Advanced Square-1 Methods/Theory

[deadalnix]

I see chat you mean. But do you think this is an objective observation ? 5I mean, it's a real question, not a rhétorical one). I feel quite insecure with my thumb on the little part, he's too big

 

[qqwref]

I'm going to agree with blade740 here - my fingers are longer than my thumb, so when I hold the Square-1 it is nice to have a longer part for my fingers so they don't get in the way of a / turn. Otherwise the grip just feels uncomfortable.

 

[blade740]

I see chat you mean. But do you think this is an objective observation ? 5I mean, it's a real question, not a rhétorical one). I feel quite insecure with my thumb on the little part, he's too big

Sure, I think it's an objective observation. I did a solve both ways and the other way requires me to hold the puzzle with more of an angle to my left hand. My thumb is a badass who stands wherever the hell he wants.

 

[jazzthief81]

To follow up on the edge orientation + corner permutation idea: I've finally worked out all cases with Jaap's solver and filtered them out. This text file lists for each case the 10 shortest solutions:
http://www.cubezone.be/square1eocp.txt

I used the letters O (no permutation), U, D, B, F, R and L to denote the permutation of the top and bottom corners. I used 0, 1, 3, 4, I and L to denote the "misoriented" edges in the top and bottom layer (should be clear what it means).

On average this would take [5.5|14.1] moves compared to normal EO ([4|9.9] moves) + normal CP ([4.6|9.2] moves).

 

[deadalnix]

It look's scary I think the numbers of case should be reduced to get more effiscient and more per case optimised. With edge permutation, a limit is reached I think.

Is theure any publication with finger friendly back to cube for all cases ?

 

[mrCage]

You can use conjugation to define parity of any shape you like.

A B A'

A takes you into cube shape. (one or many turns)
B performs some parity algorithm within cube shape
A' inverse/rverse of A - takes you back to the original shape

Note that the parity algorithm B can not possibly make A' impossible, only similar pieces are interchanged.

Per

 

[Stefan]

Per, I can always do that ABA' like you describe, so are you saying there's always (odd) parity?

 

[DavidWoner]

To follow up on the edge orientation + corner permutation idea: I've finally worked out all cases with Jaap's solver and filtered them out. This text file lists for each case the 10 shortest solutions:
http://www.cubezone.be/square1eocp.txt

I used the letters O (no permutation), U, D, B, F, R and L to denote the permutation of the top and bottom corners. I used 0, 1, 3, 4, I and L to denote the "misoriented" edges in the top and bottom layer (should be clear what it means).

On average this would take [5.5|14.1] moves compared to normal EO ([4|9.9] moves) + normal CP ([4.6|9.2] moves).

Thanks Lars that's fantastic! I'll start trying them out and hopefully find the best alg or two for each case over the next week or so, starting with 11 and LL. If I'm not too bored with it I'll even generate images for you.

 

[chowmein]

the different turn styles are very intriguing! i'm tempted to try the polish way, but i don't think my cube is loose enough to permit such smooth / turns...

 

[jazzthief81]

the different turn styles are very intriguing! i'm tempted to try the polish way, but i don't think my cube is loose enough to permit such smooth / turns...

You have a style of your own, Stephanie. It looks like you're untangling your arms and in the process you magically solve the puzzle as well.

 

[trying-to-speedcube...]

Per, I can always do that ABA' like you describe, so are you saying there's always parity?

And how do you define parity? After I do a random scramble, I don't know if I have parity or not. Of course, I can do that ABA', but that changes the parity. You never know if it makes odd even, or vice versa.

 

[joshuachan1995]

sorry for inappropriately (since this is about theVanderbergh method)bumping on this thread but I was wondering if this idea was feasible

To begin with..I am using Jason Baum's LBL method albeit I perform Corner Permutation which means I do not have to execute full PLL for the LL instead i am left with edge permutations only.

On to the idea... I could determine parity while executing Step 3B of Jason's method which is solving the last 2 or 1 edge(s) on the bottom layer http://jmbaum.110mb.com/square1pages/step3b.htm

Most of the algos listed on that page is already in its pure form, affecting no other edges and corners in the whole puzzle. With that, if there was parity during step 3b, could I execute step3b along with correcting the top layer

I would go about generating algos if this idea doesnt sound that stupid. Btw, I would really appreciate if anyone had systems or ideas to improvethe LBL method

 

[blade740]

With 2 or 4 edges in the wrong layers, parity would be a bit difficult to determine. You'd have to count the cycles of pieces to see if you had it or not.

 

[joshuachan1995]

Identifying parity isn't really actually much of a hassle. Two edges on the bottom layer are already solved. Using a single alg from step3b will solve the last two...

Take for example the top layer is orange and the bottom is red.
Since most of the algos provided by Jason Baum is already pure
I can easily determine where the red edges on the bottom will end up on the top layer, thus I can see whether there is a presence of parity

This algorithm here from Jason's website

(1,0)/(0,-3)/(-1,-4)/(1,1)/(0,3)/(0,3)/(-1,0)

actually has this effect
(UR-DR)(UB-UF)

If I identify parity on the LL, this algo comes in handy to prevent the parity from occurring,


Alternatively, if there is no parity I will use the pure form algo from Jaap's website being

(0,-3)/(0,3)/(1,0)/(2,-2)/(4,0)/(4,0)/(-4,4)/(0,1)/(0,3)/(-3,-1)/(1,-2)/(-1,0)

which has this effect
UF-DF

so yea i dunno if its possible to generate algos for this.