lecture 4: unsolvable problems

Lecture 4: Unsolvable Problems

虞台文大同大學資工所智慧型多媒體研究室

Content

Enumerable and Decidable SetsAlgorithm SolvabilityProblem ReductionDiagonalization Method


Enumerable and Decidable Sets

大同大學資工所智慧型多媒體研究室

Set Enumerability and Generability

A * is enumerable (generable) if

computable function g: N* such that

( )g N AIf g is not totally defined, then D(g)={0, 1, …, k 1}.

The Generation Process(cf )gg N A

*A Enumerable (Generable)

(0)g A(1)g A(2)g A(3)g A

A:g N =

=

=

= Generate

The Enumeration Process(cf )gg N A


(0)g A(1)g A(2)g A

A:g N =

=

=

A ?EnumerateEnumerate



(0)g A(1)g A(2)g A

A:g N =

=

=


Is such a process always

terminable?Is such a process always

terminable?



(0)g A(1)g A(2)g A

A:g N =

=

=


The Labeling Process(cf )gg N A


(0)g A(1)g A(2)g A

A:g N =

=

=

LabelingLabeling

( )g ?

2

Exercises

1. Write two algorithms to enumerate (generate) the set of all even numbers of integer in different sequences.

2. What is difference between ‘countable’ (studied in discrete math) and `enumerable’?

The Countability of Real Number

Is R countable?

Is (0,1)R countable?

Prove (0,1)R is uncountable

Assume that (0, 1) is countable.

Suppose that it is counted as:

f0

f1

f2

f3

f4

d01 d02 d03 d04d00

d11 d12 d13 d14d10

d21 d22 d23 d24d20

d31 d32 d33 d34d30

d41 d42 d43 d44d40

0.

0.

0.

0.

0.

0 1 2 3 4

……………

… … … … … … ……

…{0,1, ,9}ijd




d00

d11

d22

d33

d44

…

f0

f1

f2

f3

f4

d01 d02 d03 d04

d12 d13 d14d10

d21 d23 d24d20

d31 d32 d34d30

d41 d42 d43d40

0.

0.

0.

0.

0.

0 1 2 3 4

……………

… … … … … ……

…{0,1, ,9}ijd

Consider the following number:

0 1 2 3 40. (0,1)v h h h h h

9 {0,1, ,9}i iih d




d00

d11

d22

d33

d44

…

f0

f1

f2

f3

f4

d01 d02 d03 d04

d12 d13 d14d10

d21 d23 d24d20

d31 d32 d34d30

d41 d42 d43d40

0.

0.

0.

0.

0.

0 1 2 3 4

……………

… … … … … ……

…{0,1, ,9}ijd


0 1 2 3 40. (0,1)v h h h h h

9 {0,1, ,9}i iih d

… … … … … ……

v=fk0. h0 h1 h2 h3 h4

…




d00

d11

d22

d33

d44

…

f0

f1

f2

f3

f4

d01 d02 d03 d04

d12 d13 d14d10

d21 d23 d24d20

d31 d32 d34d30

d41 d42 d43d40

0.

0.

0.

0.

0.

0 1 2 3 4

……………

… … … … … ……

…{0,1, ,9}ijd


0 1 2 3 40. (0,1)v h h h h h

9 {0,1, ,9}i iih d

… … … … … ……

v=fk0. h0 h1 h2 h3 h4

…

0 1 20.k k k k kkv f d d d d

kk kd h 9 kkd

4.5kkd

Semidecidability

A * is semidecidable if there exists an algorithm which, when applying to any A, can decide that is in A, i.e., there exists an membership algorithm for A.

A A ?

A ?

True

False orNever Halt

Theorem 1

A is enumerable iff A is semidecidable.Pf) “”

A is enumerable g: N* such that g(N)=A.

Hence, given any A, we generate strings

g(0)=0g(1)=1

. . . . . . . .

Then, we will finally reach g(k) = k = .This allows us to concludes that A.

Theorem 1

A is enumerable iff A is semidecidable.Pf) “”What do we know?

A is semidecidableThere exists an membership algorithm for A on TM such

that( ) iff TM true A

Thinking first.

How to prove? Find g such that ( )g N A

( ) iff TM true A We have

Want ( )g N A

( ) ?g k 0,1,2,k

Theorem 1



Want ( )g N A

( ) ?g k 0,1,2,k

Preliminaries 1 2 3 4 5 6 7 8 9 10123456789

10

Numbering the followingblack cells

1 2 3 4 5 6 7 8 9 10123456789

10

Theorem 1



Want ( )g N A

( ) ?g k 0,1,2,k

0

1

2 34

5

6

7

Preliminaries

Numbering the followingblack cells

= {0, 1}

* = {, 0, 1, 00, 01, 10, 11, 000, …} 0 1 2 3 4 5 6 7

Enumerable

Theorem 1



Want ( )g N A

( ) ?g k 0,1,2,k

The algorithmto generate g(k):

Step 1. Set n = k + 1.

