Computability Decidability

1

ComputabilityDecidability

2

Unsolvable problems

Computer scientists usually deal with problems which are solvable.

We are supposed to find an algorithm to solve a particular problem.

Why do we need to know about problems which are unsolvable ?we will not waste time in finding an non-existed algorithm.we will realize that the problem needs to be simplified or altered

before we can find an algorithmic solution.unsolvable problems can stimulate our imagination and help us

gain an important perspective on computation.

3

Decidable Languages

We have seen in the last chapter an undecidable problem, the 10th problem of Hilbert.

Now we will show some other problems and consider whether they are decidable by algorithm or not.

4

Decidable Problems Concerning Regular Languages

Acceptance problem

Whether a finite automata accepts a string ?

ADFA = {<B,w>| B is a DFA that accepts input string w}

Whether a nondeterministic finite automata accepts

a string ?

ANFA = {<B,w>| B is a NFA that accepts input string w}

Whether a regular expression generates a given string.

AREX = {<R,w>| R is a regular expression that generates string w}

5

Decidable Problems concerning Regular Lang.

Empty Language problemWhether a language of a finite automata is empty ?

EDFA = {<A>| A is a DFA and L(A) = }

Equivalence problemWhether two finite automata are equivalent ?

EQDFA = {<A,B>| A and B are DFAs and L(A) = L(B)}

6

Theorem 1

ADFA is a decidable language

ADFA = {<B,w>| B is a DFA that accepts input string w}

Show that the problem of testing whether a given finite automata accepts a given string is decidable

Proof idea: Present a TM M that decides ADFA.

7

Proof

M = “on input <B,w>, where B is a DFA and w is a string:Simulate B on input w.If the simulation ends in an accept state, accept. If it ends in a

non-accepting state, reject.”

Notes: input of M is an encoded string <B,w>B can be represented reasonably as a list of its five components,

Q, , , q0 and F.before M starts its computation, M first checks on whether it

properly represents a DFA B and a string w. If not, M rejects.

8

Proof:

wInput Tape

B

Machine M

accept

rejectM keeps track of B’s current state and B’s current position in the input w by writing this information down on its tape.

9

Theorem 2

ANFA is a decidable language

ANFA = {<B,w>| B is a NFA that accepts input string w}

Proof idea: Present a TM N that decides ANFA.N has M as a subroutine.Because M is designed to work with DFAs, N first converts

the NFA B it receives as input to a DFA before passing it to M.

10

Proof

N = “on input <B,w>, where B is a NFA and w is a string:Convert NFA B to an equivalent DFA C

Run TM M from Theorem 1 on input <C,w>.

If M accepts, accept; otherwise reject.”

11

Proof:

wInput Tape

C

Machine N

accept

reject

Machine M

12

Theorem 3

AREX is a decidable language

AREX = {<R,w>| R is a regular expression that generates string w}

Proof idea: Present a TM P that decides AREX.

13

Proof

P = “on input <R,w>, where R is a regular expression and w is a string:Convert regular expression R to an equivalent DFA A.

Run TM M on input <A,w>.

If M accepts, accept; if M rejects , reject.”

14

Theorem 4

EDFA is a decidable language

EDFA = {<A>| A is a DFA and L(A) = }

Show that the problem of testing whether a given finite automata accepts any strings at all is decidable.

Proof idea: A DFA accepts some string if and only if reaching an accept state

from the start state by traveling alone the arrows of the DFA is possible.

Present a TM T that uses a marking algorithm.

15

ProofT = “on input <A>, where A is a DFA:

Mark the start state of A.

Repeat until no new states get marked:

Mark any state that has a transition coming into it from any state that is already marked.

If no accept state is marked, accept; otherwise reject.”

A is empty

16

Theorem 5

EQDFA is a decidable language

EQDFA = {<A,B>| A and B are DFAs and L(A) = L(B)}

Show that the problem of testing whether 2 DFAs recognize the same language is decidable.

Proof idea: Construct a new DFA C from A and B, where C accepts only

those strings that are accepted by either A or B but not by both.If A and B recognize the same language, C will accept nothing.Test whether L(C) is empty by the theorem 4.4

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L(C)

L(C) = (L(A) L(B)) (L(A) L(B))

L(A) L(B)

Symmetric difference of L(A) and L(B)

C is also a DFA as the class of regular language is closed undercomplementation, union and intersection

18

Proof

F = “on input <A,B>, where A and B are DFAs:Construct DFA C as described.

