Ask a New Question
Menu

1. Construct a Turing machine for L-a"b", n>0 o" er w= (a,, and simulate

1 answer

To construct a Turing machine (TM) for the language L = { a^n b^n | n > 0 }, we can outline the components of the TM, such as the states, the tape symbols, and the transition functions. This language consists of strings with an equal number of 'a's followed by 'b's, where n is a positive integer.

Components of the Turing Machine

  1. Tape Alphabet: We will use the following tape symbols:

    • Input symbols: a, b
    • Blank symbol: #
    • Marker symbols: X, Y (used to mark visited as and bs)

  2. States:

    • q0: Start state
    • q1: Mark the first a
    • q2: Move to the right to find a b
    • q3: Mark the first b
    • q4: Return to the left to find the next a
    • q5: Check for unmarked as and bs
    • q_accept: Accept state
    • q_reject: Reject state

Transition Function

We'll describe the transitions using (current state, read symbol) -> (next state, write symbol, move direction):

  1. Initialization:

    • (q0, a) -> (q1, X, R) // Mark a as X and move right
    • (q0, b) -> (q_reject, b, R) // If the first symbol is b, it's invalid
    • (q0, #) -> (q_accept, #, R) // If only # is found (empty input after marking), accept

  2. Marking b and Moving Left:

    • (q1, a) -> (q1, a, R) // Skip over unmarked as
    • (q1, b) -> (q2, Y, R) // Mark b as Y and move right
    • (q1, #) -> (q_reject, #, R) // If reached end of input and still as left, reject

  3. Finding Unmarked a:

    • (q2, a) -> (q2, a, R) // Move right past as
    • (q2, b) -> (q2, b, R) // Move right past bs
    • (q2, #) -> (q3, #, L) // If reached the end after marking, move to check for more as

  4. Returning to Mark a:

    • (q3, Y) -> (q3, Y, L) // Move back over marked bs
    • (q3, X) -> (q3, X, L) // Move back over marked as
    • (q3, a) -> (q1, X, R) // If find a, mark it and go to find b
    • (q3, #) -> (q_accept, #, R) // If find no more unmarked a, accept

  5. Completing the Cycle:

    • Continue cycling between q1 and q2 until you either accept or reject the input.
    • If at any point, you reach a configuration that cannot continue (e.g., if there's an imbalance or remaining unmarked symbols), lead to the q_reject state.

  6. Final Check:

    • Ensure no unmarked as or bs remain before transitioning to q_accept.

Summary

This Turing machine effectively checks for an equal count of as and bs, utilizing marking for both types of symbols. Once there are no unmarked as or bs left, and the steps are correctly aligned, the TM will accept the string.

Example Trace

For the input aabb, the following sequence of transitions occurs:

  1. q0 reads a, marks it as X, and moves to q1.
  2. q1 then reads a (skips), finds b, marks it (changes to Y), and then transitions to q2.
  3. Continues marking a and b until all are marked.

If the input has a perfect matching count, the TM will end in the q_accept state.

You can adjust the specific transitions and states as you implement the TM based on your formal TM specifications.