# Intuition Competition [Weekly]



## Dane man (Jun 16, 2014)

*Temporarily on Hiatus - Competition Paused*

For me, the magic of the Rubik's cube isn't in the cube being solved, it isn't even in the speed of the solution. The magic is in the intuitiveness and cleverness of the solution. It's in the challenge of having to work out a way to put it back together based on nothing but pure logic. It's in the mystery of having to find a way to fix it without the assistance of other resources. It's in the pride of knowing that you figured it out on your own, using nothing but your brain.

Sadly, that magic has been almost completely lost in a world where methods and algorithms are simply read, memorized, and executed. This kind of solving takes away the novelty of the cube, even if it increases the speed at which it can be solved.

And so, to bring the magical mystery back to solving the cube, I present...

*The Intuition Competition*

*The competition works like this:*
Once a week I will present an algorithm goal, for example "trade the position of two edges without effecting the bottom two layers or the other edges (the LL corners don't matter)". Given the algorithm goal, your goal is to create an algorithm from scratch that accomplishes that task using the fewest moves possible. Images will sometimes be included with the challenges to help explain the algorithm goal. Difficulty will be gauged each week, and the challenges will become harder as contestants present more intuitive solutions (Ex: "Using only the U and R faces, trade the corners...").

You get a week to find a solution and post it here in the thread. With your solution, you must post a move by move explanation of what it does and how it works (If you used premoves, or pseudo-blocks, please post what you used.). At the end of each week, winners (1st, 2nd, 3rd) will be chosen based on move count (Slice moves will count as single moves, not double moves, unlike FMC competitions. This is because it logically is used as a single move even though it has the same effect as two opposing side moves. The rest is HTM. The exception is when slice moves are intentionally forbidden.). And even if there are already submissions faster than yours, you're still encouraged to participate (because that's how you'll get better, and it's fun!)

*
Rules:*

_*You are not allowed to use any memorized algorithms*_ (including partial algorithms, and *known* commutators and conjugates. This means that no set of moves that is already memorized may be used).
_*You are allowed and encouraged to apply commutative and conjugative techniques*_ to your solution (but remember, nothing that is already known or memorized (see rule 1). This is so that total reliance on intuition is required).
_*You are not allowed to use any piece of software or hardware to assist you in finding a solution*_ (other than the cube, your brain, your hands, and perhaps a pen and paper).
*You are allowed and encouraged to use techniques from FMC solving* (for example premoves, pseudo-blocks, inverse scrambles (if a scramble is given with the algorithm goal), etc; you may NOT use algorithms, *known* commutators, etc. as seen in rule 1).
_*If you can't explain what everything does*_, you probably copied it from somewhere, used a memorized alg, or found a solution on accident, *therefore your solution will be discarded*. (If you've found a solution on accident, you are encouraged to study it out to see what it does so that you can explain it and submit a valid solution.)
Last but not least, *be honest about your solution*. This is for fun, not showing off.

And so we begin with our first challenge!
*
This Week's Challenge:*
6/30/14 - _Slice moves are prohibited for this challenge. Any replacement slice moves (L R', F2 B2, etc.) are also forbidden. The UF edge must go to the UB position, the UB edge to UR, and the UR edge to UF. The orientations of these edges must be maintained. Nothing else on the cube may be modified in any way, permutation or orientation._



Spoiler: Past Challenges / Winners



6/16/2014 - *1st: TDM, 2nd: Lucas Garron, HM: Lucas Garron* - _With the F2L solved, you must trade the UF edge with the UR edge, and the UL edge with the UB edge (this is equal to trading the UF and UB faces then performing U). The solved F2L must be maintained. Edges must maintain orientation. U layer corners are ignored._ 

6/23/14 - *1st: TDM, 2nd: porkynator* - _With the F2L sovled, you must flip the orientation of all four edges of the U layer without affecting their permutation. The orientation and permutation of the corners don't matter. F2L must be maintained._


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## TDM (Jun 16, 2014)

Dane man said:


> This Week's Challenge:[/B][/U]
> _With the F2L solved, you must trade the UF edge with the *UR* edge, and the UL edge with the UB edge (this is equal to trading the UF and UB faces then performing U). The solved F2L must be maintained. U layer corners are ignored._


ftfy (assuming you meant that by your description and saying F2L must be maintained)

My solution:


Spoiler



M2 (bring UL/UR edges to D layer...)
U M2 U' (... and insert them where they should go)
M2 (solve other two U layer edges)
E2 M' E2 M (solve centres; see below)


Spoiler



[E2, M']
E2 swaps F and B centres and turns the E layer 180 degrees
M' moves the other two centres which needs to be swapped, U and F, to the E layer
E2 swaps them and returns the E slice to being solved
M corrects the M slice


Solution: M2 U M2 U' M2 E2 M' E2 M (9 STM)

Edit: By the same logic as what is in the above spoiler, M' E2 M E2 has the same effect as E2 M' E2 M, just with the steps in a different order. By changing this we get M2 U M2 U' M2 E2 M' E2 M, which can be written as M2 U M2 U' M2 M' E2 M E2, which cancels to:

*Final solution: M2 U M2 U' M E2 M E2 (8 STM)*


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## Dane man (Jun 16, 2014)

TDM said:


> ftfy (assuming you meant that by your description and saying F2L must be maintained)



Oh, thank you for noticing that. I'll be more careful next time. And nice solution BTW.


