# Ultimate challenge



## mrCage (Aug 21, 2009)

Hi. Chris inspired me to come up with this variety of forced turn fewest moves challenge. Use this weeks challenge at http://fmc.mustcube.net and find a solution where exactly every second turn is a F turn. Noe tricks that i know will give a trivial solution based on a norm,al solution.

By the way the scramble is this one: *D' F D F' L U F L' U2 B' D2 B' U' F' D' L D' F2 B L2 F2 R' L D' U F' B' L' R' F'*

*Per*


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## Stefan (Aug 21, 2009)

mrCage said:


> Noe tricks that i know will give a trivial solution based on a norm,al solution.


Here's one:

1. Replace every F* turn with (U D L2 R2 U' D') B* (U D L2 R2 U' D').
2. Separate the turns with F B F B F B2 F.


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## trying-to-speedcube... (Aug 21, 2009)

You *could* do your solution with F F2 F between every single move. But that kind of throws off the challenge again.


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## Stefan (Aug 21, 2009)

trying-to-speedcube... said:


> You *could* do your solution with F F2 F between every single move.


I don't know how something can done "between a single move", plus I think your idea violates the "where *exactly* every second turn is a F turn" requirement.


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## Robert-Y (Aug 21, 2009)

Ah I think there could be some confusion.

@Per do you mean every second turn is F or do you mean every second turn is either F, F' or F2?


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## trying-to-speedcube... (Aug 21, 2009)

StefanPochmann said:


> trying-to-speedcube... said:
> 
> 
> > You *could* do your solution with F F2 F between every single move.
> ...



I'm sorry, I wasn't entirely clear. I meant you could do your normal solution, with F F2 F between every 2 moves. For example, for R U R', you would do R F F2 F U F F2 F R' F F2 F.

Why does it violate the "*exactly* every second turn is an F turn" requirement? I don't see how this implies other moves may not be F turns.


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## Tim Reynolds (Aug 21, 2009)

trying-to-speedcube... said:


> Why does it violate the "*exactly* every second turn is an F turn" requirement? I don't see how this implies other moves may not be F turns.



The word exactly used like that usually means something like if-and-only-if; that is, the moves that are F turns are exactly those that are every second turn.

Put another way: consider the algorithm F F F F F F F F. Is every second turn an F turn? Definitely. Is exactly every second turn an F turn? No, that is not an exact description of which turns are F turns; rather, every single turn is an F turn.


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## trying-to-speedcube... (Aug 21, 2009)

Sorry, I didn't understand what it actually had to imply.

I am NOT going to try this challenge anyway 

Maybe I can get a 2x2x2 block? Maybe 2x2x3, but that's where I'll give up :/


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## mrCage (Aug 21, 2009)

StefanPochmann said:


> mrCage said:
> 
> 
> > Noe tricks that i know will give a trivial solution based on a norm,al solution.
> ...


 
Ok ok!! Smartass I need to sharpen my formulation of the challenge.

Take the scramble of my first message in this thread and solve it with a sequence where exactly every second turn is exactly F and not F' or F2. Furthermore, the solution cannot also contain any subsequence that is an identity sequence - like for example F B F B F B2 F. Try to spoil this one Stefan

Per


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## trying-to-speedcube... (Aug 21, 2009)

2x2 Block: B F R' F U F U' F D F D F B' F L2 F U' F U F U' F U F U2 F D' F

Per, what exactly is an identity sequence? A sequence which can simplify or a sequence which does nothing to the cube? For example F B' F, is that forbidden?


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## mrCage (Aug 21, 2009)

trying-to-speedcube... said:


> 2x2 Block: B F R' F U F U' F D F D F B' F L2 F U' F U F U' F U F U2 F D' F
> 
> Per, what exactly is an identity sequence? A sequence which can simplify or a sequence which does nothing to the cube? For example F B' F, is that forbidden?


 
Yes, a sequence that does nothing to the cube (except possibly rotate centers ... )

Per


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## Herbert Kociemba (Aug 21, 2009)

Why should it be possible at all to find such a solution?


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## TMOY (Aug 22, 2009)

Well, here's my solution:
U' F B F L' F B F B2 F L F B F R': corner orientation
F B F L2 F B F R2 F B2 F U2: corner separation and xLL combined, 2 edges already solved
F B2 F L F B' F B2 F D': 3rd and 4th edge
F B2 F D F B2 F U2 F B2 F D: 5h edge
F D' F B2 F B' F R: 6th edge
F D2 F B F B F L F B2 F B2 F B2 F B F U2 F B2 F D F B F D2: 7th and 8th edge
F L F B2 F B" F D' F B2 F R2 F B2 F D F B2 F B2 F B2 F B F L': middle layer orientation
F B2 F D2 F U2 F B' F B2 F L2 F: middle layer permutation


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## trying-to-speedcube... (Aug 22, 2009)

Wow, Francois, that's incredible... did you come up with all those algs yourself?


