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Context-free Languages Chapter 2 Ambiguity Chomsky normal form Study Example 2.10 in the book Pushdown Automata PDA Example PDA Example What language is this? {w | w has exactly one more b than a’s where Σ = {a, b}} 1 a, ɛ a b, ɛ b a, b ɛ b, a ɛ ɛ, b ɛ 2 {w | w has exactly one more b than a’s where Σ = {a, b}} 1 a, ɛ a b, ɛ b a, b ɛ b, a ɛ ɛ, b ɛ 2 Dilemma 1: I am in State 1; The current input symbol is b and there is a b on the stack. What do you do? {w | w has exactly one more b than a’s where Σ = {a, b}} 1 a, ɛ a b, ɛ b a, b ɛ b, a ɛ ɛ, b ɛ 2 Dilemma 2: What if my input string is bbb? Dilemma 2: What if my input string is bbb? ɛ, ɛ $ 0 1 a, ɛ a b, ɛ b a, b ɛ b, a ɛ ɛ, b ɛ 2 3 ɛ, $ ɛ {w | w has exactly one more b than a’s where Σ = {a, b}} Possible Answer 1: S Sab | aSb | abS | Sba | bSa | baS | b How do you create bbaaabb? {w | w has exactly one more b than a’s where Σ = {a, b}} Possible Answer 2: S SaSbS | SbSaS | b What is the problem with this? {w | w has exactly one more b than a’s where Σ = {a, b}} Possible Answer 3: S XbX | XbX // one b surrounded by X /* X can be empty or it must contain an equal number of a’s and b’s */ X XaXbX | XbXaX | ɛ /* We need ab and ba with X’s nested anywhere inbetween */ Pumping Lemma for CFL • Can prove that a language A is NOT context free, i.e., no PDA can accept and no CFG can generate. Assume A is regular Pick string s in A • sA s has to be big enough to reach a loop in the PDA • |s| >= p p is the number of PDA transitions needed to visit the same state twice, i.e., take a loop Pumping Lemma says: s = uvxyz 1. For each i >= 0, si = uvixyiz A 2. |vy| > 0 3. |vxy| <= p Consider this CFL A = {w | w has exactly one more b than a’s where Σ = {a, b}} 1. si = uvixyiz A 2. |vy| > 0 3. |vxy| <= p s = ap bp+1 s is clearly in A Consider p = 2, s.t., s = aabbb, which we know is a reasonably small p. The adversary can choose: u = a, v = a, x = ɛ, y = b, z = bb Adversary wins s = ap bp+1 p = 2, s.t., s = aabbb The adversary can choose: u = a, v = a, x = ɛ, y = b, z = bb 2. vy = ab, i.e, |vy| > 0 3. vxy = ab, i.e., |vxy| >= 2 1. si = uvixyiz A 2. |vy| > 0 3. |vxy| <= p 1. si = uvixyiz A for all i Every time we pump up v=a, we also pump up y=b. Adversary wins 1. si = uvixyiz A 2. |vy| > 0 3. |vxy| <= p s = ap bp+1 p = 2, s.t., s = aabbb The adversary can choose: u = a, v = a, x = ɛ, y = b, z = bb 1. si = uvixyiz A for all i Every time we pump up v=a, we also pump up y=b. This proves nothing because perhaps we could have picked a better s where the adversary would not win. Consider a language that might actually be Context Free • B = {ww | w in {0,1}*} • Consider s = 0p 1 0p 1 • Consider s = 0p 1 0p 1 Non-Determinism in PDAs • All non-deterministic Finite State Automata’s can be transformed into deterministic ones (see proof of Theorem 1.39 on p55) • Many non-deterministic Pushdown Automata’s CANNOT be transformed into deterministic ones (see pp 130-135) • Non-determinism is necessary to recognize many Context-free Languages Non-Determinism in PDAs • Consider A = {wwR | w = {a, b}* } B = {wcwR| w = {a, b}* } Non-Deterministic Grammars • RS|T • S aSb | ab • T aTbb | abb • Consider the string aabb and aabbbb • There is no way to know which rules were used unless we analyze the whole string. • This indicated non-determinism. DK-Test • Essentially, a more algorithm for determining if a grammar is non-deterministic. • We will not be covering this as it requires two weeks of explanation. Is this non-deterministic • SE˧ • EE+T|T • TTxa|a • Consider the string a + a x a + a ˧ • Consider the string a x a + a x a ˧ LR(1) Grammars • SE˧ • EE+T|T • TTxa|a • Consider the string a + a x a + a ˧ • By scanning left to right (LR) and by looking ahead one symbol (1), we can determine which rule was used. (Called a forced handle). LR(k) Grammars • Grammars where you can determine which rules were used by scanning a string left to right and look ahead a constant (K) number of symbols • LR(2) would require looking ahead two symbols. • Almost all programming languages can be defined and generated using LR(k) Grammars. Big Picture • Deterministic PDAs require O(n) computation because for each symbol you deterministically move to another state and perhaps push or pop the stack. • Implementing Non-deterministic PDAs requires a lot more computation. {wwR} requires O(n2). Because you have to guess where to start popping at n possible locations. • Implementing some non-deterministic PDAs require O(2n) or O(n!) • LR(k) grammars yield non-deterministic PDAs that require n*k = O(n) computation. Cool!