Re: Defining a correct simulating halt decider

On 9/5/24 9:24 AM, olcott wrote:

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

On 2024-09-03 13:00:50 +0000, olcott said:
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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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void DDD()
{
   HHH(DDD);
   return;
}
>
New slave_stack at:1038c4
Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc
[00002172][001138bc][001138c0] 55         push ebp      ; housekeeping
[00002173][001138bc][001138c0] 8bec       mov ebp,esp   ; housekeeping
[00002175][001138b8][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ec
[00002172][0015e2e4][0015e2e8] 55         push ebp      ; housekeeping
[00002173][0015e2e4][0015e2e8] 8bec       mov ebp,esp   ; housekeeping
[00002175][0015e2e0][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
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

 DDD emulated by HHH never reaches it final halt state.
It looks like I have to repeat this 10,000 times before
anyone ever notices that I said it at least once.
 _DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
 Show the details of how DDD emulated by HHH
reaches its own machine address 0000217f.
 00002172, 00002173, 00002175, 0000217a calls HHH(DDD)
then
00002172, 00002173, 00002175, 0000217a calls HHH(DDD)...
 

Which just proves that HHH didn't correctly emulate the input.
The CORRECT emulaition would go to 000015d2, and then through the rest of the code of HHH, and finally that HHH will abort its emulation and return to 0000217f and DDD will reach its final state.
Of course, your HHH stoped its emulation before it got there, showing it doesn't actually do a CORRECT emulation of its input.

C interpretaion of allows DDD to continue forever after
HHH jas terminated.
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