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COMP-330 Theory of Computation
Fall 2017 -- Prof. Claude Crépeau Lec. 7 : Regular
Expressions & GNFA
COMP 330 Fall 2012: Lectures Schedule
14. Context-free languages15. Pushdown automata16. Parsing17. The pumping lemma for CFLs18. Introduction to computability19. Models of computation Basic computability theory20. Reducibility, undecidability and Rice’s theorem21. Undecidable problems about CFGs22. Post Correspondence Problem23. Validity of FOL is RE / Gödel’s and Tarski’s thms24. Universality / The recursion theorem 25. Degrees of undecidability26. Introduction to complexity
1-2. Introduction 1.5. Some basic mathematics2-3. Deterministic finite automata +Closure properties,4. Nondeterministic finite automata5. Minimization 6. Determinization+Kleene’s theorem7. Regular Expressions+GNFA8. Regular Expressions and Languages
+ Myhill-Nerode theorem,9-10. The pumping lemma11. Duality12. Labelled transition systems13. MIDTERM
Regular Expressions
Regular Expressions
1*(01+)*
Regular Expressions vs Regular Languages
Regular Expressions vs Regular Languages
Regular Expressions vs Regular Languages
Let RA generating LA.
If RA is a symbol “a” then use .If RA is “𝞮” then use .If RA is “∅” then use .If RA is R1 ∪ R2 then use Thm 1.45 and recursively use N1 and N2 s.t. L(N1) = L(R1) and L(N2) = L(R2).If RA is R1 ∘ R2 then use Thm 1.47 and recursively use N1 and N2 s.t. L(N1) = L(R1) and L(N2) = L(R2).If RA is R* then use Thm 1.49 and recursively use N s.t. L(N)=L(R).
Regular Expressions generate Reg. Languages
a
Reg. Expression → NFA
Two examples
(ab∪a)*
(a∪b)*aba
Automata recognize Regular Expressions
Generalized NFA
Example of GNFA
q1
q2
Generalized NFA
Generalized NFA
Definition of GNFA
1. Q={qstart,q1,q2,qaccept}
2.∑={a,b}
3.δ is given as
4.qstart is the start state
5.qaccept is the (unique) accept state
q1
q2
δ q1 q2 qaccept
qstart ab* ∅ b
q1 aa a* ab∪ba
q2 (aa)* ab b*
Let G = (Q,∑,δ,qstart,qaccept) be a generalized nondeterministic finite state automaton and let w=w1w2...wn (n≥0) be a string where each sub-string wi∈∑*.
G accepts w if ∃ s0,s1,...,sn s.t. 1. s0 = qstart 2. wi ∈ L( δ(si-1,si) ) for i = 1 ... n, and 3. sn = qaccept
Definition of GNFA
Example of GNFA
q1
q2
abbbaaaababaaba
qstart
Example of GNFA
q1
q2
abbbaaaababaaba ab*
Example of GNFA
q1
q2
abbbaaaababaaba
a*
Example of GNFA
q1
q2
abbbaaaababaaba
ab
Example of GNFA
q1
q2
abbbaaaababaaba
ab
Example of GNFA
q1
q2
abbbaaaababaaba
(aa)*
Example of GNFA
q1
q2
abbbaaaababaaba
ab∪ba
qaccept
DFA → GNFA → Reg. Exp.
Example NFA→GNFA
N1
Example NFA→GNFA
N1qstart
Example NFA→GNFA
N1𝞮qstart
Example NFA→GNFA
N1𝞮qstart
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
0∪1
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
∅
0∪1
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
0∪1
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
∅
0∪1
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
0∪1
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
0∪1
∅
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
0∪1
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
∅
0∪1
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
0∪1
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
0∪1
∅
Example NFA→GNFA
N1𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
0∪1
Example NFA→GNFA
N1
G1
𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
0∪1
Example NFA→GNFA
N1
G1
𝞮qstart
𝞮
qacc
0∪1
0∪𝞮
∅
∅ ∅
0∪1
∅
∅
DFA → GNFA → Reg. Exp.
Ripping a state
GNFA→ Reg. Exp.
