5x5x5 Tredge + Medge Flipped

[unsolved]

I can solve this in two steps but I was wondering if there was an awesome algo that solves it quickly in 1 step:

 

[natezach728]

There's no one alg that can solve that case just do edge parity on the yellow-orange edge and solve like normal. Also, in the future, post things like this in the one question answer thread

 

[unsolved]

Actually I found an awesome solution to it. Will post in a bit.

 

[IRNjuggle28]

Please stop making a new thread for each question like this. I see two threads within the last week asking for algorithms, and I'm pretty sure you've made at least one more that a moderator merged to another thread. Yes, I'm sure an algorithm for this exists. No, nobody knows of one, because most people would rather actually use an efficient, complete method than doesn't require generating a bunch of miscellaneous, good for nothing algs anytime you run into a case you don't know how to solve. If you really want to search for a bunch of useless algorithms, learn to use one of the computer programs that generates algorithms. There are many. But why do that? Why not just learn to... *gasp* solve a 5x5 properly?

If you want to solve this case, either do pure OLL parity on the orange/yellow tredge and then do an adjacent tredge pureflip of the yellow/orange and yellow/blue tredges, or do pure OLL parity on the orange/blue tredge and then do an adjacent midge pureflip on the yellow/orange and yellow/blue midges.

The main reason algs for this stuff aren't documented is that almost any alg that affects edge pairing either completely ignores the 3x3 state, or only preserves F2L. You won't find many algorithms that affect both because the reduction method completes centers and edges before solving any of the 3x3 stage, and reduction methods are the only commonly used 5x5 method. (Yau is a reduction method)

An example of a method that is not reduction and would have algs that are closer to what you're looking for is K4. You would have to adapt some stuff to make it work for 5x5, but it's worth researching if you really don't want to just use reduction.

 

[Christopher Mowla]

I can solve this in two steps but I was wondering if there was an awesome algo that solves it quickly in 1 step:

You should know there is. I found a one-step algorithm for a similar case you asked for in the past.

It just took me a few minutes to re conjugate it to solve this specific case. (26 single slice turns)
R2 U 2F2 U2 2B U2 2F' U2 2B' R2 U' 2F' 3R2 U2 3R U' R2 U 3R' U2 3R2 2F U' 2F2 U' R2

 

[unsolved]

You should know there is. I found a one-step algorithm for a similar case you asked for in the past.

It just took me a few minutes to re conjugate it to solve this specific case. (26 single slice turns)
R2 U 2F2 U2 2B U2 2F' U2 2B' R2 U' 2F' 3R2 U2 3R U' R2 U 3R' U2 3R2 2F U' 2F2 U' R2

Thanks again. Maybe it appears similar to you Chris, but other than the inversion aspect, it is quite different to me from that example from a while back.

I keep running into oddball positions at the end of my solves, like this:

Now I have seen this on the 4x4x4 cube and I know how to handle it, but that alg doesn't work for 5x5x5. Or at least the one I know doesn't. I thought I had encountered all of the algos I need to handle 5x5x5 but I guess I am not fully armed yet.

 

[Christopher Mowla]

Thanks again. Maybe it appears similar to you Chris, but other than the inversion aspect, it is quite different to me from that example from a while back.

F2 D' 2B2 3F U2 2B U2 2F' U2 2B' R2 U' 2F' 3R2 U2 3R U' R2 U 3R' U2 3R2 2F U' 3F' 2B2 D F2
___R2 U 2F2 U2 2B U2 2F' U2 2B' R2 U' 2F' 3R2 U2 3R U' R2 U 3R' U2 3R2 2F U' 2F2 U' R2

They're pretty similar.

By the way, can you post your alg? I'm curious to see the structure of your alg.

I keep running into oddball positions at the end of my solves, like this:

http://lightningcloudcomputing.com/5x5x5_split.jpg

Now I have seen this on the 4x4x4 cube and I know how to handle it, but that alg doesn't work for 5x5x5.

