Lecture on Mar 4, 2004 Undecidability of the Halting Problem Diagonalization

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Lecture on Mar 4, 2004Undecidability of the
Halting ProblemDiagonalization Reduction

• Topics
• Prove the Halting Problem is undecidable
• Introduce reduction (informally)
• Reading Sipser, pp. 165-167 in Section 4.2 ,
  pp. 171-173 Section 5.1.

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Halting Problem

• Consider programs M with 3 possible outcomes on
  an input w
• M accepts w and halts
• M rejects w and halts
• M loops (i.e. does not halt)
• Halting Problem Is there an algorithm that given
  any program M and any input w, decides whether M
  accepts w, i.e.
• outputs 1 if M accepts w
• outputs 0 if M does not accept w, i.e. M either
  rejects w, or loops.
• ATM ? ?M, w? M is a program and M accepts w.

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Halting Problem

• Does the following algorithm N decide ATM?
• On input ?M, w? , run program M on input w
• Accept if and only if M accepts w
• Answer No!
• M may not halt on w, so N may not halt on ?M, w?.
• By definition, a decider must halt.

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Halting Problem is Undecidable

• Theorem ATM is undecidable.
• Proof by Diagonalization
• Cantor, 1873
• Used to prove that real numbers are not countable

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Proof of Theorem

• Sipser, pp. 165 167
• By way of contradiction, assume that ? an
  algorithm H that decides ATM . That is, on every
  input ?M, w?
• H(?M, w?) 1 if M accepts w
• H(?M, w?) 0 if M does not accept w
• Construct algorithm/program D that on every input
  ?M?
• Run H on ?M, ?M? ?
• D accepts ?M? if H(M, ?M?) 0, i.e. M does not
  accept ?M?
• D rejects ?M? if H(M, ?M?) 1, i.e. M accepts
  ?M?

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Proof of Theorem

• But what happens when D takes ?D? as input?
• D accepts ?D? if D does not accept ?D?
• D rejects ?D? if D accepts ?D?
• Contradiction!

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Diagonalization
?M1? ?M2? ?M3? ?M4? ?D?
M1 M2 M3 M4
1 0 1 0
1 1 1 0
0 0 0 0
1 1 0 0
?
D
0 0 1 1 ?
?
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Reduction (Informal)

• Problem A reduces to Problem B if ? an algorithm
  that converts A to B s.t. a solution to B ? a
  solution to A.
• If A reduces to B B is decidable, then A is
  decidable.
• To decide A, convert A to B decide B.
• If A reduces to B A is undecidable, then B is
  undecidable.
• Contra-positive form of above
• If B is decidable, then by above A is decidable,
  contradiction.

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Halting Problem Another Variant

• Is there an algorithm that given any program M
  and any input w, decides whether M halts on w,
• outputs 1 if M halts on w (i.e. either accepts or
  rejects)
• outputs 0 if M loops on w (i.e. does not halt)
• HTM ? ?M, w? M is a program and M halts on w
• HTM Is Undecidable
• Can again prove this by diagonalization
• But have a simpler proof Reduce ATM to HTM !

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HTM Is Undecidable Reduction from ATM

• Sipser, pp. 172 173
• Assume ? algorithm R that decides HTM
• Construct algorithm S that decides ATM ,running R
  as a subroutine On input ?M, w?
• Run R on ?M, w?
• If R rejects, reject
• If R accepts, run M on w until M halts
• If M accepts, accept if M rejects, reject
• Algorithm S decides ATM. But ATM is undecidable!
  Contradiction!