1. ## Curvy Copter algs

Hi!

I challenge anyone to come up with a shortest possible way (please provide sequence or description) to flip all the 12 middle edges of the curvy copter. So far i am down to 72 turns but still improving. It is quite fun but aggrevating also ...

Per

2. One can use the following 26 turns to make up a complete 12-flip solution in 58 turns. I show part 1 only. Part 2 is same on the other half.
It is done like so: part1 + setup + part2 + unsetup.

Part1:

/*000000*/BD,
/*000001*/RB,
/*000002*/DR,
/*000003*/BD,
/*000004*/DR,
/*000005*/BD,
/*000006*/DR,
/*000007*/BD,
/*000008*/RB,
/*000009*/BD,
/*000010*/DR,
/*000011*/RB,
/*000012*/DR,
/*000013*/FU,
/*000014*/RF,
/*000015*/UR,
/*000016*/FU,
/*000017*/UR,
/*000018*/FU,
/*000019*/UR,
/*000020*/FU,
/*000021*/RF,
/*000022*/FU,
/*000023*/UR,
/*000024*/RF,
/*000025*/UR,

Copy and paste into gelatinbrain applet to verify. It flips 6 edges and twists 2 corners. Yay!!

Per

PS! I really miss a "play" feature on gelatinbrain. Should be quite easy to implement i guess ...

3. ## Curvy copter 2-flip with no jumbling.

No, this is not another "fyou have been framed" post. This is serious! It has previously been proven that 2 adjacent edges cannot be flipped without jumbling. However 2 opposite edges (FR and BL) can be flipped, and i will constructively show how. For sure not the shortest way possible, but it does the trick.

First i will flip 6 edges with 2-part sequence:
DB DR UR UF FR [UR UF]3 FR UF UR FR DR DB (17 turns)
RF RD DL BL BD [DL BL]3 BD BL DL BD RD RF (17 turns)

Now we have a 6-flip!

Next we flip 4 of those edges:
[UR UF]3 UB UL BL FR DR DF [DB DL]3 DF DR FR BL UL UB (24 turns)

Altogether a dreadful 58 turns, same as i currently have for my 12-flip (also no jumbling).

Per

PS!

Copy and paste this ugly sequence into gelatin curvy copter applet to verify my sequence(s).

/*000000*/BD,
/*000001*/RD,
/*000002*/UR,
/*000003*/FU,
/*000004*/RF,
/*000005*/UR,
/*000006*/FU,
/*000007*/UR,
/*000008*/FU,
/*000009*/UR,
/*000010*/FU,
/*000011*/RF,
/*000012*/FU,
/*000013*/UR,
/*000014*/RF,
/*000015*/RD,
/*000016*/BD,
/*000017*/RF,
/*000018*/RD,
/*000019*/DL,
/*000020*/LB,
/*000021*/BD,
/*000022*/DL,
/*000023*/LB,
/*000024*/DL,
/*000025*/LB,
/*000026*/DL,
/*000027*/LB,
/*000028*/BD,
/*000029*/LB,
/*000030*/DL,
/*000031*/BD,
/*000032*/RD,
/*000033*/RF,
/*000034*/UR,
/*000035*/FU,
/*000036*/UR,
/*000037*/FU,
/*000038*/UR,
/*000039*/FU,
/*000040*/UB,
/*000041*/LU,
/*000042*/RF,
/*000043*/LB,
/*000044*/RD,
/*000045*/DF,
/*000046*/DL,
/*000047*/BD,
/*000048*/DL,
/*000049*/BD,
/*000050*/DL,
/*000051*/BD,
/*000052*/DF,
/*000053*/RD,
/*000054*/RF,
/*000055*/LB,
/*000056*/LU,
/*000057*/UB,
Per

It can be reduced by at least 12 turns when inserting the 4-flip into the 6-flip. Please feel free to work it out ...

