Цікавий метричний простір, пов'язаний з машинами Тьюрінга


16

У цьому питанні ми розглядаємо лише машини Тьюрінга, які зупиняються на всіх входах. Якщо k N,kN то через T kTk позначаємо машину Тюрінга, кодом якої є kk .

Розглянемо наступну функцію

s(x,y)=min{k|L(Tk){x,y}|=1}

s(x,y)=min{k|L(Tk){x,y}|=1}

In other words, s(x,y)s(x,y) is the code of the smallest Turing machine that recognizes precisely one of the strings x,y.x,y. We can now define the following map

d(x,y)={2s(x,y)if xy,0otherwise.

d(x,y)={2s(x,y)0if xy,otherwise.

It can be quickly verified that d(x,y)d(x,y) induces a metric space (in fact an ultrametrics) on Σ.Σ.

Now I would like to prove that if f:ΣΣf:ΣΣ is a uniformly continuous function then for every recursive language L, f1(L)f1(L) is recursive as well.

In other words let ff be a map such that for every ϵ>0ϵ>0 there is a δ>0δ>0 such that if for strings x,yΣx,yΣ d(x,y)δ

d(x,y)δ
then d(f(x),f(y))<ϵ.
d(f(x),f(y))<ϵ.
Then we need to show that f1(L)f1(L) is a recursive language given that LL is recursive.

Now as already noted in this post one way to approach the problem is to show that there is a Turing machine that given a string xΣxΣ computes f(x).f(x).

I am stuck proving this claim and slowly wondering if there is some other approach to solve this?

Hints, suggestions and solutions are welcome!


1
Why are you trying to prove this? It reminds me of Banach-Mazur computability, which is not very well behaved.
Andrej Bauer

@AndrejBauer Homework assignment!
Jernej

Відповіді:


9

Edit: removed hints, posted my solution.

Here is my solution. We're going to pick a reference point xx where f(x)Lf(x)L and consider the universe from xx and f(x)f(x)'s points of view. It turns out that every "neighborhood" of a point corresponds to a recursive language. So LL is a neighborhood around f(x)f(x), and there will be some neighborhood around xx that maps to it; this neighborhood is a recursive language.

Lemma. In this space, a language is recursive if and only if it is a neighborhood of each of its strings.

Proof. First, fix a recursive language LL and let xLxL. Let KK be the minimal index of a decider for LL. Then we have that if yLyL, s(x,y)Ks(x,y)K, so d(x,y)1/2Kd(x,y)1/2K. Thus d(x,y)<1/2Kd(x,y)<1/2K implies that yLyL.

Second, let xx be an arbitrary string and fix ε>0ε>0; let K=log(1/ε)K=log(1/ε). Let LK={y:d(x,y)<ε}LK={y:d(x,y)<ε}; then LK={y:s(x,y)>K}LK={y:s(x,y)>K}. Then we can write

LK={y:(j=1,,K)|L(Tj){x,y}|1}.

LK={y:(j=1,,K)|L(Tj){x,y}|1}.

But LKLK is decidable: On input yy, one may simulate the first KK deciders on xx and yy and accept if and only if each either accepted both or rejected both.   

Now we're almost done:

Prop. Let ff be continuous. If LL is recursive, then f1(L)f1(L) is recursive.

Proof. Under a continuous function, the preimage of a neighborhood is a neighborhood.


Interestingly, I think that in this space a continuous function is uniformly continuous: Let ff be continuous, so for each point xx, for each εε there exists a corresponding δδ. Fix an εε and let K=log(1/ε)K=log(1/ε). There are a finite number of balls of size εε: there is L(T1)L(T2)L(TK)L(T1)L(T2)L(TK); then there is ¯L(T1)L(T2)L(TK)L(T1)¯¯¯¯¯¯¯¯¯¯¯¯L(T2)L(TK); then L(T1)¯L(T2)L(TK)L(T1)L(T2)¯¯¯¯¯¯¯¯¯¯¯¯L(TK), and so on. ff associates to each of these languages LiLi a preimage language LiLi with associated diameter δiδi. For each xLixLi, d(x,y)δid(f(x),f(y))εd(x,y)δid(f(x),f(y))ε. So we can take the minimum over these finitely many δδs to get the uniform continuity constant δδ associated with this εε.


1
Clearly d(x,y)12Kd(x,y)12K but I still miss how to show that f1(L)f1(L) is recursive!
Jernej

@Jernej OK, so first, we also have the contrapositive -- if d(x,y)>12K then either both are in L or neither is. Now let's take ϵ=12K. Then there is some δ so, if d(x,y)δ, then |L{f(x),f(y)}|=1. In particular, let's pick some x with x=f(x)L. Now we want to know where all the other elements of L lie relative to x, and therefore where must the other members of f1(L) lie relative to x?
usul

@Jernej I have posted my solution now. I hope what I posted earlier was helpful! Thanks for posting this problem, it is very cool.
usul

Thank you very much for your answer. It took me a while to digest the hints hence I haven't upvoted and accepted your answer!
Jernej

Quick question. We have shown that LK is decidable. I don't see how it follows that it is recursive? Cant it be that one of the simulated Tj never halts?
Jernej
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