Module 11

1

Module 11

• Proving more specific problems are not solvable

• Input transformation technique– Use subroutine theme to show that if one

problem is unsolvable, so is a second problem– Need to clearly differentiate between

• use of program as a subroutine and

• a program being an input to another program

2

Basic Idea/Technique

3

Proving a problem L is unsolvable

• Assume PL is a procedure that solves problem L– We have no idea how PL solves L

• Construct a program PH that solves H using PL as a subroutine– We use PL as a black box– (We could use any unsolvable problem in place of H)

• Argue PH solves H• Conclude that L is unsolvable

– Otherwise PL would exist and then H would be solvable– L will be a problem about program behavior

4

Focusing on H

• In this module, we will typically use H, the Halting Problem, as our known unsolvable problem

• The technique generalizes to using any unsolvable problem L’ in place of H.– You would need to change the proofs to work with L’

instead of H, but in general it can be done

• The technique also can be applied to solvable problems to derive alternative consequences

• We focus on H to simplify the explanation

5

Constructing PH using PL

Answer-preserving input transformations and Program PT

6

PH has two subroutines

• There are many ways to construct PH using program PL that solves L

• We focus on one method in which PH consists of two subroutines– Procedure PL that solves L– Procedure PT which computes a function f that I

call an answer-preserving (or answer-reversing) input transformation

• More about this in a moment

7

Pictoral Representation of PH *

PH

x Yes/NoPLY/NPT

PT(x)

8

Answer-preserving input transformation PT

• Input– An input to H

• Output– An input to L such that

• yes inputs of H map to yes inputs of L• no inputs of H map to no inputs of L

• Note, PT must not loop when given any legal input to H

9

Why this works *

PH

PLPT

yes input to H yes input to L yes

no input to H no input to L no

We have assumed that PL solves L

10

Answer-reversing input transformation PT

• Input– An input to H

• Output– An input to L such that

• yes inputs of H map to no inputs of L• no inputs of H map to yes inputs of L

• Note, PT must not loop when given any legal input to H

11

Why this works

PH

PLPT

yes input to H no input to L yes

no input to H yes input to L no

We have assumed that PL solves L

no

yes

12

Yes->Yes and No->No

Domain of H

Yes inputsfor H

No inputsfor H

Yes inputsfor L

No inputsfor L

Domain of L

PLPT

PH

x PT(x) Yes/No

13

Notation and Terminology

• If there is such an answer-preserving (or answer-reversing) input transformation f (and the corresponding program PT), we say that H transforms to (many-one reduces to) L

• NotationH ≤ L

Domain of H

Yes inputs No inputs

Yes inputs No inputsDomain of L

14

Examples not involving the Halting Problem

15

Generalization

• As noted earlier, while we focus on transforming H to other problems, the concept of transformation generalizes beyond H and beyond unsolvable program behavior problems

• We work with some solvable, language recognition problems to illustrate some aspects of the transformation process in the next few slides

16

Example 1

• L1 is the set of even length strings over {0,1}– What are the set of legal input instances and no inputs for

the L1 LRP?

• L2 is the set of odd length strings over {0,1}– Same question as above

• Tasks– Give an answer-preserving input transformation f that

shows that L1 LRP ≤ L2 LRP

– Give a corresponding program PT that computes f

Domain of L1


Yes inputs No inputsDomain of L2

17

Program PT

string main(string x)

{

return(x concatenate “0”);

}

18

Example 2

• L1 is the set of all strings over {0,1}

– What is the set of all inputs, yes inputs, no inputs for the L1 LRP?

• L2 is {0}– Same question as above

• Tasks– Give an answer-preserving input transformation f which

shows that the L1 LRP ≤ L2 LRP

– Give a corresponding program PT which computes f

Domain of L1



19

Program PT

string main(string x)

{

return( “0”);

}

20

Example 3

• L1 – Input: Java program P that takes as input an unsigned int– Yes/No Question: Does P halt on all legal inputs

• L2

– Input: C++ program P that takes as input an unsigned int– Yes/No Question: Does P halt on all legal inputs

• Tasks– Describe what an answer-preserving input transformation f

that shows that L1 ≤ L2 would be/do?

Domain of L1



21

Proving a program behavior problem L is unsolvable

22

Problem Definitions *

• Halting Problem H– Input

• Program QH that has one input of type unsigned int

• non-negative integer y that is input to program QH

– Yes/No Question• Does QH halt on y?

• Target Problem L– Input

• Program QL that has one input of type string

– Yes/No question• Does Y(QL) = the set of

even length strings?

