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Chilen

Post subject: vertex turning rhombidodecahedron Help needed(done) Posted: Thu May 19, 2011 12:41 am 

Joined: Tue Aug 24, 2010 8:11 am Location: Taipei, Taiwan

The situation is there are only 4 pieces incorrect. Does anyone know how to solve it.
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Last edited by Chilen on Thu May 19, 2011 4:27 am, edited 1 time in total.


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Brandon Enright

Post subject: Re: vertex turning rhombidodecahedron Help needed Posted: Thu May 19, 2011 1:14 am 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

I'm pretty sure you have solved this into the inverted color scheme. Those petal / triangle pieces should be no problem because you have duplicates. A 3cycle where A > B > C > A will work if A is brown, B is brown, and C is blue. Because of the identical pieces you can have an even permutation of the pieces that looks like and odd one.
The trouble is that the square face centers are in an odd permutation and I don't think you can resolve that without also putting the 2color edges into an odd permutation. If you did a twist of the yellowbrownlightblueblue vertex you'd put the square centers back into an even permutation which is resolvable with 3cycles. It would put the edges into an odd permutation though which can't be resolved with 3cycles.
The other possibility is that you have identical colors on the other side that we can't see.
_________________ Prior to using my real name I posted under the account named bmenrigh.


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RubixFreakGreg

Post subject: Re: vertex turning rhombidodecahedron Help needed Posted: Thu May 19, 2011 1:26 am 

Joined: Sat Jan 16, 2010 11:48 am Location: In Front Of My Teraminx (saying WTF?)


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Brandon Enright

Post subject: Re: vertex turning rhombidodecahedron Help needed Posted: Thu May 19, 2011 1:33 am 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

Actually I think there is a chance you can resolve this in the current color scheme.
Call the YellowBrownLightBlueBlue vertex A and the PeachLightBlueGreenUnknown vertex B.
If you do (B A B' A) x N where N is 2 or 3 or 4 or ...
I thin you'll eventually end up with a case where you've rotated the four edges and four petals of vertex A by 90 degrees but the centers have stayed in place. If this does happen then you can decouple the parity of the centers and edges which would allow you to solve this case.
I'm having trouble doing this in my head but can you report back what (B A B' A) x N does for N = 2, 3, 4, 5, etc? Do you ever get to a situation where you've rotated most of the vertex but not the centers around the vertex?
Edit: I followed this through in my head as best I could an unless I've made a mistake I think you need N = 5.
That is, do (B A B' A) x 5 which is 20 total moves. That should twist four edges and four petals clockwise without moving any face centers. Then you can twist the A vertex counterclockwise to put the four edges and four petals back. Now you have 3cycles to resolve everything else.
Edit2: On further thought I think my provided routine with N = 5 will actually twist the four edges and petals by 180 degrees which doesn't decouple the parity of the edges from the centers. I think the color scheme is either inverted or rotated by 90 degrees.
_________________ Prior to using my real name I posted under the account named bmenrigh.
Last edited by Brandon Enright on Thu May 19, 2011 1:55 am, edited 1 time in total.


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Chilen

Post subject: Re: vertex turning rhombidodecahedron Help needed Posted: Thu May 19, 2011 1:49 am 

Joined: Tue Aug 24, 2010 8:11 am Location: Taipei, Taiwan

this is after N=5
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Brandon Enright

Post subject: Re: vertex turning rhombidodecahedron Help needed Posted: Thu May 19, 2011 1:58 am 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

Chilen wrote: this is after N=5 Yeah see my "Edit2:". I was right that it doesn't rotate by 90 degrees but by 180. I don't see a way to resolve this without reorienting the whole color scheme by 90 degrees on some axis. If you start the solve over you can memorize the relative orientation of the colors and then solve the centers to that scheme. If they don't have a parity then you will know that you have the right orientation and you can then go about solving the edges into that orientation.
_________________ Prior to using my real name I posted under the account named bmenrigh.


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Chilen

Post subject: Re: vertex turning rhombidodecahedron Help needed Posted: Thu May 19, 2011 3:23 am 

Joined: Tue Aug 24, 2010 8:11 am Location: Taipei, Taiwan


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