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## Lucas Garron (Jun 16, 2014)

I'd basically end up recreating an <M, U> Z-perm from scratch

Bring out the opposite pairs of edges and bring them back to U in the right places: M2 U M2 U'
Fix the centers: (M2) M E2 M' E2

Gives M2 U M2 U' M' E2 M' E2 (14h, 22q, 8s, 8e)

If I were trying to make the alg easy to execute, the second half would be: U2 M' U2 M2 U2 M' U2

... which gives: M2 U M2 U M' U2 M2 U2 M' U2 (15h, 24q, 10s, 10e)
(which is of course a symmetry of the fastest <M, U> Z-perm>)

----------------

An F2L-ish solution:

Switch UB and UL: U [R, U] R
Switch UF and UR: U' [R', U'] R'
Put DFL back: [R2, L' U' L U]

Expanded, this gives: U R U R' U' R U' R' U' R U R L' U' L U R2' U' L' U L (21h, 22q, 21s, 21e)

----------------

And if we don't care about orientation:

A symmetry of [F': [R, U]] is the shortest we could do to affect LL, but it doesn't work. So let's:

Bring out an F2L pair (and leave it there): R U R'
Try to bring it back differently: F' L' U' L F

Mirror that to get: L' U' L F R U R' F' (8h, 8q, 8s, 8e)


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## Lucas Garron (Jun 16, 2014)

Dane man said:


> Lucas Garron said:
> 
> 
> > Mirror that to get: L' U' L F R U R' F' (8h, 8q, 8s, 8e)
> ...


?
It switches the correct edges (although it flips two) and:



Dane man said:


> _U layer corners are ignored._



You should also be exact and specify that edges should keep their U sticker facing up. It's possible to define orientation so that either pair of swapped edges may (both) flip while swapping (although my last alg doesn't do that).


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## Dane man (Jun 16, 2014)

Lucas Garron said:


> ?
> It switches the correct edges (although it flips two)


I meant that instead of trading UF and UR it traded UF and UL, the other pair doing the same, but your solution still counts if rotated.

And yeah, next time I'll use images to describe the challenge.


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## Lucas Garron (Jun 16, 2014)

Dane man said:


> I meant that instead of trading UF and UR it traded UF and UL, the other pair doing the same, but your solution still counts if rotated.



I'm a little confused. That animation definitely shows UF moving to UR and RU moving to UF.


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## Rocky0701 (Jun 17, 2014)

Dane man said:


> F is the green side not the red.


His solution works perfectly. You are over thinking it.


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## Dane man (Jun 17, 2014)

Rocky0701 said:


> His solution works perfectly. You are over thinking it.


Yeah, you're right. I got a little carried away with the specifics. Sorry


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## Lucas Garron (Jun 17, 2014)

Actually, this is a great way to come across regular algs. 

Switch UF and UR Domino-style: R2 U' R2 U R2
Switch UB and UL Domino-style: B2 U B2 U' B2
Fix E layer: B2 E B2' E'

Now simplify: R2 U' R2 U R2 B2 U B2 D' R2 E' y' (12h, 18q, 11s, 12e )


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## Dane man (Jun 17, 2014)

Lucas Garron said:


> Actually, this is a great way to come across regular algs.
> 
> Switch UF and UR Domino-style: R2 U' R2 U R2
> Switch UB and UL Domino-style: B2 U B2 U' B2
> ...



Wow, I like that, nice work.


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## Lucas Garron (Jun 17, 2014)

Silly idea:

Swap UF and UR: R2 U' R2 U R2
Switch with UB and UL (flipping *both*): S M
(Finish commutator)

Gives [R2 U' R2 U R2, S M] (18h, 24q, 14s, 14e )


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## Dane man (Jun 23, 2014)

This weeks winners are TDM and Lucas Garron. Both of them were able to create an algorithm of at most 8 moves to accomplish the task, though, because the two algorithms of 8 moves were the same algorithm, the winners are decided by first come first serve, and so, Lucas Garron received 2nd place. But he did win an honorable mention for coming up with many different methods to solve the problem.

*1st - TDM
2nd - Lucas Garron
HM- Lucas Garron*

Task: _With the F2L solved, you must trade the UF edge with the UR edge, and the UL edge with the UB edge (this is equal to trading the UF and UB faces then performing U). The solved F2L must be maintained. Edges must maintain orientation. U layer corners are ignored._

Onto the next challenge:
_With the F2L sovled, you must flip the orientation of all four edges of the U layer without affecting their permutation. The orientation and permutation of the corners don't matter. F2L must be maintained._


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## TDM (Jun 23, 2014)

Spoiler



*A solution:* [M' U2 M U2 L R': F] U'


Spoiler



M' U2 M U2 L R' // setup edges to F to misorient
F // flip edges
L' R U2 M' U2 M // undo setup
U' // AUF (the F move did a 4-cycle as well as changing the edges' orientations; this cycle is undone here)



L R' can be written as M' and a rotation, which we can remove for this challenge:
*Final solution: [M' U2 M U2 M', U]* (12 STM)


This is actually quite an easy alg to do quickly too. E: I might actually learn this; it's easy to remember and could be useful, especially in BLD.


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## porkynator (Jun 23, 2014)

I'll try this.


Spoiler



If you don't care about F2L, the sequence
A := M' U2 M U2 M'
flips (without permuting) the two edges on M and U layers (I don't know if I have to explain how I have found this).
Its inverse (A' := M U2 M' U2 M) obviously flips without permuting the same two edges (because "flipping without permuting two edges" is the inverse of itself) and affects the F2L in the opposite way (notice how both those algs keep all the F2L pieces, except one center, in the F2L). Therefore, the commutator 
[A,U] = A U A' U' = M' U2 M U2 M' U M U2 M' U2 M U'
Flips 2 edges (A), moves them on the side (U), flips the other 2 edges that are now on MU while also solving the F2L, since the U move didn't affect centers (A') and puts the edges back in place (U').
So my alg is:
M' U2 M U2 M' U M U2 M' U2 M U' (18 HTM, 12 STM, 22 QTM, 16 SQTM)
I know I didn't have to save the corners, but I just haven't found anything better than this (yet).