Run TM T from Theorem 4 on input <C>

If T accepts, accept. If T rejects, reject.”

19

Proof

Machine F

Machine T

C

accept

reject

accept

reject

20

Decidable Problems concerning to Context-Free Language

Acceptance problem

Whether a CFG accepts a string ?

ACFG = {<G,w>| G is a CFG that accepts input string w}

Emptiness problem

Whether a CFG generates any strings at all ?

ECFG = {<G>| G is a CFG and L(G) = }

Equivalent problem

Whether two CFGs generate the same language ?

EQCFG = {<G,H>| G and H are CFGs and L(G) = L(H)}

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Theorem 6

ACFG is a decidable language

ACFG = {<G,w>| G is a CFG that generates string w}

Proof idea: Naïve idea :

Go through every derivations to determine whether any is a derivation of w.

But this algorithm will not halt if w is not a member of G.

Better approach:We need to ensure that the algorithm tries only finitely many

derivations, so use CNF form.

any derivation of w has 2n-1 steps, where n is the length of w.

22

Proof

S = “on input <G,w>, where G is a CFG and w is a string:Convert G to an equivalent grammar in Chomsky

Normal Form.

List all derivations with 2n-1 steps, where n is the length of w.

If any of these derivations generate w, accept; if not, reject.”

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CFLs and PDAs

As we can convert back and forth between CFLs and PDAs.

Hence, everything we say about the decidability of problems concerning CFLs applies well to PDAs.

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Theorem 7

ECFG is a decidable language

ECFG = {<G>| G is a CFG and L(G) = }

Proof idea: Naïve idea : to determine whether L(G) =

try going through all possible w’s, one by one.But there are infinitely many w’s to try, so this algorithm could end

up running forever.

Better approach:We need to test whether the start variable can generate a string of

terminals.Determine for each variable whether that variable is capable of

generating a string of terminals, the algorithm keeps track of this information by placing a mark on that variable.

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ProofR = “on input <G>, where G is a CFG :

Mark all terminal symbols in G.

Repeat until no new variables get marked

Mark any variable A where G has a rule AU1 U2… Uk and each symbol U1 , …, Uk

has already been marked.

If the start symbol is not marked, accept; otherwise

reject. ”

G is empty

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EQCFG

Consider the problem of testing whether 2 CFGs generate the same language.

ECFG = {<G,w>| G is a CFG that generates string w}

Proof idea: Naïve idea :

use the same method as theorem 4.5But CFL is not closed under complementation or

intersections.

In fact, ECFG is not decidable. We’ll prove it in the next chapter.

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Theorem 8

“Every context-free language is decidable.”We show that every CFL is decidable.Proof idea:

Naïve (bad) idea : try to convert a PDA for A directly to a TM.the PDA for A may be nondeterministic, so we get an NTM

which we can convert into a deterministic TM.however, PDA can have some branches go on forever, reading

and writing the stack without coming to a halt. therefore our TM will not be a decider.

Better approach:We prove this theorem with the TM S in Theorem 6 to decide

ACFG.

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Proof

Let G be a CFG for A and design a TM MG that decides A.

We build a copy of G into MG.

It works as follow.

MG = “on input w:

Run TM S on input <G,w>

If this machine accepts, accept; if it rejects, reject.”

29

Proof

Machine MG

Machine S

accept

reject

accept

reject

1. Convert G to CNF form.

2. List all derivations with 2n-1 steps, where n is the length of w.

3. If any of these derivations generate w, accept; if not, reject.

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Relationship among classes of Language

Turing-recognizable

DecidableContext-freeRegular

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Non Recursively Enumerator language

We want to find a language that is not Recursively Enumerable

This language is not accepted by any Turing MachineIt can’t be implemented on (contemporary) computer.Example:

Verify that the program we wrote performs as specified (that it is correct).

Halting problem : Testing whether a TM accepts a given input string (ATM)

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Theorem 9

ATM = {<M,w>| M is a TM and M accepts w}.

ATM is undecidable.

Proof: observe that ATM is Turing-recognizable.

We try to find a TM (U) that recognizes ATM.

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Halting Problem

U = “On input <M,w>, where M is a TM and w is a string:Simulate M on input w.

If M ever enters its accept state, accept; if M ever enters its reject state, reject.”

Note: U loops on input <M,w> if M loops on w, that’s why U doesn’t decide ATM.

34

Universal Turing Machineproposed by Turing.

a machine is called universal if it is capable to simulate any other Turing machine from the description of that machine.

played an important early role in stimulating the development of stored-program computers.

