1. Section 12.4 Context-Free Language Topics
Algorithm. Remove L-productions from grammars for langauges without L.
Find nonterminals that derive L.
For each production A  w construct all productions A  w’ where w’ is obtained from
w by removing one or more occurrences of the nonterminals from Step 1.
Combine the original productions with those of step 2 and eliminate any L-productions.
Example. Remove L-productions from the grammar
S  ABc
A  aA | L
B  bB | L.
Solution. Step 1: The nonterminals A and B derive L.
Step 2: From the production S  ABc we construct S  Bc | Ac | c.
From the production A  aA we construct A  a.
From the production B  bB we construct B  b.
Step 3: S  ABc | Bc | Ac | c
A  aA | a
B  bB | b.
Quiz. Remove L -productions from
S  ABc | Ab | c
A  ABa | L
B  Bbc | L.
Solution.
S  ABc | Ab | c| Bc | Ac | b
A  ABa | Ba | Aa | a
B  Bbc | bc.

2. Chomsky Normal Form. Productions have one of the following forms
A  a (a a terminal)
A  BC
S  L (if L is in the language).
Advantages: Parse trees are binary, which are easy to represent. Any string of length n > 0
can be derived in 2n – 1 steps.
Algorithm. Transform to Chomsky normal form (with the additional property that no start symbol occurs on the right side of a production)
1. If the start symbol S occurs on some right side, create a new start symbol S´ and a
new production S´  S.
2. Remove A  L (if A ≠ S) by previous algorithm. (If S  L is removed, add it back.)
3. Remove unit productions (i.e., A  B): If A  B or A + B, then construct productions
A  w where B  w is not a unit production. Now remove all unit productions.
4. For each production whose right side has two or more symbols, replace all occurrrences
of each terminal a with a new nonterminal A and also add the new production A  a.
5. Replace each production B  C1…Cn with n > 2 with B  C1D where D  C2 …Cn.
Repeat this step until all right sides have length two.

3. Example. Construct a Chomsky normal form for the grammar
S  aSb | D
D  Dc | L.
Solution.
Step 1: Add the production S´  S.
Step 2: S´  S | L
S  aSb | ab | D
D  Dc | c.
Step 3: S´  aSb | ab | Dc | c | L
S  aSb | ab | Dc | c
Step 4: S´  ASB | AB | DC | c | L
S  ASB | AB | DC | c
D  DC | c
A  a
B  b
C  c.
Step 5: Replace S´ ASB and S  ASB by
S´ AE, S´ AE, and E  SB.

4. Quiz. Construct a Chomsky normal form for the grammar
S  aTbb | U | L
T  cT | c
U  Ud | d.
Solution. Step 1: No change. Step 2: No change.
Step 3: Remove unit production S  U to obtain
S  aTbb | Ud | d | L
Step 4: Transform right sides of length at least two into strings of nonterminals.
S  ATBB | UD | d | L
T  CT | c
U  UD | d.
A  a
B  b
C  c
D  d.
Step 5: Replace S  ATBB with the productions
S  AE, E  TF, F  BB.

5. Greibach Normal Form. Productions have one of the following forms
A  b (b a terminal)
A  bD1…Dk
S  L (if L is in the language).
Advantage: Any string of length n > 0 can be derived in n steps.
Algorithm (idea). Transform context-free grammar to Greibach normal form.
Perform steps 1, 2, and 3 of the Chomsky algorithm.
Remove all left-recursion, including indirect, without adding L
3. Make substitutions to transform the grammar into the proper form.
Example. Put the following grammar into Greibach normal form.
S  AB | Ac | d
A  aA | a
B Ab | c.
Solution: Steps 1 (Chomsky steps 1, 2, and 3) and 2 are non needed.
Step 3: Replace A in S  AB | Ac | d with aA | a to obtain
S  aAB | aB | aAc | ac | d.
Replace A in B Ab | c with theright side of A  aA | a to obtain B aAb | ab | c.
Now add the new productions C  c and D  b to obtain the proper form:
S  aAB | aB | aAC | aC | d
A  aA | a
B aAD | aD | c
C  c
D  b.

6. Example. Put the following grammar into Greibach normal form.
S  AB | Ac | d
A  aA | a
B Ab | c.
Solution: Steps 1 (Chomsky steps 1, 2, and 3) and 2 are not needed.
Step 3: Replace A in S  AB | Ac | d with aA | a to obtain
S  aAB | aB | aAc | ac | d.
Replace A in B Ab | c with theright side of A  aA | a to obtain
B aAb | ab | c.
Now add the new productions C  c and D  b and make appropriate replacements
to obtain the proper form:
S  aAB | aB | aAC | aC | d
B aAD | aD | c
C  c
D  b.