E: same solution as TDM 
But I have quickly found another one:


Spoiler



A := [F,R'] = F R' F' R
flips two edges; this is rather obvious if you have some ZZ/Petrus background.
C := [B,L'] = B L' B' L
does the same thing with the other edges. Now I have to solve the F2L and permute the LL edges correctly. An easy way to do so is:
L' (saves one pair before solving the other)
[U,R] = U R U' R' (standard F2L case)
L (the other pair comes back)
[U,L] = U L U' L' (solves F2L)
U2 (done).
In conclusion:
[F,R'] [B,L'] [L':[U,R]] [U,L] U2 = F R' F' R B L' B' U R U' R' L U L U' L' U2 (17 HTM, 17 STM, 18 QTM, 18 SQTM)


E2: according to the rules, only STM counts, so my first solution is way better.


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## Dane man (Jun 28, 2014)

I wanna participate as well. It's to have fun, and I did. (but I don't think I'm technically allowed to win, I am the judge after all).

Now while this isn't a very efficient algorithm, it is done without slice moves and uses only RUF moves, which was a challenge in and of itself.



Spoiler



R' - bring UR to the F face to prepare for reorientation
F - bring UF to the R face for the same reason
R - pull up the old UF piece, now oriented
F' - bring up the F2L slice to prepare to make room for the old UR piece
U' - prepare F2L slice for insertion and make room for old UR piece
F' - bring up old UR piece now oriented, aligning F2L slice
U F - insert F2L slice and bring it back down

This first sequence takes the UF and UR faces and flips them, but it also does a rotation counter-clockwise of three of the edges (UR, UF, UL). So in order to counteract that to achieve what I wanted, I applied a mirror of the same algorithm twice. Twice because I need to perform two clock-wise rotations of UF, UR, and UB in order to return the correct permutation. Also, to achieve 4 flipped edges while flipping six, I realized I needed to flip one edge three times, this is done by flipping an edge that is already flipped, and flipping one that isn't, then flipping the one that was flipped twice with one that wasn't ever flipped. Thankfully, at the end of the first alg, an unflipped edge and a flipped edge were already in position. And at the end of the second, the edges were once again already in position.

(F R' F' R U R U' R')2 - perform the mirror twice for the above reasons.

U2 - rotate the flipped pieces into position.

Done. It is 25 moves, so no gold medal for move count there but it works.


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## Dane man (Jun 30, 2014)

This weeks winners are TDM and porkynator! Gongratulations to them!

*1- TDM (12 moves STM)
2- porkynator (17 moves STM)*

The task they were given was:
_With the F2L sovled, you must flip the orientation of all four edges of the U layer without affecting their permutation. The orientation and permutation of the corners don't matter. F2L must be maintained._

The new task to be completed is:
_Slice moves are prohibited for this challenge. Any replacement slice moves (L R', F2 B2, etc.) are also forbidden. The UF edge must go to the UB position, the UB edge to UR, and the UR edge to UF. The orientations of these edges must be maintained. Nothing else on the cube may be modified in any way, permutation or orientation._


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## cmhardw (Jun 30, 2014)

I just discovered this thread, but I have a question about the rules. The way I read the rules, commutators and conjugates are strictly forbidden. I view commutators and conjugates as being 100% intuitive, as that is how they are constructed.

Are commutators and conjugates really banned for this competition? Or am I misunderstanding the rules?

--edit--
For example, I consider this solution to this week's challenge to be 100% intuitive. Would you accept this solution?


Spoiler






> *This Week's Challenge:*
> 6/30/14 - _Slice moves are prohibited for this challenge. Any replacement slice moves (L R', F2 B2, etc.) are also forbidden. The UF edge must go to the UB position, the UB edge to UR, and the UR edge to UF. The orientations of these edges must be maintained. Nothing else on the cube may be modified in any way, permutation or orientation._



First I will cycle the blocks ([UF+UFR] -> [UB+UBL] -> [UR+UBR])
I will do this by first swapping the [UF+UFR] block with the [Ub+UBL] block.

F d F' U2 F d' F' U2 F d F'

I will now swap the [UF+UFR] block with the [UR+UBR] block.

F d' F' U F d F' U' F d' F'

Notice how I started the insertion for this second cycle to go F d' F' instead of F d F'. This is because I am really "Ferris wheeling" these pieces around the top layer. However, if you break this down into it's mechanics I am performing two "double swaps" like I described above. When I put this all together, because of the "Ferris wheeling" nature of what I am doing, tons of moves cancel.

(F d F' U2 F d' F' *U2 F d F'*) (*F d' F' U* F d F' U' F d' F')

All of the bolded moves cancel, and the bolded as well as italicized moves combine. This makes the maneuver:

*F d F' U2 F d' F' U' F d F' U' F d' F'*

Now what remains to be done is to cycle the corners back to their original locations. The corners need to cycle (UFR->UBR->UBL). I perform this cycle by first getting these corners out of all being in the same layer.

Start with *R'*. This makes the cycle (FDR -> FUR -> UBL). FDR and FUR are interchangeable on the F slice. UBL can be inserted into the location of FDR by R' B2 R. I know that looking at the cycle that (FDR -> FUR -> UBL) that the corner at UBL, which I will call the "lone corner" has to cycle to FDR. I will cycle it there, without affecting any other piece on the F slice, by doing *R' B2 R*. I have now affected only one location on the F slice, namely FDR. I will now interchange such as to put the FUR corner in the FDR location by doing *F*. Next I will undo the insertion I did before, which will take the corner currently in UBL and move it to FDR. This is in affect moving the FDR corner to the FUR location based on the corners' original positions. This can be accomplished with *R' B2 R*. Lastly I need to undo the interchange maneuver with *F'* to replace the corner now at FDR to its original location at FUR. Lastly I will put all three corners back on the same face with *R*.

