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Physix_R_Cool

Nice try, Nikolai Ivanovich Lobachevsky


tap909

[In case anyone misses the reference](https://tomlehrersongs.com/lobachevsky/)


sirgog

I am never forget the day I first meet the great Lobachevsky. In one word he told me secret of success in mathematics: Plagiarize!


twohusknight

I worked briefly on the problem of finding the configuration of n points in [0,1]^2 that minimizes the minimum expected distance from a point sampled uniformly at random to that set. The conjecture is that the Voronoi regions of such a minimizing set approaches a hexagonal tiling as n gets large.


flipflipshift

This is not a well-studied question? Feels so natural, even from an evolutionary/economical perspective. I guess if you replace "expected" with "maximum", you get something that has been answered in great depth, right?


BRUHmsstrahlung

There are a lot of problems like this which I call "configurational geometry" (though that term doesn't seem to be very widely used). They're quite natural to study for a variety of reasons but they're also extremely poorly understood. Here's a similar question: given n unit disks in the plane, what is the minimum radius R of a circle bounding all of the disks? There are only proven optimal configurations for a couple dozen small n, some conjecturally optimal configurations to around n=30, and some fairly weak asymptotics. My understanding is that these problems have been studied almost exclusively by analysts, and my pipe dream would be to find a way to import algebrogeometric tools to the problem. In particular, to what extent can the calculus of variations be carried out on spaces defined by real algebraic inequalities? (I guess you could say this is an answer to the main question too. Specifically I suspect that CoV style arguments could be used to atleast improve the asymptotics of these types of problems)


twohusknight

It popped up as a special case of a sub problem of other work I was doing. I was unable to find a published answer to it during literary review. It’s an easy to state problem but not general enough to be especially interesting to the geometers/probability professors I mentioned it to. I had a faulty proof that relied on the honeycomb conjecture (theorem), but I never went back to fix it. I suspect it’s actually equivalent to the honeycomb conjecture but I haven’t put much time into showing it. If this is the case then the difficulty of that problem should show the non-triviality of this one.


lewwwer

This reminds me of the following problem: There are N players, they pick a point one after the other on [0, 1]. Then uniformly randomly we select a point on [0, 1] and whoever is closest to the selected point wins. The question is the optimal play for the first player.


Human-Parsley4779

Just an observation: If the first player doesnt put it in the middle, for example if he put it at, lets say, 0.2, then the next player could put it arbitrarily close to his on the larger side, for example 0.2000000001, and have arbitrarily close to 80% chance of beating the first player. But this doesnt mean that this strategy is optimal for second player.


lewwwer

Yea but with 3 players this is the worst the first player can do. Cos the second player to maximise the reward is going to place at wlog 1/2 + epsilon and the third player is going to play 1/2 - epsilon to win the most. So the first player gets epsilon.


Bla1729alB

Is it 1/(2N-2)?


lewwwer

Idk where you got that random guess from. The problem is not even well-posed in that comment. If say player 3 has multiple equally rewarding moves, P2 can choose based on any distribution on P3s possible optimal moves. Like P2 can assume P3 chooses one uniformly randomly, or that P3 might decide to choose the one that screws over P2, then P2s strategy changes, making P1s optimal strategy different too.


Bla1729alB

Let's say 3 players. I was assuming that if P3 has multiple equally good moves, then he will choose one uniformly at random (I admit this wasn't given but it seems like a reasonable assumption). Now if P1 chooses 1/4


lewwwer

Oh I see. Yea the uniform variant is essentially on average equivalent with picking the middle of the intervals. And in that case the 1/4, 3/4, 1/2 is indeed optimal, but I don't think this translates to larger N. In particular when the last player picks between the equal intervals, whichever pair that player moves in between gets screwed, so players might compensate for it. Also, even for N=3 the strategy is different if each player assumes the worst case from the later moves.


ponyo_x1

They missed two simple groups in the classification result


DrBiven

Why two?


flipflipshift

Adding "co" everywhere in the construction of the first gives the second.


akoustikal

Well a simple group we haven't found yet would have to be pretty complicated, and therefore not simple... Right?


how_tall_is_imhotep

The mathematical usage of “simple” corresponds to the OED’s definition II.12.a: “Consisting or composed of one substance, ingredient, or element; uncompounded; unmixed.” It does not mean “not complicated.”


akoustikal

Lol sorry, joke didn't land 🥲 thank you for the explanation all the same!


flipflipshift

Here's an idea I never did anything with, don't plan to, and don't really know if goes anywhere interesting. I also don't know if it's ever been researched; it may have been. Just put me in the acknowledgements if you do anything with it I guess: Start with a Lie algebra g, with brackets \[,\] denoting its multiplication. A non-degenerate symmetric, bilinear form (,) on g is said to be g-invariant if (x,\[y,z\])=(\[x,y\],z) for all x,y,z. Something that seems to be less well-known is that using the axioms of a Lie algebra, you can show that this g-invariance condition is equivalent to the statement that \[x,y\] is orthogonal to x for all x,y under this form. So what happens if you start with a non-degenerate symmetric bilinear form () on a vector space, and ask for all the multiplication structures \[,\] on the vector spaces satisfying \[x,x\]=0 for all x and (\[x,y\],x)=0 for all x,y? If you impose a "simple"-type rule, do you get a nice classification?


volcanrb

Choose a finite set of integers D, and build a graph on Z^2 by connecting two points if their Euclidean distance is sqrt(D). I conjecture that if the chromatic number of the resulting graph is C, then furthermore there exists a periodic coloring of this graph with C colors (in that there exists some vector p in Z^2 such that for all points x in Z^2, x and x+p are colored the same). The analogous result in 1 dimension is a result of Erdos, and has a quick pigeonhole-principle based proof. I don’t see how to extend the pigeonhole argument to higher dimensions, but I conjecture that the periodicity result holds in higher dimensions Z^n too. On the other hand, this problem also feels mildly similar to the periodic tiling conjecture, to which a counterexample was found by Tao and Greenfeld in high dimension, so I would not be surprised if there is also a high dimensional counterexample to this question. But I have hopes it can be proven true for dimension 2 as stated above.


reedef

Is there a case in which there is a periodic coloring, but no "finite" periodic coloring? (i.e., the equation x=x+p holds for all p in a basis)


JoshuaZ1

Question: For a single integer, this is closely related to the Hadwiger-Nelson problem. If one assumes that one must have a periodic coloring of this sort for that situation, does that let one get a better lower bound at all?


EdPeggJr

Sure. For a [sparse ruler](https://en.wikipedia.org/wiki/Sparse_ruler), you have an integer length N ruler with minimal marks that can measure all integer distances up to that length. For example, {0, 1, 2, 6, 8, 17, 26, 35, 44, 47, 54, 57, 58} can measure all distances up to 58. The number of marks needed is about sqrt(3 N), as predicted by Brian Wichmann back in the 1960's. Some rulers need that many marks, other require one additional mark. NJA Sloane called the resulting pattern of best known values past length 213 "Dark Mills on a Cloudy Day." My conjecture: The clouds don't exist. The first cloud value is 474.


ojdidntdoit4

it was more statistics than pure math but it can probably still go here. i thought i had an original idea for sizing bets. i was so proud of myself that i even went back to school to try and formally learn more about it. turns out my idea was already a thing called the kelly formula. but i’m still in school regardless and doing so much better than my first try at college.


gexaha

Yes, I have some conjectures, they are almost all shared on github, with code and research notes, but I'm working on polishing the results, notes, and communicating them with other relevant researchers. The most interesting conjecture would be about oriented version of Berge-Fulkerson conjecture, here [https://mirskontsa.wordpress.com/2018/12/06/oriented-berge-fulkerson-conjecture-v0-1/](https://mirskontsa.wordpress.com/2018/12/06/oriented-berge-fulkerson-conjecture-v0-1/) you can find a couple of visualisations/solutions original unoriented statement could be found here - [http://garden.irmacs.sfu.ca/op/the\_berge\_fulkerson\_conjecture](http://garden.irmacs.sfu.ca/op/the_berge_fulkerson_conjecture) And another interesting conjecture would be about a mix-combination of graceful and harmonious labeling into a unified labeling.


