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Master_Twister

Post subject: Mixup Cube Parity Posted: Sun Aug 25, 2013 12:43 pm 

Joined: Wed Jul 24, 2013 7:51 pm Location: Alberta, Canada

I've been trying to figure out a way to fix the parity of one flipped edge on the last layer of the mixup cube (solving Beginners method), and I can't seem to fix it. Does anyone have an algorithm or a method of fixing this parity?
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Brandon Enright

Post subject: Re: Mixup Cube Parity Posted: Sun Aug 25, 2013 12:46 pm 

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

Master_Twister wrote: I've been trying to figure out a way to fix the parity of one flipped edge on the last layer of the mixup cube (solving Beginners method), and I can't seem to fix it. Does anyone have an algorithm or a method of fixing this parity? Edges and centers are interchangeable on this puzzle. Do you know how to twist a center 180? Or a pair of centers 180? Put the edge in a center position and do that.
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Master_Twister

Post subject: Re: Mixup Cube Parity Posted: Sun Aug 25, 2013 12:58 pm 

Joined: Wed Jul 24, 2013 7:51 pm Location: Alberta, Canada

bmenrigh wrote: Master_Twister wrote: I've been trying to figure out a way to fix the parity of one flipped edge on the last layer of the mixup cube (solving Beginners method), and I can't seem to fix it. Does anyone have an algorithm or a method of fixing this parity? Edges and centers are interchangeable on this puzzle. Do you know how to twist a center 180? Or a pair of centers 180? Put the edge in a center position and do that. I guess you'd have to twist the edge 180 in a center location, then fix the centers without reflipping that edge.
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Pete the Geek

Post subject: Re: Mixup Cube Parity Posted: Sun Aug 25, 2013 1:18 pm 

Joined: Thu Dec 15, 2011 10:04 pm Location: Sioux Lookout, Canada

There is an algorithm for this case. If you want a video, 3x3 Mixup Plus Tutorial. See the links in the "About" tab and click "Part 9  How to Flip a Single Edge". This part of the Mixup and Mixup Plus are solved the same way.
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rline

Post subject: Re: Mixup Cube Parity Posted: Sun Aug 25, 2013 4:15 pm 

Joined: Mon Feb 28, 2011 4:54 am

You turn the middle slice containing the flipped edge, 45*, onto the front face in the center position. Then turn F2. Then turn a center from the U face down into the F face center position. Then turn F2. Then turn that center back into its U face position. Finally, return the flipped edge into its position. It's now flipped. I think you'll have 3 edges which are out of position, but not flipped. If you want to see my video of this, it's here. Skip to 11 mins 46 seconds.
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robertpauljr

Post subject: Re: Mixup Cube Parity Posted: Mon Aug 26, 2013 12:09 am 

Joined: Sun Aug 12, 2007 8:28 pm Location: Northern Central California

bmenrigh wrote: Master_Twister wrote: I've been trying to figure out a way to fix the parity of one flipped edge on the last layer of the mixup cube (solving Beginners method), and I can't seem to fix it. Does anyone have an algorithm or a method of fixing this parity? Edges and centers are interchangeable on this puzzle. Do you know how to twist a center 180? Or a pair of centers 180? Put the edge in a center position and do that. With bmenrigh's method you can flip one edge with 8 twists without disrupting anything else that is already in place. For example, hold the puzzle with the flipped edge at FL. 1. Twist the middle horizontal layer 45˚ to bring the flipped edge to the front center location. 2. Twist the middle vertical layer 90˚ to bring the flipped edge to the top center location. 3. Twist the top layer 180˚. 4. Undo twist #2. 5. Undo twist #1. 6. Repeat twist #2. 7. Repeat twist #3. 8. Repeat twist #4. You can also use something similar to this to twist one edge in the middle layer 90˚ clockwise and another edge or center in the middle layer 90˚ anticlockwise. Whichever way you go in twist #3, go the opposite way in twist #7. Master_Twister, this is simply the method I learned to twist centers on a picture cube, that can be adapted to twist edge pairs, or edgecenter pairs on the Mixup Cube.
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Puzzlemad

Post subject: Re: Mixup Cube Parity Posted: Mon Aug 26, 2013 3:47 am 

Joined: Mon Feb 27, 2012 10:57 am Location: In my study drooling over my puzzle hoard  Precioussssss!

