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CSC 363H                Lecture Summary for Week  2               Spring 2007
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More on Turing machine terminology:

 -  A "configuration" of a TM represents the current state, tape content,
    and head position as follows: "u q v", where q is current state, uv is
    current tape content (nonblank portion), and head is on leftmost symbol
    of v.  Note: because tape is infinite to the right, configuration
    "u q v" is equivalent to "u q v_", "u q v__", "u q v___", etc.

 -  For all states q_i, q_j in Q, symbols a, b, c in T, strings u, v in T*,
     .  if d(q_i,b) = (q_j,c,R), then configuration "u q_i bv" yields
        "uc q_j v" in one step of computation;
     .  if d(q_i,b) = (q_j,c,L), then configuration "ua q_i bv" yields
        "u q_j acv" in one step of computation and configuration "q_i bv"
        yields "q_j cv" (because head cannot move "off" left end).

 -  On input w, a Turing machine M computes as follows:
     .  start from initial (or "start") configuration C_1 = "q_0 w" --
        alternate convention C_1 = "_ q_0 w" to simplify computation;
     .  go from configuration to configuration following the transition
        function (i.e., C_1 yields C_2 yields C_3 yields ...);
     .  stop as soon as a "halting configuration" is reached:
        either an "accepting configuration" (where state is q_accept)
        or a "rejecting configuration" (where state is q_reject) --
        third possibility is infinite loop (never reach halting state).

Definitions:

 -  The "language of M", L(M) = { w in S* | M accepts input w }.

 -  L is "Turing-recognizable" (recognizable/semi-decidable/recursively
    enumerable) if there is some TM M such that L = L(M).

 -  L is "Turing-decidable" (decidable/recursive) if there is some TM M
    that halts on every input such that L = L(M).  Such a TM is called a
    "decider".

We know that { w#w | w in {0,1}* } is a decidable language.

Other examples for decidable laguages: 
     - the empty set
     - S* (all strings)
     - a finite set of strings

We will see many other examples.

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Variants
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How do we know TMs are the "right" model?  Consider different decisions we
made when defining TMs and how they affect the outcome.

Compare different models by the languages they recognize: models that can
recognize the same class of languages are equivalent.

Turing machines with "stay put" head movement:

 -  Allow head to stay on same square during a transition.  Formally,
    d : Q x T -> Q x T x {L,R,S}.

 -  Equivalent to TM: simulate stay put move by moving right to "extra"
    state then left to correct state (need more than one extra state).

 -  Details: tutorial exercise for next week.

Turing machines with doubly infinite tape:

 -  Similar to regular TMs except no leftmost square.  Tape initially all
    blank except for input, and head starts on first symbol of input.

 -  Equivalent to regular TM:
    .   Doubly-infinite tape TMs can simulate regular TMs: use a few extra
        states at start of computation to write a special symbol to the
        left of input; then, during computation, remain in same state and
        move right whenever special symbol is read.
    .   Regular TMs can simulate doubly-infinite tape TMs: start by placing
        special symbol on leftmost square (shift input one square right if
        necessary); during computation, whenever special symbol is read,
        shift entire tape content one square right (insert blank), then
        return to leftmost blank square to continue computation.

 -  Note that both directions are necessary (even if one is easier).

Multi-tape Turing machines:

 -  Allow more than 1 tape, each with independent head.  Initially, all
    tapes blank except tape 1 that contains input.  In one step, current
    state and symbols read on all tapes determine next state, symbols
    written and head movements for all tapes.
    Formally, transition function d : Q x T^k -> Q x T^k x {L,R}^k.

 -  Equivalent to TM: keep track of contents of k tapes separated by
    special symbol, as well as head positions by using additional symbols
    (duplicates of tape alphabet), and use additional states to remember k
    symbols to read/write (since k is fixed).  Simulating one step of
    multi-tape machine requires single-tape machine to scan over its entire
    tape twice: once to read k tapes, once to update k tapes.

 -  Details: Theorem 3.13 on p.149 (1st ed: 3.8 on p.137).