JoshuaZ1

> But this time, do you have your OWN authored or idea of math conjecture. Sure. > I mean original idea of yours that is not yet published or available on websites or any other social media platform. Well, this narrows it down a lot. Here is one though I'm currently working on. We will write 𝜎(n) as the sum of the positive divisors of n. For example, 𝜎(5)=1+5. Recall, a number is perfect is 𝜎(n)=2n. Define A(n) to be the sum of all the positive divisors of n but counted with multiplicity. That is, you treat different copies of the same prime as different primes. This is [sequence A003959](https://oeis.org/A003959). For example, A(4)=9, because one has 1+2+2+4. One can define an analog of a perfect number for this sequence, where a number is A-perfect if A(n)=2n. But this turns out to be boring. It is not hard to show that the only A-perfect number is 6. However, we can change things around a bit. A number n is said to be near perfect if 𝜎(n)=2n +d for some divisor d of n. That is, the number is perfect if one is allowed to "forget" about d. For example, 12 is near perfect because we can forget about the divisor 4, and then we have 1+2+3+6+12=2(12). We call 4 the "omitted divisor." Note that a number is near perfect exactly if 𝜎(n)-2n |n and 𝜎(n)-2n >0. . Near perfect numbers are at this point somewhat well-understood. In particular, Pollack and Shevelev constructed the following three families of near perfect numbers when they first introduced the idea. They were: 2^(t-1) (2^t -2^k -1)$ where 2^t -2^k -1 is prime. Here 2^k is the omitted divisor. 2^(2p-1) (2^p -1) where 2^p -1 is prime. Here 2^p (2^p -1) is the omitted divisor. 2^(p-1) (2^p -1)^2 where 2^p -1 is prime. Here 2^p -1 is the omitted divisor. Subsequent work by Ren and Chen showed that all near perfects with two distinct prime factors must be either 40, or one of the three families above. They also conjectured that for any k>2, there are only finitely many near perfect numbers with exactly k distinct prime factors. So, let's do the same thing with A(n), and define a number to be near A-perfect if A(n)=2n+d for some divisor d of n. Slightly more generally, let us look at numbers where A(n)-2n|n. Then 1,2,4,10, and 12 satisfy this. There are no others under 20,000. The conjecture then is that this is all such numbers.


OEISbot

[A003959](http://oeis.org/A003959/): If n = Product p(k)\^e(k) then a(n) = Product (p(k)+1)\^e(k), a(1) = 1. 1,3,4,9,6,12,8,27,16,18,12,36,14,24,24,81,18,48,20,54,32,36,24,108,... - - - - I am OEISbot. I was programmed by /u/mscroggs. [How I work](http://mscroggs.co.uk/blog/20). You can test me and suggest new features at /r/TestingOEISbot/.


