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Messages  Carca_Maker
31
« on: July 11, 2019, 09:54:25 AM »
Ooh! Meepledrone... It looks that there aren't some Quaternas branches, for example your 4221, or 1224, it's the same thing, it's not.
There are an amount of 20 Quaternas branches, remember. Although your computation worths, does the right "5040" cases, so I think that it lacks 4221, 4421, 3221, 4422 branches and their computations.
In the other hand I revised your Triplets computations and they match with mine!.Good job!. I attached here your table for clarification.
I'm using this calculations for to make an expansion in which appears "Jewels" and its combinations. You will see... Thanks. I'm waiting your wisdom words...greetings.
32
« on: July 11, 2019, 06:25:42 AM »
I will, of course!, no doubt.
33
« on: July 11, 2019, 06:23:23 AM »
Ha, ha, ha! The system says that I have to wait one hour for give you another "karma" action, "+1 merit" more.
34
« on: July 11, 2019, 06:19:01 AM »
Congratulations Meepledrone!! I intuited (intuition) that for the triplets, or quadruplas, the lowest quantity of cases were for 333, and 4444, respectively, but there were some hidden, another not obvious "combinations" which amount, worth the same lowest quantity of cases, maybe I saw, I suspected it at my triplets first trying that thanks to you, you "validated" it to me. Although I completed, I wrote that my triplets table one by one, I weren't sure at all. I'm going to compare my first trying of triplets table with yours, I suppose they match... Since I sent you my pairs table, etc, you understood that, great!. I only can say that I lift my hat!. You beated the "final big boss monster" of my maths mess!. I give you +2 merits. Now I will apply it to my variant... Thanks Meepledrone!.
35
« on: July 10, 2019, 08:35:17 AM »
It's important to say that the my triplets notation at the previous picture, means that, for example "432“ means every triplet containing that three elements. Because of it, the total amount of triplets is 720 cases, and no the other smaller result of Meepledrone's supose #2.
36
« on: July 10, 2019, 02:58:04 AM »
Good morning! As Meepledrone said at his "Case #1: Elements in different order represent different sets" the triplets, 720 cases match with my old solution, here you are a picture.
So, it's only needed by me the solution and solving for Quaternas branches, at the beginning, supposing "Case #1: Elements in different order represent different sets". Ok and thanks.
37
« on: July 09, 2019, 03:26:22 PM »
And written in the right way, I add : "Case #1: Elements in different order represent different sets", in the Meepledrone terms.
38
« on: July 09, 2019, 03:12:47 PM »
Good evening, Meepledrone! First of all, I wish to give you a "+1" merit, no doubt, you are a math sage!. Afterwards I will have to read better what you wrote, but in a first approximation I think that "90" cases for pairs, "720" for triplets , etc is the right way.
39
« on: July 09, 2019, 02:59:23 PM »
Hola DrMeeple! Efectivamente, es el que me señala el sabio de Meepledrone. Son las ramas de los árboles de mis ternas y cuaternas respecto de mis 10 elementos. Y qué reto querías compartir por aquí Carca Marker??
Enviado desde mi XT1700 mediante Tapatalk
Enviado desde mi H3311 mediante Tapatalk
40
« on: July 09, 2019, 10:46:31 AM »
Hello Meepledrone! I'm not still familiarized with "quote" action, though my smartphone, it's difficult seeing from it. Sorry. I'm going to talk from "recently" memory.
That's a good idea to identify the four kinds of elements with A, B, C, D.
I can say that your two calculations, look like, seem to "unify" some cases belonging to a "upper", category of cases. "120“ cases, and "210“ approximately in the other hand, it seems to me a nice approximation to the goal, because it's not a huge amount of rather cases. But I don't still know what does it means, so brief numbers.
Now, the question is how to FIT your first calculation, to the most simple cases, that I know and I calculated too: the PAIRS of elements. (And the trivial singles cases too in one table).
I would be glad if your mathematical model can get to, describe the same as my table of pairs. This is, how many cells are belonging to each pair, etc?. That's the question that I need for first approximation, for every trees and its branches.
Posdata: The cells of this pairs table are empty, because are an example.Thanks.
41
« on: July 08, 2019, 02:31:38 PM »
Meepledrone, I sized both two images when I uploaded them, to a A4 printable size. I don't know if that is the matter of the trouble. I'm going to souper, it's late, and disconnect. Bye.
42
« on: July 08, 2019, 02:11:37 PM »
Sorry... Maybe the two images that I accompanied are too big for a smartphone screen? Tell me if it does. Carca_Maker.
43
« on: July 08, 2019, 02:06:07 PM »
Hi everybody! I'm working on my Carcassonne variant about some "elements", but I'm stopped at this point:
To build a sets of 3, (and in the other side) , sets of 4 of this “elements". ("Triple", and "Quaterna" respectively?). Sets of one, and two elements are trivial.
So we are at this starting point:
*We have got an amount of 10 elements: one "1", two "2", three "3", and finally four "4“.
A) If we extract randomly THREE of them without replacement, how many cases are at all at every single branch?. (See my accompanying trees of three extracted elements).
I think that repeated sets of this three elements are considered as "different" cases, although they have got the same composition.
So it looks that it doesn't matter the way, the order in that this elements are considered.
Every branch of this could be ordered in any order. It only matters its composition and how many repeated times appears every case inside its own branch.
Very important: although for example, we have extracted one "2", this is DIFFERENT from the other existing "2", this means that at the end we could have got apparently some "repeated cases", but they have to be seen as "DIFFERENT", although belonging to the same branch, of course.
I estimated approximately among 720 cases at all, at all the "triplets" trees, but I'm not sure if it's true.
I followed my own describing graphical order, where I began by the most unusual, uncommon element to the most frequent elements, in my trees.
B)The same for the sets, branches, of FOUR of this extracted elements from the total amount of this 10 possible elements, (without replacement, etc).
Finally, a basic question: It would be a clarification for me to knowing how many cases are the most representative branches "1234", and "4444“, separately, for example. Thanks.
I would include the helpers and solvers of this questions at the acknowledgements of my variant, and I would give a "+1“s thanks to them.
POSDATA: I wonder also if anyone can tell me if I could write all this in Spanish, my usual language. Is it possible too?. Greetings from Carca_Maker.
44
« on: July 08, 2019, 01:57:00 PM »
Muy agradecido!, y que sirva esto para otros hispanohablantes.
45
« on: July 08, 2019, 06:16:16 AM »
Hola amigos carcasoneros!. Disculpen si les digo que me ha sorprendido gratamente verles en "quizs, puzzles and challenges", ya que pensaba que sólo se podía hablar en inglés.
He descubierto que tengo un reto pendiente por compartir aquí, y debido a mi nivel inglés "promedio", estaba atascado.
¿Se puede hacer en español en esta sección?. Gracias.
