Re: Basis for bypassing the Halting Problem ?
 From: Joshua Cranmer <Pidgeot18@xxxxxxxxxxxxxxx>
 Date: Mon, 06 Feb 2012 10:45:57 0600
On 2/6/2012 5:29 AM, Peter Olcott wrote:
On 2/5/2012 11:03 PM, Joshua Cranmer wrote:On 2/5/2012 9:40 PM, Peter Olcott wrote:I have not seen this in specific restriction within any of theirWhat I am proposing is to {add one element to Q and to transition to
this element} only upon recognizing an instance of what would otherwise
be the Halting Problem. The Halting Problem can no longer be formed
because the state that must be modified to create the Halting Problem
does not yet exist.
Turing machines are not selfmodifying code, so you cannot tell it to
"add a state."
specifications.
If this restriction is in their specification, then it would seem to
form a lack of correspondence to actual hardware machines, thus making
them less than an ideal model of computation.
Considering that Turing machines can't model random access efficiently, nitpicking on a "limited" model because it isn't selfmodifying is rather pointless.
The use of Turing machines is because they are a very simple model of computation that can emulate all known models of computation. Indeed, a selfmodifying Turing machine can be emulated with a Turing machine, via use of the Recursion Theorem (which says, informally, that "get the description of this Turing machine" is a possible operation).
Now, that said, this still does not run around the Halting problem for two reasons:
1. "Detect an instance" is impossible, as proving that the languages of two Turing machines is equivalent is undecidable.
2. Ultimately, you still get into the boolean state of either you accept it or you do not. The state of "runs forever" is not desirable (indeed, the Halting problem is recognizable but not decidable). Such a program would still be "defeatable" using the standard diagonalization argument.

Beware of bugs in the above code; I have only proved it correct, not tried it.  Donald E. Knuth
.
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