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Cubenovice

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Thank you for your feedback.

Especially on the combinations, I never realised that :)

With 5 corners I typically take the best insertion and go though the whole skeleton again.
Never gave it a second thought as finding 3 cycles with only 3 stickers is pretty straight-forward.
But as I always write down the cycle numbers anyway I should actually look into my notes to check if I already have a combination to complete the 5 cycle.
From there it is only a small extra step to check within the best "1st insertions" if there is something better.

Will check this combination-thinghy on my notes for 370 :)

BTW
 
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Kryptonite

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I prefer the thorough method, but mostly because I never go for 1-hr or less solves (for reasons just recently discussed, lol). But it's true, to be 100% thorough, you must find every insertion, and then look for each potential corresponding insertion within all the first level insertions found. Luckily the fun does terminate at the insertion within the insertion, and doesn't loop forever, haha.

I wonder however, if there's a theoretical limit to the benefit of an insertion within an insertion? For example, if the limit was 6, and I had a 6 move cancellation 5C already, I know that it really would be a waste of my time to check each 0-cancel insertion for an amazing insertion, because I could only match my previous best. I mean, theoretically the only way you could cancel 16 moves from a single insertion would be if you had added a counter-productive 3-cycle into your skeleton, lol.

Also, I'm fairly certain that you only need to look at 8f* insertions. Teemu disagreed with this when I said it earlier, and he may be right, but I'm unconvinced that it's possible to use conjugates to get an insertion fewer than 8 moves after cancellations. I'm pretty sure you would find these potential insertions by simply looking at the 8f*s.
 

kinch2002

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Just to make it clear again, when you find a nice (123) insertion, I said you have the option of a (345) before it, or a (451) after it. If you rescrambled with the new skeleton that includes the (123) insertion, you'd find that you need to cycle (451) anywhere in the scramble. So finding a (345) that comes before the (123) (having not actually applied the moves, just found them), is just the same as rescrambling and finding a (451) in the first part of the solution.
 
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kinch2002

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Also, I'm fairly certain that you only need to look at 8f* insertions. Teemu disagreed with this when I said it earlier, and he may be right, but I'm unconvinced that it's possible to use conjugates to get an insertion fewer than 8 moves after cancellations. I'm pretty sure you would find these potential insertions by simply looking at the 8f*s.

Sorry for double post. I believe that 9/10 movers that are just 8 movers with set up moves are useless. But of course if you know some 10 move algs that aren't comms for some corner cases then you could find better insertions. Personally I only ever look at 8 movers because I think 99% of the time they'd be the best ones anyway.
 

Kryptonite

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Definitely at least 99%, though it'd be nice to know either way if they are 100% or not. Are you sure there are no 3C that aren't simply conjugates of the 8f* commutators? I'm not doubting you, I just don't remember and don't have cube explorer handy.
 

Kryptonite

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Nice, that's optimal? I'm assuming so, since you can't cycle those with an 8.

It is an unconjugated commutator. The 8f*s look something like this R D R' U R D' R' U', simplifying to X Y X' Y' (our definition of commutators), where X = R D R' and Y = U.

In this case, we expand X: X = R U2 L' U2 R' and Y = U2.

Edit: Though interestingly enough, there are clearly conjugates within X in both cases, X = A B A', where A = R in the first case and R U2 in the second. This isn't directly relevant to the topic because we are only affected by conjugates outside of the commutator. But still, always interesting to examine the structure of the algs :).
 
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kinch2002

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Just asked someone because I wasn't sure, and yes it is optimal. It's known as the Per Special.

I totally didn't see that commutator form hiding in there - I forgot to look for 5 moves and 1 move for each part
 

Kryptonite

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Haha, yeah, I had to stare at it a while. Nice job though, you've eliminated the possibility of 8f*s revealing all potential optimal insertions.

Now we just have to show whether or not you need to look at conjugated commutators to find them all.
 

cuBerBruce

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Just to make it clear again, when you find a nice (123) insertion, I said you have the option of a (345) after it, or a (451) before it.

kinch2002, the valid combinations you list seem backwards to me. Maybe you number your cycles differently than I do. I'll use as an example my entry for FMC #369.

(Note, in my submitted entry I wrote the 6th move of the skeleton wrong. It's B, not B'.)

