Selective Incompatibility in sets

[HarounBoutamani]

Hello,

 

Here I share with you a paper I wrote on Selective Incompatibility in sets and an approach to the duality P vs NP.

Paper link :

 

https://­drive.google.com/­file/d/­0B2iY_1VArjmYTUttR1h5­WTRIWDg/­view?usp=drivesdk

 

Regards

 

[Phi for All]

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Moderator Note

Hello. To avoid spammers, new members aren't allowed to post links for a bit. In any case, our rules state that discussions can't require members click outside links or videos. Please post your paper here so discussion can move forward.

 

Welcome to the site.

[HarounBoutamani]

I have uploaded the file in case the link isn't working.

 

Selective Incompatibility and an approach to the duality P vs NP.pdf

Edited April 23, 2017 by HarounBoutamani

[wtf]

The first confusing remark came pretty early:

 

> We define [math]f[/math] as a function [math]f_B(x) : A \to f_B(A)[/math]

 

* You didn't define the notation [math]f_B[/math].

 

* What is [math]x[/math]?

 

* What's the difference between [math]f_B(x)[/math] and [math]f_B(A)[/math]?

 

* How am I supposed to interpret this notation?

 

Also why do you say that [math]B[/math] is a set of size [math]v[/math] then never use [math]v[/math] again

 

I mention in passing that the moderator has already asked you to express your ideas here and not solely via an external link, so I don't mean for my comment to be encouraging you in flouting the moderator's instruction.

 

But frankly your exposition is incoherent and the problems start right at the beginning.

Edited April 23, 2017 by wtf

[HarounBoutamani]

* You didn't define the notation [math]f_B[/math].

 

I don't have to define every selective function.

 

Edited April 24, 2017 by HarounBoutamani

[wtf]

In definition 1, can you define the notation [math]x|y[/math]? I'm guessing it's not the usual divisibility relation, but I can't figure out what it is.

 

Also, having spent most of the paper discussing the case of finite sets, you then pivot to the P = NP problem, which involves infinite sets. How do you justify that?

[HarounBoutamani]

Also, having spent most of the paper discussing the case of finite sets, you then pivot to the P = NP problem, which involves infinite sets. How do you justify that?

 

In infinite sets the number of elements may not be countable but I consider it a variable.

Edited April 24, 2017 by HarounBoutamani

[wtf]

You won't even answer direct questions about your ambiguous, confusing, and undefined notation. Why should I read yet another version of your paper?

Edited April 24, 2017 by wtf

[HarounBoutamani]

I point out that the paper title is "Selectivity in sets and duality P vs NP".

Minor changes in latest version.

Selectivity in sets and the duality P vs NP.pdf

[wtf]

I point out that the paper title is "Selectivity in sets and duality P vs NP".

Your entire chain of reasoning is based on the sets being finite. But the complexity classes P and NP are infinite. How do you know any of your prior reasoning still applies?

 

And why won't you answer the most basic notational questions?

[HarounBoutamani]

Your entire chain of reasoning is based on the sets being finite. But the complexity classes P and NP are infinite. How do you know any of your prior reasoning still applies?

Order(P) and Order(NP) are the number of elements of P and NP respectively, Order(P) and Order(NP) are uncountable variables but they are still comparable to the elementary properties of their respective elements.

 

For example let's define for the class P the relation R-comptability (L,K)- > { L and K have polynomial relation between their expressed complexities.}

For any element p out of P we have :

 

Compa(p) = Order(P) - 1

 

and

 

Incompa(p) = 0.

 

Let's define the same relation R-comptability defined previously but for the class NP.

For any element n out of NP we have:

 

Compa(n) < Order(NP) - 1

 

and

 

Incompa(n) > 0

 

This question of yours brings me to the subject of uncountable variables, which is a subject of research for me.

 

Regarding the notational question of x|y, it means x is different from y.

About uncountable variables, I believe the number of steps necessary to solve an NP-Hard problem to be an uncountable variable, but this is yet to be proved.

[wtf]

Order(P) and Order(NP) are the number of elements of P and NP respectively, Order(P) and Order(NP) are uncountable variables

Aren't each of P and NP countable? After all there are only countably many Turing machines.