Step 2. List the first n strings 0 1 1, , , n

Step 3. Hand simulates n steps of on TM for each 0 1 1, , , n

Theorem 1



Want ( )g N A

( ) ?g k 0,1,2,k





0 1 2 3 4 5 6 7 8 9123456789

10

0

n St

eps

n1-

Terminating Record

Theorem 1



Want ( )g N A

( ) ?g k 0,1,2,k





0 1 2 3 4 5 6 7 8 9123456789

10

0

n St

eps

n1-

Step 4. Let m= #terminating computations.Step 5. if m < k + 1

n= n + 1; goto Step 2;else return the kth string.

k

Theorem 2

A is enumerable iff A is a domain of some computable functions.

Pf)

Exercise

Theorem 2

A is enumerable iff A is a domain of some computable functions.

How about if A is not enumerable?

How about if A is not enumerable?

Discussion

Program Computability Enumeration

Enumeration Membership Determination

(Some reformation may be required)

(Semidecidability)

Complement Set

* \A A

*, A

Decidable Set

A * is decidable if there exists an computable function, say, D such that

if :

if

true AD

false A

Theorem 3

A is decidable iff and are enumerable.A A

A is decidable iff and are semidecidable.A A

Theorem 3

A is decidable iff and areA A enumerable.semidecidable.

Pf) “” Fact: * can be enumerated as 0, 1, 2, …

Since A is decidable, a computable function D st.

The computable function g to enumerate A can be:

if :

if

true AD

false A

D: 0 1 2 3 4 . . .

t f t t f . . .

g(0) g(1) g(2) . . .

A is enumerable.g:

Show that A is enumerable as follows:

Theorem 3


Pf) “” Fact: * can be enumerated as 0, 1, 2, …

Since A is decidable, a computable function D st.

The computable function h to enumerate can be:

if :

if

true AD

false A

D: 0 1 2 3 4 . . .

t f t t f . . .h:

Show that is enumerable as follows:A

is enumerable.A

A

h(0) . . .h(1)

Theorem 3


Pf) “”

Suppose 1 semidecides A, 2 semidecides .

For any *, input it to 1 and 2 simultaneously.

If 1 terminates, return true.

If 2 terminates, return false.

A


Algorithm Solvability


Enumeration, Decidability and Solvability

Enumeration:

Decidability:

Solvability:

*N

* ,true false

(Semidecidability)

To deal with the existence of algorithm (program) to solve problems.

That is, given a problem, whether there exists a program to solve (any instance of) the problem.

1

2

Example

Problem (PE: Program Equivalence):q = “Are program 1 and program 2 equivalent?”

,true falseAny

Algorithm?

PE

Example


Solvable? Computable?

1

2

:

:

string to represent 1


1 2 1 2| and are e$ quivalent A

Decidable?

PE

Example


Problem Reformulation

Solvable? Computable?

1

2

:

:



1 2 1 2| and are e$ quivalent A

Decidable?

Problem PE is solvable iff A is decidable (recursive).Problem PE is solvable iff A is decidable (recursive).

PE

Definitions

Semisolvable:

Solvable:

Totally Unsolvable:

algorithm if q is true, it answers ‘yes’.if q is false, it may not give answer (never halt).

:P a true/ f alse problemq P

algorithm if q is true, it answers ‘yes’.if q is false, it answers ‘no’.

not semisolvale.

Problem and Complement ProblemP P

Is true?qP

Is not true?P q


1 2

Is tru ?, , eTM k TM kPE

Example:

1 2

Is not tru, , e?TM k TM kPE

Turing Machine Version


( )Is tr) ue?(P x h xE g

Example:

( ) ( )Is not true?PE g x h x

Register Machine Version

Theorem 4

P is solvable iff and are semisolvable.P P

Is true?qP

Is not true?P q

Pf) “” P is solvable algorithm P such that

if q is true, it answers ‘yes’.if q is false, it answers ‘no’.To show that P is semisolvable,

we need to find a algorithm ’ such that

is true( )

is false

true qq

q

?

Theorem 4


Is true?qP

Is not true?P q


if q is true, it answers ‘yes’.if q is false, it answers ‘no’.To show that P is semisolvable,

we need to find a algorithm ’ such that

START

P

TRUEHALT

truefalse

’:

( ) if return true

(

)

else .Pq q

is true( )

is false

true qq

q

It can be

q

Theorem 4


Is true?qP

Is not true?P q


if q is true, it answers ‘yes’.if q is false, it answers ‘no’.