35

The Diagonalization method

discovered by Georg Cantor in 1873.is the proof of the undecidability of the halting

problem.using countable and uncountable set to prove.Idea is : we want to prove that

set of Turing-recognizable language is uncountablewhile set of Turing machine is countable.

36

Cantor’s problemConcerned with the problem of measuring the sizes

of infinite sets.

If we have 2 infinite sets, how can we tell whether one is larger than the other or whether they are of the same size ?

Naïve: try to count the elements of the infinite set, we will never finish!

37

Cantor’s problem

Better approach:

Observe that 2 finite sets are of the same size if the elements of the one set can be paired with the element of the other set.

We can extend this idea to infinite sets.

38

Definition: Correspondence

Consider 2 sets, A and B, and a function f from A to B.

Say that f is one-to-one if it never maps 2 different elements to the same place that is, f(a) f(b) whenever a b.

Say that A and B are the same size if there is a one-to-one, onto function f: AB.

one-to-one and onto is called CorrespondenceCorrespondence

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Example

Let N be the set of Natural numbers {1,2,3,…}

be the set of even natural numbers {2,4,6,…}

Using cantor’s method, we can see that N and have the same size.

n f(n)

1 2

2 4

3 6... ...

40

Definition : Countable

A set A is countable if either it is finite or it has the same size as N.

41

ExampleLet

N be the set of Natural numbers {1,2,3,…} Q be the set of positive rational numbers

Q = {m/n |m,n N }

Q seems the be bigger than N but actually they have the same size.

Using cantor’s method, we list all the elements of Q and then pair each element with the element of N.

We have to ensure that every member of Q appears only once on the list.

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Correspondence of N and Q

11

12

13

14

15

21

22

23

24

25

31

32

33

34

35

41

42

43

44

45

51

52

.

.

.

. . .

43

Trick

We pair the first element of Q on the list with the number 1 from N.

the second element of Q on the list with the number 2 from N.

…

After seeing the corresponding of N and Q, we can see that as each curve Q has finite number of elements which can be mapped with elements of N, thus both sets have same size.

44

Theorem 4.14R is uncountable

R is the set of real number.

Proof by the diagonalization method.

45

Proof

by contradiction : suppose that a correspondence f existed between N and R. means that we must pair all the members of N with all the members of R.

show that f fails to work as it should.

How to show ? finding an x in R that is not paired with anything in N.

46

ProofFinding x by constructing it.

choose each digit of x to make x different from one of the real numbers that is paired with an element of N

Sure that x is different from any real number that is paired.

47

Illustration of the idea

suppose we have a corresponding f:

we construct x by let the first digit of x be anything different from the first fractional digit of f(1). let the second digit of x be anything different from the second fractional digit of f(2).

…

thus we can construct x which is different from f(n) for any n.

n f(n)

1 0.123456

2 0.500000

3 0.759685... ...

n f(n)

1 0.423456

2 0.530000

3 0.756685... ...

48

Corollary 4.15

Some languages are not Turing-recognizable.

Proof idea: set of Turing-recognizable language is uncountable while set of

Turing machine is countable.

Because each Turing machine can recognize a single language and there are more languages than TMs, some languages are not recognized by any TM.

Thus, there exists some languages which are not Turing-recognizable.

49

set of all TMs is countable.Proof idea:

first of allwe have to observe that the set of all * is countable, for

any alphabet .

with only finitely many strings of each length, we may form a list of * by written down all strings of length 0, length 1, length 2 and so on.

50

set of all TMs is countable.Secondly

We can encode each TM into a string <M>

as the set of all strings is countable therefore the set of all TMs is countable.

51

set of all languages is uncountable

Proof idea:

first of allwe have to observe that the set of all infinite binary

sequence is uncountable

An infinite binary sequence is an unending sequence of 0s and 1s.

we can prove by the diagonalization method.

(imagine as the same way as R)

52

set of all languages is uncountableSecondly

Let B be the set of all infinite binary sequences.Let L be the set of all languages over alphabet .Show that L is uncountable by giving correspondence

with B.the idea is that “as B is uncountable thus L is also

uncountable.”

53


Correspondence of L and BShow that both sets have the same size.

Let * = {s1,s2,…}.

Each language A L has a unique sequence of B. (one-to-one and onto)

Characteristic sequence of A, (A)The ith bit of that sequence is a 1 if si A and is a 0 if

si A.