Ripping of a GNFA
q1
q2
Ripping q1
Ripping of a GNFA
q2
Ripping q1
R4
R1 R2
R3
Ripping of a GNFA
q2
Ripping q1
∅∪ab*(aa)*a*
Ripping of a GNFA
q2
Ripping q1
∅∪ab*(aa)*a*
R4
R1 R2
R3
Ripping of a GNFA
q2
Ripping q1
∅∪ab*(aa)*a*
b∪ab*(aa)*(ab∪ba)
Ripping of a GNFA
q2
Ripping q1
∅∪ab*(aa)*a*
b∪ab*(aa)*(ab∪ba)
R4
R1
R2
R3
Ripping of a GNFA
q2
Ripping q1
∅∪ab*(aa)*a*
b∪ab*(aa)*(ab∪ba)
ab∪(aa)*(aa)*a*
Ripping of a GNFA
q2
Ripping q1
∅∪ab*(aa)*a*
b∪ab*(aa)*(ab∪ba)
ab∪(aa)*(aa)*a*R4
R2
R3
R1
Ripping of a GNFA
q2
∅∪ab*(aa)*a*
b∪ab*(aa)*(ab∪ba)
ab∪(aa)*(aa)*a*b*∪(aa)*(aa)*(ab∪ba)
Ripping of a GNFA
q2
∅∪ab*(aa)*a*
b∪ab*(aa)*(ab∪ba)
ab∪(aa)*(aa)*a*b*∪(aa)*(aa)*(ab∪ba)
Ripping of a GNFA
q2
∅∪ab*(aa)*a*
b∪ab*(aa)*(ab∪ba)
Ripping q2ab∪(aa)*(aa)*a*b*∪(aa)*(aa)*(ab∪ba)
Ripping of a GNFA
q2
∅∪ab*(aa)*a*
b∪ab*(aa)*(ab∪ba)
Ripping q2ab∪(aa)*(aa)*a*b*∪(aa)*(aa)*(ab∪ba)
R4
R1
R2
R3
Ripping of a GNFA
q2(b∪ab*(aa)*(ab∪ba))
∪(∅∪ab*(aa)*a*)
(ab∪(aa)*(aa)*a*)*(b*∪(aa)*(aa)*(ab∪ba))
GNFA → Reg. Expression
“equivalent” means L( CONVERT(G) ) = L(G)
GNFA → Reg. Expression
Induction basis
Let G be a GNFA with exactly k=2 states. Because of the special form of our GNFA, the two states are the start and accept states. The regular expression on the transition from qstart to qaccept generates the language accepted by this GNFA.
Induction step
Let G be a GNFA with exactly k>2 states. We assume for induction hypothesis that all GNFA G’ of k-1 states accept the laguage defined by the regular expression obtained via CONVERT, i.e. L(G’)=L(CONVERT(G’)).
Since k>2 then there exists at least one state qrip which is neither qstart nor qaccept.
GNFA → Reg. Exp.
Induction step
Let G’ be, as in CONVERT, the GNFA obtained after ripping qrip from G.
Let w be a string accepted by G, w∈L(G). C o n s i d e r a n a c c e pt i n g s e q u e n c e qstart,q1,q2,...,qaccept for string w.
GNFA → Reg. Exp.
If qrip is not a state of the sequence, then the very same exact sequence will accept w in G’ because its transitions R4 contain all those R4 in G (except for qrip) in a union with new possibilities related to ripping qrip.
GNFA → Reg. Exp.
If qrip is a state of the sequence, then the same sequence (but with all qrip removed) will accept w in G’. That’s because any three elements in a row qi,qrip,qj (qi≠qrip≠qj) in G’s accepting sequence, will be processed identically through states qi,qj in G’. Remember that the transitions for qi,qj in G’ contain all those R1(R2)*R3 from G involving qrip in a union with older possibilities (R4). (we can deal with qi,qrip,...,qrip,qj similarly.)
GNFA → Reg. Exp.
This proved “if w∈L(G) then w∈L(G’)”. We should also prove “if w∈L(G’) then w∈L(G)”.
Let w be a string accepted by G’, i.e. w∈L(G’). Consider an accepting sequence qstart,q1,q2,...,qaccept for string w. Consider any two consecutive states qi,qi+1. The same portion of w is processed in G in either part of the union, R1(R2)*R3 or R4, along the transition between qi and qi+1.
GNFA → Reg. Exp.
If the portion of w is generated by R4 in G’ then it is also generated by R4 in G. If the portion of w is generated by R1(R2)*R3 in G’ then there exists an m such that it is generated by R1(R2)mR3 and it is also generated in G by R1, going through qrip m times via R2 and finally R3. Thus qi,qi+1 is replaced by qi,qrip,...,qrip,qi+1.
We conclude that if w∈L(G’) then w∈L(G).
GNFA → Reg. Exp.
GNFA → Reg. Exp.
Combining both statements we get L(G’)=L(G).
By induction hypothesis L(G’)=L(CONVERT(G’)) becau se G’ conta ins k-1 state s. By construction, CONVERT(G)=CONVERT(G’). Therefore L(CONVERT(G))=L(CONVERT(G’))=L(G’)=L(G). QED
COMP-330 Theory of Computation
Fall 2017 -- Prof. Claude Crépeau Lec. 7 : Regular
Expressions & GNFA
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