Is it possible to tile a sphere such that every point where tiles meet touch the same number of edges leading to other points, for an arbitrary number of points?
For example, a cube works for 6 points
I don't think that it is, because of the limited number of regular platonic solids, but I'm unsure
Ideally, this would form a (semi, at least) isometric grid on the sphere
You could perhaps try subdividing edges of polygons that meet that criteria indefinitely. I'm not sure if that would continue to satisfy the every-vertex-has-the-same-number-of-edges condition though.
I awoke with the following puzzle and I would like to investigate but the answer may require some programming (it may not either). I have asked on the meta site and believe the question to be suitable and hopefully interesting for the community.
I will try to explain the puzzle as best I can th...
Hmm... Thoughts on adding more points to a preexisting isometric archimedean solid: Because they are isometric, there will be a set of similar edges that touch every point once and only once between all of them (I think). If we remove this edge, the "same # of edges touching every point" condition still holds - With just 1 less edge for every point. This will in turn create a face with a larger number of edges. If we do this until we get a face with the same number of edges as the number of edges that touch each point + 1, we can place an extra point in every one of those faces, and connect…
But can it be done indefinitely?
And in all conditions?
Hmm...
Here's an example, where orange are the new points, and black are new edges. The square could be divided because it had 4 edges, and 3 edges were previously connected to every point:
Only now it can't be done further, because you can't get rid of an edge between 2 triangles, as It would create an uneven # of edges connected to each point...
Not sure the above is right, I have no clue really. The problem is doing this for a semi-arbitrary number of points
@El'endiaStarman I'll look into that tonight. Meanwhile here is up to (17,0). (13,2) is the record both in terms of distance and steps so far, and I have no clue how long 17,1 as been running for
I can't really do side-by-side stereograms. Even if I manage to lock my eyes on the two separate images, they just refuse to refocus. The onyl stereograms that work for me are those hidden images like orig14.deviantart.net/18f2/f/2017/022/5/4/…
@Zgarb Is a combinator that takes in Church numerals (which have been converted to combinators, and are all put in a single wrapped bracket) and outputs the successor of that Church numeral possible?
I'm guessing the answer is "yes"
BTW, the combinators are made from the normal 6 rules from the T thing on Wikipedia, and n-reduction
@flawr If it doesn't require additional effort, I recommend producing images that work for both cross view and parallel view, so that people can use whichever they find most natural. All it takes is to display the left image again on the right (so you have left view, right view, left view), and then people can choose to look at whichever pair suits them.
@MartinEnder Those hidden images (I think they are called autostereograms) use parallel view i.e. your eyes look to infinity (but focus on the screen), for crossview the rays cross between your eyes and the screen. It does take some practice but it is really fun. I learned both now, but prefer Crossview, as you can view bigger images. I recommend checking out reddit.com/r/CrossView and reddit.com/r/ParallelView
@Zgarb Takes one arg, which is a Church numeral converted to combinators, and outputs the successor (also in Church numeral format converted to combinators)
@MartinEnder The example image you linked to requires parallel view (looking through the image), so if that's what works for you, you might have more luck with flawr's new image, by looking at the 2nd and 3rd columns. I also find your image much easier to perceive as having depth though - I'm not sure if it's to do with having plenty of detailed texture or the fact that the repeating horizontal pattern means each eye isn't left with a "spare" image that doesn't match to anything in the other eye.
I have heard it recommended to look through two tubes for parallel view, so each eye can only see what it is intended to (to remove the ghost images).
Also, I find that for parallel view the image needs to be quite small so that the centres of the images can be closer together than my eyes (so I can line them up without my eyes having to diverge past parallel).
yeah that's the problem I'm having with parallel view side-by-side. I feel like I'd need to turn my eyes "outside" which I can't do. for cross-view I can get both images to line up fairly easily, but then I don't know how to refocus them, so while I can sort of see the 3D effect, it's all blurry.
That's what I dislike about cross view. That eye position only naturally occurs with very close objects, so my eyes try to focus too close. Focusing too far away with parallel view causes much less of a problem, since focusing "at infinity" is close enough for objects that aren't very near
I guess you could use your image as a guide - however far apart the repeating patterns are at the distance you view it comfortably, you need the parallel view images to be spaced similarly to convince your brain that it's 3D rather than just distorted
About Eisenstein primes, I wonder if there are any clusters of more than 6 connected primes (aside from the blob around the origin).
When three primes are connected around a composite number, I can never see another connected prime that isn't adjacent to that composite number (so there's partial rings of 2,3,4,5 primes and the full rings, but I can't see e.g. a 4-prime zigzag).
then again, seeing how rare the prime rings are, it's very likely that these exist but won't be found by hand/eye.
maybe I should write a sieve using small Eisenstein primes to see the resulting pattern. maybe a combination of sieving with a few small primes will already show that no such connected regions can exist.
