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Thread for Non-CA Academic Questions

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Re: Thread for Non-CA Academic Questions

Postby Macbi » January 23rd, 2019, 5:27 pm

muzik wrote:Is this identical to a sphere?

Yes.
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Re: Thread for Non-CA Academic Questions

Postby muzik » January 23rd, 2019, 5:46 pm

...and is this identical to anything? Is there anywhere I can find a list of such 2-manifolds?

x = 119, y = 113, rule = B3/S23
49bo$50b2o$52bo$53b2o$55bo$56b2o$58b2o$59b2o$60b2o$8b97o$8bo53bo41bo$
8bo52bo42bo$8bo51bo43bo$8bo49b2o44bo$8bo48bo46bo$8bo47bo47bo$8bo45b2o
48bo$8bo43b2o50bo$8bo42bo13b3o36bo$8bo41bo14bobo36bo$8bo40bo15b3o36bo$
8bo40bo15bobo36bo$8bo56bobo36bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95b
o$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo
95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo11b2o82bo13bo$8b3o9bob
o81bo12bo$8b3o9b2o82bo12bo$7b2o2bo8bobo69bo11bo11bo$6bobo2bo8b2o71bo
10bo11bo$5bo2bo3bo80bo10bo10bo$4bo3bo3bo81b2o8bo9bo$3bo4bo4bo82bo7bo8b
o$2bo5bo5b2o80bo7bo7bo$bo6bo7bo70b3o7bo6bo6bo$o7bo78bobo8bo5bo5bo$8bo
78b3o9bo4bo4bo$8b4o75bobo10bo3bo4bo$7b2o2bo75bobo11bo2bo3bo$6bobo3bo
89b3o2bo$5bo2bo4b2o88b2obo$4bo3bo6bo88b2o$4bo3bo95bo$3bo4bo95bo$3bo4bo
95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$
8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95b
o$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo95bo$8bo45b2o48bo
$8bo45bobo47bo$8bo45b2o48bo$8bo45bobo47bo$8bo45b2o48bo$8bo95bo$8bo95bo
$8bo95bo$8bo95bo$8bo95bo$8bo47bo47bo$8bo45b2o48bo$8bo45bo49bo$8bo44bo
9bo40bo$8bo43bo7b4o40bo$8bo41b2o7b2o43bo$8bo40bo8bo45bo$8bo40bo6b2o46b
o$8bo38b2o7bo47bo$8b97o$48bo6b2o$49bo6bo$50b2o5bo$52bo5b2o$53bo6bo$54b
2o$56b2o!
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Re: Thread for Non-CA Academic Questions