Putting these turns all together gives me:
(*R'*) (*R'* B2 R) (F) (R' B2 R) (F') (R)

Notice that there are some cancellations, namely that the bolded moves combine to *R2*. This gives:

*R2 B2 R F R' B2 R F' R*

Putting all of this together gives:

*F d F' U2 F d' F' U' F d F' U' F d' F' R2 B2 R F R' B2 R F' R*

And this would be a 24 move solution to this week's challenge. According to your rules I would think that you would not allow this solution as I have made use of "known commutators" in both phases of my solution, in cycling the edge+corner blocks, as well as in cycling the corners. Not only that, but you may recognize the second commutator as "The A-perm". However, to me, all of these maneuvers are completely intuitive.

I have to be honest that it sounds neat to have to try this solution without using commutators at all, but it also seems strange that something as intuitive as commutators might be banned in a competition focusing on intuition.

I will also try to come up with a solution to this case without using commutators. I don't immediately have an idea on how to do that. I think this will be fun to try


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## cmhardw (Jun 30, 2014)

Dane man said:


> *This Week's Challenge:*
> 6/30/14 - _Slice moves are prohibited for this challenge. Any replacement slice moves (L R', F2 B2, etc.) are also forbidden. The UF edge must go to the UB position, the UB edge to UR, and the UR edge to UF. The orientations of these edges must be maintained. Nothing else on the cube may be modified in any way, permutation or orientation._



Here is my shortest solution:
(clicking the first spoiler will only show the solution length)


Spoiler



This solution is 12 moves in HTM

Solution:


Spoiler



Solution: R2 D' R2 F2 R2 U2 R2 F2 R2 U2 D R2 (12 turns)

I need to create a maneuver to mimic the effect on the M slices edges that the commutator F2 M F2 M' has. Since I cannot do slice turns, I will try to replicate the effect F2 U2 has on the M slice edges (also a three cycle of M slice edges). I need to do this in such a way that destroys the rest of the cube in 2-cycles of pieces. Basically, this destroy maneuver needs to have order 2 on the non-M slice part of the cube. If I can destroy the cube with a maneuver of order 2, while also doing a three cycle of the M slice edges, then executing this maneuver twice will be *only* a three cycle of M slice edges. 

Start by swapping the 1x1x3 blocks at FL and BR using R2 F2 R2. This has done the F2 part of my F2 U2 maneuver that I am trying to do on the M slice edges. If I do U2 from here, then I swap the corners from those two 1x1x3 blocks with each other, while also swapping the edges UR and UL, and swapping the corners at UBL and UFR. So right now, on the non-M slice part of the cube, I have swapped 2/3 of the FL 1x1x3 with 2/3 of the BR 1x1x3 blocks. I have swapped the UR and UL edges. I have also swapped the UBL and UFR corners. This has left everything not on the M slice double swapped (meaning it all is destroyed in a way that has order 2). Looking only at turns that affect the M slice edges I have done F2 U2, exactly what I wanted to do, a 3 cycle on M slice edges.

Now do this maneuver, R2 F2 R2 U2, twice to destroy then solve the non-M slice stuff on the cube, while at the same time performing two three cycles on the M slice edges. Two 3 cycles is also a three cycle in the opposite direction, so I now have a three cycle on M slice edges. This 3 cycle moves the pieces as (UB DF UF).

So R2 F2 R2 U2 R2 F2 R2 U2 performs the cycle (UB DF UF).

To get the desired three cycle from this week's challenge, setup UR to DF by doing R2 D'.

This makes our three cycle:
(R2 D') (R2 F2 R2 U2 R2 F2 R2 U2) (D R2)

Final solution:
*R2 D' R2 F2 R2 U2 R2 F2 R2 U2 D R2*







A slightly longer solution, but still neat:
(clicking the first spoiler will only show the solution length)


Spoiler



This solution is 16 turns HTM

Solution:


Spoiler



*Solution:* U2 R2 U2 R2 U' R2 U' R2 U2 R2 U2 R2 U' R2 U' R2 (16 turns)

Explanation:
Let's look at the maneuver R2 U2 and see what it does to the cube. I will refer to the 1x1x3 block UFL+UL+UBL by the name "UL block". I will refer to the 1x1x3 block UFR+UR+UBR by the name "UR block". I will refer to the 1x1x3 block DFR+DR+DBR by the name "DR" block. I will also refer to the 1x1x3 block of UF+U center+UB as the "U face block", and the 1x1x3 block of FR+R center+BR as the "R face block".

Doing R2 U2 will cycle the 1x1x3 blocks as : (DR block -> UL block -> UR block). R2 U2 will also "flip" the U face and R face blocks to be misoriented. A 3 cycle has order 3, and flips have order 2. This means that if I perform R2 U2 three times, it will be an identity permutation on the 1x1x3 blocks, but it will leave the U face and R face blocks flipped since I have executed a flip on each piece 3 times. This leaves those U and R face blocks flipped.

A "flip" of the U face block is really the transposition of the UB and UF edge pieces. A "flip" of the R face block is really a transposition of the FR and BR edges.

Ok so let's build the weekly challenge 3-cycle of (UF UB UR).

Start by doing *R2 U2 R2 U2 R2 U2* to perform an identity permutation on 1x1x3 blocks, and leave the U face and R face blocks flipped.

Now place the UR edge piece at UB without affecting any of the following edges: UF, FR, BR. I can do this with:

*R2 U R2 U'*

I will now use the R2 U2 R2 U2 R2 U2 maneuver to perform an identity on 1x1x3 blocks and "flip" the U and R face blocks, but I can cancel a move if I do: *U2 R2 U2 R2 U2 R2* (the reflection) instead.

