1

Lecture 9

• Recursive and r.e. language classes– representing solvable and unsolvable problems

• Proofs of closure properties – for the set of recursive (solvable) languages– for the set of r.e. (half-solvable) languages

• Generic element/template proof technique• Relationship between RE and REC

– pseudoclosure property

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RE and REC language classes

• REC– A solvable language is commonly referred to as

a recursive language for historical reasons– REC is defined to be the set of solvable or

recursive languages

• RE– A half-solvable language is commonly referred

to as a recursively enumerable or r.e. language– RE is defined to be the set of r.e. or half-

solvable languages

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Why study closure properties of RE and REC?

• It tests how well we really understand the concepts we encounter– language classes, REC, solvability, half-solvability

• It highlights the concept of subroutines and how we can build on previous algorithms to construct new algorithms– we don’t have to build our algorithms from scratch

every time

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Example Application• Setting

– I have two programs which can solve the language recognition problems for L1 and L2

– I want a program which solves the language recognition problem for L1 intersect L2

• Question– Do I need to develop a new program from scratch

or can I use the existing programs to help?• Does this depend on which languages L1 and L2 I am

working with?

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Closure Properties of REC

• We now prove REC is closed under two set operations– Set Complement– Set Intersection

• In these proofs, we try to highlight intuition and common sense

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Quick Questions

• What does the following statement mean?– REC is closed under the set complement

operation

• How do you prove a language L is in REC?

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Set Complement Example

• Even: the set of even length strings over {0,1}• Complement of Even?

– Odd: the set of odd length strings over {0,1}

• Is Odd recursive (solvable)?• How is the program P’ which solves Odd related

to the program P which solves Even?

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Set Complement Lemma

• If L is a solvable language, then L complement is a solvable language– Rewrite this in first-order logic

• Proof– Let L be an arbitrary solvable language

• First line comes from For all L in REC

– Let P be the C++ program which solves L• P exists by definition of REC

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– Modify P to form P’ as follows• Identical except at very end

• Complement answer – Yes -> No– No -> Yes

– Program P’ solves L complement• Halts on all inputs

• Answers correctly

– Thus L complement is solvable• Definition of solvable

proof continued

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P’ Illustration

PInput x

YES

No

P’

YES

No

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Code for P’

bool main(string y)

{

if (P (y)) return no; else return yes;

}

bool P (string y) /* details unknown */

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Set Intersection Example

• Even: the set of even length strings over {0,1}• Mod-5: the set of strings of length a multiple of 5

over {0,1}• What is Even intersection Mod-5?

– Mod-10: the set of strings of length a multiple of 10 over {0,1}

• How is the program P3 (Mod-10) related to programs P1 (Even) and P2 (Mod-5)

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Set Intersection Lemma

• If L1 and L2 are solvable languages, then L1 intersection L2 is a solvable language

– Rewrite this in first-order logic

– Note we have two languages because intersection is a binary operation

• Proof– Let L1 and L2 be arbitrary solvable languages

– Let P1 and P2 be programs which solve L1 and L2, respectively

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– Construct program P3 from P1 and P2 as follows

• P3 runs both P1 and P2 on the input string

• If both say yes, P3 says yes

• Otherwise, P3 says no

– P3 solves L1 intersection L2 • Halts on all inputs

• Answers correctly

– L1 intersection L2 is a solvable language

proof continued

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P3 Illustration

P1

P2

Yes/No

Yes/No

ANDYes/No

P3

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Code for P3

bool main(string y)

{

if (P1(y) && P2(y)) return yes;

else return no;

}

bool P1(string y) /* details unknown */

bool P2(string y) /* details unknown */

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Other Closure Properties

• Unary Operations– Language Reversal– Kleene Star

• Binary Operations– Set Union– Set Difference– Symmetric Difference– Concatenation

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Closure Properties of RE

• We now try to prove RE is closed under the same two set operations– Set Intersection – Set Complement

• In these proofs– We define a more formal proof methodology– We gain more intuition about the differences

between solvable and half-solvable problems

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Quick Questions

• What does the following statement mean?– RE is closed under the set intersection

operation

• How do you prove a language L is in RE?

