Homework Homework #2 returned Context Free Languages Homework #3 returned today (for early submission) Homework #3 due today (for non-early submission) Homework #4 Pg 133 -- Exercise 1 (use structural induction) Pg 133 -- Exercise 3 Pg 134 -- Exercise 8b,c,d Pg 135 -- Exercise 22 Pg 145 -- Exercise 13 Due 10 / 10 Announcements Before We Start Exam 1 Any questions? Next class (Oct 5th) 1 hour Cover Regular Languages (up to and including today’s lecture) Closed book (1 sheet study guide okay) Remainder of next class will be a problem session Plan for today Languages 1st half Recall. Context Free Grammars and Languages What is a language? 2nd half What is a class of languages? Regular Languages and Context Free Languages / L-Systems 1
Regular Languages Last class we discovered For the past several weeks, we have Venn-diagram of languages been looking at Regular Languages: Is there something Regular Languages language: Regular Expression out here? Grammar: Regular Grammars Machine for accepting: Finite Automata YES! Finite Languages Context Free Languages Grammars Context Free Languages(CFL) is the Let’s formalize this a bit: next class of languages outside of A grammar is a 4-tuple: (V, T, P, S) where Regular Languages: V is a set of variables T is a set of terminals Language / grammar: Context Free P is a set of production rules Grammar V and T are disjoint (I.e. V ∩ T = ∅ ) Machine for accepting: Pushdown S ∈ V, is your start symbol Automata General Grammars Grammars Production Rules Let’s formalize this a bit: Production rules Of the form A → B We say that γ can be derived from α in one step: A is a string of terminals and variables A → β is a rule α = α 1 A α 2 B is a string of terminals and variables γ = α 1 β α 2 To apply a rule, replace any occurrence of A α ⇒ γ with the string B. We write α ⇒ * γ if γ can be derived from α in zero or more steps. 2
Context Free Grammars Context Free Grammars Production Rules The language generated by a grammar Of the form A → B Let G = (V, T, P, S) A is a variable The language generated by G, L(G) B is a string, combining terminals and variables L(G) = { x ∈ T * | S ⇒ * x} To apply a rule, replace an occurrence of A with the string B. A language L is a Context Free We say that the grammar is context-free since this Language (CFL) iff there is a CFG G, substitution can take place regardless of where A is. such that L = L(G) Example Example Recursive definition for palindromes Recall our friend: Palindromes (pal) over Σ A palindrome is a string that is the same 1. λ ∈ pal read left to right or right to left 2. For any a ∈ Σ , a ∈ pal First half of a palindrome is a “mirror 3. For any x ∈ pal and a ∈ Σ , axa ∈ pal image” of the second half 4. No string is in pal unless it can be obtained Examples: by rules 1-3 a, b, aba, abba, babbab. Example Back to CS Theory A CFG for palindromes over {a,b} Building the palindrome abba using Base cases grammar P → λ P ⇒ aPa (Rule 4) 1. P → a 2. P ⇒ abPba (Rule 5) P → b 3. P ⇒ ab λ ba (Rule 1) Recursion P ⇒ abba 4. P → aPa 5. P → bPb 3
Another Example Another Example Find a CFG to describe: Find a CFG to describe: L = {x ∈ {0,1} * | n 0 (x) = n 1 (x)} L = {x ∈ {0,1} * | n 0 (x) = n 1 (x)} S → λ (1) Basic idea (define recursively) S → 0S1 (2) λ is certainly in the language S → 1S0 (3) For all strings in the language, if we add a 0 S → SS (4) and 1 to the string, the result is in the language. The concatenation of any two strings in the language will also be in the language Another Example Yet Another example Let’s derive a string from L Find a CFG to describe: 00110011 L = {a i b j c k | i = k} S ⇒ SS rule 4 Number of a’s