Transcript Context-free grammars, definition & examples

Context-free Languages
Chapter 2
Ambiguity
Chomsky normal form
Study Example 2.10 in the book
Pushdown Automata
PDA Example
PDA Example
What language is this?
{w | w has exactly one more b than a’s
where Σ = {a, b}}
1
a, ɛ  a
b, ɛ  b
a, b  ɛ
b, a  ɛ
ɛ, b  ɛ
2
{w | w has exactly one more b than a’s
where Σ = {a, b}}
1
a, ɛ  a
b, ɛ  b
a, b  ɛ
b, a  ɛ
ɛ, b  ɛ
2
Dilemma 1: I am in State 1; The current input
symbol is b and there is a b on the stack. What
do you do?
{w | w has exactly one more b than a’s
where Σ = {a, b}}
1
a, ɛ  a
b, ɛ  b
a, b  ɛ
b, a  ɛ
ɛ, b  ɛ
2
Dilemma 2: What if my input string is bbb?
Dilemma 2: What if my input string is bbb?
ɛ, ɛ  $
0
1
a, ɛ  a
b, ɛ  b
a, b  ɛ
b, a  ɛ
ɛ, b  ɛ
2
3
ɛ, $  ɛ
{w | w has exactly one more b than a’s
where Σ = {a, b}}
Possible Answer 1:
S Sab | aSb | abS | Sba | bSa | baS | b
How do you create bbaaabb?
{w | w has exactly one more b than a’s
where Σ = {a, b}}
Possible Answer 2:
S SaSbS | SbSaS | b
What is the problem with this?
{w | w has exactly one more b than a’s
where Σ = {a, b}}
Possible Answer 3:
S XbX | XbX // one b surrounded by X
/* X can be empty or it must contain an equal
number of a’s and b’s */
X  XaXbX | XbXaX | ɛ
/* We need ab and ba with X’s nested anywhere
inbetween */
Pumping Lemma for CFL
• Can prove that a language A is NOT context free,
i.e., no PDA can accept and no CFG can generate.
Assume A is regular
Pick string s in A
• sA
s has to be big enough to
reach a loop in the PDA
• |s| >= p
p is the number of PDA
transitions needed to visit
the same state twice, i.e.,
take a loop
Pumping Lemma says:
s = uvxyz
1. For each i >= 0,
si = uvixyiz  A
2. |vy| > 0
3. |vxy| <= p
Consider this CFL
A = {w | w has exactly one more b than a’s
where Σ = {a, b}}
1. si = uvixyiz  A
2. |vy| > 0
3. |vxy| <= p
s = ap bp+1
s is clearly in A
Consider p = 2, s.t., s = aabbb, which we know is
a reasonably small p.
The adversary can choose:
u = a, v = a, x = ɛ, y = b, z = bb
Adversary wins
s = ap bp+1
p = 2, s.t., s = aabbb
The adversary can choose:
u = a, v = a, x = ɛ, y = b, z = bb
2. vy = ab, i.e, |vy| > 0
3. vxy = ab, i.e., |vxy| >= 2
1. si = uvixyiz  A
2. |vy| > 0
3. |vxy| <= p
1. si = uvixyiz  A for all i
Every time we pump up v=a, we also pump up y=b.
Adversary wins
1. si = uvixyiz  A
2. |vy| > 0
3. |vxy| <= p
s = ap bp+1
p = 2, s.t., s = aabbb
The adversary can choose:
u = a, v = a, x = ɛ, y = b, z = bb
1. si = uvixyiz  A for all i
Every time we pump up v=a, we also pump up y=b.
This proves nothing because perhaps we could have
picked a better s where the adversary would not
win.
Consider a language that might
actually be Context Free
• B = {ww | w in {0,1}*}
• Consider s = 0p 1 0p 1
• Consider s = 0p 1 0p 1
Non-Determinism in PDAs
• All non-deterministic Finite State Automata’s
can be transformed into deterministic ones
(see proof of Theorem 1.39 on p55)
• Many non-deterministic Pushdown
Automata’s CANNOT be transformed into
deterministic ones (see pp 130-135)
• Non-determinism is necessary to recognize
many Context-free Languages
Non-Determinism in PDAs
• Consider
A = {wwR | w = {a, b}* }
B = {wcwR| w = {a, b}* }
Non-Deterministic Grammars
• RS|T
• S  aSb | ab
• T  aTbb | abb
• Consider the string aabb and aabbbb
• There is no way to know which rules were
used unless we analyze the whole string.
• This indicated non-determinism.
DK-Test
• Essentially, a more algorithm for determining
if a grammar is non-deterministic.
• We will not be covering this as it requires two
weeks of explanation.
Is this non-deterministic
• SE˧
• EE+T|T
• TTxa|a
• Consider the string a + a x a + a ˧
• Consider the string a x a + a x a ˧
LR(1) Grammars
• SE˧
• EE+T|T
• TTxa|a
• Consider the string a + a x a + a ˧
• By scanning left to right (LR) and by looking
ahead one symbol (1), we can determine
which rule was used. (Called a forced handle).
LR(k) Grammars
• Grammars where you can determine which
rules were used by scanning a string left to
right and look ahead a constant (K) number of
symbols
• LR(2) would require looking ahead two
symbols.
• Almost all programming languages can be
defined and generated using LR(k) Grammars.
Big Picture
• Deterministic PDAs require O(n) computation because
for each symbol you deterministically move to another
state and perhaps push or pop the stack.
• Implementing Non-deterministic PDAs requires a lot
more computation. {wwR} requires O(n2). Because you
have to guess where to start popping at n possible
locations.
• Implementing some non-deterministic PDAs require
O(2n) or O(n!)
• LR(k) grammars yield non-deterministic PDAs that
require n*k = O(n) computation. Cool!