slides - Computer Science - University of San Francisco

Automata TheoryCS411-2013F-08Context-Free GrammarsGuest Lecturer: Sophie EngleDepartment of Computer ScienceUniversity of San Francisco

08-0: Context-Free GrammarsSet of Terminals (Σ)Set of Non-TerminalsSet of Rules, each of the form: → Special Non-Terminal – Initial Symbol

08-1: Generating Strings with CFGsStart with the initial symbolRepeat:Pick any non-terminal in the stringReplace that non-terminal with the right-handside of some rule that has that non-terminal asa left-hand sideUntil all elements in the string are terminals

08-2: CFG ExampleS → aSS → BbB → cBB → ǫGenerating a string:S replace S with aSaS replace S wtih BbaBb replace B wtih cBacBb replace B wtih ǫacb Final String

08-4: CFG ExampleS → aSS → BbB → cBB → ǫRegular Expression equivalent to this CFG:

08-5: CFG ExampleS → aSS → BbB → cBB → ǫRegular Expression equivalent to this CFG:a ∗ c ∗ b

08-6: CFG ExampleCFG for L = {0 n 1 n : n > 0}

08-7: CFG ExampleCFG for L = {0 n 1 n : n > 0}S → 0S1 or S → 0S1|01S → 01(note – can write:A → αA → βasA → α|β )(examples: 01, 0011, 000111)

08-8: CFG Formal DefinitionG = (V,Σ,R,S)V = Set of symbols, both terminals & non-terminalsΣ ⊂ V set of terminals (alphabet for the languagebeing described)R ⊂ ((V −Σ)×V ∗ ) Finite set of rulesS ∈ (V −Σ) Start symbol

08-9: CFG Formal DefinitionExample:S → 0S1S → 01Set theory Definition:G = (V,Σ,R,S)V = {S,0,1}Σ ⊂ V = {0,1}R ⊂ ((V −Σ)×V ∗ ) = {(S,0S0),(S,01)}S ∈ (V −Σ) = S

08-10: DerivationA Derivation is a listing of how a string is generated –showing what the string looks like after everyreplacement.S → ABA → aA|ǫB → bB|ǫS ⇒ AB⇒ aAB⇒ aAbB⇒ abB⇒ abbB⇒ abb

08-11: Parse TreeA Parse Tree is a graphical representation of aderivation.SABS ⇒ AB⇒ aAB⇒ aAbB⇒ abB⇒ abbB⇒ abba A b BεbBε

08-12: Parse TreeA Parse Tree is a graphical representation of aderivation.SABa A b BS ⇒ AB⇒ AbB⇒ aAbB⇒ aaAbB⇒ aaAb⇒ aabaAAεε

08-14: Fun with CFGsCreate a Context-Free Grammar for all strings over{a,b} which contain the substring “aba”S → AabaAA →aAA →bAA → ǫGive a parse tree for the string: bbabaa

08-15: Fun with CFGsCreate a Context-Free Grammar for all strings over{a,b} that begin or end with the substring bba(inclusive or)

08-16: Fun with CFGsCreate a Context-Free Grammar for all strings over{a,b} that begin or end with the substring bba(inclusive or)S → bbaAS → AbbaA →bAA →aAA → ǫ

08-18: L REG ⊆ L CFGWe will prove L REG ⊆ L CFG in two different ways:Prove by induction that, given any regularexpression r, we create a CFG G such thatL[G] = L[r]Given any NFA M, we create a CFG G such thatL[G] = L[M]

08-20: L REG ⊆ L CFGBase Cases:r = a, a ∈ Σ

08-21: L REG ⊆ L CFGBase Cases:S → ar = a, a ∈ Σ

08-22: L REG ⊆ L CFGBase Cases:r = ǫ

08-23: L REG ⊆ L CFGBase Cases:S → ǫr = ǫ

08-24: L REG ⊆ L CFGBase Cases:r = ∅

08-25: L REG ⊆ L CFGBase Cases:r = ∅S → SS

08-26: L REG ⊆ L CFGRecursive Cases:r = (r 1 r 2 )L[G 1 ] = L[r 1 ], Start symbol of G 1 = S 1L[G 2 ] = L[r 2 ], Start symbol of G 2 = S 2

08-27: L REG ⊆ L CFGRecursive Cases:r = (r 1 r 2 )L[G 1 ] = L[r 1 ], Start symbol of G 1 = S 1L[G 2 ] = L[r 2 ], Start symbol of G 2 = S 2G = all rules from G 1 and G 2 , plus plus newnon-terminal S, and new rule:S → S 1 S 2New start symbol S

08-28: L REG ⊆ L CFGRecursive Cases:r = (r 1 +r 2 )L[G 1 ] = L[r 1 ], Start symbol of G 1 = S 1L[G 2 ] = L[r 2 ], Start symbol of G 2 = S 2

08-29: L REG ⊆ L CFGRecursive Cases:r = (r 1 +r 2 )L[G 1 ] = L[r 1 ], Start symbol of G 1 = S 1L[G 2 ] = L[r 2 ], Start symbol of G 2 = S 2G = all rules from G 1 and G 2 , plus new non-terminal S,and new rules:S → S 1S → S 2Start symbol = S

08-30: L REG ⊆ L CFGRecursive Cases:r = (r ∗ 1)L[G 1 ] = L[r 1 ], Start symbol of G 1 = S 1

08-31: L REG ⊆ L CFGRecursive Cases:r = (r ∗ 1)L[G 1 ] = L[r 1 ], Start symbol of G 1 = S 1G = all rules from G 1 , plus new non-terminal S, andnew rules:S → S 1 SS → ǫStart symbol = S(Example)

08-35: CFG – AmbiguityA CFG is ambiguous if there exists at least onestring generated by the grammar that has > 1different parse treePrevious CFG is ambiguous (examples)S → AabaAA →aAA →bAA → ǫ

08-36: CFG – AmbiguityConsider the following CFG:E → E +E|E −E|E ∗E|NN → 0|1|2|3|4|5|6|7|8|9Is this CFG ambiguous?Why is this a problem?

