Re: Why do people here insist on denying these verified facts?

On 6/22/24 11:16 AM, olcott wrote:

On 6/22/2024 9:42 AM, Richard Damon wrote:

On 6/22/24 10:31 AM, olcott wrote:

https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/113318779X/
>
To understand this analysis requires a sufficient knowledge of
the C programming language and what an x86 emulator does. HHH0
and HHH1 have this criteria as their algorithm:

>
Which you just showed you don't have, since on comp.lang.c++ you thought that
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x *= ++f * ++f;
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had defined behavior for primative types for f.
>

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<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
   If simulating halt decider H correctly simulates its input D
   until H correctly determines that its simulated D would never
   stop running unless aborted then
>
   H can abort its simulation of D and correctly report that D
   specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

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Which used the definition of "Correct Simulation" to mean a simulation that produces the EXACT results of the direct execution of the machine being simulated, which requires a simulation that will not "abort" its simulation, EVER (except by reaching a final state).
>

 I have had enough of your deception trying to get away
with denying the semantics of the x86 programming language.
I really hope that you don't get condemned to Hell over this.

And what in the actual detail of the x86 programming language says something diffferent than what I say?
I think YOU are the one in danger, as you LIE about what the x86 assembly code says,

 

You Deciders do not do this, nor do they do this about an actual Correct Simulation per that definition of the input, so they can not use the second clause.
>

>
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
 > I don't think that is the shell game. PO really /has/ an H
 > (it's trivial to do for this one case) that correctly determines
 > that P(P) *would* never stop running *unless* aborted.
 >
>
Ben only agrees that the criteria is met for the input. He
does not agree that the criteria has been meet for non-inputs.

>
Ben only agrees that H correctly decides the exact criteria that you state, that no H can "correctly simuate" (per YOUR definition) the input to a final state.

 And you happily deny the verified facts at the possible cost
of damnation in Hell.
 

Thus, he is agreeing that you H is a POOP decider, for this input, but not that it is a HALTING decider for this input, since its criteria is not the Halting Criteria.
>

>
Computable functions are the formalized analogue of the intuitive
notion of algorithms, in the sense that a function is computable if there exists an algorithm that can do the job of the function, i.e. *given an input of the function domain*
*it can return the corresponding output*
https://en.wikipedia.org/wiki/Computable_function
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*That seems to say that non-inputs do not count*

>
But we aren't talking about "Non-Inputs",

 The D(D) that calls H(D,D) such that this call returns has
provably different behavior than D correctly simulated by H
is measured by the actual semantics of the x86 programming
language.
 How the Hell does anyone feel that they can get away with
flatly contradicting the semantics of the x86 programming
language?
 We are a herd and our first duty it to follow the herd
even if the herd leaps off a cliff.
 

and in fact, YOUR arguement needs to look at the non-inputs, as it allows the input to change when you argue about other deciders.
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int sum(int x, int y){ return x + y; }
In other words sum(3,4) must consider the sum of 5 + 6?
 

The input is the finite string.
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The MEANING of that finite string is defined by the PROBLEM.

 LIAR. You know that the meaning of the finite string
is defined by the semantics of the x86 language.

Right, by what the DIRECT EXECUTION of that set of instruction will do.
Which means, for your input, in needs the instuctions of the decider it calls.

 I might start wishing that you get a tiny taste of Hell
to set you on the correct path.
 As Christ said as ye judge ye shall be judged so I do
wish the same thing upon myself. If I am on the wrong
path then I sincerely wish for the minimum adversity
required to definitely set me on the right path.
 

 The decider gets to define the encoding, but not the meaning/behavior of the encoded items.
>

 When the x86 language is specified then the decider
has zero leeway in this.

So, what in the x86 language shows what you claim?

 

Halting DEFINES the meaning/behavior to be that of the directly run program represented by the input.

 That makes it contradict one of its own axioms, thus
conclusively proving that it is incorrect:
 Computable functions are the formalized analogue of the
intuitive notion of algorithms, in the sense that a
function is computable if there exists an algorithm
that can do the job of the function, i.e.
*given an input of the function domain it*
*can return the corresponding output*
https://en.wikipedia.org/wiki/Computable_function

Who says Halting is a Computable Function?
THAT is part of your problem.
The question actually is, "IS Halting Computable?"

 

>
*Here is the verified facts that everyone denies*
*Here is the verified facts that everyone denies*
*Here is the verified facts that everyone denies*
*Here is the verified facts that everyone denies*
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void DDD()
{
   HHH0(DDD);
}
>
int main()
{
   Output("Input_Halts = ", HHH0(DDD));
   Output("Input_Halts = ", HHH1(DDD));
}
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It is a verified fact that the behavior that finite string DDD
presents to HH0 is that when DDD correctly emulated by HH0
calls HH0(DDD) that *THIS CALL DOES NOT RETURN*
>
It is a verified fact that the behavior that finite string DDD
presents to HH1 is that when DDD correctly emulated by HH1
calls HH0(DDD) that *THIS CALL DOES RETURN*
>
>

>
>
The problem is that the "behavior" that the finite string DDD presents to HH0, is DEFINED by the problem.

 LIAR. It is defined by the semantics of the x86 language.

Right, as the results of the direfdt execution of the input, which means the input needs to contain ALL the instructions that will be executed.

 

And if that problem is the Halting Problem, that behavior is the behavior of the machine the input represents. If HH0 treats the input as having a different behavior, then HH0 just isn't a Halting Decider, but something else.
>
If HH0 is supposed to be a Halting decider, but uses a method that makes it see something other than that behavior, then it is just an incorrect Halting Decider, and its algorithm just creates an incorrect recreation of the property of the input it is supposed to be working on.
>
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A bit of a side note, the actual "Input" to HH0, is a pointer to memory, and as such it passes a reference to ALL of memory considering the starting point to be that address, so your "Input" isn't actually the few bytes of DDD, but ALL of memory and a starting point. If you actually mean that the input is just those few bytes pointed to by the address, then the input is improperly formed and is NOT a proper representation of the input machine, becuase it is incomplete.
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The fact you don't understand this, seems to imply you are lacking the basic knowledge to be talking about this sort of thing.
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 The input to HHH0(DDD) includes itself.
The input to HHH1(DDD) DOES NOT include itself.

So?

 It is stipulated that correct emulation is defined by the
semantics of the x86 programming language and nothing else.

Right, so HHH0 needs to determine what the ACTUAL BEHAVIOR of its input is to be a Halt Decider, and that means it needs to figure out what it will itself do.

 DDD correctly emulated by HHH0 correctly determines that
the call from the emulated DDD to HHH0 DOES NOT RETURN.

Nope, It just determins that it doesn't know if the call will return.

 DDD correctly emulated by HHH1 correctly determines that
the call from the emulated DDD to HHH0 DOES RETURN.