Lecture 21 Reducibility
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Transcript of Lecture 21 Reducibility
Lecture 21 Reducibility
Many-one reducibility
• For two sets A c Σ* and B c Γ*, A ≤m B if there exists a Turing-computable function
f: Σ* → Γ* such that x ε A iff f(x) ε B
Complete in r. e.
• An r. e. set A is complete in r. e. if for every r. e. set B, B ≤m A .
Halting problemTheorem. K = { <x, M> | M accepts x } is complete
in r. e. .Proof. (1) K is a r. e. set. (2) For any r. e. set A, there exists a DTM
MA such that A = L(MA). For every input x of MA, define f(x) = <x, MA>. Then
x ε A iff f(x) ε K .
Remark
• If (1) A is a r. e. set, and (2) K ≤m A, then A is complete in r. e.
Nonempty
• Nonempty = {M | L(M) ≠ Φ } is complete in r. e.
Proof. (1) Nonempty is a r. e. set. Construct a DTM M* as follows: For each M, we may try every input of M,
one by one. If M accepts an input, then M is accepted by M*.
(2) K ≤m Nonempty. Suppose M’ is a DTM accepting every
input. For each input <x, M> of K, we define f(<x, M>) is a code of the TM M* behave as follows:
on an input y, M* carry out the following computation:
Step 1. M* simulates M on input x. If M accepts x, then go to Step 2. Step 2. M* simulates M’ on input y
Therefore,
<x, M> ε K => M accepts x => M* accepts every input y => f(<x, M>) = M* ε Nonempty <x, M> not in K => M doesn’t halt on x => M* doesn’t halt on y => L(M*) = Φ => f(<x, M>) not in Nonempty
r. e. –hard
• A set B is r. e.-hard if for every r. e. set A, A ≤m B Remark• Every complete set is r. e.-hard.• However, not every r. e.-hard set is
complete. • Every r. e.-hard set is not recursive.
All = {M | M accepts all inputs}
• All is r. e. hard.• All is not r. e. • All is not complete.
Empty = {M | L(M) = Φ}
• Empty is not r. e. • We don’t know if Empty is r. e.-hard.
r. e. property
• A subset P of TM codes is called a r. e. property if
M ε P and L(M’) = L(M) imply M’ ε P.
e.g., Nonempty, Empty, All are r. e. properties. Question: Give an example which is a subsets of
TM codes, but not a r. e. property.
Nontrivial
• A r. e. property is trivial if either it is empty or it contains all r. e. set.
Rice Theorem 1
Every nontrivial r. e. property is not recursive.
Proof
• Let P be a nontrivial r. e. property. For contradiction. Suppose P is a recursive set. So is its complement.
• Note that either P or its complement P does not contains the empty set. Without loss of generality, assume that P does not contains the empty set.
• Since P is nontrivial, P contains a nonempty r. e. set A.
• Let Ma be a TM accepting A, i.e., A=L(Ma).
• We want to prove K ≤m P.• For each input <x,M> of K, we define f(<x, M>) to be a code of TM M* which
behaves as follows. For each input y of M*, M* does the
following:
Step 1. M* simulates M on input x of M. If M accepts x, then go to Step 2.
Step 2. M* simulates Ma on y. If Ma accepts y, then M* accepts y.
Therefore, if <x, M> ε K then L(M*) = L(Ma) = A ε P,and if <x, M> not in K, then L(M*) = Φ not in
P
• Since K is not recursive and K ≤m P, we obtain a contradiction.
Recursive = {M | L(M) is recursive} is not recursive.
RE = {M | L(M) is r. e.} is trivial.
Question: • Is K an r. e. property? • Is every r. e. property complete?• Is it true that for any r. e. property, either it
or its complement is complete?
Rice Theorem 2A r. e. property P is r. e. iff the following three
conditions hold:(1) If A ε P and A c B for some r. e. set B, then B ε P. (2) If A is an infinite set in P, then A has a finite
subset in P.(3) The set of finite languages in P is enumerable,
in the sense that there is a TM that generates the (possibly) infinite string code1#code2# …, where codei is a code for the ith finite languages in P.
• The code for the finite language {w1, w2, …, wn} is [w1,w2,…,wn].
• In other words, there exists an r. e. set B that is a subset of codes of finite languages in P such that for every finite language F in P, B contains at least one code of F.
Examples
• All is not r. e. because All does not satisfy condition (2).
• The complement of ALL is not r. e. because it does not satisfy condition (1).
• Empty is not r. e. because it does not satisfy (1)
• Nonempty is r. e. because it satisfies (1), (2) and (3).
• Given TMs M and M’, is it true that L(M)=L(M’)?
A = {<M,M’> | L(M) = L(M’)} is not recursive.
Proof. Empty ≤m A. Let Mo be a fixed TM such that L(Mo) = Φ. Define f(M) =<M, Mo>. Then,
M ε Empty iff <M, Mo> ε A.