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Module 10

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

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

• Generic Element proof technique

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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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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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REC and Set Complement

RECEVEN

All Languages

ODD• Even: set of even length strings

• Is Even solvable (recursive)?

• Give a program P that solves it.

• Complement of Even?– Odd: set of odd length strings

• Is Odd recursive (solvable)?

• Does this prove REC is closed under set complement?

• How is the program P’ that solves Odd related to the program P that solves Even?

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P’ Illustration

PInput x Yes/No

P’

No/Yes

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Code for P’

bool main(string y)

{

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

}

bool P (string y)

/* details deleted; key fact is P is guaranteed to halt on all inputs */

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Set Complement Lemma

• If L is a solvable language, then L complement is a solvable language

• 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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REC Closed Under Set Union

All Languages

L1

L2

L1 U L2

REC

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

• 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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REC Closed Under Set Union

All Languages

L1

L2

L1 U L2

REC

– Construct program P3 from P1 and P2 as follows

– P3 solves L1 U L2 • Halts on all inputs• Answers correctly

– L1 U L2 is solvable

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Yes/No

Yes/No

P3 Illustration

P1

P2

ORYes/No

P3

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Code for P3

bool main(string y) {

if (P1(y)) return yes;

else if (P2(y)) return yes;

else return no;

}bool P1(string y) /* details deleted; key fact is P1 always halts. */

bool P2(string y) /* details deleted; key fact is P2 always halts. */

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Other Closure Properties

• Unary Operations– Language Reversal– Kleene Star

• Binary Operations– Set Intersection– 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 Union – 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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RE Closed Under Set Union

• Expressing this closure property as an infinite set of facts– 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

– For example, L or L1 or L2

• Only use facts which apply to any relevant objects– For example, there must exist a program P1 that half-

solves L1

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

often work our way in

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RE Closed Under Set UnionL1

L2

L1 U L2

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

• L1 U L2 is an r.e. language

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

• There exists a program P that half-solves L1 U L2

• Construct program P3 from P1 and P2

• Prove Program P3 half-solves L1 U L2

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• What code did we use for P3 when we worked with solvable languages?

Constructing Program P3

L1

L2

L1 U L2

bool main(string y) {

if (P1(y)) return yes;

else if (P2(y)) return yes;

else return no;}bool P1(string y) /* details deleted; key fact is P1 always halts. */

bool P2(string y) /* details deleted; key fact is P2 always halts. */

• Will this code work for half-solvable languages?

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Proving P3 Is Correct

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

– For all x in L1 U L2, must show P3

– For all x not in L1 U L2, must show P3

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P3 works correctly?

– Let x be an arbitrary string in L1 U L2

• Note, this subproof is a generic element proof

– What are the two possibilities for x?• x is in L1

• x is in L2

– What does P3 do if x is in L1?• Does it matter if x is in or not in L2?

– What does P3 do if x is in L2?• Does it matter if x is in or not in L1?

– Does P3 work correctly on such x?• If not, what strings cause P3 a problem?

bool main(string y) {

if (P1(y)) return yes;

else if (P2(y)) return yes;

else return no;

}

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P3 works correctly?

– Let x be an arbitrary string NOT in L1 U L2

• Note, this subproof is a generic element proof

– x is not in L1 AND x is not in L2

– What does P3 do on x in this case?

• What does P1 do on x?

• What does P2 do on x?

– Does P3 work correctly on such x?

bool main(string y) {

if (P1(y)) return yes;

else if (P2(y)) return yes;

else return no;

}

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Code for Correct P3

bool main(string y){if ((P1(y) || P2(y)) return yes; /* P1 and P2 run in parallel (alternating execution) */

else return no;}bool P1(string y)

/* key fact is P1 only guaranteed to halt on yes input instances */

bool P2(string y) /* key fact is P2 only guaranteed to halt on yes input instances */

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P3 works correctly?

– Let x be an arbitrary string in L1 U L2

• Note, this subproof is a generic element proof

– What are the two possibilities for x?• x is in L1

• x is in L2

– What does P3 do if x is in L1?• Does it matter if x is in or not in L2?

– What does P3 do if x is in L2?• Does it matter if x is in or not in L1?

– Does P3 work correctly on such x?• If not, what strings cause P3 a problem?

bool main(string y){ if ((P1(y) || P2(y)) return yes; else return no;}/* P1 and P2 run in parallel */

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• First-order logic formulation for statement– RE is closed under set complement

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

• L1 complement is in RE

• L2 complement is in RE

• ...

RE and Set complement L

LcRE

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RE and Set complement

• Let L be an arbitrary r.e. language

• L complement is an r.e. language

• There exists P s.t. Y(P)=L

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

• Construct program P’ from P• Prove Program P’ half-solves L complement

LLc

RE

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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?

• Does this prove that RE is not closed under set complement?

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Other closure properties

• Unary Operations– Language reversal– Kleene Closure

• Binary operations– set intersection– concatenation

• Not closed– Set difference (on practice hw)

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Pseudo Closure Property

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

• Question: What about Lc?

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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?

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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) /* guaranteed to halt on strings in L*/

bool P2(string y) /* guaranteed to halt on strings in Lc */

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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?

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RE and REC

REC

RE

All Languages

H

So where does Hc belong?

Hc

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Closure Property Questions

• Which of the following imply L is solvable given REC is closed under set union?– L1 U L2 = L

– L1 U L = L2

– L U L2 = L1

• In all cases, L1 and L2 are known to be solvable

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Closure Property Questions

• Which of the following imply L is NOT solvable given REC is closed under set union?– L1 U L2 = L

– L1 U L = L2

– L U L2 = L1

• In all cases, L1 is solvable and L2 is NOT solvable

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Summary• Definition of REC and RE

• Proofs of some closure properties for both language classes– RE more complex

• Pseudo-closure Property

• RE is not closed under set complement

• Proving a language is or is not in a language class using closure properties