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 Post subject: 2x3x4 Edge Parity Help!!
PostPosted: Mon Apr 18, 2011 12:26 pm 
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I was solving the fully functional 2x3x4 and ended up in a situation with just 2 single edges that need swapping. I believe this is the same situation that happens on the 3x4x5. Does anyone know how to solve this parity and can help me? Algos please!


Last edited by 77mouser on Mon Apr 18, 2011 7:19 pm, edited 1 time in total.

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 Post subject: Re: 2x3x4 Edge Parity Help!!
PostPosted: Mon Apr 18, 2011 1:42 pm 
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I haven't gotten my 2x3x4 assembled yet, but I look forward to the challenge. One caveat of the 2x3x4, or any N*(N+1)*(N+2) cuboid is that the typical Domino algorithms won't work when solving the last layers. One set of layers (either the inner 2x3 slices or the outer 2x3 slices) along the 4-length can be solved like a 2x2x3 (which can mostly use unmodified Domino corner algorithms), but that leaves you with two 2x3 layers remaining which can be solved only using half-turns. Those outer parts (the four corners) are chiral and can only have one orientation when the puzzle is in a rectangular state. That said, they should be commutable, though it's not unlikely they could get a parity.

BTW, your diagram shows a 3x4x5.

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 Post subject: Re: 2x3x4 Edge Parity Help!!
PostPosted: Mon Apr 18, 2011 2:33 pm 
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The L and R sides of this 3x4x5 diagram are the same as the 2x3x4 and shows the 2 swapped edge pieces. I'm looking for how to switch those 2 edges on the 2x3x4-I'm assuming you would use the same method as the 3x4x5.


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 Post subject: Re: 2x3x4 Edge Parity Help!!
PostPosted: Mon Apr 18, 2011 5:29 pm 
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I think you have used my picture from this thread, right? My solution to this parity on the 3x4x5 cannot be translated to the 2x3x4.
I have solved the 2x3x4 (Traiphum's and Garrett's) many times and I do not recollect a similar parity situation as I understand it reading your description. Are you sure that just two edges are swapped? Can you, please, provide a picture of the real 2x3x4? I think the 2x3x4 is much easier than the 3x4x5 and it does not help very much to come from a 3x4x5 solution.

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 Post subject: Re: 2x3x4 Help!!
PostPosted: Mon Apr 18, 2011 7:24 pm 
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I see that it is an extended 2x2x3 with an inner 2x2x3 and an outer 2x2x3. My real question is how you solve the outer 2x2x3 corners after you've solved the inner withour messing up anything else. Help?


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 Post subject: Re: 2x3x4 Edge Parity Help!!
PostPosted: Tue Apr 19, 2011 6:03 am 
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I don't remeber how I did the corners, but it's pretty easy to swap two edges. On a 3x3 you can swap two pair of edges with (FR)3. If you do that on the 2x3x4 with T=2x3, F=3x4 and R=2x4, the edges in the right face are non-existent, so the net result is a single swap of the TF- and BF-edge.

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 Post subject: How to solve the fully functional 3x4x5.
PostPosted: Tue Apr 19, 2011 11:29 am 
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How to solve a fully functional 3x4x5.

Image

Step 1. x y l2;
Step 2. (l' U2)*2 F2 l' F2 r U2 r' U2 l2 (known algorithm);
Step 3. F2;
Step 4. l2 U2 r U2 r' F2 l F2 (U2 l)*2 (this algorithm is opposite to algorithm in item "Step 2.");
Step 5. F2 l2.

:D :D :D


Last edited by pytlivyj_1 on Thu Apr 21, 2011 9:42 am, edited 1 time in total.

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 Post subject: Re: 2x3x4 Edge Parity Help!!
PostPosted: Tue Apr 19, 2011 11:34 am 
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maarten wrote:
I don't remeber how I did the corners, but it's pretty easy to swap two edges. On a 3x3 you can swap two pair of edges with (FR)3. If you do that on the 2x3x4 with T=2x3, F=3x4 and R=2x4, the edges in the right face are non-existent, so the net result is a single swap of the TF- and BF-edge.
I understand that 77mouser is talking about the edges on the 2x4 faces. I have revisited the 2x3x4 (Garrett's fully functional version) and have solved it many times today. Not a single time I had a parity of just two edges in the 2x4 faces.
I had always to do swaps of two pairs or 3-cycles of those edges.
One would expect that you should get parity in 50% of the solves, if there is a parity.
Again, can you please show photos of all six faces showing the parity?
Your original post used a diagram of a 3x4x5 (I didn't mind that you had used it) and from that I made the conclusion that we are talking about the 2x4 edges (or 3x4 edges).

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 Post subject: How to solve the fully functional 3x4x5.
PostPosted: Tue Apr 19, 2011 11:46 am 
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Yesterday I have seen this task on a forum. This task very much has interested me. Today I have made sketches on a paper of known algorithm and at me it has turned out to solve this task!


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 Post subject: Re: How to solve the fully functional 3x4x5.
PostPosted: Tue Apr 19, 2011 2:45 pm 
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pytlivyj_1 wrote:
How to solve the fully functional 3x4x5.

Yes, but 77mouser has problems with the edges of a 2x3x4. The algorithm given cannot be executed on a 2x3x4, because there are no inner slices on a 2x3x4 that can be turned by 90 degree turns. That's why I assume that a real parity of two swapped edges as on a 4x4x4 is not possible at all, in this case.
EDIT: I solve the 2x3x4 by these steps
1. From shapeshifted state back to cuboid
2. Corners (with my usual Domino algorithm; shapeshifting moves may be necessary)
3. 2x4 edges (swaps of edge pairs are trivial; 3-cycle may be necessary)
4. centres (trivial)
5. 2x3 edges (trivial, see maarten's sequence)

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 Post subject: Re: How to solve the fully functional 3x4x5.
PostPosted: Tue Apr 19, 2011 5:39 pm 
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Konrad wrote:
because there are no inner slices on a 2x3x4 that can be turned by 90 degree turns.
A 2x4x1 midsection can be rotated by 90 degrees.
Attachment:
90 midsection.PNG
90 midsection.PNG [ 6.63 KiB | Viewed 2621 times ]

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 Post subject: Re: How to solve the fully functional 3x4x5.
PostPosted: Wed Apr 20, 2011 3:29 am 
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stardust4ever wrote:
Konrad wrote:
because there are no inner slices on a 2x3x4 that can be turned by 90 degree turns.
A 2x4x1 midsection can be rotated by 90 degrees.
Attachment:
90 midsection.PNG
One midsection is correct. I had said "slices", plural. Look at the algorithm above. It relies on the existence of two inner slices l and r which can be turned by 90 degrees. This is possible on a 3x4x5 but not on a 2x3x4.
I cannot prove it, but I strongly believe that there is no parity of just two edges on the 2x4 faces of a 2x3x4.
I have done at least 30 - 50 solves and have not seen this kind of parity
Image (The rUF and lUB edges are exchanged and nothing else.)
I understood the original description of 77mouser's problem exactly as in this diagram.
(BTW, in this diagram the U, E and D layer can be turned by 90 degrees, all other layers by 180 degrees, only.)
What I'm getting quite often is this:
Image
Here we need a 3-cycle of edges and I have no problem solving this.

Is nobody out there, who can prove in a mathematical sense if there is a parity on a2x3x4 or not?
If it exists, how likely is it that I have not seen it doing 30 to 50 solves?

EDIT: BTW, stardust, have you assembled yours? Yesterday, I disassembled my Garrett's puzzle and have found a single part that is glued. Mine had been assembled and stickered by Garrett, but I think the assembly should be easy for you.

EDIT2: I'm a bit confused about the title of this thread. In the sidebar my post shows up as "◦Re: How to solve the fully functional 3x4x5. by Konrad 33 minutes ago."
On top of this thread it says "2x3x4 Edge Parity Help!!" as in 77mouser's original post.
How that?
77mouser, have you changed the title?
When I click on "View your posts" I get " 2x3x4 Edge Parity Help!! in Solving Puzzles"
Strange

EDIT3: If somebody is interested, here is my algorithm for the 3-cycle of edges
[I have reused my usual corner algorithm for cuboids. Maybe there are shorter sequences, but this is, how I resolve the 3- cycle situation. r2 y (R2 U R2 D‘ R2 U R2 U‘ R2 U‘ D R2) f2 U‘ (R2 U R2 D‘ R2 U R2 U‘ R2 U‘ D R2) U‘ My diagram shows the result of this when applied to a solved 2x3x4. The notation should be obvious. r and f are the inner slices next to F and R. ]

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 Post subject: Re: 2x3x4 Edge Parity Help!!
PostPosted: Wed Apr 20, 2011 4:48 am 
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Thanks so much Konrad. I have it solved! It seemed to help if I solved the trivial centers later in the process-I was solving those first. Also the hint about the outer corners helped-sometimes you have to shapeshift to get those right.
I didn't change the title at all-don't know how.


Last edited by 77mouser on Wed Apr 20, 2011 1:57 pm, edited 1 time in total.

