Re: The philosophy of computation reformulates existing ideas on a new basis --- getting somewhere

On 11/3/24 1:33 PM, olcott wrote:

On 11/3/2024 12:20 PM, Richard Damon wrote:

On 11/3/24 9:39 AM, olcott wrote:

>
That is why I used to fully defined semantics of the x86
language to make this 100% perfectly unequivocal.
>
A few lines of x86 code express complex algorithms
succinctly enough that human minds are not totally
overwhelmed by far too much tedious detail.
>

It is not pspecified
in the usual formulation of the problem. Also note that
the behaviour exists before those strings so "describe"
should be and usually is used instead of "specify". The
use of latter may give the false impression that the behaviour
is determined by those strings.
>

>
In order for any machine to compute the mapping from
a finite string it must to so entirely on the basis
of the actual finite string and its specified semantics.

>
You have that somewhat backwards. It *CAN* only do what it can compute.
>
The mapping is not required to *BE* computable.
>

>
The finite string input to HHH specifies that HHH
MUST EMULATE ITSELF emulating DDD.

>
Right, and it must CORRECTLY determine what an unbounded emulation of that input would do, even if its own programming only lets it emulate a part of that.
>

 Yes this is exactly correct. I don't understand
why you keep disagreeing with your own self this.

Right, buyt you keep on forgetting that correct means the UNBOUNDED emulation, which isn't what you decider does.
The unbounded emulation of DDD is what HHH1 does, and that shows it halts.
HHH doesn't show a "different" behavior, as every step it sees will be identical (at least if HHH is the claimed pure function) to that seen by HHH1 to the point it decides that it sees a "necessarilty non-halting pattern", which is disproven by the emulation of HHH1.
Remember, DDD MUST inlclude the code of HHH to be validly emulated, and thus it is only the emulation of the DDD that calls the HHH that aborts and returns will show the behavior of the DDD that was given to the HHH that aborts its emulaiton to answer.

 

>
The finite string input to HHH1 specifies that HHH1
MUST NOT EMULATE ITSELF emulating DDD.

>
But the semantics of the string haven't changed, as the string needs to contain all the details of how the machine it is looking at will work.
>

 DDD emulated by HHH specifies that HHH will
emulate itself emulating DDD.

Right, and a CORRECT (and complete) emulation of HHH emulating DDD will see that it will reach a point where it aborts its emulation and returns. This will cause the DDD that is being emulated, that called that HHH to return.
THe fact that you HHH aborts its emulation before getting there doesn't change the behavior of the correct and complete emulation of that input.
Remember, you can NEVER change the code of the HHH that DDD calls and say you are looking at the same input, as it MUST contain the same code.
At least not until you define your new system and show how you can make an OBJECTIVE standard that uses some other definition.

 DDD emulated by HHH1 specifies that HHH1 will
NOT emulate itself emulating DDD.

But the question isn't "emulation itself" it is emulating the input.
Your problem is you are trying to rephrase an OBJECTIVE problem into a SUBJECTIVE one, which put it out of bounds of the theory,
That breaks your whole concept of being "isomorphic"

 

>
Unless HHH rejects its input DDD as non halting the
executed DDD never stops running. This itself proves
that HHH is correct and that DDD is not the same
instance as the one that HHH rejected.

>
You have cause and effect backwards.
>

 <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its input D
     until H correctly determines that its simulated D would never
     stop running unless aborted then...
 The conditional branch instruction criteria has been met.
 

Nope, becuase there are conditional branches in the PROGRAM D, as the H it calls is part of it, and that has conditional branches that can break the loop.
You are just confirming that you are ignorant of the definition of a "program".
Sorry, you are just proving how stupid you are.