SpeedCuber71
Member
Before you read any further, i just want to say that this method or just modification to already existing methods may not be good for speedsolving and completely stupid but i thought just to post it.
I average around 20 seconds at SQ1 and was thinking about EP. I was thinking it would be so good if i had learnt all EPs then it would just be like "PLL" in 3x3. Super fast. Just recognise the case and execute alg. But recognising on both layers was too much for me. So then i came across the roux method. In roux the bottom layer is solved intuitively, and i liked that a lot. And then just solving the last layer in one shot/quickly with few EP algs would be much better and faster. Yes, adj-adj algs can be used in Vandenbergh;s method but i just thought of this.
Roux turned out to be pretty slow for me so i decided to take some techniques from it and aplly it to Vandenbergh;s method. So here are the steps.
1. Cube shape.
2. Corner Orientation just like in Vandenbergh's method.
3. Corner Permutation.
4. We solve 2 adjacent edges in the bottom layer. Now in roux method doing M2 (which all sq1ers know) we attach the edge to two permuted corners. So i thought about this. Of course we can't solve two opposite because the second edge will always kick out the first one we solved in the same line. You will understand this if you know how M2 works. This time it isn't as slow as roux. We pick any one edge on bottom layer randomly that is of the bottom color. Then we just do an M2. If the color was green, then we just bring the green headlights/corners in the front by doing a D, D' or D2. (Yeah i was too lazy to type in sq1 notations but you get what i am talking about
) and do an M2 again so that is solved and that will kick out another edge. So as we have to solve adjacent color, we just do D or a D' and another M2. Now we have 2 adjacent edges solved. I still don't know how to solve the remaining 2 adjacent fast. Maybe someone could come up with an alg set? Because using the EO algs we all know doesn't always work as the two adjacent edges could end up switched
Once this is done the last step
5. Last layer EP. This will be super fast to recognise and exec FOR ME.
I know this method may be pretty stupid and not even worth a post but....why not try
here is an example solve
Scramble
(0, -1)/(-3, 6)/(4, 4)/(-3, -3)/(5, 0)/(6, -3)/(-3, 0)/(-5, -1)/(0, -2)
Solution
(0,2)/(-4,-5)/(3,0)/ - CS
(1,3)/(0,3)/ - CO
(-4,6)/(-3,0)/(-3,-3)/(-3,0)/ - CP
Here the green edge is already solved and we can just put the red and orange in but for the sake for explanation, lets assume its not solved. Bring the red headlights in front. You will see the blue edge. Do and M2. then quickly do a D and another M2 which will solve the blue then see which color the blue kicks out. If its an adjacent color, we can again do a D or a D' and another M2. In this case, the blue has a yellow edge. So we need to bring in an edge from the top layer.
After this the remaining two which i mentioned about above and last layer EP.
I average around 20 seconds at SQ1 and was thinking about EP. I was thinking it would be so good if i had learnt all EPs then it would just be like "PLL" in 3x3. Super fast. Just recognise the case and execute alg. But recognising on both layers was too much for me. So then i came across the roux method. In roux the bottom layer is solved intuitively, and i liked that a lot. And then just solving the last layer in one shot/quickly with few EP algs would be much better and faster. Yes, adj-adj algs can be used in Vandenbergh;s method but i just thought of this.
Roux turned out to be pretty slow for me so i decided to take some techniques from it and aplly it to Vandenbergh;s method. So here are the steps.
1. Cube shape.
2. Corner Orientation just like in Vandenbergh's method.
3. Corner Permutation.
4. We solve 2 adjacent edges in the bottom layer. Now in roux method doing M2 (which all sq1ers know) we attach the edge to two permuted corners. So i thought about this. Of course we can't solve two opposite because the second edge will always kick out the first one we solved in the same line. You will understand this if you know how M2 works. This time it isn't as slow as roux. We pick any one edge on bottom layer randomly that is of the bottom color. Then we just do an M2. If the color was green, then we just bring the green headlights/corners in the front by doing a D, D' or D2. (Yeah i was too lazy to type in sq1 notations but you get what i am talking about
) and do an M2 again so that is solved and that will kick out another edge. So as we have to solve adjacent color, we just do D or a D' and another M2. Now we have 2 adjacent edges solved. I still don't know how to solve the remaining 2 adjacent fast. Maybe someone could come up with an alg set? Because using the EO algs we all know doesn't always work as the two adjacent edges could end up switched Once this is done the last step
5. Last layer EP. This will be super fast to recognise and exec FOR ME.
I know this method may be pretty stupid and not even worth a post but....why not try
here is an example solve
Scramble
(0, -1)/(-3, 6)/(4, 4)/(-3, -3)/(5, 0)/(6, -3)/(-3, 0)/(-5, -1)/(0, -2)
Solution
(0,2)/(-4,-5)/(3,0)/ - CS
(1,3)/(0,3)/ - CO
(-4,6)/(-3,0)/(-3,-3)/(-3,0)/ - CP
Here the green edge is already solved and we can just put the red and orange in but for the sake for explanation, lets assume its not solved. Bring the red headlights in front. You will see the blue edge. Do and M2. then quickly do a D and another M2 which will solve the blue then see which color the blue kicks out. If its an adjacent color, we can again do a D or a D' and another M2. In this case, the blue has a yellow edge. So we need to bring in an edge from the top layer.
After this the remaining two which i mentioned about above and last layer EP.
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