Re: Can the 'Turing Problem' be deflated?

Marshall wrote:
On Mar 29, 5:05 pm, John Jones <jonescard...@xxxxxxx> wrote:
[snip]

If you're interested in learning about Turing machines, there are
a number of good references on the web. If you have any specific
questions about how they work, I'd be glad to discuss them with
you. It would be better to try to understand what Turing machines
are and how they work, and try to understand the Halting Problem
itself, and at least the standard proof of it, before trying to
deconstruct the issue. I am sure you are up to it.

In fact, Turing Machines are just simple state machines with
an associated "tape" as a memory. I don't think they're a
particularly good abstract computational model; they're too
low level to accomplish very much without a lot of work.
But they remain the standard for the definition of computable
function, and for questions such as the halting problem.

You understand that the question of whether a Turing machine
halts or not is not in general decidable, right?


Marshall

My claim is that "A TM always halts, and where a command/fulillment is not associated with a halt (e.g., when we do not know whether the machine will halt or not) then the TM does not engage in computation."

Briefly, because the steps a machine makes are solitary and not connected through a process, then a machine, including the TM, halts at each mooted 'step in a process'. Further, if the machine is determined by the steps, then at each step we have a new, unrelated machine.

So, there can be no distinction between a first halt and a last halt, at least as far as the machine is concerned. It is then up to us to say when a halting has fulfilled a command, for that is the stipulated definition of a halt. The machine cannot tell us. So, if for M cases there are haltings and commands that cannot be associated, such as arise in the claim that we cannot tell whether a machine will halt or not, then for M cases we have not issued a command.

Thus, for M cases (the indeterminacy of a halting event), the TM does not engage in computation.
.

Relevant Pages

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