Transcript Lecture 9

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
1
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 halfsolvable languages
2
Why study closure properties of
RE and REC?
• It tests how well we really understand the
concepts we encounter
– language classes, REC, solvability, halfsolvability
• 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
3
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?
4
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
5
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?
6
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?
7
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
8
proof continued
– 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
9
P’ Illustration
P’
P
YES
No
No
YES
Input x
10
Code for P’
bool main(string y)
{
if (P (y)) return no; else return yes;
}
bool P (string y) /* details unknown */
11
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)
12
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
13
proof continued
– 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
14
P3 Illustration
P1
Yes/No
P3
P2
AND
Yes/No
Yes/No
15
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
18
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?
19
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
...
20
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
21
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
22
Set Intersection Example
• Let L1 and L2 be arbitrary r.e. languages
• There exist P1 and P2 s.t. Y(P1)=L1 and Y(P2)=L2
– By definition of half-solvable languages
• 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
• There exists a program P which half-solves L1
intersection L2
• L1 intersection L2 is an r.e. language
23
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
24
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
25
P3 Illustration
P1
Input
Yes/No/P3
P2
AND
Yes/No/-
Yes/No/-
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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 */
27
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
28
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
29
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
30
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
• ...
31
Set complement proof overview
• Let L be an arbitrary r.e. language
• There exists P s.t. Y(P)=L
– By definition of r.e. languages
• Construct program P’ from P
– Note, we can assume very little about P
• Prove Program P’ half-solves L complement
• There exists a program P’ which half-solves L
complement
• L complement is an r.e. language
32
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
33
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
34
Other closure properties
• Unary Operations
– Language reversal
– Kleene Closure
• Binary operations
– union
– concatenation
• Not closed
– Set difference
35
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
36
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
38
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
39
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
40
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
S*
L
Lc
P1 halts
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
44
P3 Illustration
P1
Input
Yes
Yes
P3
P2
Yes
No
45
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 */
46
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
All Languages
RE
L
REC
Lc
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RE and REC
All Languages
Lc
L
Lc
RE
Lc REC
Are there any languages L in RE - REC?
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