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

Dacuba

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I doubt that there is any pattern for God's numbers. Just because the 2x2 and 3x3 and 4x4 are very different compared to each other (3x3 has no center pieces, 2x2 has no edges nor centerpieces). I don't think they fit in a scheme in that way.
 

Chrisalead

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I never say I had any justification for this ! I said "I Think" that's all. It's simply my personal feeling about this. One thing is sure : that kind of maths are really hard for human beings. This reminds me of the Syracuse conjecture.
 

asportking

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I imagine if you tried hard enough, and you already knew God's number for three or four cubes, you could find a pattern. However, I doubt it would be anything as simple as "multiply the cube size by seven," and you'd probably have to figure out a few more Gods' numbers in order to confirm the pattern.
 

AbstractAlg

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again me, with crazy examples.
find next number in row:
1, 2, 4, ...?

It could be 8 if you look this as powers of 2.
It could be 7 if you look this that increment between every two numbers rises for 1.
It can be any number actually, unless you give some rule or function to follow (it depends on previous number, it's arithmetic array, it's geometrical array, it depends on previous two numbers...)

Meaning, the pattern could exist, but as we know, even the simplest math arrays like Fibonachi (don't know how to spell) row have very big formulas, complicated with imaginary parts and square and cubic roots.
So even if pattern does exist, I believe we could never calculate it.
 

LNZ

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For the Fibonacci numbers at least, there is an easy and very accurate approximation expression:

F(n)=(1/sqrt(5))*G^n where G = Golden ratio = (1 + sqrt(5))/2

Example #1: n=5, F(5)=4.95967...... , exact value of F(5)=5

Example #2: n=2, F(2)=1.17082......, exact value of F(2)=1
 
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Well, an n x n x n cube can be optimally solved with Θ(n2 / log(n)) moves. [by Wikipedia]
But I don't know how to do this calculation :D

Don't trust Wikipedia to get everything correct. As far as I know, a method of solving an nxnxn cube which was of Θ(n^2 / log(n)) was found. That provides an upper bound of how fast God's number grows, it doesn't say how fast it grows or what the formula for God's number actually is (assuming such a formula can ever be found).

Basically, for every n, it is possible to solve a fully scrambled nxnxn cube using no more than x((n^2)/log(n)) moves, for some constant x which is the same for all n. I can't bothered checking if (n^2)/log(n) is actually correct, but it looks roughly like what I remember it being.
 

tx789

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31 is my estimate now since if you only turn two layers it reduced to a 2x2 and gods number for a 2x2 is 11 and if you only turn the outer layers it reduced to a 3x3 which gods number is 20 abd 20+11=31
 

tx789

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You need to pair up edges and put them in the right spot relative to your centers as well.
opitmal reduction isn't the shortest 4x4 solution. Still finding this out will be hard what about the 13ish quintillion solutions that look like a 3x3 but have parity can't that be solved in 11 moves

finding this out will take a while there's
7,401,196,841,564,901,869,874,093,974,498,574,336,000,000,000
7.4 quattuordecillion
7.401196841564901869874093974498574366e+45


7.4 quintillion quintillion quintillion i think
 
Last edited:

TMOY

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opitmal reduction is the shortest 4x4 solution.
Well, no. Here is a counterexample: apply the following 6-move scramble:
Lw' U R' U' Lw R2
Now perform optimal reduction:
x' Rw'' D R' D' Rw (5 moves)
If optimal reduction yielded the shortest solution, you should be able to finish the solve in 1 move or less by now. This is obviously not the case.
 

Stefan

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Don't trust Wikipedia to get everything correct. As far as I know, a method of solving an nxnxn cube which was of Θ(n^2 / log(n)) was found. That provides an upper bound of how fast God's number grows, it doesn't say how fast it grows or what the formula for God's number actually is (assuming such a formula can ever be found).

Please don't illegitimately badmouth Wikipedia, especially when it's you who doesn't know the difference between O and Θ.
 

CarlBrannen

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Hmmm. I've a 4x4x4 in my hand. If I keep the TFL corner unmoved, there are 36 possible moves. Assuming I want to eliminate repeated moves like RR = R2, I get 33 moves. But I also have to eliminate stuff like R 2R R' so let's call it about 30 available moves per turn.

The number of positions is 7.40E+45, according to wikipedia, so this gives a lower bound estimate (using Shannon information theory) of around log_30(7.40E+45) = 31.05 moves, so I'm going to guess 32 or 33.
 

tx789

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Well, no. Here is a counterexample: apply the following 6-move scramble:
Lw' U R' U' Lw R2
Now perform optimal reduction:
x' Rw'' D R' D' Rw (5 moves)
If optimal reduction yielded the shortest solution, you should be able to finish the solve in 1 move or less by now. This is obviously not the case.

I meant isn't
 

uberCuber

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31 is my estimate now since if you only turn two layers it reduced to a 2x2 and gods number for a 2x2 is 11 and if you only turn the outer layers it reduced to a 3x3 which gods number is 20 abd 20+11=31

I'm not sure how much sense it makes to separate wide turns and outer-layer turns like this because, for example, you can't even solve the centers of any scramble with only wide turns.
 
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