Let's consider a simple example of converting a CFG to an LBA. Suppose we have the following CFG:

CFG:

S → aSb | ε

This CFG generates strings of the form "a^n b^n" where 'n' is the same number of 'a's and 'b's. Now, let's convert this CFG to an LBA.

Linearly Bounded Automaton (LBA):

• Tape Alphabet: {a, b, $ } (including the end-of-tape marker $)
• State Set: {q0, q1, q_accept, q_reject}
• Transition Rules:
  1. (q0, a) → (q0, a, R) (Move right when reading 'a' in state q0)
  2. (q0, b) → (q1, $, L) (Move left when reading 'b' in state q0, replace 'b' with $)
  3. (q0, $) → (q_accept, $, R) (Accept when reaching the end of the string in state q0)
  4. (q1, a) → (q_reject, a, R) (Reject if 'a' is encountered after 'b' in state q1)
  5. (q1, b) → (q1, b, L) (Move left when reading 'b' in state q1)
  6. (q1, $) → (q0, $, R) (Move right when reaching the end-of-tape marker in state q1, replace $ with 'a')

Acceptance Criteria:

• Accept if the LBA reaches the state q_accept.
• Reject if the LBA reaches the state q_reject.

Now, let's walk through an example derivation and simulate the LBA for the input "aaabbb":

LBA Simulation Steps:

1. Initial Configuration: q0 aaabbb$ (Start in state q0 with the input string on the tape)
2. Apply Transition 1: q0 aabbb$ (Move right, read 'a')
3. Apply Transition 1: q0 abbb$ (Move right, read 'a')
4. Apply Transition 1: q0 bbb$ (Move right, read 'a')
5. Apply Transition 2: q1 $bb$ (Move left, replace 'b' with $)
6. Apply Transition 5: q1 $b$ (Move left, read 'b')
7. Apply Transition 5: q1 $b$ (Move left, read 'b')
8. Apply Transition 5: q1 $b$ (Move left, read 'b')
9. Apply Transition 3: q_accept $b$ (Move right, reach end of the string, accept)

The LBA successfully accepts the input "aaabbb" according to the language generated by the CFG.