Decidability and Partial Decidability

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Decidability and Partial Decidability
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• An alphabet is a finite set of symbols.
• Example a,b,c are symbols a,b,c is an
  alphabet.
• Example 0,1 are symbols 0,1 is an alphabet.
• ? (Sigma) represents an alphabet.
• A string over ? is a sequence of symbols from ?.
• Example 01001 is a string over 0,1
• Example aabbaaa is a string over a,b,c
• ? is the set of all finite strings over ?

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• Example 0,1 is the set of all finite binary
  strings
• Note that the input to a Turing machine is a
  string of tape symbols.
• The output to a Turing machine is also a string
  of symbols.

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• A problem is a set of yes-no questions.
• A problem consists of a set of objects (instances
  of the problem) called the base set, together
  with a right answer, either yes or no, for
  each object.
• Example Blank tape halting problem.
• Base set set of all Turing machines
• Instance of the problem A Turing machine M
• Right answer for M Yes if TM M halts on blank
  tape, no if not

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• Example Student passing problem
• Base set set of all comp006 students
• Instance of problem A particular student A
• Right answer for A Yes if student A passes,
  no if not
• If P is a problem then ?P, the complement of P,
  is P with the yes and no answers
  interchanged.

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• The complement of the blank tape halting problem
  is the blank tape non-halting problem, in which
  the answer is yes if the TM fails to halt on
  blank tape, and no if it halts.
• The complement of the passing problem is the
  failing problem.
• A problem may have one or more encodings which
  represent instances of the problem as strings of
  symbols.
• For the blank tape halting problem an encoding
  represents Turing machines as strings of symbols.

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• For the passing problem an encoding represents
  students as strings of symbols.
• Let S be the base set of a problem P and let x be
  an element of S. A Turing machine M solves the
  instance x of P if M gives the right answer on
  the encoding x of x and halts.
• M can print yes or no on the tape
• M can halt in state y or n
• M solves a problem P if M gives the right answer
  on all instances of P, and halts.

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• M solves the blank tape halting problem if M,
  given an encoding of Turing machine T, always
  halts and answers yes if T halts on blank tape,
  and no if not.
• M solves the passing problem if M, given an
  encoding of a student A, halts and answers yes
  if A passes, and halts and answers no if A
  fails.

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• A problem P is decidable (solvable) if there is a
  Turing machine T that solves P such a T always
  halts. Else P is undecidable (unsolvable).
• A problem P is semidecidable (partially
  decidable, partially solvable) if there is a
  Turing machine T that partially solves P such a
  T solves all instances of P for which the right
  answer is yes, but fails to halt for all
  instances of P for which the right answer is no.

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• The blank tape halting problem is semidecidable
  if there is a Turing machine M that, given an
  encoding of TM T, halts and says yes if T halts
  on blank tape, but M fails to halt if T fails to
  halt on blank tape
• The passing problem is semidecidable if there is
  a Turing machine M that, given an encoding of a
  student A, halts and says yes if A passes the
  course, but fails to halt if A fails the course.
• Any problem whose base set is finite, is
  decidable! Can you see why? Even if we dont
  know the answer the problem is decidable.

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• If P is decidable then P is semidecidable.
• If P is semidecidable and ? P is semidecidable
  then P is decidable.
• Let Pyes be the set of encodings of instances x
  of P such that the right answer is yes.
• Let Pno be the set of encodings of instances x of
  P such that the right answer is no.

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• For the blank tape halting problem, Pyes is the
  set of encodings of Turing machines that halt on
  blank tape. Pno is the set of encodings of
  Turing machines that do not halt on blank tape.
• For the passing problem Pyes is the set of
  encodings of students who pass the course and Pno
  is the set of encodings of students who do not
  pass the course.

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• If P is semidecidable then Pyes is called
  recursively enumerable.
• If P is decidable, Pyes and Pno are said to be
  recursive.
• Functions can also be Turing computable
• A function f from strings to strings is Turing
  computable if whenever the Turing machine starts
  with x on the tape, it ends with f(x) on the tape
• Encode integers and tuples of integers as strings

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• Thus one defines computability of functions on
  integers
• The common functions (addition, multiplication,
  subtraction) on integers are all computable by
  Turing machines
• Not all functions are computable by Turing
  machines.
• The busy beaver function is not computable
• We may discuss this later
• All primitive recursive functions are Turing
  computable

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• We may discuss primitive recursive functions
  later
• Some computable functions are not primitive
  recursive (Ackermanns function)
• Maybe we will discuss this later