Edited April 24, 2017 by wtf

[HarounBoutamani]

Aren't each of P and NP countable? After all there are only countably many Turing machines.

 

What I am referring to is that we can't tell the number of their respective elements, being infinite sets.

Edited April 24, 2017 by HarounBoutamani

[wtf]

What I am referring to is that we can't tell the number of their respective elements, being infinite sets.

Have you considered studying the very basics of computability theory and infinitary set theory before tacking P = NP? Do you understand my point about the countability of the set of Turing machines?

[HarounBoutamani]

Have you considered studying the very basics of computability theory and infinitary set theory before tacking P = NP? Do you understand my point about the countability of the set of Turing machines?

I think you may consider set theory from a selectivity perspective, or not.

[wtf]

I think you may consider set theory from a selectivity perspective, or not.

If P is infinite then what can "Compa(p) = Order(P) - 1" possibly mean?

 

Since P is infinite, is Order(P) - 1 any different than Order(P) + 47?

 

About uncountable variables, I believe the number of steps necessary to solve an NP-Hard problem to be an uncountable variable, but this is yet to be proved.

Do you understand that by definition all algorithms consist of a finite sequence of steps?

 

You are demonstrating a complete lack of basic knowledge of the subject areas you are attempting to work with. Why not just pick up the basics? Start here. https://en.wikipedia.org/wiki/Countable_set. We'll work up to basic computer science later.

Edited April 24, 2017 by wtf

[HarounBoutamani]

If P is infinite then what can "Compa(p) = Order(P) - 1" possibly mean?

 

Since P is infinite, is Order(P) - 1 any different than Order(P) + 47?

 

This exactly my point when I say that uncountable variables are an area of research for me.

 

Compa(p) and Order(P) may be uncountable but they are comparable, and they can be involved in calculus, how far can this go, I don't know yet.

Edited April 24, 2017 by HarounBoutamani

[wtf]

This exactly my point when I say that uncountable variables are an area of research for me.

 

Compa(p) and Order(P) may be uncountable but they are comparable, and they can be involved in calculus, how far can this go, I don't know yet.

I suggest starting with the link I gave in my previous post, about countable and uncountable sets. After that you might consider learning some basic computer science.

 

I'm afraid I can't add much to what I've written. You need to master the basics.

[HarounBoutamani]

Do you understand that by definition all algorithms consist of a finite sequence of steps?

 

Except for an algorithm that does not exist.

I suggest starting with the link I gave in my previous post, about countable and uncountable sets. After that you might consider learning some basic computer science.

 

I'm afraid I can't add much to what I've written. You need to master the basics.

This may be great as I may not have to answer anymore of your questions.

[wtf]

Except for an algorithm that does not exist.

 

This may be great as I may not have to answer anymore of your questions.

LOL. Nevermind. Here's the answer you were hoping for.

 

"Congratulations. Enclosed find check for US $1,000,000 for solving the P = NP problem. Your Fields medal is in the mail too."

 

All the best.

[Phi for All]

 

I don't have to define every selective function.

 

 

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Moderator Note

To have a meaningful discussion of your paper, it's necessary to clarify anything readers don't find clear. If you aren't willing to answer questions, how can discussion happen? Please address questions regarding your paper or the thread will be closed.

[HarounBoutamani]

 

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Moderator Note

To have a meaningful discussion of your paper, it's necessary to clarify anything readers don't find clear. If you aren't willing to answer questions, how can discussion happen? Please address questions regarding your paper or the thread will be closed.

I answer questions that I judge to be interesting enough, and I did. I suppose you have the moderator permissions to close the thread.

[Phi for All]

I answer questions that I judge to be interesting enough, and I did.

 

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Moderator Note

And that sounds more like blog behavior. This is a science discussion forum.

 

Do you wish to engage with others in discussion, or shall I close the thread?

[HarounBoutamani]

 

!

Moderator Note

And that sounds more like blog behavior. This is a science discussion forum.

 

Do you wish to engage with others in discussion, or shall I close the thread?

I have answered some questions that were previously posted, and I am willing to answer other questions, but I don't promise that I will answer every word directed to me.

 

 

Edited April 25, 2017 by HarounBoutamani