Final answer:
The question involves designing a Turing Machine for a language where the number of 'a's must be greater than the number of 'b's, and there is at least one 'b'. The solution outlines a strategy for the Turing Machine to iteratively mark 'a's and 'b's to ensure the language requirement is met.
Step-by-step explanation:
The subject of the given question is to design a Turing Machine for a specified language. The language L=i>j>0 indicates that the string should consist of a sequence of 'a's followed by a sequence of 'b's where the number of 'a's (i) is greater than the number of 'b's (j), and j must be greater than zero. In other words, there must be at least one 'b', and there must be more 'a's than 'b's in the string.
To construct this Turing Machine, one must follow the steps below:
- Create a start state where the machine scans from left to right.
- When the machine reads an 'a', it marks it (e.g., replacing it with an 'x') and moves to the right looking for 'b's.
- It then marks the first 'b' it finds (e.g., replacing it with a 'y') and returns to look for more 'a's to mark.
- Each time it marks an 'a', it must find a corresponding 'b' to mark. If it finds more 'a's after no more 'b's are left to match, the string is accepted.
- If the machine tries to mark a 'b' but there are no unmarked 'b's left, and unmarked 'a's remain, it moves to an accept state. If there are no unmarked 'a's left and unmarked 'b's remain, the machine moves to a reject state.
This process ensures that there is always one more 'a' than 'b's, satisfying the language requirement of i being greater than j, with j being greater than zero. The construction and transition functions of a Turing Machine need considerable detail, usually provided by state diagrams, transition tables, or formal notation.