Re: What it would take... People to address my points with reasoning instead of rhetoric -- RP

On 14/05/2025 04:38, olcott wrote:

On 5/13/2025 10:28 PM, dbush wrote:

On 5/13/2025 11:06 PM, olcott wrote:

On 5/13/2025 9:44 PM, dbush wrote:

On 5/13/2025 10:41 PM, olcott wrote:

On 5/13/2025 8:56 PM, dbush wrote:

On 5/13/2025 9:52 PM, olcott wrote:

On 5/13/2025 8:38 PM, dbush wrote:

On 5/13/2025 9:35 PM, olcott wrote:

On 5/13/2025 8:26 PM, dbush wrote:

On 5/13/2025 9:16 PM, olcott wrote:

On 5/13/2025 8:03 PM, dbush wrote:

Nope.  Russell's Paradox was derived from the base axioms of naive set theory, proving the whole system was inconsistent.
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In contrast, there is nothing in existing computation theory that requires that a halt decider exists.

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I see you made no attempt to refute the above statement. Unless you can show from the axioms of computation theory that the following requirements can be met, your argument has no basis:
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Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
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A solution to the halting problem is an algorithm H that computes the following mapping:
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(<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
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A halt decider doesn't exist
for the same reason that the set of all sets
that do not contain themselves does not exist.
*As defined both were simply wrong-headed ideas*

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There's nothing wrong-headed about wanting to know if any arbitrary algorithm X with input Y will halt when executed directly.

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Yes there is. I have proven this countless times.

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That requirements are impossible to satisfy doesn't make them wrong. It just makes them impossible to satisfy, which is a perfectly reasonable conclusion.
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It did with Russell's Paradox.
ZFC rejected the whole foundation upon which
RP was built.
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ZFC did not solve some other Russell's Paradox
it rejected the whole idea of RP as nonsense.
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Unless you can show from the axioms of computation theory that the following requirements can be met, your argument has no basis:
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Alternatively I can do what ZFC did and over-rule
the whole foundation upon which the HP proofs are build.

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You mean the assumption that the following requirements (which are *not* part of the axioms of computation theory) can be satisfied? The assumption that Linz and other proved was false and that you *explicitly* agreed with?
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The conventional halting problem proofs have your
requirements as its foundation.
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They have the *assumption* that the requirements can be met, and via proof by contradiction show the assumption to be false.
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And the fact that the requirements can't be met is fine, just like the the fact that these requirements can't be met is fine:
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A mythic number is a number N such that N > 5 and N < 2.

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We can also say that no computation can compute
the square root of a dead rabbit. In none of these
cases is computation actually limited.
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We could equally say that no whale can give
birth to a pigeon. This places no actual limit
on the behavior of whales. Whales were never
meant to give birth to pigeons.
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And as was said before:
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On 5/5/2025 5:39 PM, olcott wrote:
 > On 5/5/2025 4:31 PM, dbush wrote:
 >> Strawman.  The square root of a dead rabbit does not exist, but the
 >> question of whether any arbitrary algorithm X with input Y halts when
 >> executed directly has a correct answer in all cases.
 >>
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 > It has a correct answer that cannot ever be computed
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This qualifies as both a non-rebuttal and your confirmation you agree that Linz and others are correct that no algorithm exists that satisfies the below requirements:
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Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
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A solution to the halting problem is an algorithm H that computes the following mapping:
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(<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
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It is true that a TM either halts or does not halt.
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None-the-less the above requirements simply ignore
that some inputs specify behavior that differs
from the behavior of their direct execution.

 And some inputs can be deduced to be impossible to satisfy.
 

It is stupidly wrong to REQUIRE a correct sum()
function to require sum(3,2) to report the sum of 5 + 7.

 It is also wrong to require sum() to accurately report the sum for all inputs. Given code of int sum(int x, int y) { return x + y; } sum(INT_MAX, INT_MAX) will get the answer wrong because + can't do what is expected of it.  In such circumstances, the C standard simply refuses to specify what will happen and adopts a `que sera, sera' attitude --- can't take a joke, shouldn't have joined.
 The Halting Problem is slightly different. It proposes a specific computation, a universal computation halting decider (call it H), but in doing so demonstrates that with a single flip of H's logical tail it produces a computation that simply can't be computed. There can be no H.
 Mr Olcott's response to this appears to be that we should allow H to set aside impossible inputs in the pursuit of some kind of meaningful answer.
 What he fails to notice is that it isn't necessary to set impossible inputs aside. Why not? Because they're impossible.
 

There is no need for H to cope with itself as input for the excellent reason that H cannot be written.
 As dbush rightly points out:
 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
 It has been known for 90 years that such a solution is impossible. Attempting to dodge round this fact is as fruitless as trying to construct a square with the same area as a given circle by using only a finite number of steps with compass and straightedge.
 Of course, that doesn't stop people trying. We have a name for such people.