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 Post subject: Number of Pentultimate positions
PostPosted: Wed Mar 26, 2008 1:44 pm 
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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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 Post subject: Re: Number of Pentultimate positions
PostPosted: Wed Mar 26, 2008 2:17 pm 
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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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 Post subject: Re: Number of Pentultimate positions
PostPosted: Wed Mar 26, 2008 2:42 pm 
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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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 Post subject: Re: Number of Pentultimate positions
PostPosted: Wed Mar 26, 2008 2:58 pm 
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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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 Post subject: Re: Number of Pentultimate positions
PostPosted: Wed Mar 26, 2008 3:03 pm 
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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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 Post subject: Re: Number of Pentultimate positions
PostPosted: Wed Mar 26, 2008 3:27 pm 
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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...

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 Post subject: Re: Number of Pentultimate positions
PostPosted: Wed Mar 26, 2008 7:39 pm 
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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 positions
PostPosted: Wed Mar 26, 2008 8:08 pm 
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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! :P
(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 positions
PostPosted: Wed Mar 26, 2008 8:11 pm 
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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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 Post subject: Re: Number of Pentultimate positions
PostPosted: Thu Mar 27, 2008 2:27 am 
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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....

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