Solution

The correct answer is L = {anbn|n ≥ 1}.

Key Points

• This Turing machine recognizes the language L = {anbn|n ≥ 1} which means it accepts strings where the number of 'a's is equal to the number of 'b's, and there is at least one 'a' and one 'b'.
• The transition functions describe how the machine processes the input:
  • δ(q0,a) = (q1, x, R): When in state q0 and reading 'a', it replaces 'a' with 'x' and moves right, transitioning to state q1.
  • δ(q1,a) = (q1, a, R): When in state q1 and reading 'a', it keeps 'a' and moves right, staying in state q1.
  • δ(q1, y) = (q1, y, R): When in state q1 and reading 'y', it keeps 'y' and moves right, staying in state q1.
  • δ(q1, b) = (q2, y, L): When in state q1 and reading 'b', it replaces 'b' with 'y' and moves left, transitioning to state q2.
  • δ(q2, y) = (q2, y, L): When in state q2 and reading 'y', it keeps 'y' and moves left, staying in state q2.
  • δ(q2, a) = (q2, a, L): When in state q2 and reading 'a', it keeps 'a' and moves left, staying in state q2.
  • δ(q2, x) = (q0, x, R): When in state q2 and reading 'x', it keeps 'x' and moves right, transitioning to state q0.
  • δ(q0, y) = (q3, y, R): When in state q0 and reading 'y', it keeps 'y' and moves right, transitioning to state q3.
  • δ(q3, y) = (q3, y, R): When in state q3 and reading 'y', it keeps 'y' and moves right, staying in state q3.
  • δ(q3, ◻) = (qf, ◻, R): When in state q3 and reading blank (◻), it keeps the blank and moves right, transitioning to final state qf.

Additional Information

• In state q0, the machine starts by marking an 'a' with 'x' and then searches for the corresponding 'b' to mark it with 'y'.
• In state q1, the machine skips over 'a's and 'y's until it finds a 'b'.
• In state q2, the machine moves left to revisit and check the marked 'x' and continues the process until all 'a's and 'b's are marked.
• When all 'a's and 'b's are marked, the machine moves to state q3, where it verifies the marking and then transitions to the final state qf.