The last thing that remains is to undo the setup I did by setting down the edge at UB back to UR. I do this with *U R2 U' R2*

This makes my entire maneuver:

R2 U2 R2 U2 R2 U2 R2 U R2 *U' U2* R2 U2 R2 U2 R2 U R2 U' R2

Notice that the bolded moves combine. This makes my solution, so far:

(R2 U2 R2 U2 R2 U2) R2 U R2 U R2 U2 R2 U2 R2 U R2 U' R2 (19 turns)

Notice that if I replace the first 6 turns with it's reflection U2 R2 U2 R2 U2 R2, that I cancel moves again. Let's do that replacement:
(U2 R2 U2 R2 *U2 R2*) *R2 U* R2 U R2 U2 R2 U2 R2 U R2 U' R2

Notice that the bolded moves combine/cancel. This makes my, almost, final solution:
U2 R2 U2 R2 U' R2 U R2 U2 R2 U2 R2 U R2 U' R2

If I change the second (U2 R2 U2 R2 U2 R2) back to (R2 U2 R2 U2 R2 U2), the reflection, then this entire solution can be written as a commutator, which I think is prettier.
[U2 R2 U2 R2 U2 R2, R2 U R2 U'] = *U2 R2 U2 R2 U' R2 U' R2 U2 R2 U2 R2 U' R2 U' R2* (16 turns, after move cancellations)


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## Dane man (Jun 30, 2014)

cmhardw said:


> I just discovered this thread, but I have a question about the rules. The way I read the rules, commutators and conjugates are strictly forbidden. I view commutators and conjugates as being 100% intuitive, as that is how they are constructed.
> 
> Are commutators and conjugates really banned for this competition? Or am I misunderstanding the rules?


The rule is that if it is a memorized group of moves (like a algorithm) of which it's effects are already known, then it is not allowed. But if you create a commutator or conjugate using intuition during the challenge, then you may use that. The rules are to prevent others from using already known algorithms (or even partial algorithms) to achieve solution unless of course those algorithms are like 3 or less moves which would likely be used anyway. However the rules are interpreted, the goal should be to do everything intuitively, not relying on pre-memorized solutions to a problem.

It's not commutators and conjugates that are forbidden, it's the already memorized sets of moves that could be used in them that are not allowed. The technique of using commutators and conjugates to achieve a solution is allowed as long as it isn't done using already known move sequences.

EDIT: Rules have been updated for clarification. Thanks!


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## TDM (Jun 30, 2014)

So because of rule 1, and me not being able to find a way to explain (R2 U2)3 or (R2 F2 R2 U2)2, I can't actually find a solution for this. Even some really inefficient CFOP with commutator LLs has slice moves in.
And cmhardw, I think it's fine to use commutators if you explain them in enough detail, like I have done in past solutions (not as much as I should have done in round 2, but I used one in round 1 that I explained better).


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## 10461394944000 (Jun 30, 2014)

Dane man said:


> This weeks winners are TDM and porkynator! Gongratulations to them!
> 
> *1- TDM (12 moves STM)
> 2- porkynator (17 moves STM)*
> ...



i don't think this competition will work too well if people already know an algorithm for the case they are trying to solve, especially because some people (eg. me) understand why some algorithms do what they do (decompositions n stuff) so some standard algs can be intuitive

anyway heres an alg that does that, that I found intuitively



Spoiler



y' R U2 R' U2 R' F R U R U' R2 F' r U R U' r' y


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## cmhardw (Jun 30, 2014)

Dane man said:


> The rule is that if it is a memorized group of moves (like a algorithm) of which it's effects are already known, then it is not allowed. But if you create a commutator or conjugate using intuition during the challenge, then you may use that. The rules are to prevent others from using already known algorithms (or even partial algorithms) to achieve solution unless of course those algorithms are like 3 or less moves which would likely be used anyway. However the rules are interpreted, the goal should be to do everything intuitively, not relying on pre-memorized solutions to a problem.
> 
> It's not commutators and conjugates that are forbidden, it's the already memorized sets of moves that could be used in them that are not allowed. The technique of using commutators and conjugates to achieve a solution is allowed as long as it isn't done using already known move sequences.
> 
> EDIT: Rules have been updated for clarification. Thanks!



Thank you for the clarification. I'm pretty sure I followed the spirit of the rules when coming up with my solutions for this week's challenge. When you judge/read/grade them please let me know if I have followed the rules as you intended them. I basically went with making sure that I could fully explain the logic behind every turn I did, and I provided that explanation in my answers.

I gave two answers, only because I discovered the longer solution first and was really proud of it because I had never before seen the alg I ended up creating! However, I also found a shorter solution and would like that one to be my official answer for the challenge.


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## Dane man (Jul 1, 2014)

cmhardw said:


> Thank you for the clarification. I'm pretty sure I followed the spirit of the rules when coming up with my solutions for this week's challenge. When you judge/read/grade them please let me know if I have followed the rules as you intended them. I basically went with making sure that I could fully explain the logic behind every turn I did, and I provided that explanation in my answers.
> 
> I gave two answers, only because I discovered the longer solution first and was really proud of it because I had never before seen the alg I ended up creating! However, I also found a shorter solution and would like that one to be my official answer for the challenge.


You're welcome. And your shorter solution has been counted. Your solutions are fantastic, though the first solution you presented (24 moves) would not be counted because it includes algorithms that you already knew, but as you have discovered, this isn't a problem as there is always another way to find a solution. But brilliant job with your solutions.



10461394944000 said:


> I don't think this competition will work too well if people already know an algorithm for the case they are trying to solve, especially because some people (eg. me) understand why some algorithms do what they do (decompositions n stuff) so some standard algs can be intuitive.


Actually, the goal of the competition is to overcome that issue. The rules are built to make those who cube seek to discover a solution for themselves instead of relying on pre-memorized solutions. Of course you're allowed to understand what your memorized alg does (and that understanding can help you create a new one), but you cannot use it as your solution. You must present something else, and trust me, there are plenty of algorithms to be found for each case. 

