CS 3240 – Chapter 5
Language
Machine
Grammar
Regular
Finite Automaton
Regular Expression,
Regular Grammar
Context-Free
Pushdown Automaton
Context-Free
Grammar
Recursively
Enumerable
Turing Machine
Unrestricted PhraseStructure Grammar
CS 3240 - Introduction
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5.1: Context-Free Grammars
 Derivations
 Derivation Trees
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5.2: Parsing and Ambiguity
5.3: CFGs and Programming Languages
 Precedence
 Associativity
 Expression Trees
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S ➞ aaSa | λ
It is not right-linear or left-linear
 so it is not a “regular grammar”
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But it is linear
 only one variable
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What is it’s language?
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S ➝ aSb | λ
Deriving aaabbb:
S ⇒ aSb ⇒ aaSbb ⇒ aaaSbbb ⇒ aaabbb
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Variables
 aka “non-terminals”
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Letters from some alphabet, Σ
 aka “terminals”
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Rules (“substitution rules”)
 of the form V → s
▪ where s is any string of letters and variables, or λ
 Rules are often called productions
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ancbn
anb2n
anbm, where 0 ≤ n ≤ m ≤ 2n
anbm, n ≠ m
Palindrome (start with a recursive definition)
Non-Palindrome
Equal
anbnam
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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.
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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
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By convention, we usually use a leftmost
derivation
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<S> → <NP> <VP>
<NP> → the <N>
<VP> → <V> <NP>
<V> → sings | eats
<N> → cat | song | canary
<S> ⇒ <NP> <VP>
⇒ the <N> <VP>
⇒ the canary <VP>
⇒ the canary <V> <NP>
⇒ the canary sings <NP>
⇒ the canary sings the <N>
⇒ the canary sings the song
CS 3240 - Context-Free Languages
“sentential forms”
(aka “productions”)
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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
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Note how there was only one parse tree or
the string “the canary sings the song”
 And only one leftmost derivation
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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
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Such a grammar is ambiguous
 Not easy to process programmatically
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<exp> → <exp> + <exp> | <exp> * <exp> | (<exp>) | a | b | c
<exp> ⇒ <exp> + <exp>
⇒ a + <exp>
⇒ a + <exp> * <exp>
⇒ a + b * <exp>
⇒a+b*c
<exp> ⇒ <exp> * <exp>
⇒ <exp> + <exp> * <exp>
⇒ a + <exp> * <exp
⇒ a + b * <exp>
⇒a+b*c
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Which one is “correct”?
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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
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Top-down Parsing
 Easiest conceptually
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Bottom-up Parsing
 Most efficient (used by commercial compilers)
 We will use a simple one in Chapter 6
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Try to match a string, w, to a grammar
If there is a rule S → w, we’re done!
 Fat chance :-)
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Try to find rules that match the first character
 A “look-ahead” strategy
 This is what we do “in our heads” anyway
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Repeat on the rest of the string…
Very “brute force”
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S → SS | aSb | bSa | λ
Parse “aabb”:
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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)!
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A top-down parsing technique
Grammar Requirements:
 no ambiguity
 no lambdas
 no left-recursion (e.g., A -> Ab)
 … and some other stuff
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Create a function for each variable
Check first character to choose a rule
Start by calling S( )
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Grammar:
S -> aSb | ab
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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
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Grammar:
A -> BA | a
B -> bB | b
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See parsebstara.cpp
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Lambda rules can cause productions to shrink
 Then they can grow, and shrink again
 And grow, and shrink, and grow, and shrink…
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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?
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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
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Precedence
Associativity
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It was ambiguous because it treated all
operators equally
 But multiplication should have higher precedence
than addition
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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
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<exp> → <exp> + <mulexp> | <mulexp>
<mulexp> → <mulexp> * <rootexp> | <rootexp>
<rootexp> → (<exp>) | a | b | c
Now only one leftmost derivation for a + b * c:
<exp> ⇒ <exp> + <mulexp> ⇒ <mulexp> + <mulexp>
⇒ <rootexp> + <mulexp>
⇒ a + <mulexp>
⇒ a + <mulexp> * <rootexp>
⇒ a + <rootexp> * <rootexp>
⇒ a + b * <rootexp>
⇒a+b*c
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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
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Exercise:
 Add a right-associative power (exponentiation)
operator (^, with variable <powerexp>) to the
grammar with the proper precedence
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