Scramble: B' L R B' F R' F2 L U' L' U' L U D' F D' U' F2 B R2 D2 F2 U D' B2 D U' L' B2 U'
Skeleton
2x2x3: B2 F' L' U' R' . B R' U R' L F (11/11)
Edges: L U B L B' L .. U' L2 (8/19)

I number the cubies the way they need to be cycled, so I numbered the cubies like this:
1) UFL (WGO)
2) LDB (OYB)
3) BUL (BWO)
4) DLF (YOG)
5) UBR (WBR)

I then looked for insertions for all five 3-cycles. I found an insertion for (123) after the fifth move that cancels four moves, and an insertion for (451) that cancels two moves after the 17th move. For my way of numbering I knew that these would combine properly since the first number of the earlier 3-cycle insertion matches the last number of the later 3-cycle insertion (where the cycles are given so the "gap" in the numbering is after the last number of the triplet, of course).

Insert at ".": L2 B R' B' L2 B R B' (8-4/23) cycles (123)
Insert at "..": R D2 R' U' R D2 R' U (8-2/29) cycles (451) with respect to the original skeleton
 

Kryptonite

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Yeah, I agree with Bruce, by normal cycle notation, I don't think your list is correct.

The combos should be:
(123)(451)
(234)(512)
(345)(123)
(451)(234)
(512)(345)
 
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Cubenovice

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I then looked for insertions for all five 3-cycles. I found an insertion for (123) after the fifth move that cancels four moves, and an insertion for (451) that cancels two moves after the 17th move. For my way of numbering I knew that these would combine properly since the first number of the earlier 3-cycle insertion matches the last number of the later 3-cycle insertion (where the cycles are given so the "gap" in the numbering is after the last number of the triplet, of course).

Thank you for the detailled example.
I feel so stupid right now for never realising that I actually have to look for cycles only once (excluding the comm. inside comm. check)

For the bolded part: I think I will remind this as the 2nd cycle has to be the continuation of the first (as in successive numbers).

I have some spare time this evening so I definitely will analyse my notes for 370.
This has been a great read, thank you all for your feedback.
 

kinch2002

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Thanks for telling me I've got it wrong - I think I have! I've edited the relevant posts in case people come and read them some other time.
For example - 'Just to make it clear again, when you find a nice (123) insertion, I said you have the option of a (345) after it, or a (451) before it' should actually read 'Just to make it clear again, when you find a nice (123) insertion, I said you have the option of a (345) before it, or a (451) after it.'
 

guusrs

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My 29 move solution at Cachan open 2011:

Scramble: U2 L' U2 L' R' B2 F2 L' D2 U L2 B R D' B R B2 D2 U F2 R' (21)
Inverse scramble: R F2 U' D2 B2 R' B' D R' B' L2 U' D2 L F2 B2 R L U2 L U2

It was hard to find something matching instantly, so I decided to try something I never tried before: orienting all edges first (ZZ).
D2 R F orients all edges nicely though I couldnt find anything nice after this. Using this as premoves (F' R' D2) on the inverse scramble got me this:

Pre: F' R' D2
2x2x2 block: U2 L' U' L
2 more 1x2x2 blocks: R U R2 U2
All edges: R2 D' R' D R2 F' R' D2 leaving 5 corners.

I decided to first invert this so I have 5 corners left on the normal scramble to avoid time pressure of inverting the whole solution at the end.
Skeleton:
D2 R F . R2 D' R D R2 U2 R2 U' R' L' U L U2 (16) after 16 minutes.

After 40 minutes I found this at the . F' L2 F R' F' L2 F R which cancels nicely, leaving 3 corners after 21 moves.
Stupidly I didn't find anything for the last 3 corners (FAIL) and had to solve them with a stupid niklas....... -.-

Final solution:
D2 R L2 F R' F' L2 F R' D' R D R2 U2 R2 U' R' L' U L U2 ... B L' F' L B' L' F L (29) sub-30, but it could've probably at least be a 26 when I wouldn't have failed so much at the last 3 corners.

Hi Erik,

Pretty hopeless scramble for me, but you did excellent work with that edge orientation, NISS search and 16 move frame.
If you would practice more on insertions you could be an exc.....
he, ho wait, NOOOO! Do not practice insertions. You're already done great work on FMC so far.
It's not a shame you missed that WR 2 years ago because you didn't know about insertions
I want that NR and next: WR!

kidding (-;
The first insertion seems optimal to me. A second insertion with 3 move cancellation is possible somewhere at the end of the frame resulting in a 26-move result.

