Re: Defining a correct simulating halt decider

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:
 

A halt decider is a Turing machine that computes
the mapping from its finite string input to the
behavior that this finite string specifies.

 A halt decider needn't compute the full behaviour, only whether
that behaviour is finite or infinite.
 

 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
 Hence  HHH(DDD)==0 is correct

 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.

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