God's Number for 5x5x5?

[Herbert Kociemba]

Wow, how did you get that?

I derived this several years ago and I can't remember the details any more. The recursion for the number of positions in k moves can be written as (A^k)*x with some matrix A and some vector x and the largest eigenvalue of A then is the asymptotic branching factor. In OBTM and QTM instead of HTM, the asymtptotic branching factor is 1/(1.5^(1/(2(n-1)))-1).

 

[Stefan]

I derived this several years ago and I can't remember the details any more. The recursion for the number of positions in k moves can be written as (A^k)*x with some matrix A and some vector x and the largest eigenvalue of A then is the asymptotic branching factor. In OBTM and QTM instead of HTM, the asymtptotic branching factor is 1/(1.5^(1/(2(n-1)))-1).

Cool, thanks. I wouldn't be able to do it (at least not now), but I think I understand the rough idea. At least it's not something trivial that I just missed

 

[Herbert Kociemba]

I also found a closed formula for the number of maneuvers on the the (n+1)*(n+1)*(n+1) cube in OBTM which is not very handy as you might imagine.

With

bh[n_,j_]:=3/((3/2)^(1/n) e^((2 i j Pi)/n)-1), where e, i and Pi are the well known mathematical constants, we have for the count in k moves:

countHW[n_,k_]:=Sum[(3+bh[n,j])/(6n) bh[n,j]^k,{j,0,n-1}]

In QTM it is

bq[n_,j_]:=1/((3/2)^(1/(2n)) e^(( i j Pi)/n)-1)

countQW[n_,k_]:=Sum[(1+bq[n,j])/(4n) bq[n,j]^k,{j,0,2n-1}]

The asymptotic branching factors just are bh[n,0] and bq[n,0].

EDIT: If you ignore all terms with j>0 in the sums above you still get very good approximations with lim approxvalue/truevalue =1 for k->infinity.
This means

approxcount(n,k):=(2^(-((1 + n)/n)) 3^(k + 1/n) (1/(-1 + 1.5^(1/n)))^(1 + k))/n

is a good approximation for the number of maneuvers of length k for the (n+1)*(n+1)*(n+1) cube in OBTM.

 

[TheGermanCuber]

Does anybody know how to calculate a god's number ?

 

[unsolved]

Does anybody know how to calculate a god's number ?

The easiest way to get the exact number is also the least likely way.

Just say something like: "Hey, God, what's your number for the 5x5x5?" and wait for an answer.