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

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.

 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.

>

2) changing the input is not allowed

>
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 input D calls UTM1 then the
behavior of D simulated by UTM1 never reaches its final
halt state.
When D is simulated by ordinary UTM2 that D does not call
Then D reaches its final halt state.

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.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer