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RSS Feed for Wolfram Community showing any discussions in tag Algebra sorted by activeRandomness of the number C[10]!, a candidate for replace Pi (random basis)
http://community.wolfram.com/groups/-/m/t/1485074
Hello people of the community, I'm an enthusiast of **Mathematica** and **Wolfram|Alpha** who's been busy for the last few days with a project on possible numerical candidates to be used as random alternative bases for various applications. As I know there are amazing mathematical friends in the community, I decided to humbly expose my work and ask about opinions, etc. Only for the purpose of presenting some of my ideas and also to start an informal discussion on this kind of subject: number randomness. And maybe it can also be useful for someone here.
To begin with, I understand that the randomness that I speak in this text is not truly random, but it serves as the basis for almost-random operations and distributions that need that specific degree of trust.
In order to study the randomness of the transcendental numbers, candidates for transcendental and notable irrational numbers, I used a table base and developed a digit counting workbook. I only used numbers with 10000 decimal digits for the study (generated using **Mathematica** and the data later adapted to data workbook). Below is the example of the interface I used with the Pi number:
![enter image description here][1]
Each workbook of data like this above is a point on a chart, that is, many similar to this will result in the characteristic curve of each number review.
In this study I compared four different specific characteristics (Y-axis). Using the workbook I could detail these quantities as the digits increased to 10000 on the X-axis. I used the transcendental number Pi to start the study. In this study I made my own version of properties to study the numbers and are not necessarily the conventional way of doing it. Given:
C = Deviation of the average arithmetic between all the decimal digits in the range.
S = Deviation of the average count of different digits: how many nine, how many 8 etc...
T = It measures the difference of how many numbers are between 0 and 4 in contrast to those between 5 and 9, such as a coin toss, result between rounding up or down situations.
A = Total number of digits forming or part of doubles, triples, etc.: 11,222,5555, 333333333... (in the interval studied).
I've tested several numbers and their combinations. As for example: **E^Pi, Pi^E, E^sqrt(2), 2^sqrt(2), Zeta (3), Gamma(1/3), Ln(2), Ln(Pi), E^(1/Pi), E+Pi, GoldenRatio, EulerGamma, E+Ln(2)+EulerGamma**, etc... around 30 different numbers, preferably transcendentals, irrational and other notorious candidates. There are two types of accuracy in this project, some I made with with 31 data points and some more detailed with 91 data points.
Below is the detailed graph of **Pi** referring to the characteristics already stipulated:
![enter image description here][2]
In this graph each vertical line is one point to the curve and has a separation of 110 digits, there are 91 points from 100 to 10000 digits on the X-axis.
Below are a few more examples with other notable numbers:
E Number
![enter image description here][3]
Gamma(1/3) Number
![enter image description here][4]
Ln(2) Number
![enter image description here][5]
Each of these charts above have the space between the vertical lines of 330 digits (X-axis) and use 31 points between 100 and 10000. They represent the characteristic curve of each number (Y-axis).
Note 1: Realize that the closer to the X-axis are the curves, in all graphics, the more well distributed and favored is the number for its use in random applications.
Then the following: I calculated the **AREA** below the curve in the graphs to characterize each of its value. The method I used was to calculate the area through average trapezoids formed by the arithmetic mean, so consequently I considered its own degree of precision.
Note 2: The important point in this study **IS NOT** the absolute values that I found (because I used a specific method), **BUT** the comparison of the values between the different numbers, since I used the same process in all objects of study, making it possible to compare. Below is the table for four important numbers using the accuracy of 31 points.
![enter image description here][6]
The Pi number has the lowest frequency to form repetitions of ALL the numbers tested (..would that be the manifestation of it irrationality?).
Well, after a sequence of tests and more tests, in this quest to find candidates equal or almost good as Pi in this characteristic, I found by chance a very good candidate number: the number **C (10)!** , or **ChampernowneNumber(10)!** (! = factorial):
I used **Mathematica** to generate the test numbers (examples):
![enter image description here][7]
In this example above are the first 500 digits of the numbers C(10) and C(10)!, but in the real study I used 10000 digits (also generated by **Mathematica**).
Examples of digit count according to the amount of total digits. The left is the C (10)! And the right is Pi:
![enter image description here][8]
Below is the result of the workbook I generated for the C (10)! using 31 points of precision:
![enter image description here][9]
Full chart of Champernowne (10)! (now with 91 points, 110 in 110 digits, 100 to 10000):
![enter image description here][10]
![enter image description here][11]
Comparing the data I got for Pi e C(10)! numbers (max accuracy, chart of 91 points):
![enter image description here][12]
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Pigraph0.jpg&userId=1316061
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Pigraph1.jpg&userId=1316061
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=e.jpg&userId=1316061
[4]: http://community.wolfram.com//c/portal/getImageAttachment?filename=gamma13.jpg&userId=1316061
[5]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ln2.jpg&userId=1316061
[6]: http://community.wolfram.com//c/portal/getImageAttachment?filename=tablei.jpg&userId=1316061
[7]: http://community.wolfram.com//c/portal/getImageAttachment?filename=champernowneini.jpg&userId=1316061
[8]: http://community.wolfram.com//c/portal/getImageAttachment?filename=tabledual.jpg&userId=1316061
[9]: http://community.wolfram.com//c/portal/getImageAttachment?filename=tablei2.jpg&userId=1316061
[10]: http://community.wolfram.com//c/portal/getImageAttachment?filename=champernowne!.jpg&userId=1316061
[11]: http://community.wolfram.com//c/portal/getImageAttachment?filename=tablef.jpg&userId=1316061
[12]: http://community.wolfram.com//c/portal/getImageAttachment?filename=conclusao.jpg&userId=1316061
I conclude that: of all the numbers tested (transcendental, irrational, etc.) the number that has the characteristic of not-repeating-numeral to those of Pi is the **ChampernowneNumber (10)**! : a possible candidate to replace it in applications that need randomness and it IS NOT possible or convenient to incorporate Pi (is that a best alternative candidate? ). Currently I take 2 minutes to do a fast previous checkup on any number with the workbook, 1 hour to create and analyze completely with the chart 31 points and 3 hours for the chart of 91 points.
Please if you liked the work I did let me know giving a LIKEClaudio Chaib2018-09-29T21:29:53ZThe Octagonal Dodecahedron
http://community.wolfram.com/groups/-/m/t/1520664
On 17 October 2018, [Ivan Neretin discovered the octagonal dodecahedron](https://math.stackexchange.com/questions/2869725/), a toroid made from twelve octagons.
**4**, 6, 8 triangles can make a tetrahedron and up. The [Snub Disphenoid](http://mathworld.wolfram.com/SnubDisphenoid.html) has 12 faces.
**6**, 8, [9](https://en.wikipedia.org/wiki/Herschel_graph), 10 quadrilaterals can make a cube and up. The [Rhombic Dodecahedron](http://mathworld.wolfram.com/RhombicDodecahedron.html) has 12 faces.
**12**, 16, 18, 20 pentagons can make a [tetartoid](http://demonstrations.wolfram.com/TheTetartoid/) or dodecahedron [and up](https://math.stackexchange.com/questions/1609854/).
**7**, 8, 9, 10 hexagons can make make a [Szilassi toroid](http://demonstrations.wolfram.com/TheParametrizedSzilassiPolyhedron/) and [up](http://dmccooey.com/polyhedra/ToroidalRegularHexagonal.html).
**12**, 24 heptagons can make a [heptagonal dodecahedron](http://dmccooey.com/polyhedra/HigherGenus.html) or [Klein quartic 3-torus](http://mathworld.wolfram.com/KleinQuartic.html).
**12** octagons can make an octagonal dodecahedron.
![octagonal dodecahedron][1]
So how did I make that picture? First, I looked through the [Canonical Polyhedra](https://datarepository.wolframcloud.com/resources/Canonical-Polyhedra) resource object for the outer polyhedron. The index turned out to be "8_9".
ResourceObject["Canonical Polyhedra"]
ResourceData["Canonical Polyhedra"][["8_9"]]
It's a geometric object with constraints since it's a [canonical polyhedron](http://demonstrations.wolfram.com/CanonicalPolyhedra/). My WTC talk [Narayama's Cow and Other Algebraic Numbers](https://wtc18.pathable.com/meetings/895905) discussed how to use algebraic number fields to simplify geometrically constrained objects.
1. Get two or more points to simple fixed values.
2. Use RootApproximant[] on remaining points.
3. If remaining points have the same value for NumberFieldDiscriminant[coord^2], the object is in an algebraic number field.
