Rubik Top (Hybrid Toy Project)

[ta0]

Great work, Hydrosur!

What exactly do you mean by 1h20m time of assembly?

Have you spun it? Do the center caps stay in place or do they fly out?

[jim in paris]

Salut Henri

Oulala , tu es un vrai master

TB réalisation, avec couleurs africa

Le plan des ressorts ça doit pas être facile à
mettre en place !

Bravo encore et on se voit peut-être à Marines

Jim

[HYDRUS17R]

Bonjour Ta0, bonjour Jim

Pour le temps de montage, 1H20 c’est à partir des pièces imprimées posées sur la table le temps qu’il m’a fallu pour les assembler et obtenir la toupie rubik complètement assemblée. À deux reprises certains éléments m’ont échappé et j’ai dû recommencer.

Actuellement j’ai mélangé les éléments, pour bien vérifier le fonctionnement de l’ensemble et je contemple le désordre à résoudre pour remettre la toupie dans son aspect initial.

J’ai effectué quelques lancements, mais seulement au sol, et rien n’a été expulsé. Les capuchons restent bien en place.



Pour tenter un boomerang, il va falloir que je la remette dans son état initial.

Pour les ressorts, j’ai acheté dans un magasin de bricolage, et j’ai coupé à la longueur qui me convenait.

Malheureusement Jim nous ne pourront pas être avec vous pour Marines.

Bonne réussite à cette manifestation.

J’ai également modifié les pièces CORE, CENTER et CENTER CAP, pour pouvoir insérer les ressorts (voir photos précédentes).

[ta0]

I took advantage of the vacation time and (for the third time) developed algorithms to solve a 3x3 Rubik's cube. Now that I can solve again a traditional Rubik's cube, I can attack the spintop twisting puzzle. Alas, my first attempts have been a complete failure: I find the spintop very confusing, even though it's essentially the same puzzle

@ta0: [/b]I hope you have the cube with you and you are going to find your own algorithms again.
Please have a look at the attached article "How to solve almost any Rubik-like puzzle".
It is not a spoiler but it can enable you to find your algorithms faster.

I used the method described in the paper and in Mathologer's video to find the algorithms to solve the last layer (how to spin around 2 centers, spin around 2 corners, rotate positions of 3 centers, rotate positions of 3 corners). Finding these algorithms was straight forward but I kept making mistakes when reversing the steps (I didn't write them down). For the second layer I didn't know how to apply this method, so I just solved it by trial and error. But this one is very efficient while the others probably are not. Also by trial and error I found a way of rotating a center, what I hadn't done in my previous solves. Actually, it rotates 4 centers and leaves the other 2 still, but I'm guessing that's enough to solve any cube (but I'm not 100% sure).

[ortwin]

I took advantage of the vacation time and (for the third time) developed algorithms to solve a 3x3 Rubik's cube. Now that I can solve again a traditional Rubik's cube, I can attack the spintop twisting puzzle. Alas, my first attempts have been a complete failure: I find the spintop very confusing, even though it's essentially the same puzzle

Congratulation on that! It reminds me of that bad joke: "What is one the advantage of having Alzheimer's disease? - You meet new friends every day!"
I also have some twisty puzzles on my shelf that I need to freshly find the solution for every other year I take them down to play with. - In part that is a good feeling, in part frustrating.
Also I have some puzzles that are essentially a 3x3 cube but they confuse me every time again. For example this tetrahedron "master pyramorphix":

...I just solved it by trial and error. But this one is very efficient while the others probably are not. Also by trial and error I found a way of rotating a center, what I hadn't done in my previous solves. Actually, it rotates 4 centers and leaves the other 2 still, but I'm guessing that's enough to solve any cube (but I'm not 100% sure).

3-cycles of corners are at least 8 moves in length. Some of those can easily be found with the method you are using.If you have those, the method is not necessarily inefficient. It is even a method described for fmc (fewest moves contest) : Build a skeleton of as many parts of the cube solved by intuitive moves (not necessarily layer by layer), solve the rest of the cube with 3-cycles.

With the centers you can do better: I know a very understandable algorithm that rotates only two centers, it can be found with the method in the paper.But there is a case where you need to rotate one center by 180 degrees. I find that a bit harder to find a solution for, since I forget the algorithm for that.

[ta0]

This is an older version of a Rubik cube spintop made by Ronald Torres of Colombia. It appears to have a different setup while still being a normal 3x3. I'm trying to get Ronald to explain it to me. By the way, he is a spintop trick player who watches some of our videos.

3-cycles of corners are at least 8 moves in length. Some of those can easily be found with the method you are using . . .
With the centers you can do better: I know a very understandable algorithm that rotates only two centers, it can be found with the method in the paper.But there is a case where you need to rotate one center by 180 degrees. I find that a bit harder to find a solution for, since I forget the algorithm for that.

My 3-cycles are 12 long.
I'll work a little more in finding other algorithms.

[ortwin]

...
My 3-cycles are 12 long.
...

