Re: Can D simulated by H terminate normally? Message_ID Provided V2

On 5/19/2024 8:10 PM, Richard Damon wrote:

On 5/19/24 8:06 PM, olcott wrote:

On 5/1/2024 7:10 PM, Richard Damon wrote:
>
typedef int (*ptr)();  // ptr is pointer to int function
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.
>
For every H/D pair of the above template D correctly simulated by
*pure function* H cannot possibly reach its own final state at
line 06 and halt.
>

 Ok, so adding that H is a pure function, that means that since your outer H(D,D) is going to return 0, all logic must be compatible with the fact that EVERY call to H(D,D) will also eventually return 0.
  Remember also, THIS D is defined to call THIS H, that does exactly the same as the H that is deciding it.
 

OK, good.

>
<snip so that Message ID links to whole message>
We can use my unique time/date stamp as an alternative.
>

Remember, YOU are the one saying you are needing to change the definition from the classical theory, where we have things well defined.
>
YOU have decider that H is just whatever C code you want to write for it, and D is the input proved. (which doesn't actually match the Linz or Sipser proof, but fairly close).
>
With THAT set of definitions we have a lot of options that break your incorrectly assumed results.
>
The first method has been discussed here by Flibble. While the final answer he got to doesn't fit the requirements, the first part of the method DOES show that it is possible for an H to simulate to past line 3.
>
THe basic idea is that if H(M,d) finds that its simulation of M(d) get to a call to H(M,d) then rather that your idea of just saying it will get stuck and declair the input invalid, since there ARE a number of possible inputs that there is a "correct" answer that H can give to

>
That D is calling H does not prove recursive simulation.
That D is calling H with its same parameters does seem
to prove non-halting recursive simulation.

 Nope. Try to actuall PROVE it.
 

That is off-topic for this post.
All that we need know is that no D simulated by any H
ever reaches its own line 06 and halts.
I start ignoring everything you say as soon as you go off-topic.
If you said anything below that it relevant to some other post
I will read it when you post it there.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer