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

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

2. 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.

3. 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.

4. 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.

5. 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.

6. 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

7. Originally Posted by LNZ
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
To be precise: Fibonacci - Wolfram|Alpha

8. 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

9. Originally Posted by Marco Aurelio
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
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.

10. 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

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