P vs NP Problem

The Kissing Number Problem
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In geometry, the kissing number of a mathematical space is defined as the greatest number of non-overlapping unit spheres that can be arranged in that space such that they each touch a common unit sphere. For a given sphere packing (arrangement of spheres) in a given space, a kissing number can also be defined for each individual sphere as the number of spheres it touches. For a lattice packing the kissing number is the same for every sphere, but for an arbitrary sphere packing the kissing number may vary from one sphere to another.
When a bunch of spheres are packed in some region, each sphere has a Kissing Number, which is the number of other spheres it’s touching; if you’re touching 6 neighboring spheres, then your kissing number is 6. Nothing tricky.

(check the images on the comment to be clear.)
✅In one dimension, the kissing number is 2
✅In two dimensions, the kissing number is 6
✅In three dimensions, the kissing number is 12
✅In four dimensions, it was known for some time that the answer was either 24 or 25
✔️The kissing number in n dimensions is unknown for n > 4, except for n = 8 (where the kissing number is 240), and n = 24 (where it is 196,560).
There are several hurdles to a full solution, including computational limitations. So expect incremental progress on this problem for years to come.

Collatz conjecture
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This is the simple math problem we still can’t solve!!
Just pick a number (any number). If it's even, divide it by 2. If it's odd, multiply it by 3 and add 1. Take that new number and repeat the process, again and again. If you keep this up, you’ll eventually get stuck in a loop. At least, that’s what we think will happen.

Take 10 for example: 10 is even, so we cut it in half to get 5. Since 5 is odd, we triple it and add 1. Now we have 16, which is even, so we halve it to get 8, then halve that to get 4, then halve it again to get 2, and once more to get 1. Since 1 is odd, we triple it and add 1. Now we’re back at 4, and we know where this goes: 4 goes to 2 which goes to 1 which goes to 4, and so on. We’re stuck in a loop.

Or try 11: It’s odd, so we triple it and add 1. Now we have 34, which is even, so we halve it to get 17, triple that and add 1 to get 52, halve that to get 26 and again to get 13, triple that and add 1 to get 40, halve that to get 20, then 10, then 5, triple that and add 1 to get 16, and halve that to get 8, then 4, 2 and 1. And we’re stuck in the loop again. it is a mathematical black hole.

As simple as it sounds, it actually works. But the problem is that even though mathematicians have shown this is the case with millions of numbers, they haven't found any numbers out there that won't stick to the rules.

"It's possible that there's some really big number that goes to infinity instead, or maybe a number that gets stuck in a loop and never reaches 1," explains Thompson. "But no one has ever been able to prove that for certain."

𝐓𝐰𝐢𝐧 𝐏𝐫𝐢𝐦𝐞 𝐂𝐨𝐧𝐣𝐞𝐜𝐭𝐮𝐫𝐞
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Twin Prime Conjecture is the most famous in Number Theory or the study of natural numbers and their properties, frequently involving prime numbers. Since you've known these numbers since grade school, stating the conjectures is easy.

When two primes have a difference of 2, they're called twin primes. So 11 and 13 are twin primes, as are 599 and 601. Now, it's a Day 1 Number Theory fact that there are infinitely many prime numbers. So, are there infinitely many twin primes? The Twin Prime Conjecture says yes.
Let's go a bit deeper. The first in a pair of twin primes is, with one exception, always 1 less than a multiple of 6. And so the second twin prime is always 1 more than a multiple of 6. You can understand why, if you're ready to follow a bit of heady Number Theory.

All primes after 2 are odd. Even numbers are always 0, 2, or 4 more than a multiple of 6, while odd numbers are always 1, 3, or 5 more than a multiple of 6. Well, one of those three possibilities for odd numbers causes an issue. If a number is 3 more than a multiple of 6, then it has a factor of 3. Having a factor of 3 means a number isn't prime (with the sole exception of 3 itself). And that's why every third odd number can't be prime.

How's your head after that paragraph? Now imagine the headaches of everyone who has tried to solve this problem in the last 170 years.

The good news is that we've made some promising progress in the last decade. Mathematicians have managed to tackle closer and closer versions of the Twin Prime Conjecture. This was their idea: Trouble proving there are infinitely many primes with a difference of 2? How about proving there are infinitely many primes with a difference of 70,000,000? That was cleverly proven in 2013 by Yitang Zhang at the University of New Hampshire.

