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Decidability and Partial Decidability

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Decidability and Partial Decidability Decidability and Partial Decidability An alphabet is a finite set of symbols. Example: a,b,c are symbols; {a,b,c} is an alphabet. – PowerPoint PPT presentation

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Title: Decidability and Partial Decidability


1
Decidability and Partial Decidability
2
  • 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 ?

3
  • 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.

4
  • 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

5
  • 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.

6
  • 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.

7
  • 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.

8
  • 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.

9
  • 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.

10
  • 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.

11
  • 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.

12
  • 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.

13
  • 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

14
  • 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

15
  • We may discuss primitive recursive functions
    later
  • Some computable functions are not primitive
    recursive (Ackermanns function)
  • Maybe we will discuss this later
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