Rather than work out a specific algorithm for this, I tried to understand why it had happened in terms of the law of the cube. A 3x3 should not be able to have a single flipped edge  there MUST be 2 pieces that are flipped. Obviously none of the other edges are flipped and the corners do not play a part in edge flipping. By deduction it had to be that a centre was rotated 180*  this cannot be determined by looking at it because it is not a supercube. So my solution was to take the front centre piece and move it to the right by 45* (I.e. into the RF edge position. I then used beginners method for the F2L to take that centre out onto the top face (it ends up at UB) then bring it around to UF and use the same algorithm to put it back. This is exactly what I do with a standard 3x3 if an equatorial edge happens to be inserted flipped. Then I can turn that equator layer back 45* to put the centre back in place. After this the top face will have the either 0, 2 or 4 edges facing upwards as they are supposed to be. Hope this helps explain rather than just give an algorithm. I am impressed that I actually was able to think it through myself!
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delbert

Post subject: Re: Mixup Cube Parity Posted: Tue Feb 04, 2014 9:28 pm 

Joined: Thu May 28, 2009 6:33 pm

What about the other parity: where you end up with 2 upper edges swapped? I remember this used to happen on the void cube, but my solution for that on the void cube is not working for the mix up cube. (maybe because the centers are present and must be placed)


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Brandon Enright

Post subject: Re: Mixup Cube Parity Posted: Tue Feb 04, 2014 9:55 pm 

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

delbert wrote: What about the other parity: where you end up with 2 upper edges swapped? I remember this used to happen on the void cube, but my solution for that on the void cube is not working for the mix up cube. (maybe because the centers are present and must be placed) This is possible because it's possible to perform a 8cycle of the edges without cycling centers at all. A evencycle is an oddpermutation which creates the parity in the edges. So if you want to fix this parity, replace the 4 centers in a slice move with 4 edges and perform a 1/8th turn which does an 8cycle and will fix the parity. The next trick up is to put the edges into the center locations in the right orientation so that the 8cycle keeps the puzzle in a cube. To do that, call the standard Rubik's cube moves U, D, R, L, F, B which are quarterturns. Call the slice turns in the same direction "under" each face u, d, r, l, f, b which are eightturns. Then do: (f, u2, R, L', U, D'), r, (D, U', L, R', u'2, f') which will fix the parity and maintain the cubic shape. The parenthesized moves are setup moves that will keep the r slice move from changing the shape while fixing the parity. After that just resolve the 3x3x3 (IIRC the optimal solution at this stage is 10 moves). Edit: thank you Matt!
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Last edited by Brandon Enright on Tue Feb 04, 2014 10:35 pm, edited 2 times in total.


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Allagem

Post subject: Re: Mixup Cube Parity Posted: Tue Feb 04, 2014 10:21 pm 

Joined: Sun Oct 08, 2006 1:47 pm Location: Houston/San Antonio, Texas

Brandon Enright wrote: Then do: (f, u2, R, L', U), r, (U', L, R', u'2, f') which will fix the parity and maintain the cubic shape. (f, u2, R, L', U, D'), r, ( D, U', L, R', u'2, f') if you want to maintain the cubic shape Peace, Matt Galla


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Brandon Enright

Post subject: Re: Mixup Cube Parity Posted: Tue Feb 04, 2014 10:34 pm 

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

Allagem wrote: (f, u2, R, L', U, D'), r, ( D, U', L, R', u'2, f') if you want to maintain the cubic shape Ooops! Thank you That's what I get for trying to do it in my head.
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robertpauljr

Post subject: Re: Mixup Cube Parity Posted: Tue Feb 04, 2014 11:01 pm 

Joined: Sun Aug 12, 2007 8:28 pm Location: Northern Central California

After solving the Mixup Cube in various ways, I finally decided that my favorite way is to solve it in such a way that there is never any resolving due to parity issues! 1. Solve the white and yellow layers.
2. Look hard at the middle layer. See what needs to go where to be solved. Each swap counts as a swap. Each 3cycle as 2 swaps. Each 4cycle as 3 swaps. Each 5cycle as 4 swaps. And so on. Add together all the swaps needed to solve. If it is even, you are good to go. If it is odd, do a 45˚ turn of the middle layer, and now you are good to go.
3. Use any algorithms you like to solve the middle layer that maintain even parity. For example, commutators and conjugates for moving the pieces into place. Or things like M' U M E45˚ M' U' M E'45˚, which twists the front center 90˚ clockwise and the piece at FL 90˚ anticlockwise (if I remember correctly which way E and M go).
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cuberboy13

