# Message #1182

From: Andrew Gould <agould@uwm.edu>

Subject: number of states

Date: Tue, 28 Sep 2010 21:38:43 -0500

My initial findings Friday were that when comparing the {4,3,3} 3 as is,

there would be twice as many possible states if you allow more non-3D face

twists (specifically 3x3x1x1 twists), and yet, it should make it an easier

puzzle.

As is, I calculate 2152*377*514*79*115*134*172*192*232*29*31 possible

states,

or if the html breaks down: 2^152 * 3^77 * 5^14 * 7^9 * 11^5 * 13^4 * 17^2 *

19^2 * 23^2 * 29 * 31

, but with the other twists added the only change is that the 2^152 goes to

2^153. When I use calc to compute them I get around 1.75*10^120 and

3.5*10^120 possible states respectively.

The difference is from 16!/2 ways of currently placing the 4C pieces.but

with the other twists added there are 16! ways of placing the 4C pieces. To

refresh everyone on the ‘!’ symbol, 7! = 7*6*5*4*3*2*1.

Similarly for the {4,3,3} 2: there are twice as many states if you allow

2x2x1x1 twists.

currently 240*320*53*72*111*131 possible states

= 2^40 * 3^20 * 5^3 * 7^2 * 11^1 * 13^1

the only change is that 2^40 goes to 2^41 for the exact same 4C reason.

This makes for 3.36*10^27 and 6.72*10^27 possible states respectively.

–

Andy