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RE Closed Under Set Intersection

• First-order logic formulation?– For all L1, L2 in RE, L1 intersect L2 in RE

– For all L1, L2 ((L1 in RE) and (L2 in RE) --> ((L1 intersect L2) in RE)

• What this really means– Let Li denote the ith r.e. language

• L1 intersect L1 is in RE

• L1 intersect L2 is in RE

• ...

• L2 intersect L1 is in RE

• ...

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Generic Element or Template Proofs

• Since there are an infinite number of facts to prove, we cannot prove them all individually

• Instead, we create a single proof that proves each fact simultaneously

• I like to call these proofs generic element or template proofs

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Basic Proof Ideas• Name your generic objects

– In this case, we use L1 and L2

• Only use facts which apply to any relevant objects– We will only use the fact that there must exist P1 and

P2 which half-solve L1 and L2

• Work from both ends of the proof– The first and last lines are usually obvious, and we can

often work our way in

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Set Intersection Example

• Let L1 and L2 be arbitrary r.e. languages

• L1 intersection L2 is an r.e. language

• There exist P1 and P2 s.t. Y(P1)=L1 and Y(P2)=L2

– By definition of half-solvable languages

• There exists a program P which half-solves L1 intersection L2

• Construct program P3 from P1 and P2

– Note, we can assume very little about P1 and P2

• Prove Program P3 half-solves L1 intersection L2

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Constructing P3

• What did we do in the REC setting?

• Build P3 using P1 and P2 as subroutines

• We just have to be careful now in how we use P1 and P2

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Constructing P3

• Run P1 and P2 in parallel

– One instruction of P1, then one instruction of P2, and so on

• If both halt and say yes, halt and say yes

• If both halt but both do not say yes, halt and say no– Note, if either never halts, P3 never halts

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P3 Illustration

P1

P2

Yes/No/-

Yes/No/-

ANDYes/No/-

P3

Input

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Code for P3

bool main(string y)

{

parallel-execute(P1(y), P2(y)) until both return;

if ((P1(y) && P2(y)) return yes;

else return no;

}

bool P1(string y) /* details unknown */

bool P2(string y) /* details unknown */

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Proving P3 Is Correct

• 2 steps to showing P3 half-solves L1 intersection L2

– For all x in L1 intersection L2, must show P3 • accepts x

– halts and says yes

– For all x not in L1 intersection L2, must show P3 • rejects x or• loops on x or• crashes on x

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Part 1 of Correctness Proof

• P3 accepts x in L1 intersection L2

– Let x be an arbitrary string in L1 intersection L2

• Note, this subproof is a generic element proof

– P1 accepts x

• L1 intersection L2 is a subset of L1

• P1 accepts all strings in L1

– P2 accepts x

– P3 accepts x• We reach the AND gate because of the 2 previous facts

• Since both P1 and P2 accept, AND evaluates to YES

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Part 2 of Correctness Proof• P3 does not accept x not in L1 intersection L2

– Let x be an arbitrary string not in L1 intersection L2

– By definition of intersection, this means x is not in L1 or L2

– Case 1: x is not in L1 • 2 possibilities

• P1 rejects (or crashes on) x

– One input to AND gate is No– Output cannot be yes

– P3 does not accept x

• P1 loops on x

– One input never reaches AND gate– No output

– P3 loops on x

• P3 does not accept x when x is not in L1

– Case 2: x is not in L2

• Essentially identical analysis

– P3 does not accept x not in L1 intersection L2

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RE closed under set complement?

• First-order logic formulation?– For all L in RE, L complement in RE– For all L (L in RE) --> ((L complement) in RE)

• What this really means– Let Li denote the ith r.e. language

• L1 complement is in RE

• L2 complement is in RE

• ...

