Tuesday, September 26

At the request of one of the students we went over the proof in the book that REGULAR_TM is undecidable.  We took quite a bit of time to review the notation and to be sure that we understood what the proof said and why it was right.

This is a tricky proof, because to show that we cannot decide in general whether an encoded TM recognizes a regular language or not, we need to show that a known undecidable problem can be reduced to  REGULAR_TM.  A_TM is used for this purpose.  So, we somehow need to reduce ATM to REGULAR_TM.  The way this is done is, as usual, to first suppose that a solution exists to REGULAR_TM, a TM we call R.  We then build new TM based on R that decides A_TM.  That is, a solution to REGULAR_TM would give us a solution to A_TM.  Since A_TM is undecidable, though, that means that a solution to REGULAR_TM cannot exist.

The solution TM S we build for A_TM based on R must process input strings <M,w> and accept such strings if M would accept w and reject strings if M would not accept w.  To show how R could be used to help decide A_TM, we must build a new TM M' from the input M such that M' decides a nonregular language if M would not accept w, and M' decides a regular language if M would accept w.

Notice that at the time M' is being built, M and w are constants.  That is, we build M' based on M accepting the fixed input w, not M accepting an arbitrary w.  So, we build M' to first look at whatever string x is on the tape given to it.  If x has the form 0ⁿ1ⁿ, M' accepts this string.  Otherwise, M' acts like M on w (the fixed input M on the fixed input w, not M on x).  The accept state of M is modified to be the accept state of M'.  That is, if M accepts w, M' accepts x.  What if M would never halt w?  No problem, because then M' only accepts strings of the form 0ⁿ1ⁿ, a non-regular language.  What if M would accept w?  Then M' accepts any x given to it, so it recognizes a regular language. 

This construction of M' is easily possible.  Since M' is given in encoded form, the construction part of S just builds a new TM that checks a string to see whether it has the form 0ⁿ1ⁿ.  This is easy, because this is just a constant part that S has to put into any new M' it builds for each M and w that comes in as input.  S melds this part with the input M and w, so that M's accept state becomes the accept state in the constructed M'.

Note that S is constructing M' on S's tape.  At the end of the construction, S ensures that only the encoding of the new M' (<M'>) is on the tape.  At that time, S calls the mythical R to run on the input tape, which now contains <M'>.  R can tell us whether M' recognize a regular language or not. But M' was constructed to recognize a regular language iff the input M would accept the input w.  Hence, S is a decider for A_TM. 

Since the construction of M' is all possible given M and w, the only fallacy with S is that the R portion cannot exist.  That is, there is no decider for REGULAR_TM.