1 Prove this Language is Context-Free

\[ L = \{ a^ib^ia^jb^j | i, j \geq 0\} \]

2 Prove this Language is Not Context-Free

\[ L = \{ a^ib^ia^ib^i \} \]

(Hint: This type of problem is often addressed via the pumping lemma.)

3 Converting to CNF

Consider this grammar for expressions with properly nested parentheses & brackets.

\[ \begin{align} S &\rightarrow SA | SB | \epsilon \\ A &\rightarrow ( S ) | S | E \\ B &\rightarrow [ S ] | E \\ E &\rightarrow x \\ \end{align} \]

A discussion of what this grammar “means” and how to read it:

There are 4 steps in converting a grammar to Chomsky Normal Form.

See if you can do each of them in turn:

1. Eliminate the $\epsilon$ productions.

   Rewrite the grammar so that all productions that can derive the empty string are removed.

   Reveal

   The language as a whole does contain $\epsilon$, so our CNF will actually be for $L - \{\epsilon\}$

   Which variables are nullable?

   • S is nullable because $S \rightarrow \epsilon$.
   • A is nullable because $A \rightarrow S$ and all symbols of the RHS are nullable

   So every production with an S or A on the RHS gets a new production(s) representing what would happen if S or A or both were empty. Then we eliminate the actual $\epsilon$ production.

   \[ \begin{align} S &\rightarrow SA | S | A | SB | B \\ A &\rightarrow ( S ) | ( ) | S | E \\ B &\rightarrow [ S ] | [ ] | E \\ E &\rightarrow x \\ \end{align} \]

2. Eliminate unit productions

   Rewrite the grammar to remove all productions where the entire right-hand-side consists of a single variable.

   Reveal

   We have quite a few unit productions, many of them introduced by our eliminating the $\epsilon$ productions.

   \[ \begin{align} S &\rightarrow S | A | B \\ A &\rightarrow S | E \\ B &\rightarrow E \\ \end{align} \]

   So we have pairs of variables (S,S), (S,A), (S,B), (S,E), (A,S), (A,A), (A,B), (A,E), (B,E)

   So for each of the pairs $(G, H)$, we replace that unit production by finding every non-unit production $H \rightarrow \alpha$ and adding a new production $G \rightarrow \alpha$

   \[ \begin{align} S &\rightarrow SA | S | A | SB | B \\ A &\rightarrow ( S ) | ( ) | S | E \\ B &\rightarrow [ S ] | [ ] | E \\ E &\rightarrow x \\ \end{align} \]

   (S,A)

   \[ \begin{align} S &\rightarrow SA | S | ( S ) | ( ) | S | E | SB | B \\ A &\rightarrow ( S ) | ( ) | S | E \\ B &\rightarrow [ S ] | [ ] | E \\ E &\rightarrow x \\ \end{align} \]

   (S,B)

   \[ \begin{align} S &\rightarrow SA | S | ( S ) | ( ) | S | E | SB | [ S ] | [ ] \\ A &\rightarrow ( S ) | ( ) | S | E \\ B &\rightarrow [ S ] | [ ] | E \\ E &\rightarrow x \\ \end{align} \]

   (S,E)

   \[ \begin{align} S &\rightarrow SA | S | ( S ) | ( ) | S | x | SB | [ S ] | [ ] \\ A &\rightarrow ( S ) | ( ) | S | E \\ B &\rightarrow [ S ] | [ ] | E \\ E &\rightarrow x \\ \end{align} \]

   (A,S)

   \[ \begin{align} S &\rightarrow SA | S | ( S ) | ( ) | S | x | SB | [ S ] | [ ] \\ A &\rightarrow ( S ) | ( ) | SA | x | SB | [ S ] | [ ] | E \\ B &\rightarrow [ S ] | [ ] | E \\ E &\rightarrow x \\ \end{align} \]

   (A, B) no change

   (A,E)

   \[ \begin{align} S &\rightarrow SA | ( S ) | ( ) | S | x | SB | [ S ] | [ ] \\ A &\rightarrow ( S ) | ( ) | SA | x | SB | [ S ] | [ ] \\ B &\rightarrow [ S ] | [ ] | E \\ E &\rightarrow x \\ \end{align} \]

   (B, E)

   \[ \begin{align} S &\rightarrow SA | ( S ) | ( ) | S | x | SB | [ S ] | [ ] \\ A &\rightarrow ( S ) | ( ) | SA | x | SB | [ S ] | [ ] \\ B &\rightarrow [ S ] | [ ] | x \\ E &\rightarrow x \\ \end{align} \]

3. Eliminate useless variables and their productions.

   Rewrite the grammar to eliminate any useless variables (variables that can never be derived or that will never generate a terminal).

