I have this question five times and five times it has been copied and pasted from some previous answer:

PLEASEPLEASE ANSWER ONLY IF YOU CAN SUBMIT WORKED OUT PEN AND PAPER PICTURES. I will send a complain to chehg support if it is copied and written on a digital keyboard.

question: diagram added to the bottom of the page:

Consider the Turing Machine depicted below. It uses the input alphabet {B, X} and the tape alphabet {B, X, }. A is the accepting state, initial state is 1. Instead of having an explicit rejecting state, we use the convention that whenever a state encounters a tape character for which the state has no transition, the input is rejected. (This has been done for no other reason than to avoid clutter in the diagram.) (a) For this TM, give complete computations for the words BXBB and BBXB, i.e. se- quences of configurations ending in a final state configuration. In case of a rejection, assume a rejecting transition c/c/R to the (implicit) rejecting state R, where c is what- ever character was unaccounted for. For example, state 3 has implicit transitions / /R and X/X/R to state R. (b) The language recognized by this machine is context-free. Give a context-free grammar that describes the same language. Note: there is no general technique for doing this; but here is a hint: look at the big loop going from state 1 to state 1; this is like a recursive language description, but one loop iteration changes the tape content in a particular way. (c) Show that the language recognized by this TM is not regular. (d) In general, the halting problem for Turing Machines is undecidable. That does not mean it is always hard. Explain why this TM always halts for any input. (e) Modify the TM such that each accepted word is exactly the reverse order of an accepted word of the original machine.

picture/diagram: