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I made a default cylinder using blender and exported it into fbx file. When I looked at the content of the file after converting it into json, I found the field "PolygonVertexIndex" (under "Geometry"-"Objects") contains:"3, 1, 63, 61, 59, 57, 55, 53, 51, 49, 47, 45, 43, 41, 39, 37, 35, 33, 31, 29, 27, 25, 23, 21, 19, 17, 15, 13, 11, 9, 7" and " 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60".
These consistent positive indices series confuse me a lot since I believe fbx
PolygonVertexIndex store indices in "positive, positive, negative" (when the surface is triangular) or "positive, positive, positive, negative" (when the surface is quadrilateral). And I cannot find any information regarding consistent positive indices when I googled.
The whole list looks like:
["PolygonVertexIndex", [[0, 1, 3, -3, 2, 3, 5, -5, 4, 5, 7, -7, 6, 7,
9, -9, 8, 9, 11, -11, 10, 11, 13, -13, 12, 13, 15, -15, 14, 15, 17,
-17, 16, 17, 19, -19, 18, 19, 21, -21, 20, 21, 23, -23, 22, 23, 25, -25, 24, 25, 27, -27, 26, 27, 29, -29, 28, 29, 31, -31, 30, 31, 33, -33, 32, 33, 35, -35, 34, 35, 37, -37, 36, 37, 39, -39, 38, 39, 41, -41, 40, 41, 43, -43, 42, 43, 45, -45, 44, 45, 47, -47, 46, 47, 49, -49, 48, 49, 51, -51, 50, 51, 53, -53, 52, 53, 55, -55, 54, 55, 57, -57, 56, 57, 59, -59, 58, 59, 61, -61, 3, 1, 63, 61, 59, 57, 55, 53, 51, 49, 47, 45, 43, 41, 39, 37, 35, 33, 31, 29, 27, 25, 23, 21, 19,
17, 15, 13, 11, 9, 7, -6, 60, 61, 63, -63, 62, 63, 1, -1, 0, 2, 4, 6,
8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42,
44, 46, 48, 50, 52, 54, 56, 58, 60, -63]], "i", []],
It can be seen that the indices are fine at the start of the list but goes weird later.
The list contains negative numbers to end a polygon and looks consistent. A polygon is not restricted to be a triangle or a quad, it can contain as many vertices as it needs.
One way to create a cylinder mesh is using quads on the sides and circles on both ends. In your case, the two long series of positive numbers are the two circles encoded as two big polygons containing 32 vertices each.
(Also open to other similar non-Rails methods)
Given (0..99), return entries that are randomly picked in-order.
Example results:
0, 5, 11, 13, 34..
3, 12, 45, 67, 87
0, 1, 2, 3, 4, 5.. (very unlikely, of course)
Current thought:
(0..99).step(rand(0..99)).each do |subindex|
array.push(subindex)
end
However, this sets a single random value for all the steps whereas I'm looking for each step to be random.
Get a random value for the number of elements to pick, randomly get this number of elements, sort.
(0..99).to_a.sample((0..99).to_a.sample).sort
#⇒ [7, 20, 22, 29, 45, 48, 57, 61, 62, 76, 80, 82]
Or, shorter (credits to #Stefan):
(0..99).to_a.sample(rand(0..99)).sort
#⇒ [7, 20, 22, 29, 45, 48, 57, 61, 62, 76, 80, 82]
Or, in more functional manner:
λ = (0..99).to_a.method(:sample)
λ.(λ.()).sort
To feed exactly N numbers:
N = 10
(0..99).to_a.sample(N).sort
#⇒ [1, 5, 8, 12, 45, 54, 60, 65, 71, 91]
There're many ways to achieve it.
For example here's slow yet simple one:
# given `array`
random_indexes = (0...array.size).to_a.sample(rand(array.size))
random_indexes.sort.each { |i| puts array[i] }
Or why don't you just:
array.each do |value|
next if rand(2).zero?
puts value
end
Or you could use Enumerator#next random number of times.
Below example returns a sorted array with random entries from given range based on randomly picked true or false from array [true, false]:
(0..99).select { [true, false].sample }
=> [0, 3, 12, 13, 14, 17, 20, 24, 26, 28, 30, 32, 34, 35, 36, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 53, 54, 55, 56, 58, 59, 60, 61, 62, 65, 67, 69, 70, 71, 79, 81, 84, 86, 91, 93, 94, 95, 98, 99]
To reduce the chances of a bigger array being returned, you can modify your true/false array to include more falsey values:
(0..99).select { ([true] + [false] * 9).sample }
=> [21, 22, 28, 33, 37, 58, 59, 63, 77, 85, 86]
I'm building a simple program after learning a little bit of Ruby. I'm trying to associate values from one array to another here's what I've got so far.
ColorValues = ["Black", "Brown", "Red", "Orange", "Yellow", "Green", "Cyan", "Blue", "Violet", "Pink", "Grey"]
(0..127).each_slice(12) {|i| p i}
.each_slice returns these arrays
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
[12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]
[24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35]
[36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47]
[48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59]
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71]
[72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83]
[84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95]
[96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107]
[108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119]
[120, 121, 122, 123, 124, 125, 126, 127]
What I'm attempting to do is then take the returned arrays and associate them with each color in the ColorValues[] array
i.e.
"Black" = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
"Brown" = [12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]
I'm sure there's a simple way of doing this I'm just not sure how to go about doing it.
Using Enumerable#zip and Hash::[]
colorValues = ["Black", "Brown", "Red", "Orange", "Yellow", "Green", "Cyan", "Blue", "Violet", "Pink", "Grey"]
Hash[colorValues.zip((0..127).each_slice(12))]
# => {"Black"=>[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
# "Brown"=>[12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23],
# "Red"=>[24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35],
# "Orange"=>[36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47],
# "Yellow"=>[48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
# "Green"=>[60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71],
# "Cyan"=>[72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83],
# "Blue"=>[84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95],
# "Violet"=>[96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107],
# "Pink"=>[108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119],
# "Grey"=>[120, 121, 122, 123, 124, 125, 126, 127]}
You can use zip for that:
ColorValues.zip( (0..127).each_slice(12) )
Method #zip is convenient, but it turns out that your problem is so frequent in Ruby, that I have overloaded #>> operator on Array class to perform zipping into a hash. First, install y_support gem. Then,
require 'y_support/core_ext/array' # or require 'y_support/all'
h = ColorValues >> ( 0..127 ).each_slice( 12 ) # returns a hash like in falsetru's answer
h["Black"] #=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
I am trying to come up with an algorithm to iterate thru a list via pagination.
I'm only interested in the initial index and the size of the "page".
For example if my list is 100 items long, and the page length is 10:
1st page: starts at 0, length 10
2nd page: starts at 11, length 10
3rd page: starts at 21, length 10
...
Nth page: starts at 90, length 10
My problem is coming up with an elegant solution that satisfies these cases:
1. list has 9 elements, page length is 10
1st page: starts at 0, length 9
2. list has 84 elements, page length is 10
1st page: starts at 0, length 10
2nd page: starts at 11, length 10
3rd page: starts at 21, length 10
...
Nth page: starts at 80, length 4
I could do this with a bunch of conditionals and the modulo operation, but I was wondering if anyone could offer a better/elegant approach to this problem.
Thanks!
There follows some code doing it the long way in Python which could be used for other languages too; followed by how it could be done in a more maintainable fashion by the intermediate Pythoneer:
>>> from pprint import pprint as pp
>>> n, perpage = 84, 10
>>> mylist = list(range(n))
>>> mylist[:10]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> mylist[-10:] # last ten items
[74, 75, 76, 77, 78, 79, 80, 81, 82, 83]
>>> sublists = []
>>> for i in range(n):
pagenum, offset = divmod(i, perpage)
if offset == 0:
# first in new page so create another sublist
sublists.append([])
# add item to end of last sublist
sublists[pagenum].append(i)
>>> pp(sublists)
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74, 75, 76, 77, 78, 79],
[80, 81, 82, 83]]
>>> # Alternatively
>>> sublists2 = [mylist[i:i+perpage] for i in range(0, n, perpage)]
>>> pp(sublists2)
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74, 75, 76, 77, 78, 79],
[80, 81, 82, 83]]
>>>
I'm looking form a programatic way to take an integer sequence and spit out a closed form function. Something like:
Given: 1,3,6,10,15
Return: n(n+1)/2
Samples could be useful; the language is unimportant.
This touches an extremely deep, sophisticated and active area of mathematics. The solution is damn near trivial in some cases (linear recurrences) and damn near impossible in others (think 2, 3, 5, 7, 11, 13, ....) You could start by looking at generating functions for example and looking at Herb Wilf's incredible book (cf. page 1 (2e)) on the subject but that will only get you so far.
But I think your best bet is to give up, query Sloane's comprehensive Encyclopedia of Integer Sequences when you need to know the answer, and instead spend your time reading the opinions of one of the most eccentric personalities in this deep subject.
Anyone who tells you this problem is solvable is selling you snake oil (cf. page 118 of the Wilf book (2e).)
There is no one function in general.