START

P

TRUEHALT

falsetrue

”:

( ) if return true

no ( )

else .

t Pq q

not is true( )

is true

true qq

q

It can be

q

To show that is semisolvable, we need to find a algorithm ” such that

P

Theorem 4


Pf) “”

To run ’ and ” alternatively,

if ’ answers ‘yes’, then ‘yes’;

if ” answers ‘yes’, then ‘no’.

Corollary


P is semisolvable but not solvable

Pis totally unsolvable.

The Halting Problem



: an -progra( ) ?

m" "

: memory configurationm

mHP

MM

: a -program

" ": memory configurati

,o, n

?HP kk

MTM

T

More simple,

" " : register functio( ? n)HP g x g

START

HALT

Theorem 5

The HP (halting problem) is not solvable.

: a -program


,o, n

?HP kk

MTM

T

?

S1 S2 Sk1 Sk Sk+1 Sn1 Sn… … Blank TapeBlank Tape

Head

We will prove a simpler version.

We will prove a simpler version.會停嗎 ?

START

HALT

Theorem 5


: a -program

" ": memory configuratio

,n

1 ?,1

HP

TM

TM

S1 S2 Sk1 Sk Sk+1 Sn1 Sn… … Blank TapeBlank Tape

Head

?會停嗎 ?

Theorem 5


: a -program

" ": memory configuratio

,n

1 ?,1

HP

TM

TM

Pf) Assume that HP is solvable.

algorithm, given and <, 1>, can tell whether will halt on <, 1>.

Define @ ,1A TM

We can have an algorithm, say, decide to decide the elements of A.

( ) q

decide qq

true A

false A

That is,

Theorem 5


Pf)

@ ,1A TM

( ) q

decide qq

true A

false A

Construct strange

by making use of decide

START

Duplicate as @

HALT

decide

true falseWhat are done by strange?

What are done by strange?

Theorem 5


Pf)

@ ,1A TM

( ) q

decide qq

true A

false A

START

Duplicate as @

HALT

decide

true false

,1

,1@ ,1strange TM iff ,1 TM

Construct strange


Theorem 5


Pf)

@ ,1A TM

( ) q

decide qq

true A

false A

START

Duplicate as @

HALT

decide

true false

,1

,1@ ,1strange TM iff ,1 TM

Construct strange


The input to strange is a program’s string, say, .

strange will never halt on

if and only if

program will halt on feeding its code as input

parameter.

The input to strange is a program’s string, say, .

strange will never halt on

if and only if

program will halt on feeding its code as input

parameter.

Theorem 5


Pf)

@ ,1A TM

( ) q

decide qq

true A

false A

,1strange TM iff ,1 TM

,1strange strange TM iff ,1

strange strange TM

Assume that HP is solvable.. . . . . . . . . . . . . . . .

Theorem 5


Pf)

@ ,1A TM

( ) q

decide qq

true A

false A

,1strange TM iff ,1 TM

,1strange strange TM iff ,1

strange strange TM

Assume that HP is solvable.. . . . . . . . . . . . . . . .

START

Duplicate as @

HALT

decide

true false

,1strange

@ ,1strange strange

Theorem 6


The HP (halting problem) is semisolvable.

Pf)

Trivial(Run it)

The Non-Halting Problem (NHP)



: an -progra( ) ?

m" "

: memory configurationm

mHP

MM

: a -program


,o, n

?HP kk

MTM

T

NHP HP

Theorem 7

The NHP is not semisolvable.

The NHP is totally unsolvable.


Problem Reduction


Is a Problem Solvable?

Methods to identify the problems’ solvability:

Problem Reduction Diagonalization method Rice’s Theorem

P

Problem Reduction

Pq qr:

Let P and P’ be two problems. If there is an algorithm (r) which will map any qP into q’P st.

q is true iff q’ is true.

Then, we say that problem P reduces to P’.

P

Problem Reduction

Pq qr:

Let P and P’ be two problems. If there is an algorithm (r) which will map any qP into q’P st.


Then, we say that problem P reduces to P’.

Conceptually, P’ is larger.

P

Discussion

Pq qr:


If P’ issemisolvablesolvable How about P?

How about P?

P

Discussion

Pq qr:


If P’ issemisolvablesolvable P is

semisolvablesolvable

P

Discussion

Pq qr:


If P issemisolvablesolvable How about P’?

How about P’?

P

Discussion

Pq qr:


If P issemisolvablesolvable P’ is ?????????

P

Discussion

Pq qr:


If P’ isnot solvable

totallyunsolvable

How about P?How about P?

P

Discussion

Pq qr:


If P’ isnot solvable

totallyunsolvable

P is ????????

P

Discussion

Pq qr:


If P isnot solvable

totallyunsolvable

How about P’?How about P’?

