Re: D correctly simulated by H proved for THREE YEARS --- rewritten

Op 13.jun.2024 om 14:44 schreef olcott:

On 6/13/2024 3:15 AM, Fred. Zwarts wrote:

Op 12.jun.2024 om 21:53 schreef olcott:

On 6/12/2024 2:46 PM, Fred. Zwarts wrote:

Op 12.jun.2024 om 21:20 schreef olcott:

>
On 5/29/2021 2:26 PM, olcott wrote:
https://groups.google.com/g/comp.theory/c/dTvIY5NX6b4/m/cHR2ZPgPBAAJ
>
If that was true then you could provide every step of D correctly
simulated by H such that D simulated by H reaches its own simulated
"ret" instruction.

>
I said that each H is unable to hit its target, so how could it reach the "ret" instruction of D? Please, think before you reply.

>
It is a binary choice either D correctly simulated by H can
possibly terminate normally by reaching its "ret" instruction
or not. Your attempt to twist these words to make it look like
there is more than these two possibilities is either ignorant
or deceptive.
>

>
Please, take some more attention to what I said. Read, then think, before you reply.
I said that H is not able to reach its own "ret" when it is simulating itself.

 That has always been totally irrelevant.

So, you think that if H does not reach its "ret", D can still reach its "ret"?
Try to think. D does not reach its "ret", *because* "H" does not reach its "ret".

 

So, no disagreement with that. That proves that H misses its target. The abort is too early. The target is just some steps further. It does not mean that the target is at infinity.
>

 The outer H always has one more execution trace to base its halt
status decision on than any of the nested emulations. This means
that unless the outer H aborts its simulation then none of them do.

That is true. But it also means that H aborts one execution trace too early. And since a simulation is unable to simulate itself, it is not possible to do it correctly.
The H that simulates one execution trace is unable to see the abort of its simulated self, because it needs two execution traces to see that.
The H that simulates two execution traces is unable to see the abort of its simulated self, because it needs three execution traces to see that.
Etc.
The invariant is that they all abort one execution trace too early.
The other invariant is that there is never an infinite simulation recursion.
Conclusion is that it is impossible for such a simulator to correctly simulate itself.

 

It is like an archer who is asked to hit a target twice as far as his bow can reach. His bow reaches 50m and the target is at 100m. He misses.
Then he uses a new bow that reaches 100m, but now the target is at 200m. He is able to reach the old target, but again he misses the target for the new bow. He can continue with a stronger bow, but if the bow reaches further, the target is also further away. But note, the target is never at infinity.

 Sure just like Zeno's paradox where he "proved" that it is impossible to cross a ten foot wide room in finite time.

Strange twist of Zeno's paradox.
*You* try to "prove" similarly that there is an infinite recursion, where H is only some steps away from its target.

 

Similarly, the target of the simulator is never at infinity, but always some steps further that the simulation goes. You can make a simulator that simulates further, which can reach the target of the old simulator, but it is unable to reach its own target. So, there is no infinite recursion, but the simulation always misses the target. The simulation is never able to simulate itself up to the end. It always aborts prematurely.
So, your claim proves that it is not a good idea to simulate H by itself. It will always miss the target.

 *As soon as H sees the repeating state it stops*
*As soon as H sees the repeating state it stops*
*As soon as H sees the repeating state it stops*

Yes, it stops. But the conclusion is that there is repetition, but not an infinite repetition, but the repetition is only one more than the simulator could see. If it would simulate one execution trace more, it would see its own abort. But it is not possible to do that with this simulator. Another simulator, with more steps would see that, but than there is another target for this new simulator when it would also try to simulate itself.

 *If you don't understand the infinite recursion example then*
*You lack the required prerequisite knowledge to understand me*

If you do not understand that there is a difference between an finite number of repetitions and an infinite recursion, you miss the background to talk about infinite recursions.

 void Infinite_Recursion(u32 N)
{
   Infinite_Recursion(N);
}
 int main()
{
   H(Infinite_Recursion, (ptr)5);
}
 

If you do not understand that the simulation of Infinite_Recursion by H is totally different from H simulating itself, than you do not have the competence to think about infinite recursion.
H has the requirement to halt. Infinite_Recursion does not have that requirement. So, a correct simulation of H would also reach its "ret", but that correct simulation cannot be done by H itself, because its simulation runs one execution trace behind its simulator execution.