Re: DDD specifies recursive emulation to HHH and halting to HHH1

On 3/27/2025 1:50 PM, olcott wrote:

On 3/27/2025 2:18 AM, Fred. Zwarts wrote:

Op 27.mrt.2025 om 04:09 schreef olcott:

On 3/26/2025 8:22 PM, Richard Damon wrote:
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_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]
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Non-Halting is that the machine won't reach its final staste even if an unbounded number of steps are emulated. Since HHH doesn't do that, it isn't showing non-halting.
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DDD emulated by any HHH will never reach its final state
in an unbounded number of steps.
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DDD emulated by HHH1 reaches its final state in a finite
number of steps.
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It is not very interesting to know whether a simulator reports that it is unable to reach the end of the simulation of a program that halts in direct execution.

 That IS NOT what HHH is reporting.
HHH correctly rejects DDD because DDD correctly
emulated by HHH cannot possibly reach its own
final halt state.

In other words, HHH is not a halt decider because it is not computing the required mapping:
Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly

 

It is interesting to know:
'Is there an algorithm that can determine for all possible inputs whether the input specifies a program that (according to the semantics of the machine language) halts when directly executed?'

 It is the halts while directly executed that is impossible
for all inputs.

In other words, you *once again* explicitly agree with Linz and others that no algorithm behaves as required above.

A TM can only report on the behavior that the machine code
of another TM specifies.

And the machine code of DDD specifies that it halts when executed directly.

When it specifies a pathological
relationship then the behavior caused by the pathological
relationship MUST BE REPORTED.

False.  The behavior that must be reported is what the requirements stipulate:
Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly

 

This question seems undecidable for Olcott.
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