INHERENT LIMITATIONS OF COMPUTER PROGAMS CSci 4011

QUIZ 2 The regular pumping lemma says that… if w  L and |w| ≥ P, then Ǝ x,y,z so w = xyz and: (1) |xy| ≤ P (2) |y| > 0 (3) For every i ≥ 0, xy i z  L. Give a CFG for the language {a n b 2n : n ≥ 0 }: S → aSbb | ε A PDA transition function takes as input a state, an input symbol, and: a stack symbol If P is a PDA then L(P) is the set: { w | P accepts w } Let G = (V,Σ,R,S) be a CFG then R is the set of: Production rules

What are practical uses for PDAs? Why are pumping lemmas annoying? Buddhist temple, CFG, levitating during meditation? What will be on the upcoming midterm? WTF Implementation of PL? Context-Free Pumping Lemma? Why do I care about CNF? Where are we going with this? What does it mean for something to be context free? How does a CFG relate to a computer? What is your favorite topic in this class? Is there a rule for the CF pumping lemma like |xy|≤P? Is there a multi-stack PDA? How many more types of languages are there? PDA/CFG conversions? CNF Unit rule redundancy? Do we need to prove every case for the pumping lemma? Is this class graded on a curve?

I will be in my office: Today 4-5:30pm and Thurs. 9:00am-11:00am HW4 SOLUTIONS POSTED TOMORROW AT NOON.

An alphabet Σ is a finite set (e.g., Σ = {0,1}) A string over Σ is a finite-length sequence of elements of Σ For x a string, |x| isthe length of x The unique string of length 0 will be denoted by ε and will be called the empty or null string NOTATION A language over Σ is a set of strings over Σ

Q = {q 0, q 1, q 2, q 3 } are states Σ = {0,1} is the alphabet  : Q  Σ → Q transition function * q 0  Q is start state F = {q 1, q 2 }  Q accept states M = (Q, Σ, , q 0, F) where  01 q0q0 q0q0 q1q1 q1q1 q2q2 q2q2 q2q2 q3q3 q2q2 q3q3 q0q0 q2q2 * q2q2 0 0, q0q0 q1q1 q3q3 M NFAs have “choices” --  : Q  Σ  P (Q), and accept on input w if there is a path from q 0 to a q a ∈ F.

DFA NFA Regular Language Regular Expression DEF

SOME LANGUAGES ARE NOT REGULAR! B = {0 n 1 n | n ≥ 0} is NOT regular! Suppose B is regular and let P be the pumping length. The string 0 P 1 P  B and has length at least P. So by the pumping lemma, there should be xyz = 0 P 1 P so that |xy| ≤ P, |y| > 0, and xyyz  B. But y must be only 0s (otherwise |xy| > P), so xyyz has at least P+1 “0”s (because |y| > 0) but only P “1”s. So xyyz  B, contradicting the pumping lemma. Therefore B is not regular.

ε,ε → $ 0,ε → 0 1,0 → ε ε,$ → ε stringpoppush A PDA accepts the string x if there is a path on input x and empty stack from q 0 to some q a 2 F. Following transition “a,b  c” reads “a” from the input, pops “b” off the stack, and pushes “c” onto it.

A → 0A1 A → B B → # CONTEXT-FREE GRAMMARS A  0A1  00A11  00B11  00#11 A derives 00#11 in 4 steps. uVw yields uvw if (V → v) 2 R. variables terminals production rules start variable

PARSE TREES A  0A1  00A11  00B11  00#11 A B 01# A A 01 Each node corresponds to a rule in the derivation. A string is derived ambiguously if it has ≥ 2 parse trees. A grammar is ambiguous if it derives a string ambiguously.

THE CHOMSKY NORMAL FORM A context-free grammar is in Chomsky normal form if every rule is of the form: A → BC A → a S → ε B and C are not start variable a is a terminal S is the start variable Any variable A that is not the start variable can only generate strings of length > 0 Theorem: If G is in CNF, w  L(G) and |w| > 0, then any derivation of w in G has length ≤ 2|w| - 1

THE CONTEXT-FREE PUMPING LEMMA Let L be a context-free language Then there exists P such that For every w  L with |w| ≥ P 1. |vy| > 0 there exist uvxyz=w, where: 3. uv i xy i z  L for any i ≥ 0 2. |vxy| ≤ P { ww | w ∈ Σ* } is not regular!

USING THE PUMPING LEMMA Prove L = {w#w R | w ∈ {0,1}* and #1s = #0s} is not context-free Assume L is context-free. Then there is a pumping length P. No matter what P is, the string s = 0 P 1 P #1 P 0 P has |s| ≥ P and s ∈ L. So there should be uvxyz=s with: 1) |vy| > 0, 2) |vxy| ≤ P, 3) ∀ i, uv i xy i z ∈ L. s = 00…0011…11#11…1100…00 P P P P

USING THE PUMPING LEMMA Prove L = {w#w R | w ∈ {0,1}* and #1s = #0s} is not context-free Assume L is context-free. Then there is a pumping length P. No matter what P is, the string s = 0 P 1 P #1 P 0 P has |s| ¸ P and s 2 L. So there should be uvxyz=s with: 1) |vy| > 0, 2) |vxy| ≤ P, 3) ∀ i, uv i xy i z ∈ L. s = 00…0011…11#11…1100…00 P P P P vxy cannot be only in either half, since pumping would make one side of # longer than the other. # cannot be in v or y since pumping would add too many #s. So it must be in x. vxy since # is in x and |vxy| · P, neither v nor y can have 0s, and at least one must have a 1. Then uv 2 xy 2 z has more 1s than 0s and is not in L.

CONTEXT-FREE LANGUAGES REGULAR LANGUAGES NON-REGULAR NON-CFL FINAL JEOPARDY

A DFA for the language (ab ∪ ba)*

A regular expression for the set of strings accepted by the NFA: 0,1 1

A proof that the language { w  Σ*| w ≠ w R } is not regular.

The language accepted by the following PDA: ε,ε → $ 0,ε → 0 1,0 → ε ε,0 → ε

A CFG for the language { w # b n | |w| = n }.

Regular operations that the Context-Free Languages are closed under. DAILY DOUBLE

A class of strings in the language L = { 1 n #1 n | n ≥ 0} that contradicts the regular pumping lemma for each pumping length p.

A class of strings in the language {a n #a n #a n | n ≥ 0 } that contradicts the context-free pumping lemma for each pumping length p.

A proof that the language { a n #b m | m = n 2, n ≥ 0 } is not context-free.

TRUE