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On 10/16/2024 1:33 AM, joes wrote:Line 502 (if(Root)) wasn't changed.Am Tue, 15 Oct 2024 16:51:15 -0500 schrieb olcott:https://github.com/plolcott/x86utm/blob/master/Halt7.c#L502 shows: u32On 10/15/2024 4:24 PM, joes wrote:Nope, still there: https://github.com/plolcott/x86utm/blob/master/Am Tue, 15 Oct 2024 15:01:36 -0500 schrieb olcott:https://github.com/plolcott/x86utm Halt7.c was updated last month.On 10/15/2024 2:33 PM, joes wrote:I don't follow your repo. Can you point me to the relevant commit?Am Tue, 15 Oct 2024 13:25:36 -0500 schrieb olcott:There is some code that was obsolete several years ago.On 10/15/2024 10:17 AM, joes wrote:>Am Tue, 15 Oct 2024 08:11:30 -0500 schrieb olcott:On 10/15/2024 6:35 AM, Richard Damon wrote:On 10/14/24 10:13 PM, olcott wrote:On 10/14/2024 6:50 PM, Richard Damon wrote:On 10/14/24 11:18 AM, olcott wrote:On 10/14/2024 7:06 AM, joes wrote:Am Mon, 14 Oct 2024 04:49:22 -0500 schrieb olcott:On 10/14/2024 4:04 AM, Mikko wrote:On 2024-10-13 12:53:12 +0000, olcott said:Oh, did you take out the check if HHH is the root simulator?There are no static root variables. There never has been any "notYes! It really has different code, by way of the static RootIt explains in great detail that another different DDD (samehttps://chatgpt.com/share/6709e046-4794-8011-98b7-27066fb49f3eI did that, and it admitted that DDD halts, it just tries to
When you click on the link and try to explain how HHH must be
wrong when it reports that DDD does not terminate because DDD
does terminate it will explain your mistake to you.
justify why a wrong answer must be right.
machine code different process context) seems to terminate only
because the recursive emulation that it specifies has been
aborted at its second recursive call.
variable.
No wonder it behaves differently.
a pure function of its inputs" aspect to emulation.
It doesn't seem to have happened this year.
Halt7.c#L502
H(ptr P, ptr I) // 2024-09-15 was HH and edit on line 643
The repository indicates that it was updated: "last month"
I am talking about the referential transparency, not whether the functionTherefore if HHH even guesses that its input is non-termination then HHHSure.Sure yet only when the input is non-terminating.You seemed to not understand that a simulation may be nonterminating.You and Richard never seemed to understand this previously.Every termination analyzer that emulates itself emulating itsThat point can never come in the complete simulation of a non-
input has always been a pure function of this input up to the
point where emulation stops.
terminating input, because it is infinite.
is correct.
In other words you don't have a clue that an input to a terminationIn which way is DDD screwed up that it is both free of side effects andDDD is free to be totally screwed up every which way.Sure. How can a function without side effects have differentYou may be half right. Only the analyzer must be pure.By "pure" I mean having no side effects. You mean total vs.Non-terminating C functions do not ever return, thus cannotYou err because you fail to understand how the same C/x86Do explain how a pure function can change.
function invoked in a different process context can have
different behavior.
possibly be pure functions.
partial.
The input is free to get stuck in an infinite loop.
behaviour?
It is only HHH that must be a pure function.
referentially intransparent?
analyzer can be non-terminating thus violating the (1) criteria of pure
functions shown below.
(1) *the function return values are identical for identical arguments*Yes, this has nothing to do with termination. Your function
(no variation with local static variables, non-local variables, mutable
reference arguments or input streams, i.e., referential transparency),
and (2) the function has no side effects (no mutation of local static
variables, non-local variables, mutable reference arguments or input/
output streams).
When-so-ever any input to a termination analyzer iss/pure/total
non-terminating for any reason then this input is not a pure function.
Terminating C functions must reach their "return" statement.Which DDD does.
--I am, as of a couple months back. This is still related to the LinzFor many months now I have been talking about the termination analyzerWeren't we discussing the halting DDD(){HHH(DDD);} before?Inputs are not required to be pure functions.HHH is a pure function of its input the whole time that it isI thought DDD was fixed to only call HHH(DDD)?
emulating.
DDD has no inputs and is allowed to be any finite string of x86
code.
Inputs to HHH are by no means required to ever return AT ALL.
HHH applied to input DDD.
I am not aware of ever referring to HHH as a halt decider. When I talk
about halt deciders I talk about the Linz proof.
proof.
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