For the last six years, mathematicians have been improving that number in Zhang's proof, from millions down to hundreds. Taking it down all the way to 2 will be the solution to the Twin Prime Conjecture. The closest we've come—given some subtle technical assumptions—is 6. Time will tell if the last step from 6 to 2 is right around the corner, or if that last part will challenge mathematicians for decades longer.

Goldbach’s Conjecture
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One of the greatest unsolved mysteries in math is also very easy to write. Goldbach's Conjecture is, "Every even number (greater than two) is the sum of two primes." You check this in your head for small numbers: 18 is 13+5, and 42 is 23+19. Computers have checked the Conjecture for numbers up to some magnitude. But we need proof for all natural numbers.

Goldbach's Conjecture precipitated from letters in 1742 between German mathematician Christian Goldbach and legendary Swiss mathematician Leonhard Euler, considered one of the greatest in math history. As Euler put it, "I regard [it] as a completely certain theorem, although I cannot prove it."

Euler may have sensed what makes this problem counterintuitively hard to solve. When you look at larger numbers, they have more ways of being written as sums of primes, not less. Like how 3+5 is the only way to break 8 into two primes, but 42 can broken into 5+37, 11+31, 13+29, and 19+23. So it feels like Goldbach's Conjecture is an understatement for very large numbers.

Still, a proof of the conjecture for all numbers eludes mathematicians to this day. It stands as one of the oldest open questions in all of math.

Nigerian professor have solved 156 year old maths problem?
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One of the most important problems in mathematics - the Riemann Hypothesis - said to have finally solved by a Nigerian professor in 2015. but still it was not confirmed by clay mathematics institute. it said unsolved problem.

Dr Opeyemi Enoch claims he made a key breakthrough in 2010 which later enabled him to solve the puzzle, which is one of the seven Millennium Problems in Mathematics.

The Riemann Hypothesis was proposed by mathematician Bernard Riemann in 1859 and concerns the distribution of prime numbers.
It has become arguably the most famous problem in mathematics, since Fermat's Last Theorem was solved in the 1990s.

At its most simple, the distribution of prime numbers among all others doesn't follow a regular pattern.

However, Riemann noticed that the frequency of prime numbers is very closely related to the behavior of an elaborate function called the Riemann Zeta function.

The hypothesis asserts that all solutions of the equation ζ(s) = 0 lies on a certain vertical straight line, according to the Clay Mathematics Institute.

While this has been checked for the first 10,000,000,000 solutions, it is only now that a 'proof' explaining their distribution beyond this has been found.

However, The Clay Mathematical institute has neither confirmed nor denied that Dr Enoch has officially solved the problem, simply saying it does not comment on solutions to the Millennium Problems.

This has led to critics claiming the story is a hoax and MailOnline has contacted the professor for more information.

Dr Enoch, who teaches at the Federal University of Oye Ekiti (FUOYE) in Nigeria, said he was motivated to solve the 156-year-old problem because of his students.

He told the BBC that they wanted him to make money from the internet.

'The motivation was because my students trusted that the solution could come from me - not because the financial reward and that was why I started trying to solve the problem in the first place,' he said.

THE MILLENNIUM PRIZE PROBLEMS
The Millennium Prize Problems were launched on 24 May, 2000.

They include seven problems considered by the Clay Mathematics Institute to be 'important classic questions that have resisted solution over the years'.

These include: P versus NP, The Hodge conjecture, The Poincaré conjecture, The Riemann hypothesis, Yang–Mills existence and mass gap, Navier–Stokes existence and smoothness and The Birch and Swinnerton-Dyer conjecture.

The full details of each are available from the institute's website.

The first person to solve each of the problems will receive $1 million (£658,000).

The professor presented his proof on 11 November during the International Conference on Mathematics and Computer Science in Vienna, Nigerian news site Vanguard reported.

Join our new Telegram channel: https://t.me/unsolvedmillenniumproblems

Tatenda Kubalalika tries to solve the Riemann Hypothesis. and he wants your review on his proof and tries to contact with clay mathmatics of institute. Read and Comment your review below.
Full paper link: https://figshare.com/articles/preprint/Untitled_Item/14776146

If You Solve P vs NP Problem, You Could Steal All the Bitcoin in the World!
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If you can prove or disprove its cryptically short equation, you’d be a million dollars richer—and maybe even billions of dollars richer, depending on your scruples.

