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

Op 30.mrt.2025 om 20:27 schreef olcott:

On 3/30/2025 3:12 AM, joes wrote:

Am Sat, 29 Mar 2025 16:46:26 -0500 schrieb olcott:

On 3/29/2025 3:14 PM, dbush wrote:

On 3/29/2025 4:01 PM, olcott wrote:

>

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.

A complete simulation of a nonterminating input doesn't halt.
>

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

So not an UTM.
>

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.

Doesn't matter if it calls it, but if the UTM halts.
>

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.

You changed UTM1, which is part of the input D.
>

 UTM1 simulates D that calls UTM1
simulated D NEVER reaches final halt state
 UTM2 simulates D that calls UTM1
simulated D ALWAYS reaches final halt state
 

Since DDD specifies a finite recursive emulation according to the x86 semantics, as proven by direct execution, UTM1 fails to reach the end of this finite recursion emulation. This failing emulation, cannot be used for any conclusion about the halting behaviour of the input. So, UTM2 is the right tool for this problem. UTM1 only reports its failure to complete the emulation correctly up to the end.