Also, you need to explain your algorithm or it will not be counted.


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## 10461394944000 (Jul 1, 2014)

Dane man said:


> Actually, the goal of the competition is to overcome that issue. The rules are built to make those who cube seek to discover a solution for themselves instead of relying on pre-memorized solutions. Of course you're allowed to understand what your memorized alg does (and that understanding can help you create a new one), but you cannot use it as your solution. You must present something else, and trust me, there are plenty of algorithms to be found for each case.



but that would mean that someone who hasnt memorized an alg (lets ignore your no slice moves rule and say R2 U S' U2 S U R2) could find that alg and explain it as

R2 U: setup
S' U2 S U2: 3 cycle
U' R2: unsetup

but someone else who already knows that alg couldnt enter using the same alg with the same explanation

anyway heres a different alg:



Spoiler



F2 D R2 D' R2 U R2 U' R2 F2 R2 B2 R F R' B2 R F' R

[y':[B':[R,B'L2B]][R2:[f2,DB2D']]]'

[B':[R,B'L2B]] = corner 3 cycle which just happens to be known as an A perm
[R2:[f2,DB2D']] = 1x1x2 block 3 cycle which just happens to be known as a J perm


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## Dane man (Jul 1, 2014)

10461394944000 said:


> but that would mean that someone who hasnt memorized an alg (lets ignore your no slice moves rule and say R2 U S' U2 S U R2) could find that alg and explain it as
> 
> R2 U: setup
> S' U2 S U2: 3 cycle
> ...


You will need more explanation than that to justify an algorithm of that length. You're supposed to demonstrate that you understand the mechanisms involved in the solution process, not simply say "it has this effect, then it has that effect".

And yes, the natural effect is that, the more you have memorized, the less you are allowed to use. This is so that it can be a fair challenge. To a beginner with few things memorized it will still be a challenge, and someone who has lots of experience and memorized algorithms under his belt can enjoy the same level of challenge. A challenger won't be condemned for discovering an algorithm simply because someone, somewhere already discovered it, even if it is a really common algorithm. As long as the competitor arrived at that algorithm on his/her own, using his/her intuition without extra assistance, and can explain how/why it works, then it counts as valid.


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## 10461394944000 (Jul 1, 2014)

Dane man said:


> You will need more explanation than that to justify an algorithm of that length. You're supposed to demonstrate that you understand the mechanisms involved in the solution process, not simply say "it has this effect, then it has that effect".



I wrote it as a bunch of commutators and conjugates, how am I supposed to explain it more than that?


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## cmhardw (Jul 1, 2014)

Dane man said:


> You will need more explanation than that to justify an algorithm of that length. You're supposed to demonstrate that you understand the mechanisms involved in the solution process, not simply say "it has this effect, then it has that effect".
> 
> And yes, the natural effect is that, the more you have memorized, the less you are allowed to use. This is so that it can be a fair challenge. To a beginner with few things memorized it will still be a challenge, and someone who has lots of experience and memorized algorithms under his belt can enjoy the same level of challenge. A challenger won't be condemned for discovering an algorithm simply because someone, somewhere already discovered it, even if it is a really common algorithm. As long as the competitor arrived at that algorithm on his/her own, using his/her intuition without extra assistance, and can explain how/why it works, then it counts as valid.



You should disqualify both of my solutions to this week's challenge then. I write the rest in spoilers for those still working on the challenge:


Spoiler



I've known the effects of R2 U2 R2 U2 R2 U2 and R2 F2 R2 U2 R2 F2 R2 U2 for years, long before I ever understood why they worked. Now I understand why they work, so I would consider them intuitive. However, if I can't use sequences that I've known about, then neither of my solutions are valid by your rules.

I interpreted the rules as "Be able to explain every turn in your solution on a completely intuitive level", but based on your recent posts I think that the rules are:
1) Be able to explain every turn in your solution on a completely intuitive level
2) Derive a solution that is previously unknown to you, using component sequences that are previously unknown to you

Is that correct?


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## Dane man (Jul 1, 2014)

cmhardw said:


> You should disqualify both of my solutions to this week's challenge then. I write the rest in spoilers for those still working on the challenge:
> 
> 
> Spoiler
> ...


The second point is 100% correct. 
The first point is also correct. You should be able to explain what every move does, but you don't need to be super nit-picky about every detail if it's unnecessary. If you can explain a group of moves like "R U R'" by says something like "insert F2L slice" then you can do that because it's something obvious, but don't just gloss over the effects of the algorithm or parts of it. If a group of algorithms causes corners to rotate, then one should be able to explain _how_ it makes them rotate instead of simply saying "it rotates the corners". There is a balance. And it's up to each one to feel it out, and if I feel that someone is generalizing too much, I'll give them a heads up so they can explain a little more. I don't require a perfect explanation of every detail, but enough that one could understand "why" that algorithm works.

I don't want to be too harsh, but it's a requirement for the sake of the competitors, both for fairness, and to encourage the competitors to really get a grasp on what is happening when they are executing moves.

Your posts are perfect because you clearly demonstrate that you have a full grasp of what happens in your algorithms, and it's easy enough for someone else to see how it works.


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## Phillip1847 (Jul 1, 2014)

Because I suck at this, I'll try it:


Spoiler



z' y2 R' U2 R U' R' U2 R U2 R' U R2 U2 R' U2 B F' L F R F' L' F R' B' F' D F R' F2 R B' R' F2 R B F' D' F

z' y2
R' U2 R U' R' U2 R U2 R' U *R* - F2L solve BR Pair, intuitive,I don't use this alg
*R* U2 R' U2 - solve all edges by inserting the last one, intuitive
(B: [F' L F : R]) - commutator
(F' D F:[R' F2 R, B']) - Epic 3 move conjugate commie-tator


I feel like my signature is the way I feel about this solution.

