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On 5/22/24 5:19 PM, olcott wrote:Only specifies and no matter how many times you deny it,On 6/24/2022 2:53 AM, Malcolm McLean wrote:No. "Behavior of their inputss" MEANS for Turing Machines that are computing properties of Turing Machines (like Halt Deciders) have the "behavior of their input" defined as the Behavior of the machine their input represents/describes/specifies.He's dry-run P(P) and established that it doesn't halt. He's invoked H on it>
and H reports that it doesn't halt. He's run P(P) and it halts.
>
So something odd is going on there that needs an explanation.
*MUCH BETTER WORDS THAN ONE YEAR AGO*
*MUCH BETTER WORDS THAN ONE YEAR AGO*
*MUCH BETTER WORDS THAN ONE YEAR AGO*
>
typedef int (*ptr)(); // ptr is pointer to int function in C
00 int H(ptr p, ptr i);
01 int D(ptr p)
02 {
03 int Halt_Status = H(p, p);
04 if (Halt_Status)
05 HERE: goto HERE;
06 return Halt_Status;
07 }
08
09 int main()
10 {
11 H(D,D);
12 return 0;
13 }
>
In the above case a simulator is an x86 emulator that correctly emulates
at least one of the x86 instructions of D in the order specified by the
x86 instructions of D.
>
This may include correctly emulating the x86 instructions of H in the
order specified by the x86 instructions of H thus calling H(D,D) in
recursive simulation.
>
It is trivial to see that for every H/D pair of the infinite
set of H/D pairs that match the above template that
>
D correctly simulated by H cannot possibly reach its own final
state at line 06 and halt because D correctly simulated by
H remains stuck in recursive simulation.
>
Deciders are only accountable for the behavior of their inputs
and are thus not allowed to report on the behavior of the computation
that they themselves are contained within.
And, there is no rule that prohibits that machine given as a representation from including a copy of the machine that is deciding on it.The rule is the deciders operate on inputs and thus cannot possibly
Yes, you litterally can't phrase the question as being about "The machine you are contained within",Everyone has always implicitly assumed this false assumption.
but you can give the description of that machine and ask about it.That is one level of indirection away from the actual machine and
The point is that Turing Machines don't have that sort of "reference" to allow the use of the pronoun to reference the machine, you need to actually give the machine (by an encoding of it).When Ĥ is applied to ⟨Ĥ⟩
I already proved otherwise. That you ignored or failed to>But representations of them are, so we CAN ask Turing Machine about the behavior of a Turing Machine that includes a copy of itself.
There is no Turing machine that can possibly take its actual
self as an input because actual Turing Machines are not allowed
as inputs to other Turing Machines.
>No, because that is what DEFINES Halting.
*thus the behavior of D(D) executed from main() has always been moot*
*thus the behavior of D(D) executed from main() has always been moot*
*thus the behavior of D(D) executed from main() has always been moot*
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