Homework and Exams Homework Context Free Languages Return - PDF document

Homework and Exams • Homework Context Free Languages – Return Homework #2 – Homework #3 Due today Grammars Homework and Exams Homework and Exams – Homework #4 (due 10/13) • …and Exams • Exercise 5.1.2 b, c (pg 180) – Exam 1 is Wednesday!! • Exercise 5.1.7 a,b (pg 181) • Exercise 5.2.1 b,c (pg 191) • Design a CFG for the following languages – Set of odd length strings in {a,b} * with middle symbol a – Questions? – Set of even length strings in {a,b} * with the 2 middle symbols equal • Show that the CFG with the productions below is ambiguous: – S → a | Sa | bSS | SSb | SbS Schedule Mangling Schedule Mangling • Today • Perhaps more mangling to come in Week 8 – Chapter 5: CFGs, Parse Trees, Ambiguity – Move exam 2 to Wednesday? • Wednesday – Exam 1 – Problem Session for Chapter 5 – Stay tuned • Monday, Oct 13 th • Watch the schedule page – Homework #4 due – Homework #5 Assigned – Chapter 6: Pushdown Automata • Wednesday, Oct 15 th – Chapter 6 (cont) – Problem Session for Chapter 6

Before We Start Plan for today • Any questions? • Context Free Languages – Next class of languages in our quest! Languages Regular Languages • Recall. • For the past several weeks, we have been looking at Regular Languages: – What is a language? – Means of describing: Regular Expression – What is a class of languages? – Machine for accepting: Finite Automata Last class we discovered Context Free Languages • Venn-diagram of languages • Context Free Languages(CFL) is the next class of languages outside of Regular Is there Languages: something Regular Languages out here? – Means for defining: Context Free Grammar – Machine for accepting: Pushdown Automata YES! Finite Languages

Plan for today Introduction • Grammars • Define and describe context free grammars – Think back to your days of learning English – Rules for constructing a valid sentence • Questions? • Sentence = noun phrase + verb phrase • Noun phrase = – Name (Joe) – Article + noun (the car) • Verb Phrase = – Verb (runs) – Verb + prepositional phrase Introduction Introduction • Look at the sentence. Is this grammatically • Grammars correct? – Rules for constructing a simple sentence – Joe runs from the car. • Sentence = noun phrase + verb phrase • Noun phrase = – Sentence = noun phrase + verb phrase – Name (Joe) – = noun + verb phrase – Article + noun (the car) – = Name + verb phrase • Verb Phrase = – = Joe + verb phrase – Verb (runs) – = Joe + verb + prepositional phrase – Verb + prepositional phrase – = Joe + verb + preposition + noun phrase • Prepositional Phrase = – = Joe + verb + from + noun phrase – Preposition + noun phrase (from the car) – = Joe + verb + from + article +noun Introduction Introduction • Look at the sentence. Is this grammatically • Observations correct? – Our definition of a valid simple sentence – Joe runs from the car. includes: • A set of grammar categories (sentence, noun phrase, – Sentence = Joe + verb + from + article +noun verb phrase, noun, verb, etc) – = Joe + verb + from + the + car • Rules for defining each category – = Joe + ran + from + the + car • Set of “final strings” (Joe, ran, from, the, car) • Initial category of what we wish to create (sentence) • Valid sentence!

Back to CS Theory Back to CS Theory • CFLs are defined using Context Free • Recall our friend: Palindromes Grammars(CFG) – A palindrome is a string that is the same read left to right or right to left – Grammars, like their spoken language counterparts, provide a means to recursively – First half of a palindrome is a “mirror image” “deriving” the strings in a language. of the second half – Examples: • a, b, aba, abba, babbab. Back to CS Theory Back to CS Theory • Recursive definition for palindromes (pal) • A CFG for palindromes over {a,b} over Σ – Base cases 1. ε ∈ pal 1. P → ε 2. P → a 2.For any a ∈ Σ , a ∈ pal 3. P → b 3.For any x ∈ pal and a ∈ Σ , axa ∈ pal – Recursion 4.No string is in pal unless it can be obtained by 4. P → aPa rules 1-3 5. P → bPb Back to CS Theory Back to CS Theory • What we did: • Building the palindrome abba using grammar – At each step, we replaced the symbol P on the right with one of the definitions of P from our • P ⇒ aPa (Rule 4) rule set. • P ⇒ abPba (Rule 5) – Relating to our sentence example • P ⇒ ab ε ba (Rule 1) • Categories were replaced by Symbols • P ⇒ abba • Final strings were replaced by Symbols • Start category was replaced by a symbol • We still have our set of rules.

Context Free Grammars Context Free Grammars • Let’s redefine grammars for CS Theory use: • Production Rules 1. Terminals = Set of symbols that form the strings of – Of the form A → B the language being defined • A is a variable 2. Variables = Set of symbols representing categories • B is a string, combining terminals and variables 3. Start Symbol = variable that represents the “base • To apply a rule, replace an occurance of A with the category” that defines our language string B. 4. Production rules = set of rules that recursively define the language Context Free Grammars Context Free Grammars • Let’s formalize this a bit: • Let’s formalize this a bit: – Production rules – A context free grammar (CFG) is a 4-tuple: (V, • Of the form A → β where T, P, S) where – A ∈ V • V is a set of variables – β ∈ (V ∪ T) * string with symbols from V and T • T is a set of terminals • We say that γ can be derived from α in one step: • P is a set of production rules – A → β is a rule – α = α 1 A α 2 • V and T are disjoint (I.e. V ∩ T = ∅ ) – γ = α 1 β α 2 • S ∈ V, is your start symbol – α ⇒ γ Context Free Grammars Context Free Grammars • Let’s formalize this a bit: • The language generated by a CFG – Production rules – Let G = (V, T, P, S) • We say that the grammar is context-free since this – The language generated by G, L(G) substitution can take place regardless of where A is. • L(G) = { x ∈ T * | S ⇒ * x} • We write α ⇒ * γ if γ can be derived from α in zero or more steps. – A language L is a Context Free Language (CFL) iff there is a CFG G, such that • L = L(G)

Example 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 → ε (1) – Basic idea (define recursively) • S → 0S1 (2) • ε is certainly in the language • S → 1S0 (3) • For all strings in the language, if we add a 0 and 1 to • S → SS (4) the string, the result is in the language. • The concatenation of any two strings in the language will also be in the language Example Another example • Let’s derive a string from L • Find a CFG to describe: – L = {a i b j c k | i = k} – 00110011 – 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 to ⇒ 00 ε 11 00 ε 11 rule 1 add c to the right. = 00110011 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 → ε (4) S ⇒ aabbBcc rule 3 – Can also write as S ⇒ aabbbBcc rule 3 • S → B | aSc S ⇒ aabbb ε cc rule 4 • B → bB | ε = aabbb ε cc

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 A real practical example Practical uses for grammars • How a compiler works • Grammars for programming languages – <stmt> → … | <for-stmt> | <if-stmt> | … – <stmt> → { <stmt> <stmt> } | ε Stream Parse lexer parser codegen of tokens – <if-stmt> → if ( <expr> ) then <stmt> Tree – <for-stmt> → for ( <expr> ; <expr> ; <expr> ) Object <stmt> code Source file