Would the technique I suggested help to make the new object? Turns out it did.
I took the points from "8_9", kept the center at (0,0,0), found an EulerMatrix[] to forced the midpoints of two opposing edges to (0,0,1),(0,0,-1) and force those two edges to be parallel to the x,-y axes.
After using Chop[] in various ways to get 0, 1, and -1 values, I used RootApproximant[] on everything else, then looked at NumberFieldDiscriminant[coord^2] on all reasonable seeming values. The discriminant -104 turned out a lot, and soon I had all coordinates using the algebraic number field based on Root[#^3 - # - 2 &, 1].
I've found these functions useful for algebraic number fields.
FromSqrtSpace[root_, coord_] := Module[{ dim, degree, vector},
dim = Dimensions[coord];
degree = {1, 2}.NumberFieldSignature[root];
vector = (root^Range[0, degree - 1]);
Map[With[{k = (#).vector}, RootReduce[Sign[k] Sqrt[Abs[k]]]] &, coord, {Length[dim] - 1}]];
ToSqrtSpace[root_, coord_] := Module[{dim, order, algebraic},
dim = Dimensions[coord];
order = {1, 2}.NumberFieldSignature[root];
algebraic = Map[Function[x, ToNumberField[Sign[x] RootReduce[x^2], root]], coord,{Length[dim] - 1} ];
Map[Function[x, If[Head[x] === AlgebraicNumber, Last[x], PadRight[{x}, order]]], algebraic, {Length[dim]} ]];
The algebraic number field coordinates, actual coordinates, and faces.
valsV={{{0,1/2,-1/4},{0,0,0},{-1,0,0}},{{-2,0,1},{0,-2,1},{-1,2,-1}},{{0,2,-1},{2,0,-1},{1,-2,1}},{{0,2,-1},{-2,0,1},{1,-2,1}},{{-2,0,1},{0,2,-1},{-1,2,-1}},{{0,-1/2,1/4},{0,0,0},{-1,0,0}},{{0,0,0},{0,1/2,-1/4},{1,0,0}},{{0,0,0},{0,-1/2,1/4},{1,0,0}},{{2,0,-1},{0,2,-1},{-1,2,-1}},{{2,0,-1},{0,-2,1},{-1,2,-1}},{{0,-2,1},{-2,0,1},{1,-2,1}},{{0,-2,1},{2,0,-1},{1,-2,1}}};
p89v = FromSqrtSpace[Root[#^3 - # - 2 &, 1], valsV];
p89F={{1,2,3,4,5},{2,10,12,8,3},{4,7,11,9,5},{6,9,11,12,10},{1,5,9,6},{1,6,10,2},{3,8,7,4},{7,8,12,11}};
Code for the initial picture.
reg=RegionBoundary[RegionDifference[ConvexHullMesh[p89v],ConvexHullMesh[With[{a=.7, b=.6, c=.9},{{a,b,c}, {-a,-b,c},{-b,a,-c}, {b,-a,-c} }]]]];
DiscretizeRegion[reg,MeshCellStyle->{{2,All}->Opacity[.7]}, SphericalRegion-> True, ImageSize-> 600, ViewAngle-> Pi/10]
Showing the original polyhedron and subtracted tetrahedron.
Graphics3D[{EdgeForm[Thick], Opacity[.8], GraphicsComplex[p89v, Polygon[p89F]],
With[{a = .7, b = .6, c = .9}, Polygon[Subsets[{{a, b, c}, {-a, -b, c}, {-b, a, -c}, {b, -a, -c} }, {3}]]]}, Boxed -> False, SphericalRegion -> True, ViewAngle -> Pi/9]
![octagonal dodecahedron][2]
Might be possible to remove the canonical sub-polyhedron constraint and add a constraint that the octagons all have unit area. Or to minimize the ratio of largest/smallest edge.
If you'd like a hexagonal dodecahedron, here's a simple one.
