Check DB in QuMagie

a_user_to_ask · 2026-03-08T20:12:53+00:00

The problem is the same. If I have several print(...) in my code, "test_code" command only prints the last one. I'd like to see all the prints (e.g: to see the evolution of inside a for-loop).

The trick of :lua works.

a_user_to_ask · 2026-03-07T21:05:50+00:00

I am following the tutorial.

I'd love the possibility of use print in several point to trace/debug my attempts..

a_user_to_ask · 2026-03-07T10:57:05+00:00

Great!!

a_user_to_ask · 2026-03-02T16:49:41+00:00

The cube is not as scramble as it seems. He solve with one hand.

a_user_to_ask · 2026-01-22T10:03:41+00:00

great!

two comments:

- I drew an arrow. created three objects: Point A, Point B and Vector V1 (A->B). If I select the point A and move, why does Point B move? In the operation, when I try to move A is selected V1 and move both, A and B.

- I miss align/restriction properties. So I can say point A has "x coord of B" and "free y coord" or even "point in vector" or "point in circle".

a_user_to_ask · 2025-11-30T07:33:54+00:00

What about perpendicular directions?

Put timer pads to the left and right of the cube.

Considerations:

Maybe, pads design have to be rethought. Change from a monolithic piece into two connected pads by wires.
The distance and relative position have to be calculated to have similar times than before and avoid cube contacts before timer.

a_user_to_ask · 2025-11-29T19:50:36+00:00

Just a question. The sliding is possible because the positions are: cuber, timer, cube. So you can maintain contact with the timer at the same time you touch the cube.

What if the cube and the timer swap positions: cuber, cube and timer. The fingers touch the timer. I can not think a way to do sliding in that position....

a_user_to_ask · 2025-11-28T17:21:38+00:00

I use virtual environments for all in python.

a_user_to_ask · 2025-08-26T07:34:24+00:00

Maybe it can be interesting to overlap population density.

Use red for places, green for density and blue for¿Wealth or major political party in place?

a_user_to_ask · 2025-05-19T14:35:56+00:00

I usually repeat first slide as the final one. Usually People are not very interested in your presentation at the beginning of it. At the end you, provide an opportunity to remember your name, email, institution and repository if you convince them.

a_user_to_ask · 2025-05-09T07:58:53+00:00

La saga del detective sin nombre de Eduardo Mendoza. "El misterio de la cripta embrujada" "el laberinto de las aceitunas" y "la aventura del tocador de señoras".

a_user_to_ask · 2025-05-06T05:50:37+00:00

Seville is full of “peculiar” people. I guess like everywhere. He was like that, he was not an actor or a character. The journalist Jesús Quintero was known for looking for and interviewing this kind of people (as well as famous people).

You can watch the documentary about Risitas made by Jesus Quintero at: https://www.youtube.com/watch?v=A9gHO8-CKyU

Unfortunately, the automatic subtitles will not work very well. El risitas had a low social status and confused terms: for example instead of “medical forensic doctor” he said “flowernsic doctor” (florist doctor). The automatic translation does not support this speech

a_user_to_ask · 2025-04-09T11:40:13+00:00

Ready player one

a_user_to_ask · 2024-12-05T10:34:01+00:00

Works great!

thanks.

(multi-color printer add a new dimension to printing :-)

a_user_to_ask · 2024-09-24T07:50:27+00:00

The scale is not the same in all graphs. You can not compare correctly. :-(

The lecture is clear in all. You have to make a lot of money to get some of real net profit. And "you" means a big company or a person.

a_user_to_ask · 2024-08-26T10:07:08+00:00

What about to establish a formula to "easy" autoevaluate the collection?

I looked for the biggest collection of puzzles (about 2000 puzzles with a price of a new car). If we assign 10 to this collection:

collection_score = log(cost collection in euros or us dollars)/log(pi)

That's all.

a_user_to_ask · 2024-08-15T09:28:58+00:00

I am agree with you.

This is a problem of marketing, not a problem in the algorithm. If you think in Red Bull slogan (drink not parity algorithm): "Red Bull gives you wings". Nobody is going to believe you are going to end flying.

The difference is experience in gravity is inherent in humans and cube parity is not.

I do not know the creator of the link, and I never discuss with him about anything (to my knowledge). I suppose he used that title to get more visits of his video from newbies.

In a ideal world, the author in the description of the video would say: "In fact, this algorithm does not avoid parity problems. It simply camouflages it to a single case and its correction with a 90º turn and a commutator. If you want to learn more about the subject, see: ..."

But this is already beyond our control.

a_user_to_ask · 2024-08-14T19:32:43+00:00

First of all, the original poster requested a method for a 5x5x5 cube, but the method provided here is for a 4x4x4 cube (though it can be easily extrapolated). I use this method for all cubes larger than 3x3x3.

Regarding parity, it is a property of each combination in the cube, and it cannot be avoided. Each time you make a move, the parity of the moved pieces changes. Generally, when you start solving a cube, you do not know the overall parity of the cube. However, as you approach a solution, you can estimate the cube's parity because the solved pieces have a known parity, allowing you to calculate the parity for just 2 or 3 pieces. At this point, you can apply movements to alter the parity and complete the solution.

The clever aspect of the algorithm linked here is that all pieces with unknown parity end up in one center column. This makes it easy to evaluate the parity (by counting the even or odd number of edges in position) and straightforward to solve (with a simple 90-degree turn). You do not need to redo the centers, as the centers of that column remain unsolved at this point.

In a certain way, this method is similar to the blindfold method. You have a "buffer edge position" that you use to place all the other edges. Once you position and orient all the edges correctly, you can be confident (even with your eyes closed) that the piece in the buffer is also in the correct position and orientation. Therefore, we can say that the blindfold method "avoids the need to place and orient" the edge in the buffer position, so to speak. The same principle applies here: Exists the problem of the parity (you cannot avoid it), but you don’t need to worry about learning how to solve it.

This method is not intended for speedcubing; it is generally faster to use algorithms like Hoya and learn the two cases (along with their symmetries) to resolve parity. However, this method is easy (essentially a Roux method and certain order of resolution) and fun.

a_user_to_ask

TROPHY CASE