Re: Defining a correct simulating halt decider

On 9/7/24 9:57 AM, olcott wrote:

On 9/7/2024 3:29 AM, Mikko wrote:

On 2024-09-07 05:12:19 +0000, joes said:
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Am Fri, 06 Sep 2024 06:42:48 -0500 schrieb olcott:

On 9/6/2024 6:19 AM, Mikko wrote:

On 2024-09-05 13:24:20 +0000, olcott said:

On 9/5/2024 2:34 AM, Mikko wrote:

On 2024-09-03 13:00:50 +0000, olcott said:

On 9/3/2024 5:25 AM, Mikko wrote:

On 2024-09-02 16:38:03 +0000, olcott said:
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A halt decider is a Turing machine that computes the mapping from
its finite string input to the behavior that this finite string
specifies.

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A halt decider needn't compute the full behaviour, only whether
that behaviour is finite or infinite.
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New slave_stack at:1038c4 Begin Local Halt Decider Simulation

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Local Halt Decider: Infinite Recursion Detected Simulation Stopped
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Hence  HHH(DDD)==0 is correct

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Nice to see that you don't disagree with what said.
Unvortunately I can't agree with what you say.
HHH terminates,
os DDD obviously terminates, too. No valid

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DDD emulated by HHH never reaches it final halt state.

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If that iis true it means that HHH called by DDD does not return and
therefore is not a ceicder.

The directly executed HHH is a decider.

What does simulating it change about that?

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If the simulation is incorrect it may change anything.
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PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
 

No, the behavior was always there.
The D that calls H and does the opposite has FIXED behavior for a given H, and it doesn't matter if we ask H or H1 what that behavior is, it is always the same, because it was determined by what that PARTICULAR H that it was built on did with that input.
The fact that H can't get the right answer doesn't change what the right answer is.
You are just proving you are nothing but a stupid liar that doesn't knpw what he is talking about.
Since you can't show the actual step where HHH CORRECTLY emulates an instruction to get different behavior, you statement is just a PATHOLOGICAL LIE.
The fact that is seems you don't even TRY to find the answer, proves you are just intentionally ignorant and don't care about what is true.
The fact that you think that a "Call" instruction can do anything but enter into the function called, just proves you are totally ignorant, and the fact that you think the x86 language says that a call HHH does ANYTHING but actualy execute the instructions in HHH just shows that you are nothing but an ignorant liar.
Sorry, that is the facts, something you seem to be allergic to.