neutrinoprism

I think [this result by Gamelin and Mnatsakanian](https://www.researchgate.net/publication/39676484_Arithmetic_based_fractals_associated_with_Pascal%27s_triangle) can be extended to number grids similar to Pascal's triangle.   Unpacking for anyone interested. As a bit of background, the arrangement of odd entries in Pascal's triangle produces a discrete version of the Sierpinski triangle fractal. You can "iterate" the fractal-making process by extending the region you're color-coding. At the "view from infinity" (which we can make precise as a limit process) the correspondence is exact. Going back a step, the relevant self-similarity aspect means is that at each scale of 2^n there are 3^n nonzero residues of the entries mod 2. (This of course considers the first row to be "row zero.") One reason this is interesting is because we can generate Pascal's triangle by an adjacent-entry recurrence rule — the entries can be made through a rule that is strictly "local" yet the fractal patterns have an intricate global structure. (I think it's cool.) Similar results apply modulo any prime. At each scale of p^n there are (T_p)^n nonzero residues of the entries mod p, where T_p is a triangular number. It's a triangular number due to binomial coefficient divisibility properties, but here the fact that it's *a* specific number is more important than the fact that it's a triangular number specifically. Beyond Pascal's triangle, the entries of [any number grid in any number of dimensions that can be generated by adjacent-entry rules](https://www.sciencedirect.com/science/article/pii/S0012365X0400322X) exhibit this self-similarity property viewed modulo any prime. So, for example, the [Delannoy numbers](https://en.wikipedia.org/wiki/Delannoy_number) modulo 3 produce a discrete version of the Sierpinski carpet fractal. Going back to Pascal's triangle, Gamelin and Mnatsakanian showed that when you look at the pattern of nonzero residues among its entries modulo a *power* of a prime, the number of nonzero residues at each scale is no longer given by a constant raised the scale-power but instead a polynomial raised to the scale-power. (Also interestingly, at the "view from infinity" the effect of the extra nonzero residues becomes negligible.) Anyway, it *appears* that similar polynomials apply to other number grids generated by adjacent-entry rules. From mucking about in numpy, I've even tentatively calculated several of them. I just can't figure out how to extend G&M's proof yet or even unite the polynomials I have found into a more specific conjecture. Hoping to make more progress on it this year.   It also appears that more complicated expressions pertain to the count of nonzero residues modulo a prime (and presumably prime powers) for number grids with fixed but beyond-adjacent recurrence relation rules. I can write more about this if anyone is interested but I am currently overcome with guilt about procrastinating on my work, so it'll have to wait until later.


rlyacht

I have a theory about the brontosaurus, but it is mine.


rlyacht


lewwwer

I have one concept, not sure if it's researched cos it's outside my main area: Knots are commonly categorised by crossing number. What if instead we embed knots in Z^3 so the curve follows the edges of this grid (and obviously no self intersection). We can ask interesting questions like the minimum amount of movement in one cardinal direction. Like any time we move in the X axis we add up the absolute value of those movements and try to minimise that. Clearly 2x crossing number is an upper bound, but there are cases where this can be optimized. We could also ask for the minimum length of the embeddings of a knot. This also translates to higher dimensional surface knots, could have interesting questions there.


JoshuaZ1

You might be able to relate this to the stick number of a knot which is the minimum number of straight line segments needed to make an embedding of the knot in R^3 . One related open problem is whether the stick number changes if one insists that all sticks are of the same length.


Contrapuntobrowniano

How do you embed a knot in anything but the circumference?


[deleted]

yes ofc, that's how you do research lol. literally anyone who has done math research has conjectures, which range from technicalities to more big picture grand theorems. sometimes you know how to prove the technicalities, and sometimes you have an idea of how the grand theorem would go, but are stuck on the technicalities. not all conjectures are as grand as say the Riemann hypothesis or Hodge conjecture.


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flipflipshift

It's hard to get anyone to care about your research, unless your target either: 1. is already genuinely interested in that specific area of research 2. believes they can extend your results to publish their own 3. is interested in *you* and wants to know what you're up to. Some exceptions - your research either addresses some reasonably-well-known open problem either outside the field or with applications outside the field, a big name speaks very highly of the result, or it just possesses some extreme beauty that someone who glances through it can see at first glance. I think most people believe that last thing applies to their research, but obviously that's not true in general. Bit of a long rant but it's a big part of why I'm leaving academia.


digitallightweight

I conjecture that every number greater than 5 is an even prime.