4. My lemma. This is the only 2-flip scenario possible on the curvy copter with no jumbling sequences!!

Per

5. I'm gonna sketch a proof for my own lemma. Make complete map of the 4 center orbitals. The only place that 2 center orbitals "cross" over another edge (after first flip) is on the opposite side of the puzzle. This can be done like so (disregarding corners, that can trivially be restored):

/*000000*/UB,
/*000001*/UR,
/*000002*/LU,
/*000003*/RF,
/*000004*/FL,
/*000005*/BR,
/*000006*/LB,
/*000007*/RD,
/*000008*/DL,
/*000009*/DF,
/*000010*/DL,
/*000011*/RD,
/*000012*/LB,
/*000013*/BR,
/*000014*/FL,
/*000015*/RF,
/*000016*/LU,
/*000017*/UR,

Bolded part is setup turns! So: first swap around UB, setup, then swap around DF. Etc ....

Per

6. I completed the above sequence with solving the 4 corners also. Slight optimisations gave me this:

/*000000*/UB,
/*000001*/RF,
/*000002*/FU,
/*000003*/RF,
/*000004*/UR,
/*000005*/RF,
/*000006*/FU,
/*000007*/RF,
/*000008*/LU,
/*000009*/RF,
/*000010*/FL,
/*000011*/BR,
/*000012*/LB,
/*000013*/RD,
/*000014*/DL,
/*000015*/DF,
/*000016*/DL,
/*000017*/RD,
/*000018*/LB,
/*000019*/BR,
/*000020*/FL,
/*000021*/RF,
/*000022*/LU,
/*000023*/FU,
/*000024*/UR,
/*000025*/UB,
/*000026*/UR,
/*000027*/FU,
/*000028*/UR,
/*000029*/UB,
/*000030*/UR,
/*000031*/FU,
/*000032*/UR,
/*000033*/FU,

34 non-jumbling turns for an edge 2-flip!!

Per

7. ## Curvy copter middle edges flips (and fliptwists).

Here i will post any (best cases only) edge flipping algorithms that i have found for the curvy copter. These are invisible on the normal helicopter cube!!

First one out of the sack is the 12-flip. This one has a long story behind it, but i will spare you the details and just post the sequence:

/*000000*/UB,
/*000001*/UR,
/*000002*/RF,
/*000003*/BR,
/*000004*/RD,
/*000005*/LU,
/*000006*/FL,
/*000007*/LB,
/*000008*/DL,
/*000009*/BD,
/*000010*/DF,
/*000011*/LU,
/*000012*/FL,
/*000013*/LU,
/*000014*/FL,
/*000015*/LU,
/*000016*/FL,
/*000017*/DL,
/*000018*/LB,
/*000019*/FL,
/*000020*/LU,
/*000021*/DL,
/*000022*/LB,
/*000023*/DL,
/*000024*/LB,
/*000025*/DL,
/*000026*/LB,
/*000027*/UR,
/*000028*/RF,
/*000029*/UR,
/*000030*/RF,
/*000031*/UR,
/*000032*/RF,
/*000033*/RD,
/*000034*/BR,
/*000035*/RF,
/*000036*/UR,
/*000037*/RD,
/*000038*/BR,
/*000039*/RD,
/*000040*/BR,
/*000041*/RD,
/*000042*/BR,
/*000043*/FU,

Per

PS! More flips will follow when i get the time, and please feel free to contribute.

8. It seems like your pretty much on your own with the love for the Curvy Copter
I'm curious now though, think I'll buy one, which do you recommend?

9. Right. I guess i'm a slow solver But if you should choose between a helicopter cube and a curvy copter one. Get the curvy, unless the extra edge solving annoys you. Cheaper and better turning quality

Per

10. Ok, i will add more flip cases. Now a 2-flip from another thread.

/*000000*/UB,
/*000001*/RF,
/*000002*/FU,
/*000003*/RF,
/*000004*/UR,
/*000005*/RF,
/*000006*/FU,
/*000007*/RF,
/*000008*/LU,
/*000009*/RF,
/*000010*/FL,
/*000011*/BR,
/*000012*/LB,
/*000013*/RD,
/*000014*/DL,
/*000015*/DF,
/*000016*/DL,
/*000017*/RD,
/*000018*/LB,
/*000019*/BR,
/*000020*/FL,
/*000021*/RF,
/*000022*/LU,
/*000023*/FU,
/*000024*/UR,
/*000025*/UB,
/*000026*/UR,
/*000027*/FU,
/*000028*/UR,
/*000029*/UB,
/*000030*/UR,
/*000031*/FU,
/*000032*/UR,
/*000033*/FU,

Per

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