   Reveal

   First we look for unreachable variables.

   1. S is reachable.
   2. From S, therefore, A and B are reachable
   3. From A, we get nothing new.
   4. From B, we get nothing new.

   We are done, and have not marked E. So E is unreachable. We can eliminate all productions that mention it.

   \[ \begin{align} S &\rightarrow SA | ( S ) | ( ) | x | SB | [ S ] | [ ] \\ A &\rightarrow ( S ) | ( ) | SA | x | SB | [ S ] | [ ] \\ B &\rightarrow [ S ] | [ ] | x \\ \end{align} \]

   Next we look for non-generating variables. All three can generate x, so all variables are generating.

4. Factor the long productions to get them into the form $A \rightarrow a$ or $A \rightarrow B C$

   Reveal

   \[ \begin{align} S &\rightarrow SA | ( S ) | ( ) | x | SB | [ S ] | [ ] \\ A &\rightarrow ( S ) | ( ) | SA | x | SB | [ S ] | [ ] \\ B &\rightarrow [ S ] | [ ] | x \\ \end{align} \]

   \[ \begin{align} S &\rightarrow SA | C D | ( ) | x | SB | [ S ] | [ ] \\ A &\rightarrow ( S ) | ( ) | SA | x | SB | [ S ] | [ ] \\ B &\rightarrow [ S ] | [ ] | x \\ C &\rightarrow ( \\ D &\rightarrow S E \\ E &\rightarrow ) \\ \end{align} \]

   \[ \begin{align} S &\rightarrow SA | C D | C E | x | SB | [ S ] | [ ] \\ A &\rightarrow ( S ) | ( ) | SA | x | SB | [ S ] | [ ] \\ B &\rightarrow [ S ] | [ ] | x \\ C &\rightarrow ( \\ D &\rightarrow S E \\ E &\rightarrow ) \\ \end{align} \]

   \[ \begin{align} S &\rightarrow SA | C D | C E | x | SB | F G | [ ] \\ A &\rightarrow ( S ) | ( ) | SA | x | SB | [ S ] | [ ] \\ B &\rightarrow [ S ] | [ ] | x \\ C &\rightarrow ( \\ D &\rightarrow S E \\ E &\rightarrow ) \\ F &\rightarrow [ \\ G &\rightarrow S H \\ H &\rightarrow ] \\ \end{align} \]

   \[ \begin{align} S &\rightarrow SA | C D | C E | x | SB | F G | F H \\ A &\rightarrow ( S ) | ( ) | SA | x | SB | [ S ] | [ ] \\ B &\rightarrow [ S ] | [ ] | x \\ C &\rightarrow ( \\ D &\rightarrow S E \\ E &\rightarrow ) \\ F &\rightarrow [ \\ G &\rightarrow S H \\ H &\rightarrow ] \\ \end{align} \]

   \[ \begin{align} S &\rightarrow SA | C D | C E | x | SB | F G | F H \\ A &\rightarrow C D | C E | SA | x | SB | F G | F H \\ B &\rightarrow F G | F H | x \\ C &\rightarrow ( \\ D &\rightarrow S E \\ E &\rightarrow ) \\ F &\rightarrow [ \\ G &\rightarrow S H \\ H &\rightarrow ] \\ \end{align} \]

4 CYK parsing

Parse the string ([x])(x) using the grammar:

\[ \begin{align} S &\rightarrow SA | C D | C E | x | SB | F G | F H \\ A &\rightarrow C D | C E | SA | x | SB | F G | F H \\ B &\rightarrow F G | F H | x \\ C &\rightarrow ( \\ D &\rightarrow S E \\ E &\rightarrow ) \\ F &\rightarrow [ \\ G &\rightarrow S H \\ H &\rightarrow ] \\ \end{align} \]

5 Properties

5.1 Prove that, if $F$ is any finite set of strings, $F$ is a CFL.

5.2 Prove that, if $L$ is a CFL and $F$ is any finite set of strings, $L-F$ is a CFL.