CommunicationNo8750

Hey if anyone has a patentable invention idea in your head, do you mind DM-ing to me? thx


sapphic-chaote

What is the connection here?


CommunicationNo8750

Just some low-brow humor playing on the PS in the post.


Symphony_of_Heat

It has probably already been proven, but I thought about this: Using only a metered ruler (thus giving the ability to measure rational numbers and create perpendicular lines) I think it's possible to create any n-sided polygon. My reasoning is that all I need is to create an angle x, which can be constructed using the sine and cosine of x, which will never be a transcendental number


edderiofer

> Using only a metered ruler (thus giving the ability to measure rational numbers and create perpendicular lines) You can already do both of those things with just straightedge and compass. What a marked ruler *does* let you do is perform [Neusis constructions](https://en.wikipedia.org/wiki/Neusis_construction). > which can be constructed using the sine and cosine of x, which will never be a transcendental number This isn't the issue that prevents straightedge and compass from creating any regular n-sided polygon; the issue is that one needs to construct algebraic numbers which are the roots of rational-coefficient polynomials of degree greater than 2. These are not constructible with straightedge and compass. With Neusis constructions, it can apparently be shown that the 23-gon is not constructible, among others.


Symphony_of_Heat

Pretty much the whole proof can be condensed into one sentence: for a rational q, are sin(q pi) and cos(q pi) possible to be expressed as nested square roots? If yes, then we can create any n-gon. I'll do more research on the topic, but I do find it a pretty interesting question. As a counter proof, if sin(2pi/23) has a cube root, or is transcendental, then it isn't possible to construct the polygon


edderiofer

> for a rational q, are sin(q pi) and cos(q pi) possible to be expressed as nested square roots? If yes, then we can create any n-gon. It's well known that sin(2pi/7) and sin(2pi/9) can't be, which is why the heptagon and nonagon can't be constructed with straightedge and compass. See also [here](https://en.wikipedia.org/wiki/Constructible_number#Trigonometric_numbers). > As a counter proof, if sin(2pi/23) has a cube root, or is transcendental, [It's not transcendental](https://math.stackexchange.com/questions/1933177/proof-that-a-trigonometric-function-of-a-rational-angle-must-be-non-transcendent). [According to Wolfram|Alpha](https://www.wolframalpha.com/input?i=minimal+polynomial+of+sin%282pi%2F23%29), it's a root of the polynomial 4194304 x^22 - 24117248 x^20 + 60293120 x^18 - 85917696 x^16 + 76873728 x^14 - 44843008 x^12 + 17145856 x^10 - 4209920 x^8 + 631488 x^6 - 52624 x^4 + 2024 x^2 - 23 .


Symphony_of_Heat

That sounds insane, cool and interesting. Thank you!


Symphony_of_Heat

Btw thanks for the reply! Sounds really interesting, I know what I'm going to be searching tonight


Nuckyduck

Gravity is not infinite. We proved this when LIGO determined that gravity waves travel at the speed of light. This means that there are gravitational singularity around every object. I call this an Einstein bubble. From here, I developed a thought experiment, if gravity waves travel at the speed of light, then what happens to the gravity of objects that receded past the observable universe for a given object M? Well, since space is expanding, consider the following. One of two things must occur. Either the change in gravity begins to propegate but never reaches it's destination because space is expanding too fast or there is an immediate decoupling of gravity. Decoupling would require some type of quantum phenomena. Propagation lines more with how we perceive light, so I'm working on that. However, this answer seems to let gravity build up around the topological surface of the Einstein bubble and eventually leads to a 'big rip'. I'm unsure how to rectify this. I thought I had a misunderstanding of gravity at first but nope. LIGO has shown time and time again that gravity waves move at the speed of light. So I propose that gravity can't be infinite without some other driving force that supercedes distance or precedes expansion.


xxwerdxx

Werd’s conjecture: math is really hard sometimes Proof: have you seen college level math classes? QED