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 Post subject: Re: 2x3x4 Edge Parity Help!!
PostPosted: Wed Apr 20, 2011 5:13 am 
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77mouser wrote:
Thanks so much Konrad. I have it solved! It seemed to help if I solved the trivial centers later in the process-I was solving those first. Also the hint about the outer corners helped-sometimes you have to sahpeshift to get those right.
I didn't change the title at all-don't know how.
Congratulation! :)
I think the trivial centres do not matter so much.
May I ask again, are you sure that you have observed a "parity" where only two edges are swapped, as in my picture?
The corners are trivial, if you do no shapeshifting moves. Otherwise, I use the part of my hidden spoiler in brackets.

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 Post subject: 2x3x4 Edge Parity Help!! (3x4x5)
PostPosted: Wed Apr 20, 2011 10:54 am 
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pytlivyj_1 wrote:
How to solve the fully functional 3x4x5.
...
Step 1. x y l2;
Step 2. (l' U2)*2 F2 l' F2 r U2 r' U2 l2 (known algorithm);
Step 3. F2;
Step 4. l2 U2 r U2 r' F2 l F2 (U2 l)*2 (this algorithm is opposite to algorithm in item "Step 2.");
Step 5. F2 l2.

:D :D :D
I guess your algorithm does not a swap oft two edges but a 3-cycle.
I have done it on a solved Gelatinbrain 5x5x5 (I have no simulation of a 3x4x5 ).
l2; (l' U2)*2 F2 l' F2 r U2 r' U2 l2
F2;
l2 U2 r U2 r' F2 l F2 (U2 l)*2
F2 l2.
translates to (the &2 refers to a slice move; the first l2; l’ becomes l= L&2)
L&2,U'2,L'&2,U'2,F'2,L'&2,F'2,R&2,U'2,R'&2,U'2,L'2&2,
F'2,
L'2&2,U'2,R&2,U'2,R'&2,F'2,L&2,F'2,U'2,L&2,U'2,L&2,
F'2,L'2&2,
in Gelatinbrain Notation.
This ist he result on the 5x5x5:
Image
If I make now a 3x4x5 out of the 5x5x5 by taking out some layers I’ll get this
Image

(The grey stuff is to be removed to get a 3x4x5)

EDIT: My own solution for this 3x4x5 parity is quite complex. You have to consider that after an l turn you cannot do U (at least not on the physical puzzle I have) but only (Uu) together. Therefore, something like the algorithm above will destroy some centre pieces as well and you have to rebuild those.

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 Post subject: 2x3x4 Edge Parity Help!!
PostPosted: Thu Apr 21, 2011 9:26 am 
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2 Konrad

The cube 5х5х5 is not a fully functional cuboid 3х4х5. The principle of the decision at them differs.
At rotation of a layer U on a cube 5х5х5 it is necessary to rotate 2 top layers and after performance of algorithms to not pay attention to layers U and D of a cube 5х5х5. It is necessary to consider only layers u, 3u and d - is similar to a fully functional cuboid 3х4х5!!! Then all will be correct!!!
How I have solved this cuboid 3х4х5.


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 Post subject: Re: How to solve the fully functional 3x4x5.
PostPosted: Fri Apr 22, 2011 6:43 pm 
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Konrad wrote:
EDIT: BTW, stardust, have you assembled yours? Yesterday, I disassembled my Garrett's puzzle and have found a single part that is glued. Mine had been assembled and stickered by Garrett, but I think the assembly should be easy for you.
I got mine assembled yesterday. I colored all of the pieces black with a sharpie marker and glued two internal 3x3x3 edge parts to one face center in an "L" shape. Getting the right screws were tricky as the 3/4" 4-40 screws my local hardware store sold either had dome-shaped caps which interfered with the mech, or tapered flatheads which do not sit flush inside the face centers. I finally ended up using the tapered flat top screws, which worked oddly enough. There is hardly any clearance at all for the screw heads. My next dilemma was I attempted to assembled the puzzle by building up two thirds of it as 4x2 layers. Screwing the final screw into the puzzle did not work too well; the puzzle popped, then I realized I needed to screw the last screw through the 2x3 face (long end) of the puzzle instead. Once properly assembled, it works like a charm. I need to order a set of 4x4x4 tiles from Cubesmith though, as I thought I had some but can't find them. The shape-shifting aspect is fun for the time being.

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 Post subject: Re: 2x3x4 Edge Parity Help!!
PostPosted: Fri Apr 22, 2011 11:32 pm 
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I finally received replacement parts for mine from Shapeways (I somehow got 8 identical corners in the original shipment; kudos to Garrett and Shapeways for sorting it out).

Mine's a bit stiff, especially compared with the 3x4x4. I'll have to play with the screw tension and just twist it a lot, I guess.


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