The Halting Problem and Decidability

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The Halting Problem and Decidability

• How powerful is a TM?
• Any program in a high level language can be
  simulated by a
• TM.
• Any algorithmic procedure carried out by a human
  or a
• computer can be carried out by a TM.
• Algorithm is an informal notion but we have
  identified it with
• a TM.
• Questions
• Are there problems that cannot be solved by an
  algorithm?
• (existence)
• What are some example problems?
• (witness)
• The plan

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Universal Turing Machines

• Interpreting and running programs is algorithmic.
  It is is a
• difficult task, but nevertheless can be carried
  out by an
• algorithm (e.g. a compiler).
• SO
• We can design a (universal) TM which compiles
  or runs
• other TMs. This is the universal TM it takes
  as input a TM
• and its input tape, and runs the TM on that
  input.
• e(M)x UTM
  outputs fM(x)
• 
  iff M computes fM(x)
• 
  i.e. it halts
• 
  
• input
  loops
• 
  iff M loops on x
• If M is a TM, then encode it by e(M). x is the
  input for M, is a separator symbol.

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Encoding a Turing Machine

• Consider a TM M with states Q and alphabet S (the
  input alphabet).
• Let A a1,a2, both infinite sets
• St q1,q2,
• Such that Q C St and S C A.
• Encode every symbol ai in A by e(ai) ai, where i
  is the binary
• representation of i. For example, a5 is encoded
  by a101. Similarly, qi
• encodes qi.
• Assume a1 is the marker lt, a2 is a blank, and q1
  is the start state.
• Represent every halting state by a blank. So the
  transition (q,a),( ,a)
• is a transition into a halting state (trust me).
• The TM is represented by a sequence of
  transitions of form
• ((qi,aj),(qk,t)) where t is either ar (a
  symbol), or -gt or lt-. Note, qk
• might be blank.

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Encoding a Turing Machine

• Now, to simulate M, assume a 3-tape version for
  the UTM U.
• Tape 1 has e(M)e(x)
• Copy e(M) on to tape 2
• Shift e(x) to leftmost position of tape 1.
• Tape 1 will be used to simulate the I/O of M.
  Note that tape 1head
• does not look at a symbol from x, but an encoding
  of it, e.g. a101
• instead of a5. So, before any simulation step
  assume head is a first
• symbol of encoding. E.g. a101.
• 4. Tape 3 is used to hold qi the encoding of
  the current state qi in M.
• To simulate a step in M, assume U is in the state
  (corresponding to )
• qj in M (qj on tape 3), and tape 1 head points at
  ai (tape 1). Look up
• transition for (qj,ai) on tape 2 and simulate
  that transition, leaving new
• state on tape 3 and move tape 1 head accordingly.
  
• This concludes simulation of M by U.

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Undecidability/unsolvability

• Undecidability/unsolvability shades of the same
  thing.
• A problem is unsovable if there exists no Turing
  machine to compute
• it
• Question1 . Are there problems that cannot be
  solved by an
• algorithm? (existence)
• Yes. Proof
• Show all TMs can be enumerated
• Show all partial functions cannot be enumerated.
  (Diagonalisation)
• Encode every TM M by a unique number n(M) (Godel
  
• numbering)
• Encode M by e, as for UTMs. ie consider e(M).
  Convert every
• transition to binary, then concat all together,
  with inputs. Result is one
• long string which can be interpreted as a number.
  
• ii) Enumerate all the TMs as follows

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Undecidability/unsolvability

• 2. There exists a (partial) function fN-gtN that
  is not (partial) Turing
• computable, i.e. there is a function without a
  corresponding machine.
• Proof Diagonalisation.
• For every k, let fk be the function computed by
  Tk.
• Now construct a diagonal function f
• f(x) f(x) 1 if x is in
  dom(f(x))
• 0
  otherwise
• e.g. if f(3) 6, then f(3) 7.
• 0 1
  2 3
  
• f(0) f(0)ltgt f(0)
• f(1) f(1)ltgt f(1)
• f(2)
  f(2)ltgt f(2)
• .
• .

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The Halting Problem

• Problem Design a machine H such that
• 
  outputs 1
• e(M)e(x) H
  iff M halts given input x
• 
  
• 
  outputs 0
• Input
  iff M does not halt given
  input x
• Theorem The Halting problem is unsolvable
• Proof By contradiction. Assume H exists and
  then derive a contradiction.
• Construct the following machine H
• e(M) copier e(M)e(M)
  modified loops iff
• 
  H M halts given
  input (e(M))
• 
  outputs 0 iff
• 
  M does not halt
• H
  given input e(M))

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Implications

• There is no algorithm which can take an arbitrary
  program and arbitrary input, and tell you whether
  or not the program will halt.
• You may be able to show this for a particular
  program, or even a sub-class of programs, but
  there is no general strategy for all programs.
• E.g.
• While b do .
• For i1 to x do .
• The world of computing is littered with
  (interesting!) unsolvable problems!

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Decision problems

• Unsolvable problems yes/no solutions
• Examples
• Does z xy, (for integers x,y,z)
• Does TM M halt given x as input.
• Does TM M halt for very input x in S.
• Does TM M halt for given input e.
• Are two TMs equivalent.
• Does a diaphantine equation have a solution.
  (Hilberts 10th problem)
• A DE has form p(x1,xn) 0
• e.g. x2y2 z2
• x2-2 0
• 7. Is a predicate calculus statement valid.
  (1936, Church)

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Reduction

• Arguments by reduction
• Defn A decision problem P reduces to P, P-gtP,
  if an algorithm to solve P can be used directly
  to solve P.
• P-gtP iff
• e(I) TM e(I)
• I is instance of P, I is instance of P.
• --------------------------------------------------
  --------------------------------------------------
  
• Theorem If P-gtP and P is solvable, then so
  is P .
• Corollary If P-gtP and P is unsolvable then so
  is P.
• e(I) TM e(I) P
  1 iff I in Y(P) iff 1 in Y(P)
• P-gtP TM
  0 iff I in N(P) iff 1 in N(P)

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