CSE 105 Theory of Computation

CSE 105Theory of

Computation

Alexander TsiatasSpring 2012

Theory of Computation Lecture Slides by Alexander Tsiatas is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.Based on a work at http://peerinstruction4cs.org.Permissions beyond the scope of this license may be available at http://peerinstruction4cs.org.

CLOSURE PROOFSREVIEWING OUR EXAMPLE

Guaranteed to be on exams

From last time

• Are regular languages closed under union?

• a) True• b) False

Union of two regular languages• L1 = {w | b’s never appear after a’s}

• L2 = {w | length(w) is odd}

• L1 U L2 = {w | b’s never appear after a’s OR length(w) is odd}

Regular because there’s a DFA

Union of two regular languages

• L1 U L2 = {w | b’s never appear after a’s OR length(w) is odd}

• Does this prove that regular languages are closed under union?– (a) True or (b) False

It’s just one example

Thm. The class of regular languages is closed under the union operation.

• Proof:• Given: Two regular languages L1, L2.

• Want to show: L1 U L2 is regular.

• Because L1 and L2 are regular, we know there exist DFAs M1 = (Q1,Σ,δ1,q01,F1) and M2 = (Q2,Σ,δ2,q02,F2) that recognize L1 and L2. We construct a DFA M = (Q,Σ,δ,q0,F), s.t.:– Q = Q1 x Q2

– δ((x,y),c) = (δ1(x,c), δ2 (y,c)), for c in Σ and (x,y) in Q– q0 = (q01, q02)– F = {(x,y) in Q | x in F1 or y in F2}

• M recognizes L1 U L2.• Correctness: ___________________________________• A DFA recognizes L1 U L2, so L1 U L2 is regular, and the class of regular languages

is closed under union. Q.E.D.

Final form for your homework/test:

Could you come up with this, and write it correctly, in a short amount of time on an exam??

NONDETERMINISTIC FINITE AUTOMATANFA

They’re really good guessers!

What is deterministic?

DFA or NFA?

a) DFAb) NFAc) Both DFA and NFA

DFA or NFA?

a) DFAb) NFAc) Both DFA and NFA

Tracing in an NFA• What are the two sequences of

states on the input “100”?a) (q0,q0,q1,q2[accept]),

(q0,q1,q2[accept]) Final: Accept

b) (q0,q0,q1,q2[accept]), (q0,q1,q2[reject]) Final: Accept

c) (q0,q0,q1,q2[accept]), (q0,q1,q2[reject]) Final: Reject

d) (q0,q0,q1,q2[reject]), (q0,q1,q2[reject]) Final: Reject

“100”

b

DFAs vs. NFAs

DFAs• For each character in the

alphabet, exactly one transition leaving every state

• Computation is “deterministic,” i.e. determined by the input, i.e., the same every time for a given input

NFAs• There may be 0, 1, or many

transitions leaving a single state for the same input character

• Transition function defined on “epsilon” in addition to alphabet characters

• There may be several different ways to reach an accept state for a single string—the computation may not determined by the input (“nondeterministic”)

12

Formal Definition of an NFA

• An NFA M1 is defined as a 5-tuple as follows:

• M1 = (Q, Σ, δ, q0, F), where:– Q is a finite set of states– Σ is a finite set of characters, the alphabet– δ: Q x Σ -> P(Q), the transition function– q0, a member of Q, the start state– F, a subset of Q, the accept state(s)

NEED TO USE THIS FOR PROOFS---CANNOT MAKE GENERAL STATEMENTS BY DRAWING SPECIFIC

EXAMPLES!

WHAT IS the POWER SET?

13

Nondeterminism

• Because NFAs are non-deterministic, the outcome (accept/reject) of the computation may be different from run to run (i.e., isn’t determined by the input)

a) TRUEb) FALSE

REG. LANGS. CLOSED UNDER UNION(WHY NFAS ARE SO USEFUL IN PROOFS)

A different (easier!) way to prove what we just proved




• Because L1 and L2 are regular, we know there exist DFAs M1 = (Q1,Σ,δ1,q01,F1) and M2 = (Q2,Σ,δ2,q02,F2) that recognize L1 and L2. We construct a DFA M = (Q,Σ,δ,q0,F), s.t.:– Q = Q1 x Q2

– δ((x,y),c) = (δ1(x,c), δ2 (y,c)), for c in Σ and (x,y) in Q– q0 = (q01, q02)– F = {(x,y) in Q | x in F1 or y in F2}

• M recognizes L1 U L2.• Correctness: ___________________________________• A DFA recognizes L1 U L2, so L1 U L2 is regular, and the class of regular languages

is closed under union. Q.E.D.






• Because L1 and L2 are regular, we know there exist DFAs M1 = (Q1,Σ,δ1,q01,F1) and M2 = (Q2,Σ,δ2,q02,F2) that recognize L1 and L2. We construct an NFA M = (Q,Σ,δ,q0,F), s.t.:– Q =– δ(x,c) =– q0 =– F =

• M recognizes L1 U L2.• Correctness: ___________________________________• An NFA recognizes L1 U L2, so L1 U L2 is regular, and the class of regular

languages is closed under union. Q.E.D.



Back to our working example




• Because L1 and L2 are regular, we know there exist DFAs M1 = (Q1,Σ,δ1,q01,F1) and M2 = (Q2,Σ,δ2,q02,F2) that recognize L1 and L2. We construct an NFA M = (Q,Σ,δ,q0,F), s.t.:– Q = – δ(x,c) =– q0 is a new state not in Q1 or Q2

– F =




Discrepancy here NFA vs DFA


• Because L1 and L2 are regular, we know there exist DFAs M1 = (Q1,Σ,δ1,q01,F1) and M2 = (Q2,Σ,δ2,q02,F2) that recognize L1 and L2. We construct an NFA M = (Q,Σ,δ,q0,F), s.t.:– Q = ?– δ(x,c) =– q0 is a new state not in Q1 or Q2

– F = ?

– a) Q = Q1 X Q2, F = F1 X F2

– b) Q = Q1 X Q2, F = {(x,y) in Q | x in F1 or y in F2}

– c) Q = Q1 U Q2, F = F1 U F2

– d) Q = Q1 U Q2 U {q0}, F = F1 U F2

Why q0 in {}?




• Because L1 and L2 are regular, we know there exist DFAs M1 = (Q1,Σ,δ1,q01,F1) and M2 = (Q2,Σ,δ2,q02,F2) that recognize L1 and L2. We construct an NFA M = (Q,Σ,δ,q0,F), s.t.:– Q = Q1 U Q2

– δ(x,c) = ?– q0 is a new state not in Q1 or Q2

– F = F1 U F2






– Q = Q1 U Q2

– δ(x,c) = • δ1(x,c) if x in Q1

• δ2(x,c) if x in Q2

• {q01,q02} if x = q0 and c = ε• { } if x = q0 and c != ε

– q0 is a new state not in Q1 or Q2

– F = F1 U F2

Almost done: we have shown there is an NFA accepting L1 U L2. Does this mean it’s regular? Next class!

Zoomed in


CSE 105 Theory of Computation

Documents

regular languages l1

regular languagesl1

class of regular languages

q1 x q2x

timeare regular languages

nfa dfa

q x pq

member of q