Sujet : Re: DDD specifies recursive emulation to HHH and halting to HHH1
De : dbush.mobile (at) *nospam* gmail.com (dbush)
Groupes : comp.theoryDate : 29. Mar 2025, 23:08:43
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vs9r1b$28tqg$2@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
User-Agent : Mozilla Thunderbird
On 3/29/2025 5:46 PM, olcott wrote:
On 3/29/2025 3:14 PM, dbush wrote:
On 3/29/2025 4:01 PM, olcott wrote:
On 3/29/2025 2:26 PM, dbush wrote:
On 3/29/2025 3:22 PM, olcott wrote:
On 3/29/2025 2:06 PM, dbush wrote:
On 3/29/2025 3:03 PM, olcott wrote:
On 3/29/2025 10:23 AM, dbush wrote:
On 3/29/2025 11:12 AM, olcott wrote:
On 3/28/2025 11:00 PM, dbush wrote:
On 3/28/2025 11:45 PM, olcott wrote:
>
It defines that it must compute the mapping from
the direct execution of a Turing Machine
>
Which does not require tracing an actual running TM, only mapping properties of the TM described.
>
The key fact that you continue to dishonestly ignore
is the concrete counter-example that I provided that
conclusively proves that the finite string of machine
code input is not always a valid proxy for the behavior
of the underlying virtual machine.
>
In other words, you deny the concept of a UTM, which can take a description of any Turing machine and exactly reproduce the behavior of the direct execution.
>
I deny that a pathological relationship between a UTM and
its input can be correctly ignored.
>
>
In such a case, the UTM will not halt, and neither will the input when executed directly.
>
It is not impossible to adapt a UTM such that it
correctly simulates a finite number of steps of an
input.
>
>
1) then you no longer have a UTM, so statements about a UTM don't apply
>
We can know that when this adapted UTM simulates a
finite number of steps of its input that this finite
number of steps were simulated correctly.
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And therefore does not do a correct UTM simulation that matches the behavior of the direct execution as it is incomplete.
>
It is dishonest to expect non-terminating inputs to complete.
An input that halts when executed directly is not non-terminating
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2) changing the input is not allowed
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The input is unchanged. There never was any
indication that the input was in any way changed.
>
>
False, if the starting function calls UTM and UTM changes, you're changing the input.
>
When UTM1 is a UTM that has been adapted to only simulate
a finite number of steps
And is therefore no longer a UTM that does a correct and complete simulation
and input D calls UTM1 then the
behavior of D simulated by UTM1
Is not what I asked about. I asked about the behavior of D when executed directly.
When D is simulated by ordinary UTM2 that D does not call
Then D reaches its final halt state.
Demonstrating that UTM1 is wrong.
Changing the input is not allowed.
I never changed the input. D always calls UTM1.
thus is the same input to UTM1 as it is to UTM2.
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