What is God's number on a 4x4 Rubik's cube?

[IAssemble]

What *are* the current lower and upper bounds for 4x4x4?

67 block turns
77 single-layer turns
82 twists (face turns)

http://cubezzz.dyndns.org/drupal/?q=node/view/93

This is from a thread here on speedsolving.

My algorithm counts contiguous layers including one face turned by the same amount as a single move so this is probably close to "block turns" referenced above. So 67 looks like the the current upper bound for this approach.

I think I have only seen the lower bound quoted for "single-layer" turns...

Does anyone know of any other research?

 

[IAssemble]

Well it's at least 40 I wonder if there's a pattern between god number for 2x2-3x3 3x3-4x4 ect

How do you know it is at least 40?

BTW there is a thread currently discussing this again here ;-)

 

[Yellowsnow98]

6...

Kidding, I'd say that it could be as high as 200 but that's just me.

 

[Zoma]

I like the deck of cards analogy, but I don't get it. heheh

 

[KongShou]

42 is the answer to everything

 

[AvGalen]

I once made up this formula without any evidence.
(Amount of layers-1) * 10 + 1 for even layered cubes (or use a nice mod function if you want to get all mathy and stuff)
So
1x1x1 = (1-1) * 10 = 0
2x2x2 = (2-1) * 10 + 1 = 11
3x3x3 = (3-1) * 10 = 20
4x4x4 = (4-1) * 10 + 1 = 31
It seems really low but not improbable (and it is correct for 1, 2 and 3). I should simply see the amount of positions that can be reached on a 10x10x10 cube in 101 moves compared with the amount of total positions but I never could be bothered

 

[ben1996123]

I wonder if there's a pattern between god number for 2x2-3x3 3x3-4x4 ect

dont we all

i predict 38

 

[qqwref]

Well, we already know that God's Number is Θ(n^2 / log(n)) (see http://www.speedsolving.com/forum/showthread.php?30231). So we can guarantee that any recurrence or formula that does not also grow the same way will eventually be wrong (and probably sooner rather than later).

 

[kcl]

I once made up this formula without any evidence.
(Amount of layers-1) * 10 + 1 for even layered cubes (or use a nice mod function if you want to get all mathy and stuff)
So
1x1x1 = (1-1) * 10 = 0
2x2x2 = (2-1) * 10 + 1 = 11
3x3x3 = (3-1) * 10 = 20
4x4x4 = (4-1) * 10 + 1 = 31
It seems really low but not improbable (and it is correct for 1, 2 and 3). I should simply see the amount of positions that can be reached on a 10x10x10 cube in 101 moves compared with the amount of total positions but I never could be bothered

I was messing with something like this a few weeks ago but I did it wrong. I think you have it right..

 

[uberCuber]

I'd say that it could be as high as 200 but that's just me.

An upper bound of much less than 200 was already given in this very same thread....

67 block turns
77 single-layer turns
82 twists (face turns)

http://cubezzz.dyndns.org/drupal/?q=node/view/93

 

[MaikeruKonare]

THANK YOU applemobile!!! Atheist Speed Cubers FTW! I vote we call it "Thee Algorithm" instead of "God's Algorithm".
Therefore, thee algorithm for a 3x3x3 is 20.

 

[JasonK]

THANK YOU applemobile!!! Atheist Speed Cubers FTW! I vote we call it "Thee Algorithm" instead of "God's Algorithm".
Therefore, thee algorithm for a 3x3x3 is 20.

You realise that the concept of "God's algorithm" and "God's number" don't imply the existence of a God, right?

 

[qqwref]

I vote we call it "Thee Algorithm" instead of "God's Algorithm".

What? If we are going to call it something else, it ought to be Frank's Algorithm.

 

[cubernya]

What? If we are going to call it something else, it ought to be Frank's Algorithm.

After Frank Morris?

 

[qqwref]

Of course!

 

[AvGalen]

Well, we already know that God's Number is Θ(n^2 / log(n)) (see http://www.speedsolving.com/forum/showthread.php?30231). So we can guarantee that any recurrence or formula that does not also grow the same way will eventually be wrong (and probably sooner rather than later).

and by "we" you mean "you and a couple of other people in the speedcubing community that understand math" and actually write things like "Θ" online, right?

 

[AvGalen]

After Frank Morris?

blasphemy to even doubt!

 

[irontwig]

This is from a thread here on speedsolving.

Those are just upper bounds though, right? What type of bound would a counting argument make and how long time would it take to compute a optimal solution for a probably harder than average (analogous to the super flip) 4x4x4 scramble?

 

[AvGalen]

Those are just upper bounds though, right? What type of bound would a counting argument make and how long time would it take to compute a optimal solution for a probably harder than average (analogous to the super flip) 4x4x4 scramble?

optimal? Nobody knows but probably "centuries" so nobody is going to start such a calculation because there is no benefit to doing it.

The funny thing with that is always that if you start the calculation on a computer today it would take "100 years".
If you would start it 3 years from now computer power would have (historically) quadruppled and now it would only take "25 years".
So you will finish 78 years earlier by being lazy for 3 years

 

[Yellowsnow98]

An upper bound of much less than 200 was already given in this very same thread....

I'm sorry, don't whack me I'm brittle.