Re: Analysis of Flibble’s Latest: Detecting vs. Simulating Infinite Recursion ZFC

Op 28.mei.2025 om 18:01 schreef olcott:

On 5/28/2025 10:32 AM, Richard Heathfield wrote:

On 28/05/2025 16:17, olcott wrote:

On 5/28/2025 10:02 AM, Richard Heathfield wrote:

On 28/05/2025 15:36, olcott wrote:

On 5/28/2025 2:46 AM, Richard Heathfield wrote:

On 27/05/2025 22:25, olcott wrote:

On 5/27/2025 8:11 AM, Richard Heathfield wrote:

On 27/05/2025 11:41, Fred. Zwarts wrote:
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<snip>
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Of course HHH can be called by any other function even by DDD.

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And is. DDD's source shows this.
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But that is completely irrelevant

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Not in my view.
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I accept that that's your view and I won't dispute it because I understand your reasoning, but you and I are talking about different things. My underlying point is quite simply that Olcott made an incorrect and indeed contradictory claim about what HHH can and cannot report on. At the very, *very* least he made an insufficiently qualified claim.
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int sum(int x, int y) { return x + y; }
HHH must report on the behavior that its input actually
specifies the same way that sum(3,4) must report on the
sum of 3 + 4.

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DDD calls HHH, and you have said: "No HHH can report on the behavior of its caller" - so HHH cannot report on DDD.
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It would be wrong if HHH did report on the behavior
of its caller. Functions computed by models of computation
are only allowed to compute the mapping from their inputs.

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So you're dead in the water.
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HHH's input is DDD, and you have said: "HHH must report on the behavior that its input actually specifies" - so HHH must report on DDD.
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_DDD()
[00002192] 55             push ebp
[00002193] 8bec           mov ebp,esp
[00002195] 6892210000     push 00002192
[0000219a] e833f4ffff     call 000015d2  // call HHH
[0000219f] 83c404         add esp,+04
[000021a2] 5d             pop ebp
[000021a3] c3             ret
Size in bytes:(0018) [000021a3]
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It is a tautology that every input to a simulating
termination analyzer would never stop running unless
aborted specifies a non-terminating sequence of
configurations.

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Fails to address my point, which is that you claim that HHH cannot report on DDD and yet must report on DDD.
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HHH must report on the actual behavior that its actual
input actually specifies. Framing the problem any other
way is incorrect.

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So you're saying that it must report on its input, which means that HHH MUST report on its caller (because its caller /is/ its input). Fine. Why didn't you say that to start with?
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  *This is ALL that HHH sees*

No, HHH is given a pointer to memory and HHH can see everything in the memory. If the programmer made HHH such that it is blind for the full program specified in the input, that does not change the specification of the input, which includes DDD and all function called by DDD, in particular including the code to abort and halt the program.

 _DDD()
[00002192] 55             push ebp
[00002193] 8bec           mov ebp,esp
[00002195] 6892210000     push 00002192
[0000219a] e833f4ffff     call 000015d2  // call HHH
[0000219f] 83c404         add esp,+04
[000021a2] 5d             pop ebp
[000021a3] c3             ret
Size in bytes:(0018) [000021a3]
 That some people expect HHH to report
on things that it does not see is stupid.

That someone thinks that things do not exist if HHH is made blind for it is stupid. The code to abort and halt the program is accessible by HHH. It just needs to follow the pointer to DDD and then the addresses within DDD, etc. E.g., DDD contains the address 000015d2, so HHH could follow that and find the code which aborts and halts the program.
But HHH has a bug, so that it stops the simulation at that point and does not see the specification.
That HHH is made blind for it, does not change the specification that HHH should see.
It is childish to close your eyes and pretend that everything disappeared.