To be precise with these numbers one actually also has to state what metric one is using to count the number of moves.
A very popular metric is htm (half turn metric) where every turn of an outer layer counts as one move. So if you turn an outer layer by 90 degrees or 180 degrees or 270 degrees, it is always counted as one move. In contrast to this there is also qtm (quarter turn metric) where every 90 degree turn counts as a move. So there a 180 degree turn would count as two moves. Also some other metrices are used sometimes: stm (slice turn metric), .....
The 3-cycles of length 8 I mentioned before I counted in htm. Some of those stay at length 8 if you count in qtm but some are length 10 in qtm and some are even of length 12 in qtm.

[ta0]

I recounted and even with htm my cycling of 3 edges is still 12 and my cycling of 3 corners is actually 14! (I had counted wrong before).

[ortwin]

I recounted and even with htm the cycling 3 centers is still 12 and cycling 3 corners is actually 14! (I had counted wrong before).

You probably mean "cycling 3 edges" when you say "cycling 3 centers", centers can not cycle!
I am very tempted to send you an example of one of those corner 3-cycles of length 8, but maybe you do not want such a spoiler...
This is not a spoiler, it is save to click here, only an example of the way I would give you an algorithm, hit the play button.

[ta0]

I just spoke on the phone with Ronald. From his description this is the same setup as Patricio's, with the tip a corner and the crown another corner. The way he made it was by cutting a normal cube with a small saw (without touching the internal mechanism), sanding it, filling the voids, painting with lacquer, and attaching vinyl stickers and a wooden tip. He sold this top to somebody from the US but he will make another one. He lives in Barranquilla. I told him about this site.

You probably mean "cycling 3 edges" when you say "cycling 3 centers", centers can not cycle!

Yes, edges. I corrected the mistake while you were writing that.
At this time I'm not very interested in solving speed. I'm more interested in learning memorable algorithms that I won't forget again. Perhaps the best way is to just find them myself.
EDIT: The Hensel/Mathologer method cannot provide solutions with an even number of moves. Say you need n moves to accomplish what you want on the 1st layer. You need to do the same movements backwards to restore the other layers. And in between you need to rotate the first layer. So the total number of moves will be n+1+n = 2n+1, always odd. So those optimal solutions are not achievable. In fact, it does feel to me that my current solutions are longer than what I had accomplished before by trial and error.

[ortwin]

...
EDIT: The Hensel/Mathologer method cannot provide solutions with an even number of moves. Say you need n moves to accomplish what you want on the 1st layer. You need to do the same movements bckwards to restore the other layers. And in between you need to rotate the first layer. So the total number of moves will be n+1+n = 2n+1, always odd. So those optimal solutions are not achievable. In fact, it does feel to me that my current solutions are longer than what I had accomplished before by trial and error.

On the contrary! It provides (but not always) solutions with an even number of moves. You missed one move!
You write: "...And in between you need to rotate the first layer."
That is true, but you also have to undo that rotation at the end, that is the move you forgot and that would bring your move count to an even number.
There still are optimal algorithms for 3-cycles with uneven move count (9,11, ...).
These can be understood most of the time this way:You are in a situation where you are one move away from a position you can solve with an algorithm of even length that you know. So you first do that setup move, apply your algorithm and finally you undo your setup move. This would give of course again an even algorithm length, but there are cases where the setup move you need to do and the first move of your algorithm are on the same layer. So you combine those two moves, which reduces your move count by one, leading to an odd move count.

[ortwin]

I just spoke on the phone with Ronald. From his description this is the same setup as Patricio's, with the tip a corner and the crown another corner. The way he made it was by cutting a normal cube with a small saw (without touching the internal mechanism), sanding it, filling the voids, painting with lacquer, and attaching vinyl stickers and a wooden tip. He sold this top to somebody from the US but he will make another one. He lives in Barranquilla. I told him about this site.

...

Those setups are basically the same as in the "master pyramorphix" I put a picture of a few posts back in this thread. Since you only need to round it a bit to look like a top, I would consider modding that one a bit. Should be easier than starting from a cube.

[ta0]

That is true, but you also have to undo that rotation at the end, that is the move you forgot and that would bring your move count to an even number.

Oops, of course!

Those setups are basically the same as in the "master pyramorphix" I put a picture of a few posts back in this thread. Since you only need to round it a bit to look like a top, I would consider modding that one a bit. Should be easier than starting from a cube.

Is the master pyramorphix a 3x3 cube?

[ortwin]

...
Is the master pyramorphix a 3x3 cube?

Exactly!!
You can get it for very cheap here.
Another nice option would be this Pyraminx. It is a lot easier to solve then a 3x3 and it has magnets. So if modded into spin-top it would incorporate three of my hobbies: twisty puzzles, magnets, spin-tops.
Maybe I should start with getting a regular throw top and practice until I can do a boomerang. Any recommendations there? Is it possible to practice indoors without totally wrecking the flat?

[ta0]

Thanks. I will have to order one now. I wonder if I can just wrap it with a string and throw it. You are getting me into the time sinkhole of twist puzzles against my will!

Looking at the photos, Ronald's looks like a master pyramorphix but Patricio's doesn't. I may be all confused, but the "tip" on the pyramorphix splits into different pieces while in Patricio's it's just one piece (a corner). But Ronald said they were the same . . .