Post subject: Re: Mixup Cube Parity Posted: Tue Feb 04, 2014 11:47 pm 

Joined: Tue Dec 17, 2013 8:29 pm

An easy way to rotate a single center 180 on a 3x3x3 is to do this algorithm: (R U Ri U)x5. If you can get the edge into a center position, you can flip it.
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delbert

Post subject: Re: Mixup Cube Parity Posted: Wed Feb 05, 2014 2:33 pm 

Joined: Thu May 28, 2009 6:33 pm

Thanks guys, I can solve my mixup cube now.
Brandon, your solution was the key that my brain needed. The 1/8 slice turn. I don't want to memorize another formula. so I just do the 1/8 slice turn, then move the centers one by one back to their correct locations. then solve as normal 3x3
not efficient, but a simple concept, with no additional memorization. (this is similar to what I do for the void cube parity; 1/4 slice turn, resolve)
Robertpauljr, I like your idea solving it right the first time, but your explainations are a bit advanced for my understanding


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robertpauljr

Post subject: Re: Mixup Cube Parity Posted: Wed Feb 05, 2014 4:10 pm 

Joined: Sun Aug 12, 2007 8:28 pm Location: Northern Central California

delbert wrote: Robertpauljr, I like your idea solving it right the first time, but your explainations are a bit advanced for my understanding Can you solve the white and yellow layers of a Mixup Cube before attempting to solve anything in the layer between them? If not, forget it. Just do what you do and have fun doing it! But if you can— I will try to simplify what I was trying to say in step 2. Parity has to do with how many swaps it takes to solve the cube. With a normal 3x3x3 cube 1 twist of an outer layer does a 4cycle of corners and a 4cycle of edges. Each ncycle is equivalent to (n1) swaps. So each 4cycle mentioned above is equivalent to 3 swaps. 3 + 3 = 6. 6 is an even number. So a 1/4 turn of an outer layer on a cube does an even number of swaps. The sum of any number of even numbers is an even number. With an inner slice layer, instead of an outer layer, it isn't a 4cycle of corners, it is a 4cycle of centers along with the 4cycle of edges. So the same thing applies. No matter how a normal 3x3x3 cube is scrambled there have been an even number of swaps made. So when solving it, there must be an even number of swaps made. That is why you can not get to the end of a solve of a normal 3x3x3 cube and have just 1 swap of corners, or 1 swap of edges, or 1 swap of centers to perform. One is an odd number. But with the Mixup Cube you can twist an inner layer by 45°, or 1/8 turn. That does an 8cycle, which is equivalent to 7 swaps, which is an odd number of swaps. Let's say that if the middle layer requires an odd number of swaps to solve, we will call the middle layer odd. My step 2 involves figuring out if the middle layer is odd or even. If it is even you go to step 3. If it is odd, you give it a 45° or 1/8 turn, and then go to step 3. But how do you tell if it is odd or even? Some pieces may be solved already. Ignore them. Some pieces may need to 3cycle, or 5cycle. Ignore them, because a 3cycle = 2 swaps, and a 5cycle = 4 swaps. 2 and 4 are even, so can be ignored. If you notice two pieces need to swap, and 2 more need to swap, that is two swaps, which is even. Here is an example: say 1 piece is solved, and you need a 3cycle for 3 of the pieces and a 4cycle for the other 4. Solved is even, and a 3cycle is even, but the 4cycle = 3 swaps, which is odd, so the middle layer is odd. So you would give it a 45° twist and go to step 3. Does that make sense? Want more on step 3?
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Brandon Enright

Post subject: Re: Mixup Cube Parity Posted: Wed Feb 05, 2014 4:16 pm 

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

delbert wrote: your solution was the key that my brain needed. The 1/8 slice turn. I don't want to memorize another formula. so I just do the 1/8 slice turn, then move the centers one by one back to their correct locations. then solve as normal 3x3 I'm glad it helped. I agree with you on the memorization. I don't memorize stuff, I only do moves that I understand. I don't have the sequence I listed memorized, I just visually look at what setup moves are needed to keep all of the edges in the right orientation for the 8cycle. The first time I ran into the parity I too just did the minimal setups without regard to orientation and then fixed the resulting mess piecebypiece. After a while though I think you'll be comfortable enough to spend more time thinking about the setups so that you save a lot of time not having to fix so much later.
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