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## TMOY (Aug 22, 2009)

Well, with CF it's really not very difficult once you master some little tricks for putting the F and B layers in the position you want


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## trying-to-speedcube... (Aug 22, 2009)

I found a shorter 2x2x2 block, in ULB, so it also gives nicer continuation because the F layer is free.

U F U F B' F R F U F R F B' F U2 F B F R' F B

I haven't found a nice continuation yet though...


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## mrCage (Aug 22, 2009)

Herbert Kociemba said:


> Why should it be possible at all to find such a solution?


 
I think TMOY has clearly shown that it is possible. A hidden part of the challenge was to alternatively prove it's impossible

Per


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## calekewbs (Aug 22, 2009)

are cube turns legal?


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## Stefan (Aug 22, 2009)

mrCage said:


> the solution cannot also contain any subsequence that is an identity sequence


This is impossible in most cases. You probably mean substring, not subsequence.

http://en.wikipedia.org/wiki/Subsequence#Substring_vs._subsequence


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## mrCage (Aug 22, 2009)

StefanPochmann said:


> mrCage said:
> 
> 
> > the solution cannot also contain any subsequence that is an identity sequence
> ...


 
Hairsplitter, nitpicker !!  Is it ok if i call it a subsequence of turns?? This should be the same as a substring defining a subsequence .... You know semantics right??

Per


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## cmhardw (Aug 22, 2009)

I will try this challenge, but I must first also do my own surprise challenge (which I haven't had time to finish yet)! Nice idea Per, it sounds difficult, but I am excited to try it!

And yes I will follow the spirit of your challenge, just as I understand the spirit of my challenge as well.

Try looking at the list of stipulations I made for the final variation in my surprise challenge thread, I am curious if these are sufficient to precisely describe what exactly we mean by the "spirit" of this kind of forced turns challenge.

Chris


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## qqwref (Aug 22, 2009)

I tried a harder version of the challenge where you cannot put a U or D face turn between consecutive U's. I could only get halfway though.



Spoiler



x [to make the F turns be U turns ]
L2 U B2 U R' U F2 U R U R U B' U B2 U [block]
F' U L U F U B U B' U L' U F U R2 U R U R2 U F' U [opposite block, finish cross]



Hope this works. Can anyone continue from here?


EDIT: I have a proof that we can ALWAYS create a solution, without any U or D turns except for the U turn that happens every other turn.
Consider a sequence such as "R U". This sequence has some order x; so (R U)^(x-1) will be equivalent to the sequence U' R'. (I'll call this sequence (R U)'.) Thus, the sequence (for instance) F U (R U)' has the effect F R'. We can actually construct any 2-qtm sequence (not counting the D face) using the following sequences and their obvious variants:
- We can create F2 by (F U)(F' U)' and U2 by (F2 U)'(F2)(F2 U)'(F2), so we can reverse any turn.
- F R' = (F U)(R U)', F B' = (F U)(B U)', F U is obvious, U' F' = (F U)'.
If the solution is not of even qtm, just add (F2 U)'(F2) for a U move or (F U)(F2 U)'(F2) for an F move.
So just think of any solution, then write each 2-qtm part (and the extra 1-qtm part if there is one) as above.


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## Stefan (Aug 23, 2009)

mrCage said:


> StefanPochmann said:
> 
> 
> > You probably mean substring, not subsequence.
> ...


That doesn't change anything, as that sounds just like what we meant already (don't know what else we could mean). I suspect you misunderstand the two terms and their difference. To make it clear: If you forbid identity subsequences, then you forbid RUR' because it contains RR'.



qqwref said:


> - We can create F2 by (F U)(F' U)'


I think you're just hiding your identity parts, not avoiding them, so this seems invalid.


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## mrCage (Aug 23, 2009)

StefanPochmann said:


> mrCage said:
> 
> 
> > StefanPochmann said:
> ...


 
So then it's forbidden by two reasons, so what??!!

Per


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## qqwref (Aug 23, 2009)

StefanPochmann said:


> qqwref said:
> 
> 
> > - We can create F2 by (F U)(F' U)'
> ...



If there's an identity, couldn't we just remove it? I can't think of any way this procedure would create an identity that isn't of the form <move> U <move> U ... <move> U.


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## Stefan (Aug 23, 2009)

qqwref said:


> If there's an identity, couldn't we just remove it? I can't think of any way this procedure would create an identity that isn't of the form <move> U <move> U ... <move> U.


You said: F R' = (F U)(R U)'
Remove the U U' identity and you get F R'.


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## TMOY (Aug 24, 2009)

Which U U' identity ? (R U)' is here R U repeated 104 times, not U' R'.


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