equals the number of c’s with any number of b’s between them ⇒ 0S1 0S1 rule 2 Use variable B to represent b j ⇒ 00S11 00S11 rule 2 Every time you add a to the left of B you need ⇒ 00 λ 11 00 λ 11 rule 1 to add c to the right. = 00110011 Yet Another example Another example Find a CFG to describe: Let’s derive a string from L: aabbbcc L = {a i b j c k | i = k} S ⇒ aSc rule 2 S → B (1) S ⇒ aaScc rule 2 S → aSc (2) S ⇒ aaBcc rule 1 B → bB (3) S ⇒ aabBcc rule 3 B → λ (4) S ⇒ aabbBcc rule 3 Can also write as S ⇒ aabbbBcc rule 3 S → B | aSc S ⇒ aabbb λ cc rule 4 B → bB | λ = aabbbcc 4
One more example One more example Defining the grammar for algebraic Defining the grammar for algebraic expressions: expressions: G = (V, T, P, S) Let a = a numeric constant Set of binary operators = {+, -, *, /} V = {S} Expressions can be parenthesized T = { a, -, +, *, /, (, ) } S = S P = see next slide One more example One more example Defining the grammar for algebraic Show derivation for a + (a * a) / a expressions – Production rules S ⇒ S + S rule 1 S → S + S (1) S ⇒ a + S rule 6 S → S – S (2) S ⇒ a + S / S rule 4 S ⇒ a + (S) / S rule 5 S → S * S (3) S ⇒ a + (S * S) / S rule 3 S → S / S (4) S ⇒ a + (a * S) / S rule 6 S → (S) (5) S ⇒ a + (a * a) / S rule 6 S → a (6) S ⇒ a + (a * a) / a rule 6 Parse Trees Another example Find a CFG to describe: Graphical means to illustrate a derivation of a string from a grammar L = {a i b j c k | i = k} S → B (1) Root of the tree = start variable S → aSc (2) Interior nodes = other variables B → bB (3) Children of nodes = application of a production B → λ (4) rule Can also write as Leaf nodes = Terminal symbols S → B | aSc B → bB | λ 5
Another example Parse Tree S Let’s derive a string from L: aabbcc An inorder traversal Rule 2 of the tree will give S ⇒ aSc rule 2 a S c the the string S ⇒ aaScc rule 2 Rule 2 derived. a S c S ⇒ aaBcc rule 1 B Rule 1 S ⇒ aabBcc rule 3 S ⇒ aabbBcc rule 3 b B Rule 3 S ⇒ aabb λ cc rule 4 Rule 3 b B = aabbcc Rule 4 λ Recall our example from last time One more example Defining the grammar for algebraic Show derivation for a + a * a expressions – Production rules S ⇒ S + S rule 1 S → S + S (1) S ⇒ a + S rule 6 S → S – S (2) S ⇒ a + S * S rule 3 S → S * S (3) S ⇒ a + a * S rule 6 S → S / S (4) S ⇒ a + a * a rule 6 S → (S) (5) S → a (6) Parse Tree One more example Another derivation for a + a * a S ⇒ S + S S rule 1 S ⇒ S * S rule 3 S ⇒ a + S S + S S ⇒ S * a rule 6 rule 6 S ⇒ S + S * a rule 1 S ⇒ a + S * S S * S a S ⇒ a + S * a rule 6 rule 3 S ⇒ a + a * a rule 6 S ⇒ a + a * S a a rule 6 S ⇒ a + a * a rule 6 6
Parse Tree Parse trees S ⇒ S * S S rule 3 S S S ⇒ S * a S * S S + S S * S rule 6 a S + S S ⇒ S + S * a a S + S rule 1 S * S a a a S ⇒ a + S * a a a rule 6 a a S ⇒ a + a * a Same string, 2 derivations rule 6 Ambiguity Ambiguity Showing a grammar is ambiguous is A CFG is said to be ambiguous if there easy is at least 1 string in L(G) having two or Find a string x in the L(G) that has two more distinct derivations. derivations Showing a particular grammar is not Some grammars are inherently ambiguous is usually difficult. ambiguous. Showing that any grammar is not ambiguous is not possible. Derivations Derivations a + a + a Leftmost derivations S A leftmost derivation is one where the leftmost S variable in the current string is always the first to S + S get replaced via a production rule. S + S A rightmost derivation is one where the rightmost a variable in the current string is always the first to S + S S + S a get replaced via a production rule. a a a a rightmost derivation leftmost derivation 7
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