DiscretizeRegion[RegionBoundary[RegionDifference[Region[Cuboid[{0, 0, 0}, {3, 3, 3}]],
RegionUnion[Region[Cuboid[{0, 0, 0}, {2, 2, 2}]], Region[Cuboid[{1, 1, 1}, {3, 3, 3}]]]]],
MaxCellMeasure -> {"Area" -> 0.001}, AccuracyGoal -> 8, PrecisionGoal -> 8]
![hexagonal dodecahedron][3]
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=octagonaldodecahedron.jpg&userId=21530
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=octagonaldodecbuild.jpg&userId=21530
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=hexagonaldodecahedron.jpg&userId=21530Ed Pegg2018-10-18T21:43:37ZTwelve Prisms
http://community.wolfram.com/groups/-/m/t/1517102
My friend Gianni Sarcone recently built a 12 prism construction.
![twelve prisms][1]
So I had to build one myself. I also built in in Mathematica. To my surprise, I was able to simplify it down to three points.
base={{4,4,7},{3,6,6},{2,5,8}};
prismp=Join[base,-(Reverse/@base)];
prismf={{1,2,3},{4,5,6},{1,2,4,5},{1,3,6,5},{2,3,6,4}};
tetrahedralGroup ={{{-1,0,0},{0,-1,0},{0,0,1}},{{0,-1,0},{0,0,1},{-1,0,0}},{{0,0,1},{-1,0,0},{0,-1,0}},{{0,0,-1},{1,0,0},{0,-1,0}},{{0,1,0},{0,0,-1},{-1,0,0}},{{1,0,0},{0,1,0},{0,0,1}},{{0,-1,0},{0,0,-1},{1,0,0}},{{-1,0,0},{0,1,0},{0,0,-1}},{{0,0,1},{1,0,0},{0,1,0}},{{1,0,0},{0,-1,0},{0,0,-1}},{{0,0,-1},{-1,0,0},{0,1,0}},{{0,1,0},{0,0,1},{1,0,0}}};
Graphics3D[{Opacity[.8],Table[Polygon[prismp[[#]].tetrahedralGroup[[n]]]&/@prismf,{n,1,12}]}, Boxed-> False, SphericalRegion->True,ImageSize-> {800,800},ViewAngle-> Pi/9]
![twelve prisms][2]
Maybe try out ViewAngle -> Pi/600, ViewPoint -> {200, 0, 0}
![12 prisms from far away][3]
Sweet.
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=twelveprisms.jpg&userId=21530
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=12prismWL.jpg&userId=21530
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=12prismfar.jpg&userId=21530Ed Pegg2018-10-16T22:24:27ZOuter Billiards. How to create a test to skip triangle corner in for loop?
http://community.wolfram.com/groups/-/m/t/1487490
Hey Everyone! I'm currently working on a problem in a course on mathematical modeling concerning Outer Billiards. I'm supposed to write a program that does the following:
> Start with a ball (a point particle) somewhere outside an equilateral triangle with side length equal to 1. You have two possibilities here and you select one of them. When it arrives at the corner it has traveled a distance d1. Then the particle continues in the same direction as before the same distance d1 There it changes direction momentarily and moves towards the other corner. Then the procedure is repeated, at the second corner it has traveled a distance d2 and it continues in the same direction the same distance d2, etc.
One problem that I've encountered is that for some points the trajectory of the point particle crosses the interior of the triangle, which is not allowed. Therefore I would like to create a test inside of my for loop which says that: "IF the trajectory towards a corner crosses the interior of the triangle, move instead to the next corner." Now, my lecturer gave me a hint that one can use determinants in order to make a pretty easy test, but I find it somewhat hard to understand intuitively. So I would like to make another test, but I don't know how exactly. Here is the program that I'm working on:
corner = {{1/2, Sqrt[3] /2}, {0, 0}, {1, 0}};
ourtriangle = Triangle[{corner[[1]], corner[[2]], corner[[3]]}];
p0 = {2, 2};
plotpoints = {p0};
cornerindex = 1;
n = 3;
For[i = 1, i < n, i++,
p1 = 2*corner[[cornerindex]] - p0;
p0 = p1;
AppendTo[plotpoints, p1];
cornerindex = Mod[i, 3] + 1;
]
traj = Table[plotpoints[[i]], {i, n}];
plot1 = Graphics[{Dashed, Line[traj]}];
Show[plot1, Graphics[ourtriangle], Axes -> False]
This yields the following graph:
![enter image description here][1]
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Project_1_FINAL.jpg&userId=1487470
So, in this case I would like the trajectory to instead move towards the right-most corner but I really dont know how. Could someone please give me at least a hint?
Thank you all.Victor Galeano2018-10-01T14:08:05Z