Chapter 8 Properties of CFLs
Theorem 8.1: Let L be an infinite context-free language. Then  a
positive integer m such that any w  L with |w|  m can be
decomposed as w = uvxyz with |vxy|  m and |vy| 1 such that
uvixyiz  L for all i = 0, 1, 2, …
Proof sketch:
Finishing the Pumping Lemma Proof
Using the pumping lemma
Example 1 is in the online notes
Example 1: Proof that L = {0i1i2i | i  0} is not a CFL
Case 1: v and/or y contains two different symbols.
Case 1a:
Case 1b:
Case 2:
v  0*, y  0+ if i = 0, uxz = 0k1m2m, is not in L since k < m
v  0*, y  1+
v  1*, y  1+
v  1*, y  2+
v  2*, y  2+
Example 2: L = {0i | i is prime} is not a CFL. (# 2 p. 212)
Example 3: Prove that L = {anbnci | i  n} is not context-free.
Case 1: vy contains both a's and b's.
Case 2: vy contains both b's and c's.
Case 3: vy contains only a's.
Case 4: vy contains only b's.
Case 5: vy contains only c's.
Example 4: L = {anbjck | k > n, k > j }
See proof in the online notes
Example 5: L = {aibjck | j = max{i, k} and i k }
Read Example 8.2—L = {ww | w  {a, b}*} is not context-free
Section 8.2 Closure Properties and decision algorithms for CFL’s
Theorem 8.3 The family of context-free languages is closed under
union, concatenation and star-closure.
Proof
Let L1 and L2 be context-free languages generated by grammars,
G1 = (V1, T1, S1, P1) and G2 = (V2, T2, S2, P2). Without loss of
generality, assume V1  V2 = .
Closure under concatenation:
Star closure:
Theorem: If L is a CFL, then so is LR.
Theorem 8.4 The family of context-free languages is not closed
under intersection and complementation.
Nonclosure under complementation
Assume that the CFLs are closed under complementation
Other operations under which the context-free languages are
closed:
Theorem 8.5 Let L1 be a context-free language and L2 be a regular
language. Then L1  L2 is context-free.
Let M1 = (Q, , , 1, q0, z, F1) be an NPDA that accepts CFL L1 and
let M2 = (P, , ,2, p0, F2) be a DFA that accepts L2.
A natural question now arises, why won’t this work for two CFL’s?
Examples of using closure properties
Example 1: Prove or disprove L = {x  {a, b, c}* | na(x) = nb(x) =
nc(x)} is a CFL.
Example 2: Show that L = {ww | w  {a, b}+} is not a CFL.
Example 3: Prove that if L1 and L2 are CFLs then L1 – L2 is not
necessarily a CFL.
Some Decidable Properties of Context-Free Languages
Theorem 8.6 Given a context-free grammar G = (V, T, S, P) there
exists an algorithm for deciding whether or not L(G) is empty.
Theorem 8.7 Given a context-free grammar G = (V, T, S, P) there
exists an algorithm for determining whether or not L(G) is infinite.
Example: A  BC, B  DC, C  DA.
The CYK algorithm
wij denotes
Vij denotes
How the algorithm works
for i  1 to n
Vii  {A | A  a is in P and the ith symbol of w is a}
for j  2 to n
for i  1 to n-j+1
Vij  Ø
for k  i to i+j-1
Vij  Vij  {A | A  BC is in P and B  Vik and C  Vk+1,j}
CYK Algorithm Example
S  CB | FA | FB
B  FS | CE | b
D  AA
Fb
A  CS | FD | a
Ca
E  BB
Some Undecidable CFL Questions
Let G1 and G2 denote arbitrary CFG’s and let R be a regular
language.
1.
2.
3.
4.