Can you find it? (tip: use stickers!)

See ya

Gus
 
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Cubenovice

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Hey... I missed that post as it was at the bottom of the page
Nice solution, too bad about the missed insertion though.

I typically take a look at EO at the start, if it is short enough (as in your solution) I will explore it and take the shortest EO of normal and inverse, also checking it as premove like you did.

Regarding that NR: Spols open in Brussels has 2 rounds of FMC.
I'm coming after you :)
 

Kryptonite

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Okay guys, I'm trying to prove (or disprove) the following hypothesis:
All corner insertions that cost less than 8 moves after cancelations can be found by only looking for unconjugated corner 3-cycles.

I've got a couple "proofs" done, you guys'll have to check these, I haven't done proofs in a long time, and I've never done them in the context of the cube, which is pretty tricky.

The first case is completely canceling (eliminating rather than combining moves) both sides of a conjugated 8f*
A B * C D
Where * = X Y X^-1
then
B = X^-1 and C = X
So our skeleton = A D
and our insertion = A Y D
Meaning in actuality we inserted an unconjugated 8f*.

Let's look at completely canceling only one side (proving one should prove the other)
A B * C D
where * = X Y X^-1 and B = X^-1
This simplifies to
A B X Y X^-1 C D
A X^-1 X Y X^-1 C D
A Y X^-1 C D
Which can also be expressed as
A Y B C D
So we had no need for the conjugated commutator, we already found this result after A!

Combining both sides (canceling one move on both sides) cannot yield better than an 8 move insertion for fairly obvious reasons.
A B * C D
where * = X Y X^-1
A B X Y X^-1 C D
Where B X = E and X^-1 C = F
A E Y F D
The combined moves E and F essentially road block any additional cancellations, and B and C could have been combined in the first place.

If a move of the commutator cancels with the conjugate we have the following
A B * C D
where * = X Y X^-1
and Y = X^-1 Z
so * = Z X-1
since * is 8 moves (Y is 8, so X + Z = 8), it must be an unconjugated commutator, since all 8f* algs are unconjugated commutators.

The next case is when the commutator combines with the conjugate, but I haven't solved this one yet.

I can't think of any other cases...

Okay, that's where I'm at right now. Let me know what you think, there may be some flaws, gaps, or logical jumps, and it's incomplete. I'll keep updating as I go or as other people figure it out. It'd be really interesting to have this one answered.
 
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Erik

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Thx Guus, I'll try to find some stuff later. I did use stickers, but I'm horrible at recognizing the right commutator nonetheless. I always use a different cube to set it up so I can recognize it properly, but that takes time...

Would you come to a competition with 6 FMC attempts? Would be a fun Twente Open thing ;)
 

Cubenovice

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Erik, do you write the full commutator before making any move, or execute the comm "while" you write?
I find the first approach works best as it allows me to write down all possible commutators while working trough my solution.
I will only execute the (best) commutator after checking the full skeleton, never during.

For identifying the best commutators I only glance at the start and end move, if it doesn't cancel anything I move further trough the skeleton.
(I do write down the very first 8-mover even if it does not cancel, just in case there is nothing better.)
 

marco.garsed

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Hey... I missed that post as it was at the bottom of the page
Nice solution, too bad about the missed insertion though.

I typically take a look at EO at the start, if it is short enough (as in your solution) I will explore it and take the shortest EO of normal and inverse, also checking it as premove like you did.

Regarding that NR: Spols open in Brussels has 2 rounds of FMC.
I'm coming after you :)

this is a very good trick to start solves!

stupid question:

If in normal scramble I have n bad edges I have same n bad edges in the inverse scramble right?

another stupid question...

If I find EO for example R L F (too lucky XD) my premoves will be F' L' R' ?
 

Cubenovice

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If in normal scramble I have n bad edges I have same n bad edges in the inverse scramble right?
If I find EO for example R L F (too lucky XD) my premoves will be F' L' R' ?

Yes, same number of bad edges but in different locations so different moves to solve
Yes, starting moves found on the inverse should be inversed if you use them as premoves for the regular scramble

Same thing for switching the other way: starting moves found on the normal scrambles should be inversed if you want to use them as premoves for the inverse scramble.
 
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