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Lower bound using the Edges antipode
Submitted by mdlazreg on Mon, 01/29/2007 - 15:01.One way of solving the cube is using two phases:
1) First solve all the edges [less than 18 moves as proved by Tom Rokicki ].
2) Take it to the cube identity [less than 22 moves as proved by Silviu Radu]. Which gives a maximum of 40 moves.
Is it possible to follow the same logic but use the Edges Antipode instead?. So the two phases become:
1) Solve all the edges by taking them to the Antipode edges instead of the identity edges. [this is guaranteed to be 18 moves].
2) From the edges antipode position take it to the cube identity. [Does anyone know the maximum moves needed for this phase?]. To find this maximum it would take the same computational effort to prove the 22 moves I guess. Let's call this maximum X.
1) First solve all the edges [less than 18 moves as proved by Tom Rokicki ].
2) Take it to the cube identity [less than 22 moves as proved by Silviu Radu]. Which gives a maximum of 40 moves.
Is it possible to follow the same logic but use the Edges Antipode instead?. So the two phases become:
1) Solve all the edges by taking them to the Antipode edges instead of the identity edges. [this is guaranteed to be 18 moves].
2) From the edges antipode position take it to the cube identity. [Does anyone know the maximum moves needed for this phase?]. To find this maximum it would take the same computational effort to prove the 22 moves I guess. Let's call this maximum X.
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