Message #1828

From: Andrey <andreyastrelin@yahoo.com>
Subject: Re: [MC4D] God’s number for 3^N
Date: Thu, 07 Jul 2011 06:11:13 -0000

I just understood that we want not maximal, but minimal values of coefficients to use the formula. They are for N=5: 0.70295 for FTM and 0.22201 for QFTM.

Andrey

— In 4D_Cubing@yahoogroups.com, "Andrey" <andreyastrelin@…> wrote:
>
> Andrew,
> You are right - I forgot to mention a couple of things:
> - formula works only for N>=5 (where we haven’t additional invariants for corners orientations)
> - and even in that cases it’s only asymptotic: (2/9+o(1))*N*3^N and (3/4+o(1))*3^N. I’m not sure in constant 3/4: for N=170 computations give something like 0.739.. and this coefficient slowly decreases. Maximal coefficient for QFTM is 0.222674992 for N=71, and for FTM - 0.7642 for N=17.
>
> Andrey
>
>
> — In 4D_Cubing@yahoogroups.com, "Andrew Gould" <agould@> wrote:
> >
> > A couple problems I’ve noted for FTM:
> >
> > When N = 3: your lower bound gives 20.25 when we know it’s actually 20.
> >
> > When N = 3: your possible twists is 24 when I count 18.
> >
> > When N = 4: your possible twists is 192 when I count 184.
> >
> >
> >
> > I’m wondering if your FTM lower bound equation is a bit high.mainly because
> > it beat my lower bound :-)
> >
> > 3^4: your FTM lower bound is 60.75, mine is 56.
> >
> >
> >
> > –
> >
> > Andy
> >
> >
> >
> >
> >
> > From: 4D_Cubing@yahoogroups.com [mailto:4D_Cubing@yahoogroups.com] On Behalf
> > Of Andrey
> > Sent: Wednesday, July 06, 2011 11:27
> > To: 4D_Cubing@yahoogroups.com
> > Subject: [MC4D] God’s number for 3^N
> >
> >
> >
> >
> >
> > My estimates show that lower counting limit L for God’s number for 3^N is
> > 2/9*N*3^N for QFTM (as implemented in MC5D and MC7D - with 2*N*(N-1)*(N-2)
> > possible twists) and 3/4*3^N for FTM (where any twist of face is counted as
> > 1, so we have N!*2^(N-1) possible twists). Actual God’s number is probably
> > between L and 2*L.
> > By the way, if we take puzzle 2*1^N (with only one twisting face), its God’s
> > number in QFTM is N. But counting limit gives something like
> > N*(log(2*N)/(2*log(N)) that is N/2*(1+o(N)). So lower limit is almost the
> > half of the actual number.
> >
> > Andrey
> >
>