• Assume program PL solves L

23

Construction review

PH

x Yes/No

•We are building a program PH to solve the halting problem H

PTPT(x)

•PH will use PT as a subroutine, and we must explicitly construct PT using specific properties of H and L

PLY/N

•PH will use PL as a subroutine, and we have no idea how PL accomplishes its task

24

P’s and Q’s• Programs which are PART of program PH and

thus “executed” when PH executes– Program PT, an actual program we construct– Program PL, an assumed program which solves

problem L

• Programs which are INPUTS/OUTPUTS of programs PH, PL, and PT and which are not “executed” when PH executes– Programs QH, QL, and QYL

• code for QYL is available to PT

25

Two inputs for L *


• Program Q that has one input of type string

– Yes/No question• Does Y(Q) = the set of


• Program PL

– Solves L– We don’t know how

• Consider the following program Q1

bool main(string z)

{while (1>0) ;}

– What does PL output when given Q1 as input?

• Consider the following program Q2

bool main(string z)

{ if ((z.length %2) = = 0) return (yes)

else return (no); }

– What does PL output when given Q2 as input?

26

Another input for L *


• Program Q that has one input of type string

– Yes/No question• Does Y(Q) = the set of


• Program PL

– Solves L– We don’t know how

• Consider the following program QL with 2 procedures Q1 and QYL

bool main(string z) {

Q1(5); /* ignore return value */

return(QYL(z));}

bool Q1(unsigned x) {if (x > 3) return (no); else loop;

}

bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);

else return(no);}

• What does PL output when given QL as input?

27

Input and Output of PT *

• Input of PT

• (Also Input of H)– Program QH

• one input of type unsigned int

– Non-negative integer y

• Program QL that is the output of PT

• (Also input of L)bool main(string z) {

QH(y); /* QH and y come left-hand side */ /* ignore return value */

return(QYL(z));}

bool QH(unsigned x) {/* comes from left-hand side

}

bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);

else return(no);}

QH,y PT QL

28

Example 1 *

Input to PT

Program QH

bool main(unsigned y) { if (y ==5) return yes; else if (y ==4) return no; else while (1>0) {};}

Input y5

Output of PT

Program QL

bool QH(unsigned y) { if (y ==5) return yes; else if (y ==4) return no; else while (1>0) {};}bool QYL(string z) { if ((z.length % 2) == 0) return (yes) else return (no);}bool main(string z) { unsigned y = 5; QH(y); return (QYL(z));}

QH,y PT QL

29

Example 2

Input to PT

Program QH

bool main(unsigned y) { if (y ==5) return yes; else if (y ==4) return no; else while (1>0) {};}

Input y3

Output of PT

Program QL

bool QH(unsigned y) { if (y ==5) return yes; else if (y ==4) return no; else while (1>0) {};}bool QYL(string z) { if ((z.length % 2) == 0) return (yes) else return (no);}bool main(string z) { unsigned y = 3; QH(y); return (QYL(z));}

QH,y PT QL

30

PT in more detail

31

Declaration of PT

• What is the return type of PT?– Type program1 (with one input of type string)

• What are the input parameters of PT?– The same as the input parameters to H; in this case,

• type program2 (with one input of type unsigned int)• unsigned int (input type to program2)

program1 main(program2 QH, unsigned y)

PLPT

PH

QH,y QL Yes/No

32

program1 main(program2 P, unsigned y) {

/* Will be viewing types program1 and program2 as STRINGS over the program alphabet P */

program1 QL = replace-main-with-QH(P);/* Insert line break */

QL += “\n”;

/* Insert QYL */

QL += “bool QYL(string z) {\n \t if ((z.length % 2) == 0) return (yes) else return (no);\n }”;

/* Add main routine of QL */

QL += “bool main(string z) {\n\t”; /* determined by L */

QL += “unsigned y =”

QL += convert-to-string(y);

QL += “;\n\t QH(y)\n\t return(QYL(z));\n}”;

return(QL);}

program1 replace-main-with-QH(program2 P) /* Details hidden */string convert-to-string(unsigned y) /* Details hidden */

Code for PT PLPT

PH

QH,y QL Yes/No

33

PT in action

PT

code for QYL

QH

unsigned ystart QYL

Y/Nz

Y/N

QL

haltQH

y

Program QHbool main(unsigned y) { if (y ==5) return yes; else if (y ==4) return no; else while (1>0) {};}

Input y5

Program QLbool QH(unsigned y) { if (y ==5) return yes; else if (y ==4) return no; else while (1>0) {};}bool QYL(string z) { if ((z.length % 2) == 0) return (yes) else return (no);}bool main(string z) { unsigned y = 5; QH(y); return (QYL(z));}

PT

QYL

PLPT

PH

QH,y QL Yes/No

34

Constructing QL (and thus PT)

How to choose QYL or QNL

35

Start with no input for H

• If QH, y is a no input to the Halting problem

• Program QL bool main(string z) {

QH(y); /* ignore return value */

return(Q?L(z)); /* yes or no? */}


}

bool Q?L(string y) {

}

– Thus Y(QL) = {}

– QH loops on y

– Determine if this makes QL a no or yes input instance to L

36

Answer-preserving input transformation


– Thus Y(QL) = {}

– QH loops on y




return(QYL(z)); /* yes */}


}

bool QYL(string y) {

}

– Now choose a QYL (or QNL) that is a yes (or no) input instance to L

37

Make yes for H map to yes for L


– Thus Y(QL) = {}

– QH loops on y







}

bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);else return (no);

}

38

Possible shortcutProgram QL bool main(string z) {

QH(y); /* ignore return value */if ((z.length( ) % 2) = = 0) return (yes);else return (no);

}


}

Program QL bool main(string z) {




}


}

39

Another Example

40

Problem Definitions

• Halting Problem H– Input

• Program QH that has one input of type unsigned int

• non-negative integer y that is input to program QH

– Yes/No Question• Does QH halt on y?


• Program QL that has one input of type string

– Yes/No question• Is Y(QL) finite?

• Assume program PL solves L

41

Start with no input for H




return(Q?L(z)); /* yes or no? */}


}

bool Q?L(string y) {

}

– Thus Y(QL) = {}

– QH loops on y


42

Answer-reversing input transformation




return(QNL(z)); /* no */}


}

bool QNL(string y) {

}

– Thus Y(QL) = {}

– QH loops on y



43

Make yes for H map to no for L




return(QNL(z)); /* no */}


}

bool QNL(string y) {if ((y.length( ) % 2) = = 0) return(yes);else return(no);

}

– Thus Y(QL) = {}

– QH loops on y



44

Analyzing proposed transformations

4 possibilities

45

Problem Setup• Input of Transformation

• Program QH, unsigned x• Output of Transformation



return(QYL(z)); /* yes or no */}

bool QH(unsigned x) {}


}

• Problem L• Input: Program P• Yes/No Question: Is Y(P) =

{aa}?

• Question: Is the transformation on the left an answer-preserving or answer-reversing input transformation from H to problem L?

46

Key Step• Input of Transformation

• Program QH, unsigned x• Output of Transformation






}

• Problem L• Input: Program P• Yes/No Question: Is Y(P) =

{aa}?

• The output of the transformation is the input to the problem.• Plug QL in for

program P above• Is Y(QL) = {aa}?

47

Is Y(QL) = {aa}?• Problem L

• Input: Program P• Yes/No Question: Is Y(P) =

{aa}?• Analysis

• If QH loops on x, Y(QL)={}• No input to H creates a QL that

is a no input for L• If QH halts on x, Y(QL) =

{even length strings}• Yes input to H creates a QL

that is a no input for L• Transformation does not work

• All inputs map to no inputs

• Input of Transformation• Program QH, unsigned x

• Output of Transformation• Program QL bool main(string z) {





}

48

Three other problems• Problem L1

• Input: Program P• Yes/No Question: Is Y(P)

infinite?

• Problem L2• Input: Program P• Yes/No Question: Is Y(P)

finite?

• Problem L3• Input: Program P• Yes/No Question: Is Y(P) =

{} or is Y(P) infinite?







}

49

Is Y(QL) infinite?• Problem L1


infinite?• Analysis

• If QH loops on x, Y(QL)={}• No input to H creates a QL

that is a no input for L• If QH halts on x, Y(QL) =


that is a yes input for L• Transformation works

• Answer-preserving







}

50

Is Y(QL) finite?• Problem L2


finite?• Analysis


that is a yes input for L• If QH halts on x, Y(QL) =


that is a no input for L• Transformation works

• Answer-reversing







}

51

Is Y(QL) = {} or is Y(QL) infinite?• Problem L3

• Input: Program P• Yes/No Question: Is Y(P) =

{} or is Y(P) infinite?• Analysis


that is a yes input for L• If QH halts on x, Y(QL) =


that is a yes input for L• Transformation does not work

• All inputs map to yes inputs







}

Documents

Module 11