Postby Macbi » January 23rd, 2019, 6:04 pm

That's a Klein bottle. You can see by cutting it and rearranging it like this:
x = 325, y = 120, rule = B3/S23
49bo219bo$50b2o218b2o$52bo219bo$53b2o218b2o$55bo219bo$56b2o218b2o$58b
2o218b2o$59b2o218b2o$60b2o218b2o$8b97o123b97o$8bo53bo40b2o123bo53bo40b
2o$8bo52bo40b3o122b2o52bo40bo$8bo51bo40b2obo121b3o51bo40b2o$8bo49b2o
41bo2bo121bobo49b2o41bo$8bo48bo42bo3bo120b2obo48bo42bo$8bo47bo43bo3bo
119b2o2bo47bo43bo$8bo45b2o43bo4bo119bo3bo45b2o43bo$8bo43b2o44bo5bo118b
2o3bo43b2o44bo$8bo42bo13b3o28b2o6bo118bo4bo42bo13b3o28b2o$8bo41bo14bob
o27b2o7bo116b2o5bo41bo14bobo27b2o$8bo40bo15b3o27bo8bo115b2o6bo40bo15b
3o27bo$8bo40bo15bobo26bo9bo115bo7bo40bo15bobo26bo$8bo56bobo25bo10bo
115bo7bo56bobo25bo$8bo82b2o11bo114bo8bo82b2o$8bo80b2o13bo114bo8bo80b2o
$8bo80bo14bo113bo9bo80bo$8bo79bo15bo112bo10bo79bo$8bo78b2o15bo111bo11b
o78b2o$8bo77bo17bo110bo12bo77bo$8bo76bo18bo108b2o13bo76bo$8bo75b2o18bo
107b2o14bo75b2o$8bo74bo20bo107bo15bo74bo$8bo73bo21bo106bo16bo73bo$8bo
73bo21bo104b2o17bo73bo$8bo72bo22bo103b2o18bo72bo$8bo72bo22bo101b2o20bo
72bo$8bo70b2o23bo99b2o22bo70b2o$8bo69bo25bo98bo24bo69bo$8bo68bo26bo96b
2o25bo68bo$8bo67bo27bo95b2o26bo67bo$8bo65b2o28bo93b2o28bo65b2o$8bo63b
2o30bo92b2o29bo63b2o$8bo11b2o30b3o16b2o31bo13bo78bo30bo11b2o30b3o16b2o
$8b3o9bobo29bo18bo32bo12bo78bo31b3o9bobo29bo18bo$8b3o9b2o29bo17b2o33bo
12bo78bo31b3o9b2o29bo17b2o$7b2o2bo8bobo27bo17bo23bo11bo11bo78bo31b2o2b
o8bobo27bo17bo$6bobo2bo8b2o28bo16b2o24bo10bo11bo77bo31bobo2bo8b2o28bo
16b2o$5bo2bo3bo37bo15b2o25bo10bo10bo77bo31bo2bo3bo37bo15b2o$4bo3bo3bo
37b5o10b2o27b2o8bo9bo76b2o31bo2b2o3bo37b5o10b2o$3bo4bo4bo50b2o30bo7bo
8bo72bo3bo32bo3b3o3bo50b2o$2bo5bo5b2o48bo31bo7bo7bo73bo2bo32bo2b2obobo
3b2o48bo$bo6bo7bo46bo23b3o7bo6bo6bo74bobo32bo2bo3bo2bo4bo46bo$o7bo45b
11o22bobo8bo5bo5bo69bo5b2o32bo3bo3bo2bo42b11o$8bo51bo3bo22b3o9bo4bo4bo
70bo5b2o35bo4bo3bo47bo3bo$8b4o47bo4bo22bobo10bo3bo4bo70bo5b2o34bo5b4ob
o45bo4bo$7b2o2bo47bo4bo22bobo11bo2bo3bo66bo4bo3b2o2b5o27b2o5b2o2bo2bo
44bo4bo$6bobo3bo43bobo5bo37b3o2bo67bo4bob2o36bo5bob2o2bobo41bobo5bo$5b
o2bo4b2o36b7o6bo38b2obo68bo3b3o37bo5b6o2b3o35b7o6bo$4bo3bo6bo40b3o5b2o
38b2o69bo4b2o3b6o33bobobo2bo3bo40b3o5b2o$4bo3bo46bo2bo6bo38bo70bo3bob
4o39b2o2bo3bo42bo2bo6bo$3bo4bo46bo2bo6bo38bo70bob2o44b2o3bo3bo42bo2bo
6bo$3bo4bo45bo3bo5bo39bo70b2o46b2o3bo4bo40bo3bo5bo$8bo45bo2bo46bo70bo
47bo4bo45bo2bo$8bo39b6o3bo46bo70bo47bo4bo39b6o3bo$8bo42b3o3bo46bo69b2o
46b2o4bo42b3o3bo$8bo41bo3bo2bo46bo67b2ob7o2bo5b2o36bo41bo3bo2bo$8bo40b
o4bo2bo46bo66b2o8b3o5bo38bo40bo4bo2bo$8bo39bo5bo2bo46bo65b2o17bo38bo
39bo5bo2bo$8bo38bo6bo49bo64b2o18bo38bo38bo6bo$8bo38bo6bo49bo63b2o20b2o
36bo38bo6bo$8bo37bo57bo62bo60bo37bo$8bo36bo58bo62bo60bo36bo$8bo35bo59b
o60b2o61bo35bo$8bo33b2o60bo59b2o62bo33b2o$8bo31b2o62bo59bo63bo31b2o$8b
o30b2o63bo58bo64bo30b2o$8bo30bo64bo58bo64bo30bo$8bo29bo65bo57bo65bo29b
o$8bo28b2o65bo56bo66bo28b2o$8bo28bo66bo55bo67bo28bo$8bo27bo67bo54bo68b
o27bo$8bo26bo68bo54bo68bo26bo$8bo26bo68bo52b2o69bo26bo$8bo25bo69bo50b
2o71bo25bo$8bo25bo69bo49bo73bo25bo$8bo24bo70bo48bo74bo24bo$8bo23b2o20b
2o48bo47b2o74bo23b2o$8bo23bo21bobo47bo46bo76bo23bo$8bo22bo22b2o48bo45b
o23b3o51bo22bo$8bo20b2o23bobo47bo44b2o23bobo51bo20b2o$8bo19b2o24b2o48b
o42b2o25b3o51bo19b2o$8bo19bo75bo42bo26bobo51bo19bo$8bo18bo76bo41bo27bo
bo51bo18bo$8bo17bo77bo41bo21bo59bo17bo$8bo16bo78bo40bo23b2o57bo16bo$8b
o15bo79bo39bo26bo56bo15bo$8bo13b2o32bo47bo37b2o27bo56bo13b2o$8bo10b3o
32b2o48bo36b2o29b2o54bo10b3o$8bo9b2o34bo49bo36bo32bo53bo9b2o$8bo9bo34b
o9bo40bo35bo34bo52bo9bo$8bo7b2o34bo7b4o40bo34bo36bo51bo7b2o$8bo5b3o33b
2o7b2o43bo32b2o38bo50bo5b3o$8bo4b2o34bo8bo45bo30b2o41bo49bo4b2o$8bo2b
3o35bo6b2o46bo29b2o43bo48bo2b3o$8bob2o35b2o7bo47bo29bo44b2o47bob2o$8b
97o27b100o$48bo6b2o124bo$49bo6bo123bo$50b2o5bo121bo$52bo5b2o118bo$53bo
6bo115b2o$54b2o119bo$56b2o116bo$173bo$172bo$171bo$170bo$168b2o$166b2o$
165bo!
For a classification of surfaces, see here: https://en.wikipedia.org/wiki/Surface_(topology)#Classification_of_closed_surfaces
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Re: Thread for Non-CA Academic Questions

