I came across this question:
A position-based Turing machine model is the same as the normal model except for the following change - the transition function also depends on the current position of the head. Prove/disprove: A position-based Turing machine model is equivalent to the standard model.
I understand that in the position-based model, the transition function depends on the current position of the head, and is defined by:
$\delta : (Q \setminus F) \times \Gamma \times \mathbb{N}^+ \rightarrow Q \times \Gamma \times {L,S,R}$
However, I'm not sure how to proceed with the proof or disproof Thank you in advance for any insights or guidance you can provide!