CS 3240 – Chapter 8
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Is anbncn context-free?
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

anbncn is not context-free
Neither is ww
 although wwR is!

We will develop a pumping lemma for
context-free languages
 (oh joy! :-)
 as before, can only be used to show that a
language is not CF
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
How can you tell by looking at a CFG whether
its language is infinite or not?
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S → aaB
A → bBb | λ
B → Aa
Consider the derivation:
S ⇒ aaB ⇒aaAa ⇒ aabBba ⇒ aabAaba
⇒ aabbBbaba ⇒ aabbAababa …
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

Grammars for infinite CFL must reuse a variable in some
derivation
S ⇒* uAz ⇒* uvAyz
 That is, A ⇒* vAy
⇒
 u,v,y,z,are derivable strings of characters and variables

We can repeat the same choices for A again:
 S ⇒* uAz ⇒* uvAyz ⇒* uvvAyyz

And again and again… (finally stopping with x)
 S ⇒* uvnAynz ⇒* uvnxynz

So for any sufficiently long string, s, we have
 s = uvnxynz
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
Hint: think of derivation trees, and the
relationship between the depth and the
number of leaves in a tree
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
If a complete binary has depth d, how many
leaves does it have?
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Consider a path from the root of a tree (S) to
a leaf.

It is all variables, except for the leaf
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The longer the string, the deeper the path

Eventually a variable must be repeated!
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
Based on a repeated variable (a type of loop)
 For sufficiently-long strings (≥ p = 2v), some
variable will be a descendant of itself in the parse

Every string of sufficient length from an
infinite CFL can be written as uvxyz, and
pumped as uvixyiz, i ≥ 0:
 |v| + |y| > 0
 |vxy| <= p
Intuitively: We’ve already used up the
stack to coordinate the anbn prefix
 Must consider all cases for a proof

 CFLPL-8.PDF

Use the pumping lemma with apbpapbp

(N)CFLs are closed under:
 Union
 Concatenation
 Kleene Star
 “Regular Intersection” (CF ∩ R = CF)

(N)CFLs are not closed under:
 intersection
 complement
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Let S1 be the start symbol for L1, and S2 for L2
Just have a new start symbol point to the OR
of the old ones:
S => S1 | S2
S1 => …
S2 => …

S => S1S2
S1 => …
S2 => …


Rename the old start variable to S1
S => S1S | λ
S1 => …

Let L1 = anbncm
 The concatenation of anbn and cm

Let L2 = anbmcm
 The concatenation of an and bmcm
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These are both context-free
L1 ∩ L2 = anbncn
 We already showed this is not context free
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Let R be a regular language and L a contextfree language
The R∩L is context-free
Why?
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a,λ
b,X
±s
s/X
f/λ
+f
Φ
f/λ
a
b
±A
B
C
B
A
D
C
D
A
D
C
B
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a,λ
b,X
± sA
s/X; B
f/λ; C
sB
s/X; A
f/λ; D
fC
;D
f/λ; A
fD
;C
f/λ; B
+ fA
;B
f/λ; C
fB
;A
f/λ; D
CS 3240 - Properties of Context-Free Languages
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jail
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
Proof by contradiction, derived from the
formula for intersection:
L1 ∩ L2 = (L1' + L2')'
Since the intersection is not closed, but union
is, then the complement cannot be.
 (Otherwise we could compute the intersection,
which in general is not CF)
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Non-determinism is the problem
Remember NFAs?
 To find the complement, we needed to first
convert to a DFA, then flip the states


Since some CFLs are inherently nondeterministic, they have no deterministic
equivalent to “flip”
But… what does this say about deterministic
CFLs?
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
DCFLs are closed under:
 Complement!
 Concatenation
 Kleene Star
 “Regular Intersection” (CF ∩ R = CF)

DCFLs are not closed under:
 Intersection
 Union!
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
Consider:
L1 = {aibjck | i = j}
L2 = {aibjck | j = k}
Each of these is DCF
 (Easy to show – your 7.1 homework was similar)

The union is not!
 It requires non-determinism
 It’s still context-free, but not Deterministic CF

DCFLs always have an associated CFG that is
unambiguous

Closed under Union, Concatenation, Kleene
Star
Not closed under intersection, complement
CFL ∩ Regular = CFL

DCFLs are closed under complement

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 But not union!
 Swap those two
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Unanswerable questions
Answerable questions

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Do 2 arbitrary CFGs generate the same
language?
Is a CFG ambiguous?
Is a given NCFL’s complement also CF?
Is the intersection of 2 given CFLs CF?
Do 2 CFLs have a common word?

Is a non-terminal ever used in a productive
derivation?
 Draw the connectivity graph ✔

Does a CFG generate any words?
 Substitute each “terminating production” (RHS is
all terminals) throughout and see what happens
▪ “back substitution method”

Is a CFL finite or infinite?
 Procedure to detect useful, repeated variables
S → aA | bB | λ
A → a | aCA | bDA | bBa | aAa
B → b | aAb | aCB | bDB | bBb
C → aCC | bDC
D → aCD | bDD
First remove useless variables…
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S → aA | bB | λ
A → a | bBa | aAa
B → b | aAb | bBb
Pick a non-empty, terminal rule: A → a
Back-substitute that rule:
S → aa | bB | λ
A → a | bBa | aaa
B → b | aab | bBb
Keep going until we have S → <terminal string>, or we can’t continue.
We have S → aa. STOP.
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S → aA | bB | λ
A → a | bBa | aAa
B → b | aAb | bBb
All variables are useful. Let’s see if A is repeated, for instance.
First, mark all A’s on the right:
S → aA | bB | λ
A → a | bBa | aAa
B → b | aAb | bBb
Now mark all variables affected on the left:
S → aA | bB | λ
A → a | bBa | aAa
B → b | aAb | bBb
Since A was marked on the left, it is repeated.
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S ➞ aA | SB
A ➞ baB | λ
B ➞ bB | bA
Mark on left:
S ➞ aA | SB
A ➞ baB | λ
B ➞ bB | bA
Mark A’s on right:
S ➞ aA | SB
A ➞ baB | λ
B ➞ bB | bA
A was marked on left. DONE.
Now mark corresponding variables on left:
S ➞ aA | SB
A ➞ baB | λ
B ➞ bB | bA
Repeat marking on right:
S ➞ aA | SB
A ➞ baB | λ
B ➞ bB | bA
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