The importance of P versus NP is mainly in its consequences for computing. It happens to be one of the seven Millennium Prize Problems, meaning The Clay Mathematics Institute of Cambridge, Massachusetts will award $1 million to whoever manages to prove or disprove the statement. But should you prove that P in fact does equal NP, you wouldn’t even need the $1 million prize. As theoretical computer scientist Scott Aaronson explained last week at a lecture in a stuffy auditorium at Los Alamos National Lab in New Mexico, proving that P=NP would open up some intriguing possibilities.

“If someone proves P=NP, the first thing they should do is steal $200 billion in bitcoin. The second thing they should do is solve all of the other Millennium Prize Problems,” Aaronson said.

To understand this, you need to know that computers are devices that solve problems, abstracted into code readable by the physical computing device, based on the principles put forth by Alan Turing. Solving problems takes a number of steps and a certain amount of time, with the amount of time required increasing as the problem grows larger.

So go out there and try and prove that P does, or does not, equal NP. If you’re successful, you’ll make at least a million dollars, and perhaps much, much more. If you’re unsuccessful, well, hopefully you will have led a meaningful life researching computational theory.

Travelling salesman problem (TSP)
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TSP asks "from given list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city exactly once and returns to the origin city?" it is an NP-hard problem in combinatorial optimization, important in theoretical computer science and operations research.

For example: what is the shortest possible route to start from City A and visit each city once and come back to city A?

- To solve this 6 cities symmetric TSP problem with brute force (try each path), it takes (n-1)!/2. then 5!/2 =60 possibilities. if we increase the number of cities the path possibilities increase exponentially. you may solve TSP with small number of cities with brute force, but when the number of cities become over 150 it may take many years to solve even with computers. that is why we call it np hard problems.
Of course there are some algorithms to solve TSP with approximation. but there is no exact algorithm for TSP yet. current TSP algorithms works well for small number of cities. but when the number of cities increase they start to approximate.
Can you find efficient algorithm for TSP?
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Try to solve this 6 cities symmetric TSP.

AI and P vs NP problem
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Artificial intelligence is a system that perceives its environment and takes actions that maximize its chance of success at a specific goal. that means, AI is a cognitive system that has the capability to make decisions by solving a set of optimization problems and have a real time response to its environmental signals.

Many AI optimization problems fall into Non-deterministic polynomial time (NP) problems. Though easy to understand and check the correctness of a solution; NP problems are difficult to solve due to their exponential complexity. No efficient solution has been found to date. Indeed, P vs NP is one of the 7 unsolved Millennium Problems as defined by Clay Mathematics Institute. A very known NP problem is the Traveling Salesman Problem, first introduced in 1800s, the Travelling Salesman Problem describes a salesman who must travel between N cities. The order of cities doesn't matter, as long as he visits each one during his trip, and finishes where he started.

If P=NP, everything gets much faster: computers get faster, shipping gets faster, networks get faster, etc. as the optimal solutions to the day-to-day NP-hard problems are solved. There is a number of different reasons for this speed increase, including improved routing, improved searching, and improved compression.

Next, a large number of the cryptographic methods currently used become vulnerable. Everything any government has ever done becomes public knowledge overnight. You can imagine the upshot.

What do you think if P=NP?

Why does P vs NP problems matter?
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Another way to ask whether P = NP is to ask whether every hard problem actually contains an easy, but hidden, solution.
Are these two flavor of problems irrevocably separate from one another? Are some problems simply complex by their fundamental nature?

If P does equal NP, then it would have some major implications for our way of life.
One major benefit is that many NP problems are referred to as being NP-complete, which means that their solutions can be quickly adapted to any other NP-complete problem.
So, developing a way to quickly solve one NP-complete problem would make significant strides towards completing all other NP-complete problems.

What are some examples of NP problems? Many researchers focus on one major concern.
The majority of modern cryptography relies on codes that are hard to crack but easy to check. As an example, consider the passwords or PINs to your various accounts.
Checking that they are correct is straightforward, but brute-force guessing every permutation of letters and numbers would take forever.
The encryption behind securing your credit card number when ordering something on Amazon, too, is an example of NP cryptography. If P = NP,
then cracking nearly every kind of encryption would suddenly become much, much easier.

While losing any semblance of internet security would be disastrous, there would be many beneficial consequences if P = NP.
Lance Fortnow, a computer scientist and author of The Golden Ticket: P, NP and the Search for the Impossible,
summed up some of the major consequences in an article for Communications of the ACM:

Transportation of all forms will be scheduled optimally to move people and goods around quicker and cheaper.
Manufacturers can improve their production to increase speed and create less waste. And I'm just scratching the surface.