Postby muzik » January 23rd, 2019, 6:10 pm

Probably getting a bit ridiculous here. This would result in something like a cone, which would end up being a hemisphere, which can be flattened out into a circle, which is a plane, right?

x = 111, y = 106, rule = B3/S23
41bo$42b2o$44bo$45b2o$47bo$48b2o$50b2o$51b2o$52b2o$97o$o53bo41bo$o52bo
42bo$o51bo43bo$o49b2o44bo$o48bo46bo$o47bo47bo$o45b2o48bo$o43b2o50bo$o
42bo52bo$o41bo53bo$o40bo54bo$o40bo54bo$o95bo$o48b3o44bo$o48bobo44bo$o
48b3o44bo$o48bobo44bo$o48bobo44bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$
o95bo$o95bo$o95b2o$o95b3o$o94b2obo$o93b3o2bo$o93bobo3bo$o92b2obo3bo$o
92bo2bo4bo$o91bo3bo4b2o$o91bo3bo5bo$o90bo4bo5b2o$o90bo4bo6bo$o89bo5bo
7bo$o79b3o7bo5bo7bo$o79bobo6bo6bo7b2o$o79b3o6bo6bo8bo$o79bobo5bo7bo8bo
$o79bobo4bo8bo9bo$o86bo8bo10bo$o86bo8bo10bo$o85bo9bo10b2o$o85bo9bo11bo
$o84bo10bo11bo$o95bo11b2o$o95bo12b2o$o95bo$o95bo$o95bo$o95bo$o95bo$o
95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo
$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o
95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo$o95bo
$o95bo$o95bo$o95bo$o95bo$o95bo$97o!
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Re: Thread for Non-CA Academic Questions

Postby Macbi » January 23rd, 2019, 6:33 pm

Yeah.
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Re: Thread for Non-CA Academic Questions

Postby pcallahan » January 24th, 2019, 8:29 pm

I noticed this thread and I hope that at least through affinity, somebody here might be able to help me. Is there a name for the kind of oriented rhombus tiling shown in the picture below?
49726547_2255109781479246_6600225566891180032_o (2).jpg
49726547_2255109781479246_6600225566891180032_o (2).jpg (30.83 KiB) Viewed 1242 times


To start with, it's just a rhombus with 60° acute corners, composed of two equilateral triangles. This was known to the ancient Romans among others who used it in mosaics that give the illusion of 3D cubes, and it was used in the same way much later in the video game Q*bert. Irregular rhombus tilings are analogous to domino packings on a rectangular grid and, like them, reducible to bipartite matching (each covers one upsidedown and one rightsideup triangle). So this is not my question. (I am also not asking about Penrose tilings; these rhombuses are the 60° kind.)

But what interested me was to replace the flat sides with alternating convex and concave arcs that match up. In this case, they are circular arcs with a radius of curvature equal to the triangle side, but they could be any shape as long as they match up, such as jigsaw nubs and voids.

Adding curvature, we get a "z" orientation, shown here in a dark color, and an "s" orientation, shown here in a light color. You can make one of these out of two-sided material, so that you flip it over to change the color. In fact, I made some with paper and cardboard sides:
49899078_2256694581320766_2710552348677111808_o.jpg
49899078_2256694581320766_2710552348677111808_o.jpg (25.07 KiB) Viewed 1242 times


These kinds of tilings are pretty easy to analyze, but it is fun, at least for small tilings to see how the orientations impose a coloring. Colors need to invert back and forth to go in a straight line but stay the same color around a turn.

(In fact, this reminds me of something I found very confusing (years back) about Madachy's method for constructing flexagons, which have a similar alternating rule when making the paper strips of triangles. Is there a name for this property of flipping sign/color going in a straight line?)

The shapes themselves lend themselves to decorative tiling. Surely I cannot be the first to have played around with this kind of rhombus tiling. Any leads?
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Re: Thread for Non-CA Academic Questions

Postby Hunting » February 21st, 2019, 5:21 am

Which kind of curves(not functions) has the following property:
All tangent lines intersects with the curves only once?

An example is circle.

Is it possible to have one or multiple(but finite number of) formulas that can describe all of them?
Plz correct my grammar mistakes. I'm still studying English.

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Re: Thread for Non-CA Academic Questions

Postby Apple Bottom » February 21st, 2019, 6:31 am

Hunting wrote:Which kind of curves(not functions) has the following property:
All tangent lines intersects with the curves only once?

An example is circle.

Is it possible to have one or multiple(but finite number of) formulas that can describe all of them?


My gut feeling is that those curves are precisely the boundaries of convex sets.

My gut feeling is also that you can't describe them all using a finite set of formulae (even allowing for parametrization[1]), for you can take any concave, monotonically increasing function on an interval [a, b], and mirror it four-fold (think D4 symmetry) to yield such a curve (and set). Since there's more than 2^{\aleph_0} concave, monotonically increasing functions (proof left as an exercise for the reader), these cannot be described by a finite number of formulae, and as such neither can the curves you're interested in.

But I'm no mathematician, and no doubt what I wrote above is all wrong, for reasons that will be pointed out by the more competent denizens of this forum at most two minutes after I post this. ;)
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Re: Thread for Non-CA Academic Questions

Postby Macbi » February 21st, 2019, 7:00 am

I also thought that the answer would be the boundaries of convex sets, but I think the question is impossible to answer without a more precise definition of "tangent line". If the definition was "line with the same derivative at a point where the curve is differentiable" then something like the Weierstrass function would work. It's not differentiable anywhere, so it has no tangents, so all of its tangents intersect it at one point!

My gut feeling is also that you can't describe them all using a finite set of formulae (even allowing for parametrization[1]), for you can take any concave, monotonically increasing function on an interval [a, b], and mirror it four-fold (think D4 symmetry) to yield such a curve (and set). Since there's more than 2^{\aleph_0} concave, monotonically increasing functions (proof left as an exercise for the reader), these cannot be described by a finite number of formulae, and as such neither can the curves you're interested in.
This part is definitely correct. Even if we limit ourselves to D8 symmetric infinitely differentiable convex curves there will still be 2^ℕ of them, which is Too Many.
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Re: Thread for Non-CA Academic Questions

Postby Hunting » February 21st, 2019, 9:08 am

Apple Bottom wrote:
Hunting wrote:Which kind of curves(not functions) has the following property:
All tangent lines intersects with the curves only once?

