3, q 8, q 11, q 7, q 2, q 0 } Wht we just worked through is just frgment of the DFA enoding this NFA! {q 1, q 8, q 11, q 7, q 2, q 0 }!34

35 Models of regulr lnguges Converts to GNFA Converts to RE DFA Minimizes to Converts to NFA Converts to!35

36 The suset lgorithm!36

37 The suset lgorithm Input: NFA (Q, Σ, δq, q0, FQ) Output: DFA (D, Σ, δd, d0, FD) Let d0 = ε-losureδ({q0}) mrked = {} D = {d0} δd = {} While d D suh tht d mrked: mrked = mrked {d} For eh s Σ: Let dmv = moveδ(d,s) Let dmv+ε = ε-losureδ(dmv) If dmv+ε D: D = D {dmv+ε} δd = δd {(d,s,dmv+ε)} FD = { d D q d. q FQ}!37

38 Try n exmple: L(( )( )*) L is the lnguge. q0 q1 q11 q7 q8 q6 q9 q4 q5 q10!38

39 Try n exmple: L(( )( )*) L is the lnguge. {q 5, q 10, q 11, q 7, q 2, q 0 } {q 1, q 8, q 11, q 7, q 2, q 0 } d1 d0 d2 d3 {q 9, q 4, q 6, q 10, q 11, q 7, q 2, q 0 } {q 3, q 8, q 11, q 7, q 2, q 0 }!39