# 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.

Now let's look at Tom's calculation. Is is possible to reproduce them by adding one generator which is the one that takes the identity to the edges antipode immediatly? If we do this calculation the maximum moves will be for sure less than 18moves and more than 9 moves. Let's call it Y.

This means that any cube position can be solved in either:

Y + X moves [taking it first to the edges antipode than to the cube identity]

Y + 22 moves [taking it first to the edges identity than to the cube identity]

Silviu has proved a minimum limit of 35q. So if X < 22 and Y < 13 we might lower the limit a little bit...

Please let me know what you think of this approach, if it has been attempted or if it has some logic hole.

Thanks

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.

Now let's look at Tom's calculation. Is is possible to reproduce them by adding one generator which is the one that takes the identity to the edges antipode immediatly? If we do this calculation the maximum moves will be for sure less than 18moves and more than 9 moves. Let's call it Y.

This means that any cube position can be solved in either:

Y + X moves [taking it first to the edges antipode than to the cube identity]

Y + 22 moves [taking it first to the edges identity than to the cube identity]

Silviu has proved a minimum limit of 35q. So if X < 22 and Y < 13 we might lower the limit a little bit...

Please let me know what you think of this approach, if it has been attempted or if it has some logic hole.

Thanks