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devel / comp.theory / Re: ℙ!=ℕℙ proof

SubjectAuthor
* ℙ!=ℕℙ proofwij
`* Re: ℙ!=ℕℙ proofimmibis
 `* Re: ℙ!=ℕℙ proofwij
  `- Re: ℙ!=ℕℙ proofimmibis

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ℙ!=ℕℙ proof

<3be71d1fd92725898b57e2e0af5b4458b908e527.camel@gmail.com>

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From: wynii...@gmail.com (wij)
Newsgroups: comp.theory
Subject: ℙ!=ℕℙ proof
Date: Fri, 02 Feb 2024 20:21:03 +0800
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 by: wij - Fri, 2 Feb 2024 12:21 UTC

This is the P!=NP proof revised from the previous one.

-------------
ANPC::= Set of decision problems that additional information c must
be provided to compute the problem in P-time (including 
processing the information c (certificate)).

Since the information c can be generated in O(2^N) time, the upper
bound of ANPC is O(2^N) (checking c is in P-time).

Corollary: ANPC ⊆ ℕℙ
(from ℕℙ definition: "... ℕℙ is the set of decision 
problems verifiable in polynomial time by a
deterministic Turing machine.
https://en.wikipedia.org/wiki/NP_(complexity)")

If we try to prove or show that an ANPC problem can be solved in P-
time, we would also be trying to prove/show that the requirement of
the additional information c is not necessary. Thus, such a proof
will just prove itself not a valid proof, IOW, ANPC==P is unprovable.

Conclusion: ANPC is not P-time provable/solvable (implied by
ANPC definition)

Corollary: ℕℙ!=ℙ (ANPC ⊆ ℕℙ AND ANPC!=ℙ)

------------
I had just read something about the P-NP problem again, the P-
NP problem is actually quite tricky and highly 'academic'. That is
why the proof was changed to using ANPC. This version of P!=NP
proof is easy to understand. But, sadly, it does not provide direct
insight as to why TSP (or other NPC graph problems) cannot be solved in
P-time. I think this part may be (I am not sure) explained by Cook-
Levin Theorem (The CNF-SAT problem might be provable in O(2^N)).

Re: ℙ!=ℕℙ proof

<uprenc$eahh$5@dont-email.me>

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From: new...@immibis.com (immibis)
Newsgroups: comp.theory
Subject: Re: ℙ!=ℕℙ proof
Date: Mon, 5 Feb 2024 20:59:40 +0100
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 by: immibis - Mon, 5 Feb 2024 19:59 UTC

On 2/02/24 13:21, wij wrote:
> This is the P!=NP proof revised from the previous one.
>
> -------------
> ANPC::= Set of decision problems that additional information c must
> be provided to compute the problem in P-time (including
> processing the information c (certificate)).
>
> Since the information c can be generated in O(2^N) time, the upper
> bound of ANPC is O(2^N) (checking c is in P-time).
>
> Corollary: ANPC ⊆ ℕℙ
> (from ℕℙ definition: "... ℕℙ is the set of decision
> problems verifiable in polynomial time by a
> deterministic Turing machine.
> https://en.wikipedia.org/wiki/NP_(complexity)")
>
> If we try to prove or show that an ANPC problem can be solved in P-
> time, we would also be trying to prove/show that the requirement of
> the additional information c is not necessary. Thus, such a proof
> will just prove itself not a valid proof, IOW, ANPC==P is unprovable.

If you prove the additional information c is not necessary, then you
prove the problem is not an ANPC problem.

Re: ℙ!=ℕℙ proof

<9428b7fb17c89e5c549208c2a66fc4e6a3a70032.camel@gmail.com>

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From: wynii...@gmail.com (wij)
Newsgroups: comp.theory
Subject: Re: ℙ!=ℕℙ proof
Date: Sun, 18 Feb 2024 17:51:12 +0800
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 by: wij - Sun, 18 Feb 2024 09:51 UTC

On Mon, 2024-02-05 at 20:59 +0100, immibis wrote:
> On 2/02/24 13:21, wij wrote:
> > This is the P!=NP proof revised from the previous one.
> >
> > -------------
> > ANPC::= Set of decision problems that additional information c must
> > be provided to compute the problem in P-time (including
> >          processing the information c (certificate)).
> >
> > Since the information c can be generated in O(2^N) time, the upper
> > bound of ANPC is O(2^N) (checking c is in P-time).
> >
> > Corollary: ANPC ⊆ ℕℙ
> >            (from ℕℙ definition: "... ℕℙ is the set of decision
> >            problems verifiable in polynomial time by a
> > deterministic Turing machine.
> >            https://en..wikipedia.org/wiki/NP_(complexity)")
> >
> > If we try to prove or show that an ANPC problem can be solved in P-
> > time, we would also be trying to prove/show that the requirement of
> > the additional information c is not necessary. Thus, such a proof
> > will just prove itself not a valid proof, IOW, ANPC==P is
> > unprovable.
>
> If you prove the additional information c is not necessary, then you
> prove the problem is not an ANPC problem.
>

That is correct. But, the way ANPC described does not look right,
maybe I should revert to using ANP.

The basic is the same: Devise a class, e.g. ANP (or ANPC, to avoid the
difficulty of 'conventional NP definition'). You may see the 
P-NP problem is actually a 'word game'.

Re: ℙ!=ℕℙ proof

<uqtqd8$1eebh$1@dont-email.me>

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From: new...@immibis.com (immibis)
Newsgroups: comp.theory
Subject: Re: ℙ!=ℕℙ proof
Date: Sun, 18 Feb 2024 21:47:36 +0100
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 by: immibis - Sun, 18 Feb 2024 20:47 UTC

On 18/02/24 10:51, wij wrote:
> On Mon, 2024-02-05 at 20:59 +0100, immibis wrote:
>> On 2/02/24 13:21, wij wrote:
>>> This is the P!=NP proof revised from the previous one.
>>>
>>> -------------
>>> ANPC::= Set of decision problems that additional information c must
>>> be provided to compute the problem in P-time (including
>>>          processing the information c (certificate)).
>>>
>>> Since the information c can be generated in O(2^N) time, the upper
>>> bound of ANPC is O(2^N) (checking c is in P-time).
>>>
>>> Corollary: ANPC ⊆ ℕℙ
>>>            (from ℕℙ definition: "... ℕℙ is the set of decision
>>>            problems verifiable in polynomial time by a
>>> deterministic Turing machine.
>>>            https://en.wikipedia.org/wiki/NP_(complexity)")
>>>
>>> If we try to prove or show that an ANPC problem can be solved in P-
>>> time, we would also be trying to prove/show that the requirement of
>>> the additional information c is not necessary. Thus, such a proof
>>> will just prove itself not a valid proof, IOW, ANPC==P is
>>> unprovable.
>>
>> If you prove the additional information c is not necessary, then you
>> prove the problem is not an ANPC problem.
>>
>
> That is correct. But, the way ANPC described does not look right,
> maybe I should revert to using ANP.
>
> The basic is the same: Devise a class, e.g. ANP (or ANPC, to avoid the
> difficulty of 'conventional NP definition'). You may see the
> P-NP problem is actually a 'word game'.
>

All mathematical problems are 'word games'.

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