Re: Halting Problem: What Constitutes Pathological Input

On 5/6/25 4:46 PM, olcott wrote:

On 5/6/2025 3:31 PM, dbush wrote:

On 5/6/2025 4:25 PM, olcott wrote:

On 5/6/2025 2:35 PM, dbush wrote:

On 5/6/2025 2:47 PM, olcott wrote:

On 5/6/2025 7:14 AM, dbush wrote:

On 5/6/2025 1:54 AM, olcott wrote:

On 5/6/2025 12:49 AM, Richard Heathfield wrote:

On 06/05/2025 00:29, olcott wrote:
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<snip>
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It is the problem incorrect specification that creates
the contradiction.

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Not at all. The contradiction arises from the fact that it is not possible to construct a universal decider.
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Everyone here insists that functions computed
by models of computation can ignore inputs and
base their output on something else.

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I don't think anyone's saying that.
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Maybe you don't read so well.
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What are the exact steps for DD to be emulated by HHH
according to the semantics of the x86 language?
*Only an execution trace will do*

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The exact same steps for DD to be emulated by UTM.
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_DD()
[00002133] 55         push ebp      ; housekeeping
[00002134] 8bec       mov ebp,esp   ; housekeeping
[00002136] 51         push ecx      ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call HHH(DD)
[00002141] 83c404     add esp,+04
[00002144] 8945fc     mov [ebp-04],eax
[00002147] 837dfc00   cmp dword [ebp-04],+00
[0000214b] 7402       jz 0000214f
[0000214d] ebfe       jmp 0000214d
[0000214f] 8b45fc     mov eax,[ebp-04]
[00002152] 8be5       mov esp,ebp
[00002154] 5d         pop ebp
[00002155] c3         ret
Size in bytes:(0035) [00002155]
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Machine address by machine address specifics
that you know that you cannot provide because
you know that you are wrong.
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HHH and UTM emulate DD exactly the same up until the point that HHH aborts,

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When you trace through the actual steps you
will see that this is counter-factual.
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Then what is the first instruction emulated by HHH that differs from the emulation performed by UTM?
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 HHH1 is exactly the same as HHH except that DD
does not call HHH1. This IS the UTM emulator.
It does not abort.

Rigth, and it shows that the CORRECT emulation of this input will halt.

 With HHH1(DD) the call from DD to HHH(DD) returns.
With HHH(DD) the call from DD to HHH(DD) cannot possibly return.

Sure it does, just not in the part that HHH emulated.

 THIS IS BECAUSE
HHH1(DD) DOES NOT HAVE a pathological relationship to DD.
HHH(DD) DOES HAVE a pathological relationship to DD.
 

No, it is because HHH gives up due to bad logic.
The bad logic that you gave it, because you believe in bad logic.
Your logic thinks that an incorrect emulation can be just called correct because you think it should be.,
Sorry, you are just proving your stupidity.