Re: Simulating termination analyzers by dummies --- What does halting mean?

Liste des GroupesRevenir à s logic 
Sujet : Re: Simulating termination analyzers by dummies --- What does halting mean?
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory sci.logic
Date : 19. Jun 2024, 15:31:24
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v4umjc$1vpm0$6@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
User-Agent : Mozilla Thunderbird
On 6/19/2024 4:08 AM, joes wrote:
Am Tue, 18 Jun 2024 21:51:56 -0500 schrieb olcott:
On 6/18/2024 9:41 PM, Richard Damon wrote:
On 6/18/24 10:30 PM, olcott wrote:
On 6/18/2024 9:16 PM, Richard Damon wrote:
On 6/18/24 1:25 PM, olcott wrote:
On 6/18/2024 12:06 PM, joes wrote:
 
When such a UTM has been adapted to only simulate the first ten states
of its input TMD, then every simulated TMD with more than ten states
did not terminate normally.
 
Terminating is a property of the actual machine, and not a simulation
of it.
This.
Thus according to your faulty reasoning when the source-code of a C
program is simulated by interpreter this is mere nonsense gibberish
having nothing to do what the behavior that this source-code specifies.

YOUR partial decider makes everything halt, even that which doesn't.
So yes, it can't simulate infinite loops.
 
My partial decider makes everything stop running this is not
at all the same as halting. Novices get very confused about this.

You could say the SIMULATION didn't terminate normally, but you can't
say the machine didn't or even the Turing Machine Description, as you
could give that exact same TMD to a real UTM and find out the actual
behaviof or the input.
Sure you can otherwise interpreters of source-code would be a bogus
concept.

When I write an infinite loop, I want it to be interpreted as an
infinite loop. Your H0 is bogus.
 
void Infinite_Loop()
{
   HERE: goto HERE;
}
int main()
{
  Output("Input_Halts = ", H0(Infinite_Loop));
}
_Infinite_Loop()
[00002092] 55         push ebp
[00002093] 8bec       mov ebp,esp
[00002095] ebfe       jmp 00002095
[00002097] 5d         pop ebp
[00002098] c3         ret
Size in bytes:(0007) [00002098]
H0: Begin Simulation   Execution Trace Stored at:11376f
Address_of_H0:1aa2
[00002092][0011375f][00113763] 55         push ebp
[00002093][0011375f][00113763] 8bec       mov ebp,esp
[00002095][0011375f][00113763] ebfe       jmp 00002095
[00002095][0011375f][00113763] ebfe       jmp 00002095
H0: Infinite Loop Detected Simulation Stopped

You just have lost track of the defintions of what is REALITY (the
actual behavior of the machine) and what is just imagination.
Not I but you.

The map is not the territory. The simulation is not the machine.
 
(a) If the correct simulation of the x86 machine language of
the function does not prove the actual behavior that this
finite string specifies then source-code interpreters
are a bogus concept.
(b) source-code interpreters are NOT a bogus concept.

Every C programmer that knows what an x86 emulator is knows that when
H0 emulates the machine language of Infinite_Loop, Infinite_Recursion,
and DDD that it must abort these emulations so that itself can
terminate normally.
Which doesn't mean the program DDD needs to be abort to have it halt.
The verified that that it does need to be aborted contradicts your
nonsense to the contrary.

[The verification that it ... ?]
If H0 halts, so does DDD (which only calls it).
 
My partial decider makes everything stop running this is not
at all the same as halting. Novices get very confused about this.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Date Sujet#  Auteur
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