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Set complement proof overview

• Let L be an arbitrary r.e. language

• L complement is an r.e. language

• There exists P s.t. Y(P)=L– By definition of r.e. languages

• There exists a program P’ which half-solves L complement

• Construct program P’ from P– Note, we can assume very little about P

• Prove Program P’ half-solves L complement

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Constructing P’• What did we do in recursive case?

– Run P and then just complement answer at end• Accept -> Reject• Reject -> Accept

• Does this work in this case?– No. Why not?

• Accept->Reject and Reject ->Accept ok• Problem is we need to turn Loop->Accept

– this requires solving the halting problem

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What can we conclude?

• Previous argument only shows that the approach used for REC does not work for RE

• This does not prove that RE is not closed under set complement

• Later, we will prove that RE is not closed under set complement

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Other closure properties

• Unary Operations– Language reversal– Kleene Closure

• Binary operations– union– concatenation

• Not closed– Set difference

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Closure Property Applications

• How can we use closure properties to prove a language LT is r.e. or recursive?

• Unary operator op (e.g. complement)– 1) Find a known r.e. or recursive language L

– 2) Show LT = L op

• Binary operator op (e.g. intersection)– 1) Find 2 known r.e or recursive languages L1 and L2

– 2) Show LT = L1 op L2

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Closure Property Applications

• How can we use closure properties to prove a language LT is not r.e. or recursive?

• Unary operator op (e.g. complement)– 1) Find a known not r.e. or non-recursive language L

– 2) Show LT op = L

• Binary operator op (e.g. intersection)– 1) Find a known r.e. or recursive language L1

– 2) Find a known not r.e. or non-recursive language L2

– 2) Show L2 = L1 op LT

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Example

• Looping Problem– Input

• Program P

• Input x for program P

– Yes/No Question• Does P loop on x?

• Looping Problem is unsolvable– Looping Problem complement = H

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Closure Property Applications

• Proving a new closure property

• Theorem: Unsolvable languages are closed under set complement– Let L be an arbitrary unsolvable language

– If Lc is solvable, then L is solvable• (Lc)c = L

• Solvable languages closed under complement

– However, we are assuming that L is unsolvable

– Therefore, we can conclude that Lc is unsolvable

– Thus, unsolvable languages are closed under complement

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Pseudo Closure Property

• Lemma: If L and Lc are half-solvable, then L is solvable.

• First-order logic?– For all L in RE, (Lc in RE) --> (L in REC)

– For all L, ((L in RE) and (Lc in RE)) --> (L in REC)

• Question: What about Lc?– Also solvable because REC closed under set

complement

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High Level Proof

– Let L be an arbitrary language where L and Lc are both half-solvable

– Let P1 and P2 be the programs which half-solve L and Lc, respectively

– Construct program P3 from P1 and P2

• Argue P3 solves L

– L is solvable

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Constructing P3

• Problem– Both P1 and P2 may loop on some input strings,

and we need P3 to halt on all input strings

• Key Observation– On all input strings, one of P1 and P2 is

guaranteed to halt. Why?• Nature of complement operation

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Illustration

*

L

P1 halts

Lc

P2 halts

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Construction and Proof

• P3’s Operation

– Run P1 and P2 in parallel on the input string x until one accepts x

• Guaranteed to occur given previous argument

• Also, only one program will accept any string x

– IF P1 is the accepting machine THEN yes ELSE no

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P3 Illustration

P1

P2

Yes

Yes

P3

Input

Yes

No

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Code for P3

bool main(string y)

{

parallel-execute(P1(y), P2(y)) until one returns yes;

if (P1(y)) return yes;

if (P2(Y)) return no;

}

bool P1(string y) /* details unknown */

bool P2(string y) /* details unknown */

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Question

• What if P2 rejects the input?

• Our description of P3 doesn’t describe what we should do in this case.– If P2 rejects the input, then the input must be in

L

– This means P1 will eventually accept the input.

– This means P3 will eventually accept the input.

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RE and REC

REC

RE

All Languages

L Lc

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RE and REC

REC

RE

All Languages

LLc

Lc

Lc

Are there any languages L in RE - REC?