For the sequence you specified, The On-Line Encyclopedia of Integer Sequences finds 133 matches in its database of interesting integer sequences. I've copied the first 5 here.
A000217 Triangular numbers: a(n) = C(n+1,2) = n(n+1)/2 = 0+1+2+...+n.
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431
A130484 Sum {0<=k<=n, k mod 6} (Partial sums of A010875).
0, 1, 3, 6, 10, 15, 15, 16, 18, 21, 25, 30, 30, 31, 33, 36, 40, 45, 45, 46, 48, 51, 55, 60, 60, 61, 63, 66, 70, 75, 75, 76, 78, 81, 85, 90, 90, 91, 93, 96, 100, 105, 105, 106, 108, 111, 115, 120, 120, 121, 123, 126, 130, 135, 135, 136, 138, 141, 145, 150, 150, 151, 153
A130485 Sum {0<=k<=n, k mod 7} (Partial sums of A010876).
0, 1, 3, 6, 10, 15, 21, 21, 22, 24, 27, 31, 36, 42, 42, 43, 45, 48, 52, 57, 63, 63, 64, 66, 69, 73, 78, 84, 84, 85, 87, 90, 94, 99, 105, 105, 106, 108, 111, 115, 120, 126, 126, 127, 129, 132, 136, 141, 147, 147, 148, 150, 153, 157, 162, 168, 168, 169, 171, 174, 178, 183
A104619 Write the natural numbers in base 16 in a triangle with k digits in the k-th row, as shown below. Sequence gives the leading diagonal.
1, 3, 6, 10, 15, 2, 1, 1, 14, 3, 2, 2, 5, 12, 4, 4, 4, 13, 6, 7, 11, 6, 9, 9, 10, 7, 12, 13, 1, 0, 1, 10, 5, 1, 12, 8, 1, 1, 14, 1, 9, 7, 1, 4, 3, 1, 2, 2, 1, 3, 4, 2, 7, 9, 2, 14, 1, 2, 8, 12, 2, 5, 10, 3, 5, 11, 3, 8, 15, 3, 14, 6, 3, 7, 0, 4, 3, 13, 4, 2, 13, 4, 4, 0, 5, 9, 6, 5, 1, 15, 5, 12, 11, 6
A037123 a(n) = a(n-1) + Sum of digits of n.
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 46, 48, 51, 55, 60, 66, 73, 81, 90, 100, 102, 105, 109, 114, 120, 127, 135, 144, 154, 165, 168, 172, 177, 183, 190, 198, 207, 217, 228, 240, 244, 249, 255, 262, 270, 279, 289, 300, 312, 325, 330, 336, 343, 351, 360, 370, 381
If you restrict yourself to polynomial functions, this is easy to code up, and only mildly tedious to solve by hand.
Let , for some unknown
Now solve the equations
…
which simply a system of linear equations.
If your data is guaranteed to be expressible as a polynomial, I think you would be able to use R (or any suite that offers regression fitting of data). If your correlation is exactly 1, then the line is a perfect fit to describe the series.
There's a lot of statistics that goes into regression analysis, and I am not familiar enough with even the basics of calculation to give you much detail.
But, this link to regression analysis in R might be of assistance
The Axiom computer algebra system includes a package for this purpose. You can read its documentation here.
Here's the output for your example sequence in FriCAS (a fork of Axiom):
(3) -> guess([1, 3, 6, 10, 15])
2
n + 3n + 2
(3) [[function= -----------,order= 0]]
2
Type: List(Record(function: Expression(Integer),order: NonNegativeInteger))
I think your problem is ill-posed. Given any finite number of integers in a sequence with
no generating function, the next element can be anything.
You need to assume something about the sequence. Is it geometric? Arithmetic?
If your sequence comes from a polynomial then divided differences will find that polynomial expressed in terms of the Newton basis or binomial basis. See this.
There is no general answers; a simple method can be implemented bu using Pade approximants; in two words, assume your sequence is a sequence of coefficients of the Taylor expansion of an unknown function, then apply an algorithm (similar to the continued-fraction algorithm) in order to "simplify" this Taylor-expansion (more precisely: find a rational function very close to the initial (and truncated) function. The Maxima program can do it: look at "pade" on the page: http://maxima.sourceforge.net/docs/manual/maxima_28.html
Another answer tells about the "guess" package in the FriCAS fork of Axiom (see previous answer by jmbr). If I am not wrong; this package is itself inspired from the Rate program by Christian Krattenthaler; you can find it here: http://www.mat.univie.ac.at/~kratt/rate/rate.html Maybe looking at its source could tell you about other methods.