Re: Every sufficiently competent C programmer knows --- Truthmaker Maximalism

On 3/14/2025 8:56 PM, dbush wrote:

On 3/14/2025 9:49 PM, olcott wrote:

On 3/14/2025 8:34 PM, dbush wrote:

On 3/14/2025 9:27 PM, olcott wrote:

On 3/14/2025 8:00 PM, dbush wrote:

On 3/14/2025 8:45 PM, olcott wrote:

On 3/14/2025 12:54 PM, dbush wrote:

On 3/14/2025 12:33 PM, olcott wrote:

On 3/14/2025 11:01 AM, wij wrote:

On Fri, 2025-03-14 at 10:51 -0500, olcott wrote:

On 3/14/2025 10:04 AM, wij wrote:

On Fri, 2025-03-14 at 09:35 -0500, olcott wrote:>>

void DDD()
{
     HHH(DDD);
     return;
}
>
DDD correctly simulated by HHH cannot possibly reach
its own "return" instruction in any finite number of
correctly simulated steps.
>
That you are clueless about the semantics of something
as simple as a tiny C function proves that you are not
competent to review my work.
>

>
https://en.wikipedia.org/wiki/Halting_problem
In computability theory, the halting problem is the problem of determining, from a description of
an
arbitrary computer program and an input, whether the program will finish running, or continue to
run
forever.
>
That means: H(D)=1 if D() halts and H(D)=0 if D() does not halt.
>
But, it seems you don't understand English, as least as my level, ....
>
>
>

>
void DDD()
{
    HHH(DDD);
    return;
}
>
The only difference between HHH and HHH1 is that they are
at different locations in memory. DDD simulated by HHH1
has identical behavior to DDD() directly executed in main().
>
The semantics of the finite string input DDD to HHH specifies
that it will continue to call HHH(DDD) in recursive simulation.
>
The semantics of the finite string input DDD to HHH1 specifies
to simulate to DDD exactly once.
>
When HHH(DDD) reports on the behavior that its input finite
string specifies it can only correctly report non-halting.
>
When HHH(DDD) is required to report on behavior other than
the behavior that its finite string specifies HHH is not
a decider thus not a halt decider.
>
All deciders are required to compute the mapping from
their input finite string to the semantic or syntactic property
that this string specifies. Deciders return true when this
string specifies this property otherwise they return false.
>

>
Are you solving The Halting Problem or not? Yes or No.
>
>

>
I have only correctly refuted the conventional halting
problem proof.

>
And what exactly do you think this proof is proving?  More specifically, what do you think the Linz proof is proving?

>
All of the proofs merely show that there cannot
possibly exist any halt decider that returns a
value corresponding to the behavior of any input
that is actually able to do the opposite of whatever
value is returned.
>

Not exactly.  What they prove is that no H exists that satisfies these requirements:
>
>
Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
>
A solution to the halting problem is an algorithm H that computes the following mapping:
>
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly
>

>
The executed directly part is bogus as I have
shown and your indoctrination blindly ignores.
>

>
But I want to know if any arbitrary X with input Y halts when executed directly,

>
Even when some inputs are BOGUS.
>

 Did I stutter?
 I want to know if any arbitrary X with input Y halts when executed

If you reject "ls;dlfm skdofdfn 894&49.8244bewr" as a syntactically
incorrect input then you are being inconsistent when you fail to reject
semantically incorrect inputs.

directly.  If I had an H that could tell me that in *all* possible cases, I could solve the Goldbach conjecture, among many other unsolved problems.
 Does an H exist that can tell me that or not?

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