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 Post subject: vertex turning rhombidodecahedron Help needed(done)
PostPosted: Thu May 19, 2011 12:41 am 
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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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 Post subject: Re: vertex turning rhombidodecahedron Help needed
PostPosted: Thu May 19, 2011 1:14 am 
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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 3-cycle 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 2-color edges into an odd permutation. If you did a twist of the yellow-brown-lightblue-blue vertex you'd put the square centers back into an even permutation which is resolvable with 3-cycles. It would put the edges into an odd permutation though which can't be resolved with 3-cycles.

The other possibility is that you have identical colors on the other side that we can't see.

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 Post subject: Re: vertex turning rhombidodecahedron Help needed
PostPosted: Thu May 19, 2011 1:26 am 
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This is a parity error, because this is equivaent to a dino 3x3, and the edges can only 3 cycle so... the "hidden corners" are misplaced, meaning you have to change the position of all the pices by 90┬░ on a 3x3 axis

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 Post subject: Re: vertex turning rhombidodecahedron Help needed
PostPosted: Thu May 19, 2011 1:33 am 
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Actually I think there is a chance you can resolve this in the current color scheme.

Call the Yellow-Brown-LightBlue-Blue vertex A and the Peach-LightBlue-Green-Unknown 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 counter-clockwise to put the four edges and four petals back. Now you have 3-cycles 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.

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Last edited by Brandon Enright on Thu May 19, 2011 1:55 am, edited 1 time in total.

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 Post subject: Re: vertex turning rhombidodecahedron Help needed
PostPosted: Thu May 19, 2011 1:49 am 
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this is after N=5


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 Post subject: Re: vertex turning rhombidodecahedron Help needed
PostPosted: Thu May 19, 2011 1:58 am 
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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 re-orienting 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.

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 Post subject: Re: vertex turning rhombidodecahedron Help needed
PostPosted: Thu May 19, 2011 3:23 am 
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Joined: Tue Aug 24, 2010 8:11 am
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Thank both you.
I try the method RubixFreakGreg said.
It works.

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