54


Example : Characteristic sequence

The function f: L B, where f(A) equals the Characteristic sequence of A, is one-to-one and onto and hence a correspondence. Therefore, as B is uncountable, L is uncountable as well.

* = { , 0 , 1 , 00 , 01 , 10 ,11

, 000 , 001 , … } ;

A = { 0 , 00 , 01 000 , 001 , … } ;

(A) = { 0 1 0 1 1 0 0 1 1 … } ;

55

The Halting Problem

We have proved that there are some languages which are not Turing-recognizable.

Now, we want to show that such a TM which can such a TM which can recognize the non Turing-recognizable recognize the non Turing-recognizable language doesn’t existlanguage doesn’t exist.

56

Proof of Theorem 9

ATM is undecidable.Proof:

Consider = {a}.

The set of all strings is * = {,a, aa, aaa, aaaa…}

= {,a1, a2, a3, a4…}

the set of all TMs is countable, thus we can order them as M1, M2, M3,…

to indicate which language is accepted by what machine, we can again use the characteristic sequence of that language.

57

Proof of Theorem 9

Example:assume a machine Mi accepts a language L(Mi)

L(Mi) = {a2, a4, a6}

we can write

* = { a1 , a2 , a3 , a4 , a5 , a6 , a7 , a8 ,,,

… } ;

L(Mi) = { 0 1 0 1 0 1 0 0 … } ;

58

1a 2a 3a 4a

)1(ML

0 1 10

)2(ML

)3(ML

01 0 1

0 1 11

)4(ML 0 10 0

Proof of Theorem 9

For each machine, we can write its characteristic sequence

59

Consider the language

)}(:{ iii MLaaL

L consists from the 1’s in the diagonal

Proof of Theorem 9

60

1a 2a 3a 4a

0 1 10

01 0 1

0 1 11

0 10 0

},,{ 43 aaL

)1(ML

)2(ML

)3(ML

)4(ML

61

)}(:{ iii MLaaL


)}(:{ iii MLaaL

Consider the language L

Proof of Theorem 9

Since L is a properly defined language, we can consider its complement

62

1a 2a 3a 4a

0 1 10

01 0 1

0 1 11

0 10 0

},,{ 21 aaL

)1(ML

)2(ML

)3(ML

)4(ML

63

Claim 1:

Language is not recursively enumerableL

Proof of Theorem 4.9

If we can prove that it is not recursively enumerable, we can now find a language which is not Turing recognizable.

If we can prove that it is not recursively enumerable, we can now find a language which is not Turing recognizable.

64

is recursively enumerable

LAssume for contradiction that

There must exist some machine

that accepts kM

L

LML k )(

Proof: Claim 1

65

1a 2a 3a 4a

0 1 10

01 0 1

0 1 11

0 10 0

Question: ?1MMk

)1(ML

)2(ML

)3(ML

)4(ML

66

1a 2a 3a 4a

0 1 10

01 0 1

0 1 11

0 10 0

)1(ML

)2(ML

)3(ML

)4(ML

Answer: 1MMk )(

)(

11

1

MLa

MLa k

)(

)(

11

1

MLa

MLa k

67

1a 2a 3a 4a

0 1 10

01 0 1

0 1 11

0 10 0

)1(ML

)2(ML

)3(ML

)4(ML

Question: ?2MMk Answer: 2MMk )(

)(

22

2

MLa

MLa k

)(

)(

22

2

MLa

MLa k

68

1a 2a 3a 4a

0 1 10

01 0 1

0 1 11

0 10 0

)1(ML

)2(ML

)3(ML

)4(ML

Question: ?3MMk Answer: 3MMk )(

)(

33

3

MLa

MLa k

)(

)(

33

3

MLa

MLa k

69

Similarly: ik MM for any i

Because either:

)(

)(

ii

ki

MLa

MLa

)(

)(

ii

ki

MLa

MLa

)(

)(

ii

ki

MLa

MLa

)(

)(

ii

ki

MLa

MLa

or

Proof: Claim 1

70

Therefore, the machine cannot exist kM

Therefore, the language

is not recursively enumerable

L

End of ProofEnd of Proof

Proof: Claim 1

71

There is no There is no algorithmalgorithm that describes that describes L

(otherwise would be accepted by

some Turing Machine)

L

Observation:

72

Recursive

Recursively Enumerable

Non Recursively Enumerable

L

73

Testing whether TM accepts a given string : ATM

Try to construct a TM (H) to decide whether another TM (M) accepts a string (w) or not.

H is decider on (M,w)

accept if M accepts w

H(M,w) =

reject if M rejects w

74


If w = the encoded string <M>

H is decider on (M,<M>)

accept if M accepts <M>

H(M,<M>) =

reject if M rejects <M>

75


Now we construct a new TM (D) with H as a subroutine.

D does opposite to what its subroutine does.D = “On input <M>, where M is a TM,

Run H on input <M,<M>)Output the opposite to what H outputs; that is, if H

accepts, reject and if H rejects, accept.”

76


accept if H rejects <M>

D(<M>) =

reject if H accepts <M>

accept if M rejects <M>

D(<M>) =

reject if M accepts <M>

77


D(<M>)

H (M,<M>)

M (<M>)if M accepts <M> then H acceptsif M rejects <M> then H rejects

accept

reject accept

reject

78


running M on <M>

1M

1 1

1

1

1

11

1

1M

2M

3M

4M

2M 3M 4M

1

79

output of H


1M

0

1

10

0

1

0

1

1

0

1

0

0

1M

2M

3M

4M

2M 3M 4M

1

1

1

80

Output of D


1M

0

1

10 0

1

0

1

1

0

1

0

0

1M

2M

3M

4M

2M 3M 4M

1

1

1

D 1 10 0

D

1

1

0

?

1

81

A Language which is Recursively Enumerable and not Recursive

82

We want to find a language which

There is a

Turing Machine

that accepts

the language

The machine

doesn’t halt

on some input

Is recursively

enumerable

But not

recursive

83

We will prove that the language

)}(:{ iii MLaaL

Is recursively enumerable

but not recursive

Proof

84

1a 2a 3a 4a

0 1 10

01 0 1

0 1 11

0 10 0

},,{ 43 aaL

)1(ML

)2(ML

)3(ML

)4(ML

85

Theorem:

The language )}(:{ iii MLaaL


Proof:

We will give a Turing Machine that accepts L

86

Turing Machine that accepts L

For any input string w

• Find Turing machine iM

(using the enumeration procedure for Turing Machines)

• Simulate on inputiM ia

iM• If accepts, then accept w

• Compute for whichiaw i

End of Proof

87

)}(:{ iii MLaaL

Recursively enumerable

)}(:{ iii MLaaL

Not recursively enumerable

(Thus, also not recursive)

Observation:

88


is not recursive

Theorem:

Proof:

Assume for contradiction that

L is recursive

89

Proof

But we know:

L is not recursively enumerable

thus, not recursive

CONTRADICTION!

Therefore, is not recursiveL

End of Proof

Then is recursive L

90

L

Recursive



L

91

Theorem

A language is decidable (recursive) if and only if it is both Turing-recognizable and Co-Turing-recognizable.

Co-Turing-recognizable language is the the complement of a Turing-recognizable language.

92

Corollary

is not Turing-recognizable.

Proof:

ATM is Turing-recognizable.

if is Turing-recognizable,

ATM would be decidable.

But ATM is not decidable, thus

must not be Turing-recognizable.

ATM

ATM

ATM

93

Recursively Enumerable and

Recursive

Languages

94

Definition:

A language is recursively enumerable(rekursivt uppräknelig)

if some Turing machine accepts it

95

For string :

Let be a recursively enumerable languageL

and the Turing Machine that accepts itM

Lw

w

if then halts in a final state M

Lwif then halts in a non-final stateM

or loops forever

96

Definition:

A language is recursive

if some Turing machine accepts it

and halts on any input string

In other words:

A language is recursive if there is

a membership algorithm for it

97

For string :

Let be a recursive languageL

and the Turing Machine that accepts itM

Lw

w

if then halts in a final state M

Lwif then halts in a non-final stateM

98

We will prove:

1. There is a specific language

which is not recursively enumerable

(not accepted by any Turing Machine)

2. There is a specific language

which is recursively enumerable

but not recursive

99

Recursive



100

We will first prove:

• If a language is recursive then

there is an enumeration procedure for it

• A language is recursively enumerable

if and only if


101

Theorem:

if a language is recursive then


L

102

Proof:

MM~

Enumeration Machine

Accepts LEnumerates all

strings of input alphabet

103

If the alphabet is then

can enumerate strings as follows:

},{ baM~

abaaabbabbaaaaab......

104

Enumeration procedure

Repeat:

M~ generates a string

M

w

checks if Lw

YES: print to output w

NO: ignore w

End of Proof

105

Example:

M~

abaaabbabbaaaaab......

)(ML

b

ab

bbaaa

......

