Re: DDD correctly emulated by H0 -- Ben agrees that Sipser approved criteria is met

On 6/26/24 10:13 PM, olcott wrote:

On 6/26/2024 8:52 PM, Richard Damon wrote:

On 6/26/24 9:30 PM, Mike Terry wrote:

On 27/06/2024 02:15, Mike Terry wrote:

On 27/06/2024 01:42, Richard Damon wrote:

On 6/26/24 8:20 PM, olcott wrote:

On 6/26/2024 6:55 PM, Richard Damon wrote:

On 6/26/24 7:46 PM, olcott wrote:

On 6/26/2024 6:41 PM, Richard Damon wrote:

On 6/26/24 9:42 AM, olcott wrote:

On 6/26/2024 6:02 AM, Richard Damon wrote:

On 6/25/24 11:42 PM, olcott wrote:

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That is not the way that it actually works.
That the the way that lies are defined.

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Source for you claim?
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Where is you finite set of steps from the truthmakers of the system to that claim?
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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 H0(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
>
The call from DDD to H0(DDD) when DDD is correctly emulated
by H0 cannot possibly return.

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Sure it can. I have shown an H0 that does so.
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I already told you that example does not count.
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I can't keep repeating those details or others
that so far have no idea what an x86 emulator is
will be baffled beyond all hope of comprehension.
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WHy not?
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We have already been over that you know that you cheated.
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Nope, since you didn't put in the rule, and if you had it would have shown that you lied, as if H0 is a pure function then the call to H0 emulated by H0 needs to have the same behaivor as the direct call to H0 by main.

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Incidentally, the nonconformance you're referring to is shown explicitly in the "195 page trace" that PO linked to.  [I.e. the simulated H does not correctly track the code path of the outer H.]

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I suppose I should have made clear, that's not simply due to the simulated H being aborted.  There is an instruction in H:   [actually, in Init_Halts_HH()]
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[000012e4] 753b jnz 00001321
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and in outer H control proceeds to 000012e6  [i.e. branch not taken],
whilein simulated H control proceeds to 00001321  [i.e. branch taken]
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Mike.
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Would need to look closer at the code, but I bet that the simulated machine is looking into the trace buffer to see if it is simulated or not.
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In effect, it is misusing static memory just like he says isn't allowed.
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>

 The slaves to a UTM either use a portion of the UTMs tape
or they can't possibly exist. There is probably a more pure
way to encode this.
 

Nope, they use virtual memory provided by the UTM.
They write to what they consider to be their tape, and the UTM figures out how to store that on its tape to be able to give it back when requested.
Of course, you never understood the need for putting the simulated machine in its own virtual memory space.