An example is circle.

Is it possible to have one or multiple(but finite number of) formulas that can describe all of them?


My gut feeling is that those curves are precisely the boundaries of convex sets.

My gut feeling is also that you can't describe them all using a finite set of formulae (even allowing for parametrization[1]), for you can take any concave, monotonically increasing function on an interval [a, b], and mirror it four-fold (think D4 symmetry) to yield such a curve (and set). Since there's more than 2^{\aleph_0} concave, monotonically increasing functions (proof left as an exercise for the reader), these cannot be described by a finite number of formulae, and as such neither can the curves you're interested in.


Thanks!
Plz correct my grammar mistakes. I'm still studying English.

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* Rule Y Orthogonoid

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x = 4, y = 6, rule = B2e3i4at/S1c23cijn4a
o2bo$4o3$4o$o2bo!
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Re: Thread for Non-CA Academic Questions

Postby BlinkerSpawn » February 21st, 2019, 6:04 pm

Hunting wrote:Which kind of curves(not functions) has the following property:
All tangent lines intersects with the curves only once?

An example is circle.

Is it possible to have one or multiple(but finite number of) formulas that can describe all of them?

This set contains the set of closed convex curves as a subset.
If we require the curves to be closed, this is the entire set, since the tangent line will sweep out the entire plane during a circumnavigation of the curve.
Otherwise, we can also use sets of open segments of convex curves, arranged such that no piece touches any sector swept out by the tangent of any other (for example, two halves of a circle separated by one unit perpendicular to the slice).

P.S. I'm fairly sure the set of convex closed curves has cardinality on the order of \aleph_1^{\aleph_1}, since unrolling the curve gives a segment which can be bijected onto the positive reals and each point on the curve can have any(?) positive real curvature.
P.P.S. This doesn't count curves with cusps, but I don't think adding the possibility of jump discontinuities in the first derivative changes the final answer.
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Re: Thread for Non-CA Academic Questions

Postby Hunting » February 22nd, 2019, 12:43 am

BlinkerSpawn wrote:
Hunting wrote:Which kind of curves(not functions) has the following property:
All tangent lines intersects with the curves only once?

An example is circle.

Is it possible to have one or multiple(but finite number of) formulas that can describe all of them?

This set contains the set of closed convex curves as a subset.
If we require the curves to be closed, this is the entire set, since the tangent line will sweep out the entire plane during a circumnavigation of the curve.
Otherwise, we can also use sets of open segments of convex curves, arranged such that no piece touches any sector swept out by the tangent of any other (for example, two halves of a circle separated by one unit perpendicular to the slice).

P.S. I'm fairly sure the set of convex closed curves has cardinality on the order of \aleph_1^{\aleph_1}, since unrolling the curve gives a segment which can be bijected onto the positive reals and each point on the curve can have any(?) positive real curvature.
P.P.S. This doesn't count curves with cusps, but I don't think adding the possibility of jump discontinuities in the first derivative changes the final answer.


Alright. Another example(I forgot to mention) is any exponential curve which is not closed.

Is there a proof for the "closed convex curves as a subset"?
Plz correct my grammar mistakes. I'm still studying English.

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* Rule Y Orthogonoid

Favorite gun ever:
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x = 4, y = 6, rule = B2e3i4at/S1c23cijn4a
o2bo$4o3$4o$o2bo!
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Re: Thread for Non-CA Academic Questions

Postby Macbi » February 22nd, 2019, 3:24 am

BlinkerSpawn wrote:P.S. I'm fairly sure the set of convex closed curves has cardinality on the order of \aleph_1^{\aleph_1}, since unrolling the curve gives a segment which can be bijected onto the positive reals and each point on the curve can have any(?) positive real curvature.
P.P.S. This doesn't count curves with cusps, but I don't think adding the possibility of jump discontinuities in the first derivative changes the final answer.