LCM of a list of Eisenstein primes might be a fun challenge.
I'm not familiar with Eisenstein integers (and staring at the Wikipedia page isn't doing much to change that...). Do they factor uniquely into Eisenstein primes, up to choice of unit?
Ah I see - the image really helps! So even if it is possible to have full hexagons further out, they will never have any adjacent primes not in the hexagon?
@trichoplax not sure if you were following the transcript yesterday, but we established that these full hexagon rings actually exist (and there's a good chance that there's infinitely many of them), but they are very rare
well not very, I guess. but a lot rarer than one might think
@MartinEnder You mean that you haven't seen any instances where three primes form half a ring and a fourth prime is opposite them? 'cause I just found one of those.
@El'endiaStarman no I mean, three primes that form half a ring and then either a fourth one connected to them to form a Y-shape, or to form a zigzag shape
and the sieve image above should prove that those can't exist
@MartinEnder adding (4,1) into that does not bring down the distance further. the period of the grid increases to (32,34) but now there are several hexes from the previous sieve left in each period.
I wonder at what point they get separated by multiples of 14
sorry, in case my last comment seemed random, we found yesterday that the components of all prime rings except the first one are multiples of 14 for whatever reason.
@flawr Took me a while to work out what that means, but I think I see now. Thanks. Is 23 where the uniqueness fails or has it just not been proven unique beyond that?
@MartinEnder I saw that in what you linked to today, so it didn't seem random
I haven't tried proving it, no. Although I have a feeling that if I extend the sieve sufficiently, it can be shown quite easily.
fun observation while collection some small primes: the hexagonal ring around the origin that goes through (8,0) is the first one without any primes (apart from the trivial ones through (1,0) and (0,0))
Eh, I think it's more an indication that Martin was too lazy to ensure that the variants of (2,0) and (2,1) weren't kept (and he didn't bother to take out (0,0) and variants of (1,0).
- Are there prime-free hex rings beyond a = 8? - What's the distribution of prime rings/hexes? - Are all prime rings centred on coordinates that are divisible by 14? - What's the distribution of distances you can reach by jumping between primes with fixed step sizes?
^ some random questions that have cropped up so far
If we use the representation a+b * exp(2 * pi * i/6) of the ironstone integers, is it correct that if a ring has center (a,b), then the points in that ring have coordinates {(a,b+1),(a,b-1),(a-1,b),(a+1,b), (a-1,b+1),(a+1,b-1)}?
Whoops, my statement was wrong. I didn't sieve with (2,0) and (7,0) (although I was going to). That sieve was generated with (2,0), (2,1), (3,1) and its conjugate (2,-1) which I forgot to add in earlier.
Anyway, that ticks one of the questions off the list. :)
(14,-14) is probably a better representative of the period because it minimises the absolute value of both components
Does anyone object to a policy in this room of clearing stars on stupid jokes so we can keep the starboard interesting with diagrams/animations/links to interesting posts or videos?
Well "policy" is probably the wrong word. What I meant is would anyone mind if a mod occasionally cleared some stars on the chatty stuff in favour of more interesting posts with mathematical content.
On another note: I have an algorithm for this challenge (it's a golf, but I got interested in optimizing runtime) that runs in O(k*d(k)) time, where k is the size of the input and d is the divisor function (number of divisors). I'd really like to know if this can be improved.
The idea is that given an m*n matrix with p ones, we know that if it can be divided into grid, the number a of horizontal rows of grid cells divides p and is at most m, and the number of vertical columns of grid cells is then b = p/a. So we loop through those numbers (this is the d(k) factor). Once a has been fixed, we know that between any two vertical lines there are exactly a ones, and between two horizontal lines there are b ones.
Thus we can choose (greedily) any positions for the grid lines that satisfy these conditions. Finally, we check that each grid cell actually contains exactly one 1 (this is the k factor in the runtime).
Somehow it feels wrong to just loop through the entire array as the last step in the inner loop, but I don't know what to precompute to make it faster. It's of course possible this approach is already optimal (it's asymptotically faster than O(k^(1+e)) for any positive e, and no algorithm can be faster than O(k)).
yes, of course. that was still under the assumption that this check wasn't necessary with early exit during the greedy assignment.
@Zgarb the good thing about this is example is that it shows that row and column sums don't add any useful information for the check, because I can construct a valid input with the same sums.
One can find necessary splits in O(k) but that won't help the asymptotic complexity (although I'm sure it would bring down the average runtime considerably for random inputs).
