1. CS 3240 – Chapter 5

2. LanguageMachineGrammar RegularFinite AutomatonRegular Expression, Regular Grammar Context-FreePushdown AutomatonContext-Free Grammar Recursively Enumerable Turing MachineUnrestricted Phrase- Structure Grammar 2CS 3240 - Introduction

3.  5.1: Context-Free Grammars  Derivations  Derivation Trees  5.2: Parsing and Ambiguity  5.3: CFGs and Programming Languages  Precedence  Associativity  Expression Trees CS 3240 - Context-Free Languages3

4.  S ➞ aaSa | λ  It is not right-linear or left-linear  so it is not a “regular grammar”  But it is linear  only one variable  What is it’s language? CS 3240 - Context-Free Languages4

5. 5 S ➝ aSb | λ Deriving aaabbb: S ⇒ aSb ⇒ aaSbb ⇒ aaaSbbb ⇒ aaabbb

6.  Variables  aka “non-terminals”  Letters from some alphabet, Σ  aka “terminals”  Rules (“substitution rules”)  of the form V → s ▪ where s is any string of letters and variables, or λ  Rules are often called productions CS 3240 - Context-Free Languages6

7.  a n cb n  a n b 2n  a n b m, where 0 ≤ n ≤ m ≤ 2n  a n b m, n ≠ m  Palindrome (start with a recursive definition)  Non-Palindrome  Equal  a n b n a m CS 3240 - Context-Free Languages7

8. CS 3240 - Pushdown Automata8 S → aSbSbS | bSaSbS | bSbSaS | λ Trace ababbb When building CFGs, remember that the start variable (S) represents a string in the language. So, for example, if S has twice as many b’s as a’s, then so does aSbSbS, etc.

9.  A derivation is a sequence of applications of grammatical rules, eventually yielding a string in the language  A CFG can have multiple variables on the right-hand side of a rule  Giving a choice of which variable to expand first  By convention, we usually use a leftmost derivation CS 3240 - Context-Free Languages9

10. 10 → → the → → sings | eats → cat | song | canary ⇒ ⇒ the ⇒ the canary ⇒ the canary sings ⇒ the canary sings the ⇒ the canary sings the song “sentential forms” (aka “productions”)

11.  A graphical representation of a derivation  The start symbol is the root  Each symbol in the right-hand side of the rule is a child node at the same level  Continue until the leaves are all terminals CS 3240 - Context-Free Languages11

12. CS 3240 - Context-Free Languages12

13.  Note how there was only one parse tree or the string “the canary sings the song”  And only one leftmost derivation  This is not true of all grammars!  Some grammars allow choices of distinct rules to generate the same string  Or equivalently, where there is more than one parse tree for the same string  Such a grammar is ambiguous  Not easy to process programmatically CS 3240 - Context-Free Languages13

14. CS 3240 - Context-Free Languages14 → + | * | ( ) | a | b | c ⇒ + ⇒ a + ⇒ a + * ⇒ a + b * ⇒ a + b * c ⇒ * ⇒ + * ⇒ a + * <exp ⇒ a + b * ⇒ a + b * c

15. CS 3240 - Context-Free Languages15 Which one is “correct”?

16.  The process of determining if a string is generated by a grammar  And often we want the parse tree  So that we know the order of operations  Top-down Parsing  Easiest conceptually  Bottom-up Parsing  Most efficient (used by commercial compilers)  We will use a simple one in Chapter 6 CS 3240 - Context-Free Languages16

17.  Try to match a string, w, to a grammar  If there is a rule S → w, we’re done!  Fat chance :-)  Try to find rules that match the first character  A “look-ahead” strategy  This is what we do “in our heads” anyway  Repeat on the rest of the string…  Very “brute force” CS 3240 - Context-Free Languages17

18. CS 3240 - Context-Free Languages18 S → SS | aSb | bSa | λ Parse “aabb”:

19. CS 3240 - Context-Free Languages19 S → SS | aSb | bSa | λ Parse “aabb”: Candidate rules: 1) S → SS, 2) S → aSb: 1)SS ⇒ SSS, SS ⇒ aSbS 2)aSb ⇒ aSSb, aSb ⇒ aaSbb Answer: S ⇒ aSb ⇒ aaSbb ⇒ aabb (2) Not a well-defined algorithm (yet)!

20.  A top-down parsing technique  Grammar Requirements:  no ambiguity  no lambdas  no left-recursion (e.g., A -> Ab)  … and some other stuff  Create a function for each variable  Check first character to choose a rule  Start by calling S( ) CS 3240 - Context-Free Languages20

21.  Grammar: S -> aSb | ab  Function S:  if length == 2, check to see if it is “ab”  otherwise, consume outer‘a’ and ‘b’, then call S on what’s left  See parseanbn.py, parseanbn2.py CS 3240 - Context-Free Languages21

22.  Grammar: A -> BA | a B -> bB | b  See parsebstara.cpp CS 3240 - Context-Free Languages22

23.  Lambda rules can cause productions to shrink  Then they can grow, and shrink again  And grow, and shrink, and grow, and shrink…  How then can we know if the string isn’t in the language?  That is, how do we know when we’re done so we can stop and reject the string? CS 3240 - Context-Free Languages23

24.  A rule of the form A → B doesn’t increase the size of the sentential form  Once again, we could spend a long time cycling through unit rules before parsing |w|  We prefer a method that always strictly grows to |w|, so we can stop and answer “yes” or “no” efficiently  So, we will remove lambda and unit rules  In Chapter 6 CS 3240 - Context-Free Languages24

25.  Precedence  Associativity CS 3240 - Context-Free Languages25

26.  It was ambiguous because it treated all operators equally  But multiplication should have higher precedence than addition  So we introduce a new variable for multiplicative expressions  And place it further down in the rules  Because we want it to appear further down in the parse tree CS 3240 - Context-Free Languages26

27. CS 3240 - Context-Free Languages27 → + | → * | → ( ) | a | b | c Now only one leftmost derivation for a + b * c: ⇒ + ⇒ + ⇒ + ⇒ a + ⇒ a + * ⇒ a + b * ⇒ a + b * c

28. CS 3240 - Context-Free Languages28

29.  Derive the parse tree for a + b + c …  Note how you get (a + b) + c, in effect  Left-recursion gives left associativity  Analogously for right associativity  Exercise:  Add a right-associative power (exponentiation) operator (^, with variable ) to the grammar with the proper precedence CS 3240 - Context-Free Languages29