Theory of Computation

Theory of Computation Theory of Computation Peer Instruction Lecture Slides by Dr. Cynthia Lee, UCSD are licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.Based on a work at www.peerinstruction4cs.org.

Extra Credit Quiz • If there is somebody in your group who seems really smart, it is a good idea to just let that person do all the talking like a “mini-teacher,” so that you can learn from them. • TRUE • FALSE

They’re really good guessers! Nondeterministic Finite AutomataNFA What is deterministic?

DFA or NFA? DFA NFA Both DFA and NFA

Tracing in an NFA “100” • What are the two sequences of states on the input “100”? • (q0,q0,q1,q2[accept]), (q0,q1,q2[accept]) Final: Accept • (q0,q0,q1,q2[accept]), (q0,q1,q2[reject]) Final: Accept • (q0,q0,q1,q2[accept]), (q0,q1,q2[reject]) Final: Reject • (q0,q0,q1,q2[reject]), (q0,q1,q2[reject]) Final: Reject

DFAs vs. NFAs DFAs 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”) • 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

Formal Definition of an NFA • A DFA 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!

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) • TRUE • FALSE NEED TO USE THIS FOR PROOFS---CANNOT MAKE GENERAL STATEMENTS BY DRAWING SPECIFIC EXAMPLES!

A different (easier!) way to prove what we just proved Reg. Langs. Closed Under Union(Why NFAs are So Useful in proofs)

Final form for your homework/test: 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 = Q1x 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 F1or 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. Could you come up with this, and write it correctly, in a short amount of time on an exam??

Final form for your homework/test: 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 an NFAM = (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. Could you come up with this, and write it correctly, in a short amount of time on an exam??

Back to our working example

Another construction proof Equivalence of Finite AutomataNFA & DFA

Tracing in NFA (Fig. 1.27 in your book) Run this NFA on input 010110 Fill in the missing row of states: • q2, q3, q4, q4, q4 • q1, q2, q4, q4, q4 • q1, q2, q3, q4, q4 • None of the above • I don’t understand this at all

Tracing in NFA (Fig. 1.27 in your book) Run this NFA on input 010110 • Each row is a set of states that we are in at the “same time” • {q1} • {q1,q2,q3} • {q1,q3} • {q1,q2,q3,q4} • {q1,q3,q4} • Recall that when we did the union closure proof with DFAs, we were always in a pair of states at the “same time”—same concept • What are all the possible unique sets of states? (a) QxQ (b) |Q|2 (c) P(Q) (d) |Q|! (e) I don’t understand this at all

Thm. 1.39: Every NFA has an equivalent DFA. Talk about sigma • Given: NFA M = (Q,Σ,δ,q0,F) • Want: DFA M’ = (Q’,Σ,δ’,q0’,F’) s.t. L(M) = L(M’). • Construction: //need to make a DFA that simulates nondeterminism within the constraints of determinism (!!) • Q’ = • Σ • δ’ • q0’ = • F’ = • A DFA recognizes L(M), then every NFA has an equivalent DFA. Q.E.D. P(Q) d’(R,a) = {q | q in d(r,a) for some r in R OR q can be reached by following 1+ epsilon edges from some r in R} q’0 = {q0} U {q in Q | q can be reached by following 1+ epsilon edges from q0} { R in Q’ | R contains at least one accept state of M}

Thm. 1.39: Every NFA has an equivalent DFA. • We also know every DFA isan NFA. • Corollary of these two facts: • The class of languages recognized by DFAs and the class of languages recognized by NFAs are the same class • The Class of Regular Languages • It may be surprising that adding something as powerful and magic-seeming as instantaneous, 100% accurate guessing to the DFA model could turn out to not increase the power of the model! • This course is full of surprises!

Extremely useful Regular ExpressionsPattern Matching

From the Reading Quiz • Let L be the language of this regular expression: 1*0 • Which of the following is NOT in L? • 10 • 100 • 110 • They’re all in L

Regular Expressions and UNIX/Linux Shell Wildcard • 1*0 = “Any number of 1’s (including no 1’s), followed by a single 0” • “100” not accepted by that RE • This may be confusing to those who are used to *’s usage in, say, UNIX/Linux shell—VERY DIFFERENT • In UNIX/Linux * means “replace with anything here” • 1*0 matches “100” and “1blahblahblah0” • Typical shell usage: “ls *.jpg” (to list all JPEG files) • This is very useful to know! • Just don’t confuse it with REs in this course

Regular Expressions • Let L be the language of this regular expression: ((a U Ø)+b*)* • Which of the following is NOT true of L? • Some strings in L have equal numbers of a’s and b’s • All strings in L have more b’s than a’s • L contains “aaaaaa” • a‘s never follow b’s in any string in L • None or more than one of the above

Regular Expressions • Let L be the language of this regular expression: aØb* • Which of the following is NOT true of L? • L is the empty set • L contains “a” • aaab is not in L • None or more than one of the above

Theory of Computation