,....},,,{ aaabbabbL Enumeration

Output

b

ab

bbaaa

......

106

Theorem:

if language is recursively enumerable then


L

107

Proof:

MM~

Enumeration Machine

Accepts LEnumerates all

strings of input alphabet

108

If the alphabet is then

can enumerate strings as follows:

},{ baM~

abaaabbabbaaaaab

109

Enumeration procedure

Repeat: M~ generates a string

M

w

checks if Lw

YES: print to output w

NO: ignore w

NAIVE APPROACH

Problem:

If

machine may loop forever

LwM

110

M~

M

1w

executes first step on

BETTER APPROACH

1w

M~ Generates second string 2w

M executes first step on 2w

second step on 1w

Generates first string

111

M~ Generates third string 3w

M executes first step on 3w

second step on 2w

third step on 1w

And so on............

112

1w 2w 3w 4w

1

Step

in

string

1 1 1

2 2 2 2

3 3 3 3

113

If for any string

machine halts in a final state

then it prints on the output

Miw

iw

End of Proof

114

Theorem:

If for language

there is an enumeration procedure

then is recursively enumerable

L

L

115

Proof:

wInput Tape

Enumerator

for LCompare

Machine that

accepts L

116

Turing machine that accepts L

Repeat:

• Using the enumerator,

generate the next string of L

For input string w

• Compare generated string with w

If same, accept and exit loop

End of Proof

117

We have proven:

A language is recursively enumerable

if and only if


118

A Language which is not


119

We want to find a language that

is not Recursively Enumerable

This language is not accepted by any

Turing Machine

120

Consider alphabet }{a

Strings: ,,,, aaaaaaaaaa

1a 2a 3a 4a

121

Consider Turing Machines

that accept languages over alphabet }{a

They are countable:

,,,, 4321 MMMM

122

Example language accepted by

},,{)( aaaaaaaaaaaaML i

},,{)( 642 aaaML i

iM

Alternative representation

)( iML

1a 2a 3a 4a 5a 6a 7a

0 1 1 10 0 0

123

1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

124


)}(:{ iii MLaaL


125

1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

},,{ 43 aaL

126


)}(:{ iii MLaaL


)}(:{ iii MLaaL

L

127

1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

},,{ 21 aaL

128

Theorem:

Language is not recursively enumerableL

129

Proof:

is recursively enumerableL

Assume for contradiction that

There must exist some machine

that accepts kM

L

LML k )(

130

1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Question: ?1MMk

131

1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Answer: 1MMk )(

)(

11

1

MLa

MLa k

132

1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Question: ?2MMk

133

1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Answer: 2MMk )(

)(

22

2

MLa

MLa k

134

1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Question: ?3MMk

135

1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Answer: 3MMk )(

)(

33

3

MLa

MLa k

136

Similarly: ik MM

)(

)(

ii

ki

MLa

MLa

)(

)(

ii

ki

MLa

MLa

for any i

Because either:

or

137

Thus the machine cannot exist kM

Therefore, the language

is not recursively enumerable

L

End of Proof

138

Observation:

There is no algorithm that describes L

(otherwise would be accepted by

some Turing Machine)

L

139

Recursive



L

140

A Language which is Recursively Enumerable

and not Recursive

141

We want to find a language which

There is a

Turing Machine

that accepts

the language

The machine

doesn’t halt

on some input

Is recursively

enumerable

But not

recursive

142

We will prove that the language

)}(:{ iii MLaaL

Is recursively enumerable

but not recursive

143

1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

},,{ 43 aaL

144

The language

Theorem:

)}(:{ iii MLaaL


145

Proof:

We will give a Turing Machine that

accepts L

146

Turing Machine that accepts L

For any input string w

• Compute , for whichiaw

• Find Turing machine iM(using the enumeration procedure for Turing Machines)

• Simulate on inputiM ia

• If accepts, then accept iM w

i

End of Proof

147

Observation:

)}(:{ iii MLaaL

)}(:{ iii MLaaL

Recursively enumerable

Not recursively enumerable

(Thus, also not recursive)

148

Theorem:


is not recursive

149

Proof:

Assume for contradiction that is recursiveL

Then is recursive L

Take the Turing Machine that accepts M L

halts on any input:M

If accepts then reject

If rejects then accept

MM

150

L is recursive

But we know:

L is not recursively enumerable

thus, not recursive

CONTRADICTION!

Therefore, is not recursiveL

End of Proof

151

L

Recursive



L

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Computability Decidability