Re: Infinite set of HHH/DDD pairs --- truisms

On 7/23/2024 1:13 PM, Fred. Zwarts wrote:

Op 23.jul.2024 om 19:54 schreef olcott:

On 7/23/2024 11:46 AM, Fred. Zwarts wrote:

Op 23.jul.2024 om 17:03 schreef olcott:

On 7/23/2024 3:17 AM, Fred. Zwarts wrote:

Op 22.jul.2024 om 21:57 schreef olcott:

On 7/22/2024 2:30 PM, Fred. Zwarts wrote:

Op 22.jul.2024 om 20:31 schreef olcott:

On 7/22/2024 12:45 PM, Fred. Zwarts wrote:

Op 22.jul.2024 om 17:08 schreef olcott:

On 7/22/2024 9:32 AM, joes wrote:

Am Mon, 22 Jul 2024 09:13:33 -0500 schrieb olcott:

On 7/22/2024 3:01 AM, Mikko wrote:

On 2024-07-21 13:50:17 +0000, olcott said:

On 7/21/2024 4:38 AM, Mikko wrote:

On 2024-07-20 13:28:36 +0000, olcott said:

On 7/20/2024 3:54 AM, Mikko wrote:

On 2024-07-19 14:39:25 +0000, olcott said:

On 7/19/2024 3:51 AM, Mikko wrote:

>

Anyway you did not say that some HHHᵢ can simulate the
corresponding DDDᵢ to its termination. And each DDDᵢ does
terminate, whether simulated or not.

>
>

Then DDD correctly simulated by any pure function HHH cannot possibly
reach its own return instruction and halt, therefore every HHH is
correct to reject its DDD as non-halting.

That does not follow. It is never correct to reject a halting
comoputation as non-halting.

In each of the above instances DDD never reaches its return instruction
and halts. This proves that HHH is correct to report that its DDD never
halts.

It can't return if the simulation of it is aborted.
>

Within the hypothetical scenario where DDD is correctly emulated by its
HHH and this HHH never aborts its simulation neither DDD nor HHH ever
stops running.

In actuality HHH DOES abort simulating.
>

This conclusively proves that HHH is required to abort the simulation of
its corresponding DDD as required by the design spec that every partial
halt decider must halt and is otherwise not any kind of decider at all.

Like Fred recognised a while ago, you are arguing as if HHH didn't abort.
>

That HHH is required to abort its simulation of DDD conclusively proves
that this DDD never halts.

You've got it the wrong way around.
>

>
I am talking about hypothetical possible ways that HHH could be encoded.
(a) HHH(DDD) is encoded to abort its simulation.
(b) HHH(DDD) is encoded to never abort its simulation.
>
Therefore (a) is correct and (b) is incorrect according to the
design requirements for HHH that it must halt.

>
Both are incorrect. An HHH, when encoded to abort does not need to be aborted when simulated, because it already halts on its own.

>
You must have attention deficit disorder.
(a) At least one HHH aborts.
(b) No HHH ever aborts.
>
Every X has property Y or not, there is no inbetween.

>
Do you have difficulty reading and writing English?
>
If every X has property Y or not, then it is clear that every HHH abort or not.

>
Sure and when we start a race with a single file line of
people that are 15 feet apart and everyone goes the same
speed then everyone will reach the finish line, eventually.
>
When the first HHH that reaches the finish line stops
simulating its input then no other HHH can possibly reach
the finish line because nothing is simulating them.

>
Exactly! That is the error in HHH. It stops simulating before the other HHH could reach the finish line.

>
So you don't even know how foot races work.
>
The winner of the race is not supposed to wait
so that everyone crosses the finish line at once.

>
But do you know how foot races work? When the winner reaches the finish, the other players do not disappear. They do not halt either. They continue until they also reach the finish line.
>

>
Also with HHH(DDD) there are infinite instances
in the race. Waiting for the last one to finish
waits forever. The first one to cross the finish
line ends the race. Every other instance is
immediately incapacitated

You don't have to wait for the last one of an infinite number. It is sufficient to wait for the first simulated HHH, because that one aborts and halts

>
Since HHH is the exact same machine code then when the first one waits
for the second one the second one waits for the third on and on forever.
>

 Indeed! Exactly! You are almost at the finish.

*THIS SEEMS PERMANENTLY TOO DIFFICULT FOR YOU TO UNDERSTAND*
Unless the first one aborts none of them ever aborts and HHH
itself never stops running. Therefore is is correct and necessary
for the first HHH to abort.

If HHH aborts after N cycles then only HHH that aborts after M cycles does a correct simulation, when M > N.
No HHH can possibly simulate *itself* correctly. Other HHH *can* do a correct simulation when M > N.
What HHH should do, is different from what it does, because that is how HHH is coded.
When you change the code, also the requirement changes when it has to simulate *itself* correctly. No matter how far you increase N, it never reaches N+1. It always aborts one cycle too early when simulating itself.
The HHH that aborts after N cycles can only simulate correctly the HHH that abort after K cycles when K < N. This shows that HHH cannot possibly simulate *itself* correctly.

When N instructions of DDD are simulated according to the
x86 semantics that they specify then N instructions of been
simulated correctly.
When DDD specifies non-halting behavior then it is stupidly
wrong to require a correct simulation to be a complete simulation.

That there is no way to do it correctly, does not mean that there must be another way to do it correctly.
 

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer