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 Post subject: multicoloured megaminx questionPosted: Fri Jan 03, 2014 11:58 pm

Joined: Fri Sep 27, 2013 4:27 pm
Hello If you would to sticker a megaminx like Oliver Nagy's 30 colour megaminx but only use 20 colour and repeat each colour twise so you can get 2 completley white edges and so on with 20 colours. Would the solve be really difficult because there are 2 edges wich are the same colour for 20 colours and if those edges where exchanged would a parity occur and you would need to swap all the edges or will it just be really hard because there can be up to 20 or less swaped edges and you never know wich ones are swapped and wich are not.
Thanks.
-Naz

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 Post subject: Re: multicoloured megaminx questionPosted: Sat Jan 04, 2014 12:34 am

Joined: Mon Apr 20, 2009 5:24 am
Shouldn't it be 15 colors?
Anyway, difficulty is somewhat subjective. Would you call a longer solve time more difficult, or a novel situation like parity? Some call 4x4x4 more difficult than 5x5x5, and 2x2x2 more difficult than 3x3x3 because of this reason.
I think a 15 color edge-stickered megaminx would have parity while the others won't, so arguably it would be harder. So it's your call. Imo it shouldn't be too hard to fix it. If you want to know why, highlight over this next part. If you want to try for yourself, then don't read it. You can also try making a 6 color megaminx, which would solve like a normal megaminx but would also have parity.
Quote:
You don't need to swap all 30 edges. Because pairs of edges look identical, you aren't supposed to know which ones go where anyway. If you come across parity, you only need to swap one extra pair of identical edges and the parity will be gone. But it's another challenge to figure out how to do this ; )

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 Post subject: Re: multicoloured megaminx questionPosted: Sat Jan 04, 2014 12:39 am

Joined: Fri Sep 27, 2013 4:27 pm
Thanks that cleared out alot.
-Naz

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