Unless you are using my 16q/15h algorithm:
z r' U 2L' u' r u 2L' f' 2L' f 2L2 u' r' 2U r z', which can be translated to the nxnxn like this:

4x4x4 (special case) (16q,15h)
[d' B b' u f 2L': [2L, f' u' f] [2L' u': 2L'] ]

5x5x5 (special case) (18q,17h)
[d' B b' u f 2L': [2L, f' 3u' f] [2L' u': 2L'] ]

6x6x6 (general big even cube representation) (inner orbit) (17q,16h)
[3d' B 3b' 3u 3f 3L': [3L, 3f' 3u' 3f] [2L2 3L' 3u': 3L'] ]

6x6x6 (general big even cube representation) (outer orbit) (17q,16h)
[3d' B 3b' 3u 3f 2L': [2L, 3f' 3u' 3f] [3L2 2L' 3u': 2L'] ]

7x7x7 (general big odd cube representation) (inner orbit) (19q,18h)
[3d' B 3b' 3u 3f 3L': [3L, 3f' 4u' 3f] [2L2 3L' 3u': 3L'] ]

7x7x7 (general big odd cube representation) (outer orbit) (19q,18h)
[3d' B 3b' 3u 3f 2L': [2L, 3f' 4u' 3f] [3L2 2L' 3u': 2L'] ]

, an algorithm like 2R U2 2R U2 x U2 2R U2 2L' x' U2 2L U2 2R2 should be very simple to translate to larger cubes.

EDIT:
Just in case you didn't know how, my 26 move alg can be translated to any odd cube size larger than the 5x5x5.

For example, here's the 7x7x7 translation:
R2 U 2-3f2 U2 2-3b U2 2-3f' U2 2-3b' R2 U' 2-3f' 4R2 U2 4R U' R2 U 4R' U2 4R2 2-3f U' 2-3f2 U' R2

In addition, if you just want an alg which flips a tredge and middle edge, just omit the outer two most setup moves of my 26 mover to have a 22 mover:
2F2 U2 2B U2 2F' U2 2B' R2 U' 2F' 3R2 U2 3R U' R2 U 3R' U2 3R2 2F U' 2F2

Lastly, I just realized that if you remove the central slice turns on the even cube, you get a single dedge flip alg (cmowlaparity).
2F2 U2 2B U2 2F' U2 2B' R2 U' 2F' U2 U' R2 U U2 2F U' 2F2

(I didn't make this algorithm from cmowlaparity, but it turns out that it's pretty efficient to build off of it, eh?)

EDIT2:
I "cheated" and merged the 18 single slice move <2L,2R,U2> single dedge flip alg with the two middle edge flip algorithm (which can be extracted from the above two algorithms), and I get a 28 single slice turn solution in <2L,2R,3R,R2,U2>. (Only two single slice moves cancelled\merged, as the combo is 18 + 12.)
2R U2 2-3r2 2L' U2 2R U2 2R' U2 2L 3R U' R2 U 3R' U2 3R2 U R2 U' 2R' U2 2R U2 2R' U2 2R'

 

[unsolved]

By the way, can you post your alg? I'm curious to see the structure of your alg.

Sure.

r2 B2 r- F2 r+ F2 B2 r+ B2 l+ F2 l- F2 B2 r2 R- T2 R2 T+ R- T- R- T2 L- F+ R+ F- L+

My B = your D.
My T = your U.

T+ = U
T- = U'

B+ = D'
B- = D

L+ = L'
L- = L

R+ = R
R- = R'

So it's a right-handed only coordinate system. I hated how L' and R turn the cube in the same direction, so my L and R are signed when spinning those faces in the same direction. Ditto for T and B which is your U and D. I use K for my Back designation.

5x5x5 (special case) (18q,17h)
[d' B b' u f 2L': [2L, f' 3u' f] [2L' u': 2L'] ]

This is the type of algo I am hunting for. I am looking to compile all such cases of 2-cubie trades.

This one you showed was for UFl and UBr.

I am wondering if there is a way to swap only:

UBr and URb
UBr and URf
(UBr and UFr I know how to do)
(UBr and UFl already shown)

UBl and URb
UBl and URf
UBl and UFr
UBl and UFl

I suppose some of these are not possible or just simply not known.

 

[Christopher Mowla]

Sure.

r2 B2 r- F2 r+ F2 B2 r+ B2 l+ F2 l- F2 B2 r2 R- T2 R2 T+ R- T- R- T2 L- F+ R+ F- L+

With your translation key, this translates to (2R2 D2 2R' F2 2R F2 D2 2R D2 2L' F2 2L F2 D2 2R2)(R' U2 R2 U R' U' R' U2 L' F R F' L), which, as I have shown, = (one dedge flip alg)(3x3x3 2 composite edge flip alg) = two steps. Everyone solves this case in two steps (including me) in K5. In addition, your alg cycles 3 corners? I probably can find a shorter sequence than 22 moves to flip a tredge and middle edge if I'm allowed to cycle 3 corners.

---

5x5x5 (special case) (18q,17h)
[d' B b' u f 2L': [2L, f' 3u' f] [2L' u': 2L'] ]

This is the type of algo I am hunting for.

I'm surprised that you prefer my alg over something like 2R U2 2R U2 x U2 2R U2 2L' x' U2 2L U2 2R2 for that case. My alg for this case serves more of a theoretical purpose than a practical solving purpose.

I am looking to compile all such cases of 2-cubie trades.

This one you showed was for UFl and UBr.

I am wondering if there is a way to swap only:

UBr and URb
UBr and URf
(UBr and UFr I know how to do)
(UBr and UFl already shown)

UBl and URb
UBl and URf
UBl and UFr
UBl and UFl

I suppose some of these are not possible or just simply not known.

They're all very well-known.

I found direct algorithms to solve all adjacent cases 4+ years ago (it was not known how to solve them directly before that), but everyone who uses K5 conjugates the above alg to solve all adjacent cases, unless they would like to learn my algs.

This one you showed was for UFl and UBr.

No, it was UFl and UBl.

However, algorithms I personally use to solve all of these cases for the 5x5x5 can be grabbed from the 4x4x4 parity algorithms wiki page and applied to the 5x5x5.
r2 F2 U2 2R U2 x U2 r2 U2 2R' U2 2L 2R U2 2L' U2 x' (directly translatable) (You found around 10 sequences which solve this case with your solver, why are debating if it's possible?)

2L2' U 3l U2 2R' U2 r' U2 x U' 2R' U x' U2 x U' 2L' 2R 2L' (where the bold moves are converted for the 5x5x5 from the 4x4x4 version of my alg)

2L2' U' 3l U2 2R' U2 r' U2 x U 2R' U' x' U2 x U 2L' 2R 2L' (where the bold moves are converted for the 5x5x5 from the 4x4x4 version of my alg)

L' B' 2L' U2 2L' U2 x U2 2L' U2 x' 2R U2 2R' U2 2L2 B L (directly translatable)

Why don't you just use alg.cubing.net to apply the algorithms on the 4x4x4 parity algorithms wiki page to the 5x5x5 yourself? There are all types of tools available.

---

Lastly, clearly we are not thinking in the same realm, and therefore, I'm going to finally learn my lesson and ignore your posts regarding this topic in the future because I feel that I am just producing unnecessary clutter on this site. Unless you reconsider your posting process, that is exactly what I'm going to do. I expect more from you than most because you have proven to be an intelligent person, so I take your repeated actions as being "lazy" or "careless".

Here is a list of things you should think about. It's a "free forum", and so doing the following will not get you banned, but you will get a lot of frowns from people like me who actually take the time to answer your same questions...over and over.

1. Taking this thread, for example, you made a claim that you had an algorithm to solve a specific case. It turns out your algorithm didn't solve the case specified, nor was its structure as you described. Should I mention that you forced me to translate your algorithm into the standard notation so that I could show the world your sequence?
   
   If you have your own notation, great. I have mine. However, I do not write algorithms in my own notation on the forums because it causes unneeded confusion (I'm mostly using standard notations on my own time out of habit now, however).
2. You frequently post that something is unknown when it is almost always just unknown to you. In fact, I don't know why these things are unknown to you because I and others have answered the same questions repeatedly in the past to you.
3. And most of all, you seem to either ignore or disregard any past responses. We all are busy, but at least everyone is responsible for remembering, learning, and documenting all information which was directly given to us when we asked for help in the past.

 

[unsolved]

With your translation key, this translates to (2R2 D2 2R' F2 2R F2 D2 2R D2 2L' F2 2L F2 D2 2R2)(R' U2 R2 U R' U' R' U2 L' F R F' L), which, as I have shown, = (one dedge flip alg)(3x3x3 2 composite edge flip alg) = two steps. Everyone solves this case in two steps (including me) in K5. In addition, your alg cycles 3 corners? I probably can find a shorter sequence than 22 moves to flip a tredge and middle edge if I'm allowed to cycle 3 corners.

Something is awry with my formula if it is cycling edges. The problem is, I flip the pair of tredges with an algo I have known since 1983 on the 3x3x3 cube, so I do it without thinking. And, when I try and do it slowly, I forget where I am in the middle So I undoubtedly have something off in there. I will examine it later and try to correct it.

I gave my 5x5x5 program the position to solve, and it came out with a much shorter sequence. However, as we have seen before, it is not exactly without its bugs, and it leaves one pair of centers on adjacent faces unsolved. This is from the TFS-database being stored in the move generation order and not the solving order, which Tom identified as an earlier bug. Version control in the spinoff for the 5x5x5 brute solver. I will track it down.

I'm surprised that you prefer my alg over something like 2R U2 2R U2 x U2 2R U2 2L' x' U2 2L U2 2R2 for that case. My alg for this case serves more of a theoretical purpose than a practical solving purpose.

I like the new one better. I am just moving from 4x4x4 solving to 5x5x5 solving so I am not sure yet which apply and which do not. Some of the 4x4x4 algos make mish-mosh out of wing edge solves. Example:

r2 T2 r2 [Tt]2 r2 t2 works on the 4x4x4 but you need [Rr]2 F2 T2 r2 T2 F2 [Rr]2 when solving moving wing edges from the back to the front.

They're all very well-known.

Well, not to me. I also can't find a way to move URf to UFl on the 5x5x5.

I found direct algorithms to solve all adjacent cases 4+ years ago (it was not known how to solve them directly before that)...

And wasn't that a worthwhile pursuit? I am not sure why people are harassing me about my layer-by-layer solve. I am just doing this for fun! Solving these remaining edges without perturbing the centers is of interest to me. Plus I am building a database of known-solves to assist with the brute force solver I am writing. The more positions is has in the database, the faster it can find forced solves from further away. There is a reason why I am engaged in these pursuits. The end result will be very fast forced solvers for the speed solving community.

No, it was UFl and UBl.

I stand corrected.

You found around 10 sequences which solve this case with your solver, why are debating if it's possible?

You are reading too much into my commentary. I was unsure if ALL of these conditions can arise on the 5x5x5 because I have not yet experienced all of them.

Why don't you just use alg.cubing.net to apply the algorithms on the 4x4x4 parity algorithms wiki page to the 5x5x5 yourself? There are all types of tools available.

I have not seen every case listed on those pages which describe the swapping of just 2 cubes. And one or two of the algos I tried either do not translate to the 5x5x5 or I made the same mistake several times in trying to reproduce the equivalent moves on the 5x5x5. I never claimed to be infallible, nor to be an expert. Hint: look at my forum name.

.... I'm going to finally learn my lesson and ignore your posts regarding this topic in the future... so I take your repeated actions as being "lazy" or "careless".

I am not denying being either at times. But, since my last series of posts, I have:

1) Figured out how to build 5.60 GHz servers that fit into 1.75 inches of rack space.
2) Finished writing a 550 page book, received an advance for turning it into a trilogy, and received one offer for the rights to turn it into a screenplay
3) Successfully deposited one child into college
4) Survived crashing my sports car at 85 MPH, no seat belt, convertible top down.

Item #4 actually required most of my time. Not for the crash, for the being able to get to the point where I can chew my own food again. So, yeah, I suppose part of me is still a little forgetful.

 

[goodatthis]

Survived crashing my sports car at 85mph, no seatbelt, convertible top down.

Something tells me that would lead to instant death. Crashing anything at 85mph is hard to survive, but in a sports car, with the top down, and without a seatbelt?! Come on man.

 

[Christopher Mowla]

And wasn't that a worthwhile pursuit? I am not sure why people are harassing me about my layer-by-layer solve. I am just doing this for fun! Solving these remaining edges without perturbing the centers is of interest to me.

Yeah it was worth the pursuit...I can tell many others think so as well, considering that the 4x4x4 parity algorithms page, which is relatively new, is already the #11 most viewed page in the wiki (#1 is the main page). I worked hard to make that page as easy to understand as possible, and I hope my work will be appreciated in years to come!