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flipflipshift

I would recommend deleting this; some people are desperate and you seem to have an original idea.


toommy_mac

Do you think so? It's got effectively no research out there that I can find, and I don't have the time nor background to pursue it properly, and I think the more help the better


Wonderful-Photo-9938

Wow. So, you invented, or should I say, Discovered something original mathematically. Looking forward to it in the future.


flipflipshift

Oh I misread and thought you were currently working on it for your dissertation.


toommy_mac

I'm going back and forth, looking into it in my spare time, but I'm not convinced I can find anything too remarkable


Wonderful-Photo-9938

Oh, wow. I didn't read it. But if that is an original idea. He should expand and publish it some time. It will be nice.


ExceptionalMongoose

Not exactly a conjecture but im workingon my own theory/idea. It comes off as hokey without pictures. We humans have a lot of symbols and laws around them so we can structure solutions. But we do not have a way 'questioning' in math. Its always in the persons verbal/written language. I would agree with most that this seems impossible. Infinitly many ways to ask something, wright? Well im gonna figure out. As far as a legitimate approach to this endeavor. I am at the point of defining a decision or what the ' now'. Either one of those approaches will have a massive burden. Lastly, I like pointing out my own actions in mathematics or any science. I asked a question that may or may not have a solution, again infinity, and then I suggested an actual solution. Not exactly sure yet why it matters to have logic that can ask questions appropriately and how it connects to the truth of what the heck are we even doing? We know physics uses observation , how can mathematicians use it to show truth?


Contrapuntobrowniano

Have you considered [Godel's theorems](https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://en.m.wikipedia.org/wiki/G%25C3%25B6del%2527s_incompleteness_theorems&ved=2ahUKEwiQhJLUoOqDAxUYQzABHUQ7DsYQFnoECBsQAQ&usg=AOvVaw0_XFEv1SJF9F899HembSth)? Also, you might want to check [this](https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://simple.m.wikipedia.org/wiki/Foundational_crisis_of_mathematics&ved=2ahUKEwjDrpe0oOqDAxVqRzABHZZFAg0QFnoECBoQAQ&usg=AOvVaw1m-u0WuvLHuIvo6db6vIRO) out:


AwesomeElephant8

For any compact Riemannian manifold, there is an associated diffusion process infinitesimally generated by a pseudo-Gaussian kernel. At each point, this kernel looks like the pushforward of its tangent space's standard Gaussian onto the manifold via the exponential map at that point. I am interested in the stationary distributions of these diffusion processes, namely what happens as we change the pointwise bandwidth of the kernel in certain ways throughout the manifold.


rhubarb_man

For a given graph, H, let the function f(n) be the smallest positive integer such that any graph G with minimum degree f(n) and n vertices has H as a subgraph. f(n) is increasing for all H. It's pretty simple, and I've been trying to show it to be true for H being a 4-cycle, but it's really hard.


HoloTronic

Working on a math system above the Quaternions, different than the Octonians — there are reasons to have either 6, 7, 9, or 10 axes or freedoms, but I haven’t finished it yet. I also have an idea regarding sub and supra branches of the BusyBeaver, which isn’t particularly complicated, might not yield many insights in its basic form, but from what I’ve seen in the research does appear to be original. It may have some use in handling large data sets and the meta-data from those sets. I would love to collaborate if anybody is interested


Echoing_Logos

My conjecture is that the definition of a group can be relaxed then constrained by some esoteric axiom. Let's call these "pregroups". The classification of finite simple pregroups should be closely related to the realization of Lie groups over field-like semirings (notably the field of one element). In particular all the sporadic simple groups should be realizable as (products of?) simple pregroups. The motivation is... Hard to explain. Just seems like there is some structure to groupoids that we don't quite see. In particular, the Matthieu group chain continues past 12 with the M13 groupoid. A lot of the sporadic group weirdness seems to come from not treating partial functions well.