7. Properties of Context-Free Languages
When we know some properties of context-free languages they can help us argue, BWOC,
that certain languages are not context-free.
The Pumping Lemma
If L is an infinite context-free language, then any grammar for L must be recursive, so there
must be derivations of the the following form where u, v, w, x, and y are terminal strings.
S + uNy
N + vNx (where v and x are not both L)
N + w.
These derivations lead to derivations like
S + uNy + uvNxy + uv2Nx2y + uvkNxky + uvkwxky  L for all k  N.
This is the basis for the Pumping Lemma:
There is an integer m > 0 such that if z  L and | z | ≥ m, then z has the form z = uvwxy where
1 ≤ | vx | ≤ | vwx | ≤ m and uvkwxky  L for all k  N.
Note: The number m depends on the grammar as we’ll see in the following example.

8. Example. Suppose we have the following
grammar for {L, bbc}  {abcnd | n  N}.
S  aNd | bbc | L
N Nc | b.
Here are a few derivations:
S  aNd  abd
S  aNd  aNcd  abcd
S  aNd  aNcd  aNccd  abccd
S + abckd for any k in N.
For this grammar m = 4 can be used in the pumping lemma because any derivation of a
string z with | z | ≥ 4 must use the nonterminal N. For example, if | z | = 8 and z = abcccccd,
then the pumping lemma factors z = abcccccd = uvwxy where 1 ≤ | vx | ≤ | vwx | ≤ 4 and
uvkwxky  L for all k  N. In this case let u = a, v = L, w = b, x = c, and y = ccccd.
Example. The language L = {anbncn+k | k, n  N} is not context-free.
Proof: Assume, BWOC, that L is context-free. L is infinite, so pumping lemma applies.
Choose z = ambmcm where m is the positive integer from the lemma.
Then z = ambmcm = uvwxy where 1 ≤ | vx | ≤ | vwx | ≤ m and uvkwxky  L for all k  N.
Observe neither v nor x can contain distinct letters. For example, if v = …a…b…, then
v2 = …a…b……a…b…, which can’t appear as a substring of any string in L.
So v and x must be strings of repeated occurrences of a single letter.
Now since | vwx | ≤ m, there are two possible places in ambmcm where v and x must occur:
(1) v and x occur in ambm.
(2) v and x occur in bmcm.
But we obtain the following contradictions because v and x are not both L.
(1) Let k = 2 to obtain uv2wx2y = am+ibm+jcm, where i > 0 or j > 0. So uv2wx2y  L
(2) Let k = 0 to obtain uwy = ambm-icm–j, where i > 0 or j > 0. So we have uwy  L.
These contradictions imply that L is not context-free. QED.

9. Example/Quiz. Prove that the language L = {ss | s  {a, b}
Example/Quiz. Prove that the language L = {ss | s  {a, b}*} is not context-free.
Proof: Assume, BWOC, that L is context-free. L is infinite, so pumping lemma applies.
Choose z = ambmambm where m is the positive integer from the lemma.
Then z = ambmambm = uvwxy where 1 ≤ | vx | ≤ | vwx | ≤ m and uvkwxky  L for all k  N.
Now since | vwx | ≤ m, there are three possible places in ambmambm where v and x must occur:
(1) v and x occur in ambm (on the left of z).
(2) v and x occur in bmam (in the center of z).
(3) v and x occur in ambm (on the right of z).
Notice that v and x can consist only of repetitions a single letter. For example, in case (1)
suppose v = aibj for some i > 0 and j > 0 and x = bn for some n ≥ 0. Then, letting k = 0, we
would obtain uwy = am–ibm–j–nambm, which cannot be in L. The argument is similar for the
other cases. So v and x must consist only of repetitions of a single letter.
We need to find a contradiction in each of the three cases. We’ll do it by using k = 0.
This tells us that uwy  L. But we obtain the following contradictions because v and x are not
both L.
(1) uwy = am–ibm–jambm where either i > 0 or j > 0 So uwy  L,
(2) uwy = ambm–iam–jbm where either i > 0 or j > 0. So uwy  L.
(3) uwy = ambmam–ibm–j where either i > 0 or j > 0. So uwy  L.
Therefore L is not context-free. QED.
Remark: Be careful that the choice of z is not in a context-free sublanguage of L. For example, if we chose z = (ab)m(ab)m in the preceding example, we would not get any contradictions.