From: Tim Tyler on
polymedes wrote:

> 1) I'm trying to understand the following problem: let M_1 and M_2
> two turing machines with common input alphabet and a given string x. Is
> there a particular step in which the two machines will write the same
> symbol on the tape?

These TMs will probably write the same symbol on the tape at
/every/ step - unless you explain in what way they differ.
--
__________
|im |yler http://timtyler.org/ tim(a)tt1lock.org Remove lock to reply.
From: Harri Haanpaa on
On 2008-05-06, polymedes <polymedes(a)gmail.com> wrote:
> 1) I'm trying to understand the following problem: let M_1 and M_2
> two
> turing machines with common input alphabet and a given string x. Is
> there a particular step in which the two machines will write the same
> symbol on the tape?
> is it undecidable? can somebody give an idea on how can we reduce
> this
> problem to an undecidable problem?

Well *surely* it must be undecidable. You have not only one but
*two* Turing machines that can be arbitrarily messy. And *one* messy
Turing machine is usually enough to make a problem undecidable,
isn't it. I'm absolutely convinced that there are really simple M_1
for which this problem remains undecidable...

Harri H.

From: Tim Tyler on
polymedes wrote:
> On May 6, 9:26 pm, Tim Tyler <seemy...(a)googlemail.com> wrote:
>> polymedes wrote:

>>> 1) I'm trying to understand the following problem: let M_1 and M_2
>>> two turing machines with common input alphabet and a given string x. Is
>>> there a particular step in which the two machines will write the same
>>> symbol on the tape?
>> These TMs will probably write the same symbol on the tape at
>> /every/ step - unless you explain in what way they differ.
>
> in general, they are not the same TM. but the problem is to _decide_
> if there is a (i.e. at least one) natural number n for which the two
> (different) TMs will write the same symbol on the tape on step n.
>
> In other words,let L = {<M,N,w> | There is n such that M and N (with
> input w) write the same symbol on tape on step n}
>
> Is L decidable?

If the machines are different, then the problem of whether they
write the same symbol at the same time can be transformed into
a halting problem - e.g. consider a third machine that simulates
the first two machines and halts iff the simulated machines
output the same symbol on their tapes at the same time.
--
__________
|im |yler http://timtyler.org/ tim(a)tt1lock.org Remove lock to reply.
From: polymedes on
On May 6, 10:27 pm, Tim Tyler <seemy...(a)googlemail.com> wrote:
> polymedes wrote:
> > On May 6, 9:26 pm, Tim Tyler <seemy...(a)googlemail.com> wrote:
> >> polymedes wrote:
> >>> 1) I'm trying to understand the following problem: let M_1 and M_2
> >>> two turing machines with common input alphabet and a given string x. Is
> >>> there a particular step in which the two machines will write the same
> >>> symbol on the tape?
> >> These TMs will probably write the same symbol on the tape at
> >> /every/ step - unless you explain in what way they differ.
>
> > in general, they are not the same TM. but the problem is to _decide_
> > if there is a (i.e. at least one) natural number n for which the two
> > (different) TMs will write the same symbol on the tape on step n.
>
> > In other words,let L = {<M,N,w> | There is n such that M and N (with
> > input w) write the same symbol on tape on step n}
>
> > Is L decidable?
>
> If the machines are different, then the problem of whether they
> write the same symbol at the same time can be transformed into
> a halting problem - e.g. consider a third machine that simulates
> the first two machines and halts iff the simulated machines
> output the same symbol on their tapes at the same time.
> --
> __________
>   |im |yler  http://timtyler.org/ t...(a)tt1lock.org  Remove lock to reply.

This is a great answer! thanks for helping. Any ideas for the PCP?

From: polymedes on
On May 6, 10:25 pm, Harri Haanpaa <har...(a)yahoo.com> wrote:
> On 2008-05-06, polymedes <polyme...(a)gmail.com> wrote:
>
> > 1) I'm trying to understand the following problem: let M_1 and M_2
> > two
> > turing machines with common input alphabet and a given string x. Is
> > there a particular step in which the two machines will write the same
> > symbol on the tape?
> > is it undecidable? can somebody give an idea on how can we reduce
> > this
> > problem to an undecidable problem?
>
> Well *surely* it must be undecidable. You have not only one but
> *two* Turing machines that can be arbitrarily messy. And *one* messy
> Turing machine is usually enough to make a problem undecidable,
> isn't it. I'm absolutely convinced that there are really simple M_1
> for which this problem remains undecidable...
>
> Harri H.

You cannot be absolutely convinced, unless you provide a proof. See
above the one provider by Tim Tyler. Anyway, thanks for you time.