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Tripod Finish

Lofty

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I read about tripod finishes quite some time ago and for some reason just started thinking about them a little the past few days. Does anyone know any numbers for this kind of finish? Move counts cases, etc. I know the number of algs to do this in one step is massive but if it was broken up into corners and edges or orient then permute it doesn't seem like it can be that many cases. With something like that all the cube but 7 pieces could be solved with intuition then only the last 7 pieces would need to be done with algs.
One would probably do edges then corners since the corners could be done with commutators/conjugates.
idk I just think its interesting.
 
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Lucas Garron

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I read about tripod finishes quite some time ago and for some reason just started thinking about them a little the past few days. Does anyone know any numbers for this kind of finish? Move counts cases, etc. I know the number of algs to do this in one step is massive but if it was broken up into corners and edges or orient then permute it doesn't seem like it can be that many cases. With something like that all the cube but 6 pieces could be solved with intuition then only the last 6 pieces would need to be done with algs.
One would probably do edges then corners since the corners could be done with commutators/conjugates.
idk I just think its interesting.
I think I get what you mean...
CP: 4*3*2
EP: 3*2
Parity: 1/2
CO: 3*3*3
EP: 2*2

Make 7776 states for L7P (last 7 pieces).
Splitting into 3 steps gives a lower bound of about 57 algs (175 for 2).
A bit high... It's very appealing up to L7P, though...
 

Stefan

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(not considering symmetric cases)

Rotate the tripod: 1/3
Mirror: 1/2
Inverse: 1/2

648 cases => about 51 algs for two-step

But wait, there's more! The three reductions can be used for both steps!

sqrt(7776) = 88 cases per step = 7.35 reduced cases per step = 15 algs
 

Lofty

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ok that is more what i was thinking...
I didn't think there could be more algs then OLL and PLL since you have one edge already solved.
anyway interesting but since I just saw Tomy's recent OH progress there is no time to look into it!
 

cuBerBruce

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I assume the tripod means a cube that's solved except for UFL,URF,UBR,DFR,UF,UR, and FR or some symmetrically equivalent set. In other words, it would basically be a solved cube except a keyhole in the first two layers and 2x2x1 block in the last layer (LL). The 2x2x1 block wouldn't really have to be the block opposite the keyhole position, since the LL could be rotated to move whatever 2x2x1 block you have to that position. Technically the corner and edge in the first two layers wouldn't have to be a pair either since the layer opposite the LL could be rotated to line up the corner with the edge.

I wrote a little gap code to calculate the number of symmetrically unique cases of the tripod group. I got 1317 patterns for symmetry reduction, and the number went down to 794 patterns when using inverses as well. I also note that rotation by 120 degrees about tripod's central axis does not preserve orientation of cubies using the usual orientation schemes.
 

Stefan

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Bruce, there might be some (incomplete) first two layers for this, but there's no last layer. So your thoughts can be generalized further.

Not sure this would be a good method for speedsolving, but it could be very good for fewest moves competitions.
 

cuBerBruce

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Bruce, there might be some (incomplete) first two layers for this, but there's no last layer. So your thoughts can be generalized further.

I assumed somebody would set up a tripod finish by first solving two layers except a corner/edge pair (probably using either Fridrich or Petrus style solving). Then, they would need to create a 2x2x1 block in the final layer. I saw no need to constrain which of the four possible 2x2x1 blocks you would create (except for possibly making recognition of tripod cases easier). But if you set up a "pure" tripod finish, then the concept of a last layer becomes moot, of course.
 
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