08-37: CFG – AmbiguityE → E +E|E −E|E ∗E|NN → 0|1|2|3|4|5|6|7|8|9EEE + EE*EN E * E3 N NE + E NN N 54534

08-38: CFG – AmbiguityE → E +E|E −E|E ∗E|NN → 0|1|2|3|4|5|6|7|8|9If all we care about is removing ambiguity, there isa (relatively) easy way to make this unambiguous(make all operators right-associative)

08-39: CFG – AmbiguityE → E +E|E −E|E ∗E|NN → 0|1|2|3|4|5|6|7|8|9Non-ambiguous:E → N|N +E|N −E|N ∗EN → 0|1|2|3|4|5|6|7|8|9If we were writing a compiler, would this be a goodCFG?How can we get correct associativity

08-40: CFG – AmbiguityAmbiguous:E → E +E|E −E|E ∗E|NN → 0|1|2|3|4|5|6|7|8|9Unambiguous:E → E +T|E −T|TT → T ∗N|NN → 0|1|2|3|4|5|6|7|8|9Can add parentheses, other operators, etc. (More inCompilers)

08-41: Fun with CFGsCreate a CFG for all strings over {(,)} that formbalanced parenthesis()()()(()())((()()()))((((()))))

08-42: Fun with CFGsCreate a CFG for all strings over {(,)} that formbalanced parenthesisS → (S)S → SSS → ǫIs this grammar ambiguous?

08-43: Fun with CFGsCreate a CFG for all strings over {(,)} that formbalanced parenthesisS → (S)S → SSS → ǫIs this grammar ambiguous?YES! (examples)

08-44: Fun with CFGsCreate an unambiguous CFG for all strings over{(,)} that form balanced parenthesis

08-45: Fun with CFGsCreate an unambiguous CFG for all strings over{(,)} that form balanced parenthesisS → ASS → ǫA → (S)

08-46: Ambiguous LanguagesA language L is ambiguous if all CFGs G thatgenerate it are ambiguousExample:L 1 = {a i b i c j d j |i,j > 0}L 2 = {a i b j c j d i |i,j > 0}L 3 = L 1 ∪L 2L 3 is inherently ambiguous(Create a CFG for L 3 )

08-47: Ambiguous LanguagesL 1 = {a i b i c j d j |i,j > 0}L 2 = {a i b j c j d i |i,j > 0}L 3 = L 1 ∪L 2S → S 1 |S 2S 1 → ABA → aAb|abB → cBd|cdS 2 → aS 2 d|aCdC → bCc|bcWhat happens when i = j?

08-48: (More) Fun with CFGsCreate an CFG for all strings over {a, b} that havethe same number of a’s as b’s (can be ambiguous)

08-49: (More) Fun with CFGsCreate an CFG for all strings over {a, b} that havethe same number of a’s as b’s (can be ambiguous)S → aSbS → bSaS → SSS → ǫ

08-50: (More) Fun with CFGsCreate an CFG for L = {ww R : w ∈ (a+b) ∗ }

08-51: (More) Fun with CFGsCreate an CFG for L = {ww R : w ∈ (a+b) ∗ }S → aSaS → bSbS → ǫ

08-52: (More) Fun with CFGsCreate an CFG for all palindromes over {a,b}.That is, create a CFG for:L = {w : w ∈ (a+b) ∗ ,w = w R }

08-53: (More) Fun with CFGsCreate an CFG for all palindromes over {a,b}.That is, create a CFG for:L = {w : w ∈ (a+b) ∗ ,w = w R }S → aSaS → bSbS → ǫS → aS → b

08-54: (More) Fun with CFGsCreate an CFG for L = {a i b j c k : j > i+k}

08-55: (More) Fun with CFGsCreate an CFG for L = {a i b j c k : j > i+k}HINT: We may wish to break this down into 3 differentlangauges ...

08-56: (More) Fun with CFGsCreate an CFG for L = {a i b j c k : j > i+k}S → ABCA → aAbA → ǫB → bBB → bC → bCc|ǫ

08-57: (More) Fun with CFGsCreate an CFG for all strings over {0, 1} that havethe an even number of 0’s and an odd number of1’s.HINT: It may be easier to come up with 4CFGs – even 0’s, even 1’s, odd 0’s odd 1’s,even 0’s odd 1’s, odd 1’s, even 0’s – andcombine them ...

08-58: (More) Fun with CFGsCreate an CFG for all strings over {0, 1} that havethe an even number of 0’s and an odd number of1’s.S 1 = Even 0’s Even 1’sS 2 = Even 0’s Odd 1’sS 3 = Odd 0’s Even 1’sS 4 = Odd 0’s Odd 1’sS 1 → 0S 3 |1S 2S 2 → 0S 4 |1S 1S 3 → 0S 1 |1S 4S 4 → 0S 2 |1S 3