It's not quite that large. Since it's continuous, you can specify a curve (a map ℝ ⟶ ℝ^2, or from some interval in ℝ) by specifying where it sends the rationals. So it's only size \aleph_1^{\aleph_0}, which is the same as 2^{\aleph_0}.
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Re: Thread for Non-CA Academic Questions

Postby BlinkerSpawn » February 22nd, 2019, 9:13 am

Hunting wrote:
BlinkerSpawn wrote:This set contains the set of closed convex curves as a subset.
If we require the curves to be closed, this is the entire set, since the tangent line will sweep out the entire plane during a circumnavigation of the curve.
Otherwise, we can also use sets of open segments of convex curves, arranged such that no piece touches any sector swept out by the tangent of any other (for example, two halves of a circle separated by one unit perpendicular to the slice).
[...]


Alright. Another example(I forgot to mention) is any exponential curve which is not closed.

Is there a proof for the "closed convex curves as a subset"?

Follows from the definition of "convex".
And the exponential is another interesting case; an open convex curve of any (finite or infinite) length works.
It should be straightforward to show that infinite open curves can only exist in isolation.
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Re: Thread for Non-CA Academic Questions

Postby Moosey » February 26th, 2019, 2:06 pm

X-post from random posts:
Moosey wrote:1
11
21
1211
111221
312211
13112221
What rate does the sequence grow at?
It seems erratic but around exponential.
(The code is n X’s in a row is nX. Thus 333 becomes 33. 22 becomes 22.)

If you haven’t heard of it before, I definitely did not make up the sequence.

EDIT:
See the answer here:
http://www.conwaylife.com/forums/viewtopic.php?f=12&t=2703&start=1925#p71827
My rules:
They can be found here

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Re: Thread for Non-CA Academic Questions

Postby muzik » March 20th, 2019, 6:38 am

Can E6, E7 and E8 polytopes be stellated, and, if so, have there been any attempts so far to enumerate these?
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Re: Thread for Non-CA Academic Questions

Postby danny » March 23rd, 2019, 4:14 pm

Can every number besides powers of 2 be expressed as a sum of at least two consecutive integers? I know all odds besides 1 can be, but here's a Lua program that checks the evens
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Re: Thread for Non-CA Academic Questions

Postby Moosey » March 23rd, 2019, 4:35 pm

danny wrote:Can every number besides powers of 2 be expressed as a sum of at least two consecutive integers? I know all odds besides 1 can be, but here's a Lua program that checks the evens

0 is an integer, so one is also an odd that can be expressed as the sum of two consecutive integers.
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Re: Thread for Non-CA Academic Questions

Postby toroidalet » March 23rd, 2019, 4:42 pm

danny wrote:Can every number besides powers of 2 be expressed as a sum of at least two consecutive integers? I know all odds besides 1 can be, but here's a Lua program that checks the evens

If negative numbers are allowed, any positive integer n can be written as (1-n)+(2-n)+...0+1+2+...(n-1)+n.
x = 4, y = 2, rule = B3/S23
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Re: Thread for Non-CA Academic Questions

Postby Macbi » March 23rd, 2019, 4:44 pm

danny wrote:Can every number besides powers of 2 be expressed as a sum of at least two consecutive integers? I know all odds besides 1 can be, but here's a Lua program that checks the evens
Yep.

Suppose k = n+1 + n+2 + ... + m. Then since n+1 + n+2 + ... + m = m(m+1)/2 - n(n+1)/2 = (m^2 - n^2 +m - n)/2 = (m - n)(m + n + 1)/2 we have 2k = (m - n)(m + n + 1). So if we pick a factorisation PQ of 2k then we can find m and n by solving the simultaneous equations m - n = P and m + n + 1 = Q. These have integer solutions so long as P and Q have different parities. So we need to write 2k as a product of an odd number and an even number. We can do this so long as k has an odd factor, so we'll succeed so long as k isn't a power of 2. (Of course any k divides by 1 and -1, but if you substitute these in you find the solution where k is the sum of one consecutive number, namely itself, and the solution where k = -(k-1) + ... + (k-1) + k.)
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