« first day (227 days earlier)      last day (997 days later) » 

flawr
flawr
08:25
@El'endiaStarman It clearly depends on the ring you're supposed to be using :D
But +1 for elliptic curves :D
 
6 hours later…
Nathan Merrill
Nathan Merrill
14:17
kickstarter.com/projects/1812249267/…
@MartinEnder you probably got the email about this too, but just in case: ^
Martin Ender
Martin Ender
14:34
Yeah I did, but I was more interested in the implication that they're going to announce their next game in a month or so :P
 
1 hour later…
Nathan Merrill
Nathan Merrill
15:41
@MartinEnder woah, how did I miss that sentence?
 
3 hours later…
PhiNotPi
PhiNotPi
18:36
Math problem/puzzle: consider triangles on a lattice (each vertex of the triangle is on a grid point). Find a triangle such that the difference between the longest side length and the shortest side length is as small as possible.
El'endia Starman
El'endia Starman
In other words, as equilateral as possible?
PhiNotPi
PhiNotPi
The best I've done by hand is (0,0) (4,1) (1,4) which has side lengths sqrt(17), sqrt(18), sqrt(17).
@El'endiaStarman yes, though not exactly in that a triangle can be more equilateral (angle-wise) but end up having to be much larger (due to the lattice), thus have a greater side-length difference.
El'endia Starman
El'endia Starman
Hmm. Looks like $\lim_{n \to \infty} \sqrt{n+1} - \sqrt{n} = 0$, which would mean that there's no maximum size to the triangle.
PhiNotPi
PhiNotPi
That doesn't mean a triangle of side lengths sqrt(n) and sqrt(n+1) exists.
El'endia Starman
El'endia Starman
Well, the unspoken assumption here is that one can always find a larger triangle with side lengths sqrt(n), sqrt(n+1), and sqrt(n), like you did. Given the increasing numbers of Pythagorean triples as the distance from the origin increases, I think that's plausible.
El'endia Starman
El'endia Starman
19:14
>>> foo = lambda n: abs(n - round(2*(round(math.sqrt(n)*math.cos(math.pi/12)) - round(math.sqrt(n)*math.sin(math.pi/12)))**2)) == 1
>>> foo(17)
True
>>> foo(18)
False
>>> filter(foo, range(20))
[1, 7, 9, 17, 19]
>>> filter(foo, range(200))
[1, 7, 9, 17, 19, 31, 49, 51, 71, 73, 97, 99, 127, 129, 161, 163, 199]
Ah, scratch that, the function isn't quite right. Gimme a minute. 
>>> C = lambda n: [round(math.sqrt(n) * math.cos(math.pi/12)), round(math.sqrt(n) * math.sin(math.pi/12))]
>>> C(17)
[4.0, 1.0]
>>> foo = lambda n: abs(C(n)[0]**2 + C(n)[1]**2 - 2*(C(n)[0] - C(n)[1])**2) == 1
>>> foo(17)
True
>>> filter(foo, range(200))
[1, 2, 14, 15, 16, 17, 18, 19, 20, 21]
>>> C(21)
[4.0, 1.0]
>>> filter(foo, range(2000))
[1, 2, 14, 15, 16, 17, 18, 19, 20, 21, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252, 253, 254, 255, 256, 257]
There, found another one, @PhiNotPi: (0, 0), (15, 4), (4, 15) with side lengths of sqrt(241), sqrt(242), and sqrt(241), where the difference between max and min lengths is ~0.032, which is < ~0.120.
Now, I think you're right in that these are quite rare, but I don't think they're bounded.
Certainly a nice puzzle nonetheless!
orlp
orlp
20:08
@El'endiaStarman they're not
El'endia Starman
El'endia Starman
@orlp Not rare or not bounded?
orlp
orlp
optimal solutions are found by iterating a(0) = 0, a(1) = 1, a(n) = 4*a(n-1)-a(n-2)
flawr
flawr
Which oeis entry? :D
orlp
orlp
giving [1, 4, 15, 56, 209, 780, 2911, 10864, 40545, 151316]
oeis.org/A001353
El'endia Starman
El'endia Starman
Ah, very nice.
orlp
orlp
20:10
so you can get arbitrarily close
and this is just the triangles of the form (0, 0), (a, b), (b, a), you could probably converge even faster than that
I found the sequence by solving a^2 + b^2 = 2(a-b)^2 - 1
which gives b = 2a - sqrt(3a^2 + 1)
and that sequence is precisely the integers for which 3a^2 + 1 is square
@PhiNotPi so I think that conclusively answers your puzzle :P
using a(n) = ((2+sqrt(3))^n - (2-sqrt(3))^n)/(2*sqrt(3)) you can get a closed form as well
 
2 hours later…
Nathan Merrill
Nathan Merrill
21:53
bah, I missed the puzzle

« first day (227 days earlier)      last day (997 days later) »