Also, if it's a commutator, I don't see why it needs explanation. Ben's solution is fine


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## cmhardw (Jul 1, 2014)

Dane man said:


> The second point is 100% correct.
> The first point is also correct. You should be able to explain what every move does, but you don't need to be super nit-picky about every detail if it's unnecessary.
> 
> ...
> ...



Ok, thank you for taking the time to clarify my questions. I think I understand now. Allow me to paraphrase your explanation of the rules via an example, and let me know if this is correct:
It would never be allowable to use a sequence of moves that "twists the corners". Instead you should always derive a sequence from basic principles, and be able to explain clearly how that sequence twists the corners.

If my solutions are ok by the rules, then in the future I will definitely shoot for the same level of clarity and intuitiveness of the sequences I use, and I definitely plan to be a regular participant in this competition! Very cool idea!

Thanks again!


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## TDM (Jul 1, 2014)

Phillip1847 said:


> Because I suck at this, I'll try it:
> 
> 
> Spoiler
> ...


That solution isn't allowed; see the parts I bolded.


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## Phillip1847 (Jul 1, 2014)

TDM said:


> That solution isn't allowed; see the parts I bolded.



Oh come on seriously? Those aren't even used for the edges.
Okay, I'll try and find another


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## TDM (Jul 1, 2014)

Phillip1847 said:


> Oh come on seriously? Those aren't even used for the edges.
> Okay, I'll try and find another


Well from what he says they aren't, as they are equivalent to a slice move and a rotation. I'd still keep your solution though, just in case for some reason it is allowed.


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## Phillip1847 (Jul 1, 2014)

TDM said:


> Well from what he says they aren't, as they are equivalent to a slice move and a rotation. I'd still keep your solution though, just in case for some reason it is allowed.


Would B2 F' still count?


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## cmhardw (Jul 2, 2014)

Phillip1847 said:


> Would B2 F' still count?



Yes, B2 F' is fine.


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## 10461394944000 (Jul 2, 2014)

cmhardw said:


> Yes, B2 F' is fine.



doesn't that imply that B2 F' B' is also fine?


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## Lucas Garron (Jul 2, 2014)

Dane man said:


> The new task to be completed is:
> _Slice moves are prohibited for this challenge. Any replacement slice moves (L R', F2 B2, etc.) are also forbidden. The UF edge must go to the UB position, the UB edge to UR, and the UR edge to UF. The orientations of these edges must be maintained. Nothing else on the cube may be modified in any way, permutation or orientation._



That's a rather weird restriction. You should also number your challenges, and you should *still* define what maintaining orientation means. You should really just give a picture or give us an exact cycle/Singmaster state description.

Anyhow, here's an obvious one that avoids slices: [x2 B2: [U R2 U', l2] [F' R' F, L2]] (18h)

EDIT: Actually, here's another fun one: [R2 D': [R' F2 R, R U2 R']] (13h simplified)
EDIT 2: Oh, interesting. I tried to cyclic shift that last one to remove one more move, and guess what? Our friend, the 12h: [R2 D': (R2 F2 R2 U2)2]

EDIT 3: And of course, there's the ever-popular one: [R U': [R U R U R: U'] R] (simplifies to 11h). Intuitive enough, but the fact that the core part it only affects edges is something I would probably find by experimentation rather than design, if I didn't already know how it works.


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## cmhardw (Jul 2, 2014)

10461394944000 said:


> doesn't that imply that B2 F' B' is also fine?



Probably not:



> 6/30/14 - Slice moves are prohibited for this challenge. Any replacement slice moves (L R', F2 B2, *etc.*) are also forbidden.



I think the rules leave it open ended enough that any form of slice move replacement gets filtered out.


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## elrog (Jul 2, 2014)

I found it hard not to do something easy to see that used M moves or the algorithm (R2 U2)3 in it somewhere, but I finally came up with a good solution. O also had the idea of solving the edges while messing up the corners, then solving the corners with commutators that don't use slice moves, but that seemed kind of cheap.

y' (L' U' L U) (R U R' U') (U' L' U L) (U R U' R')

I am pretty sure I've found this series of moves through messing around before, but I didn't have it committed to memory and was not thinking of it while doing this.

I started out thinking that I need to do 4 2-swaps with the edges rather than 2 because doing only 2 2-swaps as a commutator would involve slice moves. So I needed to swap UB and UR, then UF and UR, then UB and UR again, and finally, UF and UR for the finish. After doing a y', the moves (L' U' L U) do the first swap and (R U R' U') do the second. Doing these backwards does the third and forth swaps. If I do each of them backwards I also fix the cube excluding the pieces affected by both series of moves. Besides the edges being cycled, the (after a y') ULB and URB corners are the only two affected. Both series of moves affect the 2 corners in question in opposite ways, so they corners must be solved. It should also be pointed out that the corners permutation must be solved because you cannot swap only 2 pieces.


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## Dane man (Jul 2, 2014)

elrog said:


> y' (L' U' L U') (R U R' U') (U' L' U L) (U R U' R')



I am about to post the winners and the next challenge, but I'd like to verify that this is your algorithm. If you could verify that this is your algorithm, or correct my understanding of it, I can put you in the winners list.


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## elrog (Jul 2, 2014)

I am terribly sorry. I mistyped the algorithm.

It should be: y' (L' U' L *U*) (R U R' U') (U' L' U L) (U R U' R')

I will correct it in my post.


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## Dane man (Jul 2, 2014)

Current standings. These are not the final results.

*1st - elrog* with 15 moves
*2nd - cmhardw* with 12 moves. The solution works and there was a wonderfully clear explanation, but used algorithms that were already known. Was simply moved down instead of DQ'd because he explained _why_ the algorithm works.