P

Discussion

Pq qr:


If P isnot solvable

totallyunsolvable

P’ isnot solvable

totallyunsolvable

Discussion

PP

q qr:: an -program

" ": memory confi

(gura n

) ?tio

mm

HP

M MM

: a -program

" ": memory c

, ?, onfiguration

kk

HP

TM TMTM

HPM HPTM

HPTM HPM

Theorem 8

PP

q qr:: an -program

" ": memory confi

(gura n

) ?tio

mm

HP

M MM

: a -program

" ": memory c

, ?, onfiguration

kk

HP

TM TMTM

HPM HPTM

HPTM HPM

The halting-problems for TM(), R, SR, SR2,

… are semisolvable but not solvable.

The halting-problems for TM(), R, SR, SR2,

… are semisolvable but not solvable.

Theorem 9

The problem of TM equivalence (PE) is totally unsolvable.

" ?"PE TM TM

TM

TM ,true false

AnyAlgorithm?

Theorem 9


" ?"PE TM TM

Pf) Fact: NHP is totally unsolvable.

Method of problem reduction: Find r st.

: r NHP PE , ?k TM ? TM TM

true trueiff

How?

Theorem 9


Pf) Fact: NHP is totally unsolvable.

: r NHP PE , ?k TM ? TM TM

Input = <, k>?

START

HALT

true

false

?START

doesn’t halt on <k, 1>


Diagonalization Method


Example Show that HP is not solvable using diagonalization method.

Fact: The set of all programs is enumerable. (why?) Let be the string representation of all

programs. Suppose that HP is solvable. Then, we can have the

following termination record: Construct strange such that:

0 1 2, , ,...

0

1

2

3...

001...

101...

000...

011...

0 0 1 1 …………. . .

0 1 2 3 …

1ji

0ji

,1 iff ,1strange j j j TM TM

Can you write such a

program?Can you write such a

program?

... ... ... ... ...k 1 0 1 0 …

…





0 1 2, , ,...


0

1

2

3...

001...

101...

000...

011...

0 0 1 1 …………. . .

0 1 2 3 …

... ... ... ... ...k 1 0 1 0 …

…





0 1 2, , ,...


0

1

2

3...

001...

101...

000...

011...

0 0 1 1 …………. . .

0 1 2 3 …

,1 iff ,1strange k k k TM TM

,1 iff ,1k k k k TM TM

Theorem 10

The total function problem is totally unsolvable.

Pf)

totally unsolvable not semisolvable

• Assume it is semisolvable.• Then, we can enumerate the all total function as:

0 1 2, , ,

• Let x1 be the input/output register.

• Then,0 0

1 1

1 2

:

:

:

f N N

f N N

f N N

Totally defined functions

Theorem 10


Pf)


• Assume it is semisolvable.

0 0 0 0

1 1 1 1

2 2 2 2

3

0

1

3

2

33 3

(0)

(1)

0 1 2 3

(1) (2) (3)

(0) (2) (3)

( (2)

(3

0) (1) (3)

(0) (1) ) )(2

f f f f

f f f f

f

f

f

f f f

f

f

f f f f

. . . . . . . . . . . . . . . . . . . . .

• Define : ( ) 1ng n f n

• Clearly, g is total.

Theorem 10


Pf)



0 00

1

2

0 0

1 1 1 1

2 2 2 2

3 3 3 3 3

(0)

(1)

(2

0 1 2 3

(1) (2) (3)

(0) (2) (3)

(0) (1) (3)

(0) (1) (2)

)

(3)

(0) (1) (2) (3)i i i i i

f f f f

f f f f

f f f f

f f f f

f f f

f

f

f

f f

f

. . . . . . . . . . . . . . . . . . . . .



g

Theorem 10


Pf)



0 00

1

2

0 0

1 1 1 1

2 2 2 2

3 3 3 3 3

(0)

(1)

(2

0 1 2 3

(1) (2) (3)

(0) (2) (3)

(0) (1) (3)

(0) (1) (2)

)

(3)

(0) (1) (2) (3)i i i i i

f f f f

f f f f

f f f f

f f f f

f f f

f

f

f

f f

f

. . . . . . . . . . . . . . . . . . . . .



g

( ) ( )ig i f i ( ) 1if i

While programs

TotalFunctions

Theorem 11

There is a total function which is not the associate function of any count program.

Count programs

Exercises

1. Special halting problem (SHP) and general halting problem (GHP) are defined as follows:

Prove:a) If GHP is not solvable, then SHP is not solvable.b) If SHP is not solvable, then GHP is not solvable.

" ˆ ?"SHP TM

," ?"HP kG TM

Exercises

2. Special equivalence problem (SE) is defined as:

Show that SE is not solvable.

3. General equivalence problem (GE) is defined as:

Show that GE is not solvable.

" " : registe( ) 1 r fu ion? nctS ggE x

" " : register func) i( t? ong x yGE g

Exercises

4. Define

Show that MEMBER is not solvable.

: register function

" " : range of ? ff

f

MEMBER R f

z N

z R

lecture 4: unsolvable problems

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