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SC 363H                Lecture Summary for Week  6             Spring 2007
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readings: section 5.3, problem 5.28.

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Mapping Reducibility (cont'd)
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Examples:

 -  EQ_TM is neither recognizable nor co-recognizable.
    (A language is "co-recognizable" if its complement is recognizable.
    Similar definition for decidability unnecessary -- do you see why?)

    Show A_TM <=m EQ_TM (equiv. to A_TM^c <=m EQ_TM^c):
        On input <M,w>, construct <M1,M2> as follows:
            M1 = "On input x: run M on w (ignore x) and do the same."
            M2 = "On input x: accept."
        Then, <M,w> in A_TM iff M accepts w iff M2 accepts every string iff
        L(M1) = L(M2) iff <M1,M2> in EQ_TM.
    This proves that EQ_TM is not co-recognizable.

    Show A_TM <=m EQ_TM^c (equiv. to A_TM^c <=m EQ_TM):
        On input <M,w>, construct <M1,M2> as follows:
            M1 = "On input x: run M on w (ignore x) and do the same."
            M2 = "On input x: reject."
        Then, <M,w> in A_TM iff M accepts w iff M2 accepts every string iff
        L(M1) =/= L(M2) iff <M1,M2> not in EQ_TM iff <M1,M2> in EQ_TM^c.
    This proves that EQ_TM is not recognizable.

 -  INF = { <M> : L(M) contains infinitely many strings }
    is neither recognizable nor co-recognizable.

    HALT_TM <=m INF:
        On input <M,w>, construct <M'> as follows:
            M' = "On input x:
                 1. Run M on w.
                 2. Accept if M halts."
        If M halts on w, then M' accepts all strings so L(M') is infinite.
        If M loops on w, then M' accepts no string so L(M') is finite.
    This shows INF^c is unrecognizable.

    HALT_TM <=m INF^c:
        On input <M,w>, construct <M'> as follows:
            M' = "On input x:
                 1. Run M on w for |x| many steps.
                 2. Accept if M has not halted."
        If M halts on w, then M' accepts only the strings up to a certain
        length, so L(M') is finite.
        If M loops on w, then M' accepts all strings so L(M') is infinite.
    This shows INF is unrecognizable.

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Rice's Theorem
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If P is a nontrivial property of recognizable languages, then P is
undecidable.  ("Nontrivial" means there is at least one language that
satisfies the property and at least one language that does not; a property
of recognizable languages means a property of L(M) for TMs M -- it must be
the case that the property depends only on L(M) and on no other aspect of
M.)

Examples (Rice's theorem applies to all of these languages, so each one is
undecidable):

 -  E_TM = { <M> : L(M) = {} }
    (depends only on L(M) and is non-trivial: L(M) = {} for some M, and
    L(M) != {} for others)

 -  REGULAR_TM = { <M> : L(M) is regular }

 -  DECIDABLE_TM = { <M> : L(M) is decidable }

 -  INF_TM = { <M> : L(M) contains infinitely many strings }

 -  { <M> : M accepts 00101 }
    (because if L(M1) = L(M2), then <M1> and <M2> are either both in or
    both out of the language, so this depends only on L(M) and the property
    is non-trivial)

 -  ...

Languages that do NOT fall under Rice's theorem (so we cannot conclude
anything about their decidability/undecidability -- anything is possible):

 -  { <M> : L(M) is recognizable }
    (trivial: L(M) is recognizable for every TM M, by definition)

 -  { <M> : M has a useless state }
    (not a property of L(M))

 -  { <M> : M is a 2-tape TM that writes on its 2nd tape for some input }
    (not a property of L(M))

 -  ...

Proof sketch:  Selected solution to problem 5.30 on p.215 of textbook
(not in 1st edition).