Yes. There are many examples in the textbook.
S ↠ C ↠ LCR ↠ aCR ↠ aLCRR ↠ aaCRR ↠ aaCCRR ↠ aaLCRCRR ↠ aaaCRCRR ↠ aaaRCRR ↠ aaabCRR ↠ aaabLCRRR ↠ aaabaCRRR ↠ aaabaRRR ↠ aaababRR ↠ aaababbR ↠ aaababbb
Like ((()())) where ( is a and ) is b.
S → TT | U T → 0T | T0 | # U → 0U00 | #
S0 → S S → TT | U T → 0T | T0 | # U → 0U00 | #
This step isn’t necessary because S does not appear on the LHS of any rule, but this isn’t mentioned in the book.
Remove ε-rules.
There aren’t any. See CFS to CNF for a example.
Remove unit rules. There are only two unit rules: S0→S and S→U. Replace the targets of the unit rules with the LHS of rules where the single terminal variable is the RHS. In particular, copy the targets of S to S0 and of U to S. Also, remove the target of the unit.
S0 → TT | 0U00 | # S → TT | 0U00 | # T → 0T | T0 | # U → 0U00 | #
Add variables as needed to get rid of the rules that are not in the proper Chomsky form. We will use Z for 0, M for 0U, and N for M0 (0U0). See the book (page 111) for more information
S0 → TT | NZ | # S → TT | NZ | # T → ZT | TZ | # Z → 0 M → ZU N → MZ
S → 0A A → 0A | B B → 1C C → 1D D → ε | 0D
S → aSa | bSb | C C → a | b | ε
S0 → S S → aSa | bSb | C C → a | b | ε
Remove ε-rules.
First remove the C ε-rule. And fix S to deal with the problem.
S0 → S S → aSa | bSb | aa | bb | ε C → a | b
Next remove the S ε-rule. This will slightly complicate the S0 rules. But, it is OK for S0 to have an ε-rule.
S0 ⇒ S | ε S ⇒ aSa | bSb | aa | bb C ⇒ a | b
Deal with the unit rule S0 ⇒ S.
S0 ⇒ aSa | bSb | aa | bb | ε S ⇒ aSa | bSb | aa | bb C ⇒ a | b
Add variables as needed to get rid of the rules that are not in the proper Chomsky. We will use A for a, B for b, X for AS and Y for BS.
S0 → XA | YB | AA | BB | ε S → XA | YB | AA | BB C → a | b X → AS Y → BS A → a B → b
S0 ⇒ YB ⇒ BSB ⇒ bSB ⇒ bXAB ⇒ bASAB ⇒ baSAB ⇒ baaaAB ⇒ baaaaB ⇒ baaaab
Palindrome of letters a and b.
PDA for Problem 8 — plain CFG or Chomsky CFG. Maybe draw on board.
Omit.
Omit.