By the way, I just changed my template to not link to alg.cubing.net in "Play View" so that you can easily change which cube an algorithm is applied to. This way, you can easily apply any of the algorithms on the 4x4x4 parity algorithms page to the 5x5x5...I should have done this as soon as I went from alg.garron.us to alg.cubing.net, but I didn't. So go view the 4x4x4 parity algorithms page again and click on any algorithm to see a different view with a cube size drop-down box.

I don't like when people tell you to use reduction, but I guess not everyone on this site knows that some people just solve, like you said, for fun and don't care about speed.

I'm not sure what the move optimal sequence for flipping 4 middle edges on a 5x5x5 is, but, just in case you're curious, I just found a 30 single slice turn sequence which flips a tredge and 3 middle edges.

2R2 U' r U' F2 U r' 3B U2 3F2 U2 R U' F2 U R' U2 3F2 U2 3B' U' 2L' U2 2R' U2 2L U2 2R2

In addition, your alg cycles 3 corners? I probably can find a shorter sequence than 22 moves to flip a tredge and middle edge if I'm allowed to cycle 3 corners.

I don't believe I can get fewer moves than 22 slice turns to flip a tredge and middle edge, even if I allow 3 corners to be cycled (in fact, it took 26 single slice moves for me to do this: 2R2 U2 2L F2 2L' F2 l' F2 L U' l' 3F2 U2 3B' U' F2 U 3B U2 3F2 l U' 2R2).

I gave my 5x5x5 program the position to solve, and it came out with a much shorter sequence. However, as we have seen before, it is not exactly without its bugs, and it leaves one pair of centers on adjacent faces unsolved. This is from the TFS-database being stored in the move generation order and not the solving order, which Tom identified as an earlier bug. Version control in the spinoff for the 5x5x5 brute solver. I will track it down.

How short was the sequence? I know off hand that, if you don't care about centers, then flipping a tredge and middle edge requires at most 19 single slice turn moves:
2B2 U' 2F' 3R2 U2 3R U' R2 U 3R' U2 3R2 2F U R2 U2 2F U2 2B2


Lastly, I am sorry to hear you got into a lethal accident, but I'm glad to see you are alright given the circumstances. However, you can't expect anyone on the forums to know anymore than what they see you post.

Good luck with your book!

 

[qqwref]

Well, not to me. I also can't find a way to move URf to UFl on the 5x5x5.

R B setup to a UFl/UBl swap.

I was unsure if ALL of these conditions can arise on the 5x5x5 because I have not yet experienced all of them.

What? If you think this way, you REALLY need to read more about cube theory and permutations. For instance, on an NxNxN cube, any 3-cycle of similar pieces is possible, therefore if any 2-cycle is possible in an orbit, all 2-cycles are possible in that orbit. (And even without knowing about 3-cycles, you can use setup moves.) This should be immediately obvious.

2) Finished writing a 550 page book, received an advance for turning it into a trilogy, and received one offer for the rights to turn it into a screenplay
4) Survived crashing my sports car at 85 MPH, no seat belt, convertible top down.

And I ran a 3-minute mile.

 

[goodatthis]

Being thrown from the car was the life saving event.

That's still extremely surprising that you survived after being thrown at close to 85 mph out of a car. That's a lot of force.
What did the car/you crash into?

Jupiter Rising.

In movie theaters February 2015.

Thats a very short time for your recently written book to be made into a screenplay, and then a movie.

also interesting that the book was released in 2004

 

[unsolved]

I worked hard to make that page as easy to understand as possible, and I hope my work will be appreciated in years to come!

Well I have always appreciated it and continue to appreciate it though I don't pretend to understand all of the inner workings of the cube. I wish I had a better understanding of it.

By the way, I just changed my template to not link to alg.cubing.net in "Play View" so that you can easily change which cube an algorithm is applied to. This way, you can easily apply any of the algorithms on the 4x4x4 parity algorithms page to the 5x5x5

Ahhh, this will no doubt be an awesome feature! I have to check this out after more of the non-believers shoot more arrows at me.

How short was the sequence?

Something is telling me it was 17 total moves if you count the face + slice moves as 1 even if it is a double turn. I will have to double check and post it. I will run it again.

Lastly, I am sorry to hear you got into a lethal accident

Correction: nearly lethal accident. I now have more titanium in my skull than a Big Bertha 1-iron.