COSC 3340: Introduction to Theory of Computation

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COSC 3340 Introduction to Theory of Computation

• University of Houston
• Dr. Verma
• Lecture 27

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Diagonalization Principle

• Let S be any nonempty set and R any relation on
  S.
• The principle states that
• the complement of the diagonal is different from
  each row.
• Can be generalized to relations on two sets,
  etc.
• Example
• Let S a, b, c, d.
• R (a, a), (b, c), (b, d), (c, a), (c, c), (c,
  d), (d, a), (d, b)

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Example (contd.)

• R (a, a), (b, c), (b, d), (c, a), (c, c), (c,
  d), (d, a), (d, b)

a b c d

a
b
c
d
a b c d
a X
b X X
c X X X
d X X
X
X
complement of the diagonal
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Applications of Diagonalization.

• The principle is very useful to show that
  enumerations of some sets cannot be exhaustive.
• For example, one can show
• 2N is uncountable.
• The set of real numbers in the interval 0,1 is
  uncountable. (Exercise)
• The set ltM, wgt M accepts w is not Turing
  decidable.

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2N is Uncountable.

• Suppose that it is countably infinite. Then there
  is an enumeration 2N S0, S1, S2, ... using
  the bijection between N and 2N .
• Let D i ? N i ? Si.
• D is a subset of N so it belongs to 2N .
• So, D Sk for some k ? N.

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2N is Uncountable (contd.)

• We ask whether k ? Sk.
• Case 1. k ? Sk. By definition of D, k ? D, but D
  Sk, so contradiction.
• Case 2. k ? Sk. By definition of D, k ? D, but D
  Sk, again contradiction.
• Hence our hypothesis that 2N is countable is
  false (q.e.d.).

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Where was Diagonalization used?

• Consider N X S0, S1, .....
• D is the complement of the diagonal of this
  infinite relation.

0 1 2 3 . .
s0 X X
s1 X X
s2
s3 X X
. X
.
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Proof for ATM ltM, wgt M accepts w

• Suppose that H is a decider for ATM.
• On input ltM, wgt, H halts and accepts if M accepts
  w and H halts and rejects if M rejects w.
• H(ltM, wgt) accept if M accepts w
• reject if M does not accept w.
• Now we construct a new Turing machine D with H as
  a subroutine. D calls H to determine what M does
  when input to M is its own description ltMgt.

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Proof for ATM (contd.)

• D On input ltMgt, where M is a TM
• Run H on input ltM, ltMgtgt.
• Output the opposite of what H outputs that is,
  if H accepts, reject and if H rejects, accept.
• D(ltMgt) accept if M does not accept ltMgt
• reject if M accepts ltMgt.
• What happens when we run D with its own
  description ltDgt as input? In that case we get
• D(ltDgt) accept if D does not accept ltDgt
• reject if D accepts ltDgt.

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Proof for ATM (contd.)

• Basically,
• Assume that a TM H decides ATM.
• Then use H to build a TM D that when given input
  lt M gt accepts exactly when M does not accept
  input lt M gt.
• Finally, run D on itself.
• The machine take the following actions, with the
  last line being the contradiction.
• H accepts ltM, wgt exactly when M accepts w.
• D rejects lt M gt exactly when M accepts lt M gt.
• D rejects lt D gt exactly when D accepts lt D gt.

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Proof for ATM (contd.)

• Examine tables of behavior for TMs H and D.

lt M1 gt lt M2 gt lt M3 gt lt M4 gt . . .
M1 accept accept
M2 accept accept accept accept
M3 . . .
M4 accept

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Another Perspective on Halting Problem

• If L is a Turing-decidable language then L
  (complement of L) is also Turing-decidable.
• If it were possible to predict for any TM M and
  any input string whether or not M would halt on
  that input, then every Turing-acceptable language
  would also be Turing decidable.

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Another Perspective

• K0 ?(M) ?(w) Turing machine M accepts input
  string w.
• If K0 is decidable then every Turing acceptable
  language is also Turing decidable.
• K0 is Turing acceptable.
• Every Turing acceptable language is Turing
  decidable iff K0 is Turing decidable.

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Another Perspective

• If K0 is decidable,
• K1 ?(M) TM M accepts input string ?(M)
  would also be Turing decidable.
• If K1 is decidable then so is complement of K1,
  K1, is not even Turing acceptable
• K1 w ? I, c w is not the encoding of any
  TM M, or w ?(M) for some M that does not accept
  its own encoding