*HM - Lucas Garron* with 12, 13, and 18 moves. The solution works, but there was no sufficient explanation of the origin or workings of any algorithm. In doing so, hinted that the solution of 11 moves was "popular" and therefore already known. Given an honorable mention because clearly understood what the algorithms did and why, despite the lack of sufficient explanation. (If he can give a sufficient explanation of these algorithms, then the rankings will be adjusted to include this entry).

Final standings and new challenge coming soon!


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## 10461394944000 (Jul 3, 2014)

Dane man said:


> Current standings. These are not the final results.
> 
> *1st - elrog* with 15 moves
> *2nd - cmhardw* with 12 moves. The solution works and there was a wonderfully clear explanation, but used algorithms that were already known. Was simply moved down instead of DQ'd because he explained _why_ the algorithm works.
> ...



what about me?


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## Dane man (Jul 3, 2014)

10461394944000 said:


> what about me?


You can still get in there if you explain your algorithms.


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## Phillip1847 (Jul 4, 2014)

Dane man said:


> You can still get in there if you explain your algorithms.



they are commutators? They don't really need explanation


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## elrog (Jul 4, 2014)

Just because it is a commutator does not mean you understand it. Commutators are algorithms as well. You have to show that you do understand. That's the point in this.


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## Dane man (Jul 4, 2014)

New challenge!

_You must create the following pattern from this angle on the cube. It doesn't matter what the rest of the cube looks like._


Yeah yeah, patriotic I know. Have fun!

PS: If you're not particularly fond of this color scheme, you may choose your own, but it must achieve this pattern.


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## Lucas Garron (Jul 5, 2014)

Hmm. I can't use Stefan's alg, then. Here's something very simple:

We're going to swap four pieces around UBR with FDR, and then do the same with FDR and LBD

Setup onto U: U' R D R F' D' R'
Swap with U2. Full commutator: [x y', [U' R D R F' D' R': U2]]

Flip a corner on D: R U2 R' F' U2 F
Full commutator to flip rotate corners: [R U2 R' F' U2 F, D]
Conjugated to handle opposite corners with a cancellation: [R U2 R' F' U2 F, D]

Invert both of the parts, and rotate the cube to align the color scheme:
y2
[x' R2: [D, R U2 R' F' U2 F]]
[[U' R D R F' D' R': U2], x y']
y'

It's easy to take off another 2-3 moves using the right rotations, but I'm obviously not going to movecount here.

EDIT: There's also an obvious setup into 3-cycles of pairs around the rotated ring:

Setup to move DFR next to RUB: F'
Commutator to swap pairs: [U L2 U', r']]
Repeat twice and combine with corner twist from previous alg: y2[x' R2: [D, R U2 R' F' U2 F]][F': [U L2 U', r']][D: [R' B2 R, f]]y'


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## elrog (Jul 5, 2014)

I started off holding blue on top, red on right, and white on front.

First off, I started this challenge trying to solve the corners as they were harder. The corners have only 1 position they can be where as the edges have multiple edges not shown. The corners also require you to orient, but not permute, them.

I realized that I could swap the URF corner with the DLB corner and then swap them back into the correct orientation.

To swap the corners the first time, I applied B2 followed by the commutator J(b)-perm: R U2 F' R' F U' F' R F U' R' U'



Spoiler: It works like this



Because commutators have to be in an even permutation, the commutator J-perm is actually a 3 cycle of pairs followed by an AUF.

A commutator requires that you insert something into a layer or space on the cube, then change it out, and then fix the rest of the cube by undoing the insertion, and then undoing the swap. I have to move a pair out of the U layer so it can be inserted in place of another. The move R does this. Doing a U2 makes the insertion easier.

You could insert the pair in place of another pair in the U layer with R', but you mess other parts of the U layer up. You must first move the URF corner out of the way with F'. This gives you F' R' F for the insertion.

To swap the inserted pair out with another, you can do U'. You then undo the insertion giving you F' R F. You now undo the swap giving you a U. You then undo the premoves giving you U2 R'. The U and U2 cancel to give you U' instead. At the end you AUF with a U.


 To swap the corners back into the correct orientation, do a z' rotation and apply the mirror of the above J-perm: L' U2 F L F' U F L' F' U L U. You can then undo the roatation with a z and fix the back side with another B2.

At this point, you need to cycle the UF, RF, and RU edges with the commutator U M U' M' F' M U M' U' F.



Spoiler: This commutator works like this



The UR ewdge needs to got to UF. I You can place it there with a U, but you need your URF and RLF corner to stay where they were. You can put them back with a U' but first, hide the edge with an M. After the U', undo the M with an M'.

Next, do an F' to swap the RF edge with the UR edge so that the RF edge will end up in the RF position and the UF edge will end up in the RF position after undoing moves. 

Now you undo the insertion with M U M' U' To finish the commutator, undo the F' with an F.

You know that the first edge you insert will be oriented correctly if you insert it that way. If you make the first insertion into a slot that is already oriented, (the UF edge) that slot will stay in the orientation it was in. The last edge must be correctly oriented because you cannot flip 1 edge.


 Finally, you need to fix the centers with the commutator E' M E M'.



Spoiler: This short commutator works like this



The E' moves the R center to the F centers position. The M places the U center in the F position taking the place of the R center.

When you undo the E' and M with E and M' the U centers goes back to where the R center was and the F center goes back to where the U center was.

The centers in back where cycled as well, but other than that, nothing else moved because you can only move the pieces where the part one and two of the commutator intersect.


 The whole solution: B2 R U2 F' R' F U' F' R F U' R' U' z' L' U2 F L F' U F L' F' U L U z B2 U M U' M' F' M U M' U' F E' M E M'.


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