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 Post subject: Number of Pentultimate positionsPosted: Wed Mar 26, 2008 1:44 pm

Joined: Sat Apr 21, 2007 11:21 pm
Location: Marin, CA
Searching the web, this site, and Jaap's page, I wasn't able to come up with a published number of positions for the Pentultimate. Ignoring permutation parities, it should be just a matter of figuring out the possible positions of each piece and dividing out duplicate positions based on symmetry. Right? I've never done this, so any help making sure this is accurate would be nice.

//pentagons first
>>> 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
479001600

//triangles (posiiton and orientation)
>>> 60 * 57 * 54 * 51 * 48 * 45 * 42 * 39 * 36 * 33 * 30 * 27 * 24 * 21 * 18 * 15 * 12 * 9 * 6 * 3
8483004771271882804592640000L

Now all together, dividing out for symmetry (12 * 5)
>>> (479001600 * 8483004771271882804592640000L) / 60
67722880970781098306872698470400000L

So I get 67,722,880,970,781,098,306,872,698,470,400,000 as the possible positions for the Pentultimate.

Anyone care to correct my undergrad level math?

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Jason Smith posted here as 'io' through 2012.
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 Post subject: Re: Number of Pentultimate positionsPosted: Wed Mar 26, 2008 2:17 pm

Joined: Sun Jun 04, 2006 10:05 am
Location: Minneapolis, Minnesota, USA
We'll assume 1 center is stationary. Can't have just 2 swapped.

11!/2 = 19958400

Position of 20 corners. Can't have just 2 swapped.

20!/2 = 1216451004088320000

Orientation of corners. Restricted to rotations of 3.

(3^20)/3 = 1162261467

Compiled.

(11!)(20!)(3^20)/(2)(2)(3) = 2.82 x 10^34

You got 6.77 x 10^34, so at least our numbers are close.

Correct me if I'm wrong, which is entirely possible.

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Fridrich
3x3 PB 22.63
3x3 Av 30.57

25, Male
Started cubing Oct 15 '05

Out of the game, but not completely.

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 Post subject: Re: Number of Pentultimate positionsPosted: Wed Mar 26, 2008 2:42 pm

Joined: Sat Apr 21, 2007 11:21 pm
Location: Marin, CA
Oh, so these are the permutation parities! Great. I don't know the puzzle well enough to know these restrictions yet.

You accounted for the fixed face positional symmetry by using 11! instead of 12!. But I think you left out the symmetry of rotation around the fixed face. Even assuming the face is fixed, you want to discard the variations that are the same under rotation around that face.

So, seems like we've got this with your terms in blue and mine in red:

(facePositions * cornerPositions * cornerOrientations) / (facePositionRestriction * cornerPositionRestriction * cornerOrientationRestriction * symmetryOfFacePosition * symmetryOfFixedFaceOrientation)

Could that be right?

If so, I get:
(12! * 20! * 3^20) / (2 * 2 * 3 * 12 * 5)

(479001600 * 2432902008176640000 * 3486784401) / (2 * 2 * 3 * 12 * 5)

3762382276154505461492927692800000

or

3.76 x 10^33

Could that be right?

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Jason Smith posted here as 'io' through 2012.
Visit Jason Smith's PuzzleForge on Shapeways!
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 Post subject: Re: Number of Pentultimate positionsPosted: Wed Mar 26, 2008 2:58 pm

Joined: Sun Jun 04, 2006 10:05 am
Location: Minneapolis, Minnesota, USA
I don't think you need the "symmetryOfFacePosition". That is covered already in it. We are just choosing a random face to remain stationary while moves are applied. The same position could still be achieved regardless of stationary center.

However, I did forget the "symmetryOfFixedFaceOrientation", so thank you for correcting that.

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Fridrich
3x3 PB 22.63
3x3 Av 30.57

25, Male
Started cubing Oct 15 '05

Out of the game, but not completely.

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 Post subject: Re: Number of Pentultimate positionsPosted: Wed Mar 26, 2008 3:03 pm

Joined: Sat Apr 21, 2007 11:21 pm
Location: Marin, CA
I think "symmetryOfFacePosition" is covered if you use 11! as you did, but I allowed all face positions initially with 12!, and then divided the twelve (symmetryOfFacePosition) back out in the symmetry terms. So I think we're on the same page, actually!

So it seems like we're in buisness. It has

3.76 x 10^33

positions!

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Jason Smith posted here as 'io' through 2012.
Visit Jason Smith's PuzzleForge on Shapeways!
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 Post subject: Re: Number of Pentultimate positionsPosted: Wed Mar 26, 2008 3:27 pm

Joined: Sun Jun 04, 2006 10:05 am
Location: Minneapolis, Minnesota, USA
Ah, yes. That will do it. Sorry, I didn't catch that.

So let's see how that compares to other puzzles. It's less than a 3x3x3...

_________________
Fridrich
3x3 PB 22.63
3x3 Av 30.57

25, Male
Started cubing Oct 15 '05

Out of the game, but not completely.

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 Post subject: Re: Number of Pentultimate positionsPosted: Wed Mar 26, 2008 7:39 pm

Joined: Mon Oct 18, 2004 11:06 pm
Location: Portland, ME
io wrote:
So it seems like we're in buisness. It has

3.76 x 10^33

positions!

Also known as 3.7 Decillion. Wow!

Chris

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 Post subject: Re: Number of Pentultimate positionsPosted: Wed Mar 26, 2008 8:08 pm

Joined: Sat Aug 12, 2006 6:40 pm
Location: California
pelley wrote:
io wrote:
So it seems like we're in buisness. It has

3.76 x 10^33

positions!

Also known as 3.7 Decillion. Wow!

Chris

Also known as 3.7 Quintilliard. Wow!
(Actually, isn't it 3.8 rounded? Also, I haven't computed myself, but qqwref gets a different number. I'll let him explain)

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 Post subject: Re: Number of Pentultimate positionsPosted: Wed Mar 26, 2008 8:11 pm

Joined: Sat Jan 22, 2005 12:12 pm
Location: NY, USA
I think you're all wrong.

Google says (12! * 20! * 3^20) / (2 * 2 * 3 * 12 * 5) = 5.644 * 10^33 (which is incidentally exactly what I got)

Here's the way I did it. Fix one specific corner to fix orientation. Then we have:
- 12 centers with no orientation (12!/2 because each turn makes 5-cycles)
- 19 corners with 3 orientations each (19!/2 * 3^18, again because of 5-cycles)
So I had 12! * 19! * 3^18 / 4 which is the same as yours, or again 5.644 * 10^33.

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Puzzle Solving Service! - a puzzle that has never been scrambled and solved has been wasted.

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 Post subject: Re: Number of Pentultimate positionsPosted: Thu Mar 27, 2008 2:27 am

Joined: Sat Apr 21, 2007 11:21 pm
Location: Marin, CA
Woah, you're right. Must be a rogue copy/paste that got me.

I get your same answer just pasting my formula into google.

So it's:

5.644 * 10^33

Going once.... going twice....

_________________
Jason Smith posted here as 'io' through 2012.
Visit Jason Smith's PuzzleForge on Shapeways!
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