My algorithm maker software

[Stefan]

They work for me. He just solve the first layer, as he mentioned (white / yellow)

Ah yes, thanks. I had missed that, sorry. This thread is a bit confusing. And given that I'm used to color-neutral first face, those solutions looked very long.

of course EG is a really fast method!!! I just mean my method may be better because:
in EG we just make one face with irregular sides and then we use some longer algorithms to fix it again.
but in my method we don't need to make that irregular. so we don't use longer algorithms.

I have already told you that irregular means shorter algorithms, not longer. Counter-intuitive maybe, but still true (unless I messed up the computations). You have thus just given a reason why your method is worse than EG.

 

[Mohammadahmadi]

Although I don't understand much more... Why is the R in the the "Build a case" phase?

before R that isn't one of my method cases. for example sometimes you do U to make an F2L and then you do the alg.

I have already told you that irregular means shorter algorithms, not longer. Counter-intuitive maybe, but still true.

eeeee. so i failed in my method!! but same 108 alg may be useful. I'll make that soon! i want to hear your ideas about it!

I'm working on an other method too. something just like F2L!!

Ah yes, thanks. I had missed that, sorry. This thread is a bit confusing.

NP! I know that my speech is confusing. because I'm Persian!

 

[Renslay]

how do you do the first face? anyway it's better than nothing!

Take the scramble:
F2 R2 F' R U' F'

Then with EG, I would do:
x y' U R U' R' for EG2 first layer (misaligned: opposite case),
then R2 F2 R2 for CLL. That's 7 moves (6 with the cancellation).

I doubt that there is any shorter solution with aligned first layer unless you solve the whole cube in one step, like the inverse scramble.

 

[Stefan]

Using your three example scrambles:

Scramble: D B' D' L2 U' R2 B' D2 L2 B'
Orient: z y'
Solve blue face: R U' R2'

Scramble: F U2 R2 U2 R D' L' D' F L
Orient: z2 y'
Solve green face: R2' U' R

Scramble: F' D2 L' B D2 R D2 R B2 R
Orient: z' y2
Solve red face: R2' U' F

(I'm no good at 2x2, though, use poor man's Ortega)

 

[qqwref]

before R that isn't one of my method cases. for example sometimes you do U to make an F2L and then you do the alg.

I see. So not only you have to learn 300+ algs, but you may have to do setup moves at the start? It doesn't look like a very good strategy. Why did you choose these cases, anyway? Couldn't you choose more cases to eliminate setup moves? Or, couldn't you have less algs but still minimize the number of setup moves you need? Your method has so much to remember but not very much benefit.

 

[Mohammadahmadi]

i designed one another method with my program again! HEHE! and i named it FF2L. it's mean First F2L
(First First 2 layer?! omg )

in F2L you solve FR and FRD without any effect on BR and BRD, but sometimes there is a faster way to solve FR but it can move BR too. so when BR isn't true, you can use FF2L to solve FR. so you can use it 1-3 times in each solve

you can see my method here
please give your comments and ideas about my method

qqwref! what do you think?

 

[Mohammadahmadi]

it's time to see the power of my program
After weeks of trying I made a method for 2x2 first layer.
you can see my method here
give your comments

I changed my method and I fixed bugs

we have 90 cases for the first face (not first layer). we just do the first face and then we use EG method to solve the cube
Notice that sometimes you should make this cases with one move
the average of moves is 4.05 so we can solve the first layer in average with 4 moves

Please give your comments and ideas about my method

you can see my method here

 

[mark49152]

This is quite useful. I've been trying to capture a list of optimal solutions to first face for my own use.

Could you perhaps filter it so that only cases with two adjacent stickers are included? For example, always have DBL and DFL solved. Generally when choosing which face to solve, I'll go for one with two adjacent stickers already.

For some of the 5-long cases the alg list is a bit long. In such cases you could perhaps filter so that only options that turn two faces are included, or maybe three faces.

Oh - and this isn't really a method, it's an algorithm set.

 

[Mohammadahmadi]

This is quite useful. I've been trying to capture a list of optimal solutions to first face for my own use

if you do this, please send us what algorithms did you choose. as you know it's too hard to search and choose a good one. tnx

Could you perhaps filter it so that only cases with two adjacent stickers are included? For example, always have DBL and DFL solved. Generally when choosing which face to solve, I'll go for one with two adjacent stickers already.

these are all the possible cases for the first layer so i don't want to omit some of them because if you want to find a fast way to solve the first layer you should know bad cases too! so if you don't like to memorize long algorithm you can just start from the top. because they are arranged according to number of moves

Oh - and this isn't really a method, it's an algorithm set.

It really doesn't matter. because I cant speak English very well and i just say something to tell the story.

I'm too happy to hear from you
if you have any other ideas please feel free to ask me.

 

[mark49152]

these are all the possible cases for the first layer so i don't want to omit some of them because if you want to find a fast way to solve the first layer you should know bad cases too! so if you don't like to memorize long algorithm you can just start from the top. because they are arranged according to number of moves

Most people are CN on 2x2, so the first step is to choose which case looks easiest from the six choices. It's relatively unusual to find only hard cases. My own approach of choosing a case with two stickers already adjacent doesn't always give the shortest solution; I'm sure some cases without that characteristic are more efficient. But it does mean I have a reliable and well-practiced solution to each case I recognize.

 

[ben1996123]

or you could just use intuition

 

[mark49152]

or you could just use intuition

Obviously, for many cases intuition would give the shortest solution, but not always. As with F2L, there are always neat tricks to solve some cases quicker, and that's why I'm interested in a list of optimal solutions.

 

[Mohammadahmadi]

Most people are CN on 2x2, so the first step is to choose which case looks easiest from the six choices. It's relatively unusual to find only hard cases. My own approach of choosing a case with two stickers already adjacent doesn't always give the shortest solution; I'm sure some cases without that characteristic are more efficient. But it does mean I have a reliable and well-practiced solution to each case I recognize.

1 - why do you choose one face that have two stickers already adjacent? maybe some none adjacent cases are faster.
2 - my algs have all possible cases so it contain two sticker adjacent too

or you could just use intuition

of course you can!! but every body know that one memorized alg is faster that think and make a new alg. sometime our mine cant take everything into consideration.

 

[mark49152]

1 - why do you choose one face that have two stickers already adjacent? maybe some none adjacent cases are faster.

Because it's easier to recognise. I wouldn't bother with a non-adjacent case unless there were no adjacent cases or all the ones available had only slow solutions.

 

[Mohammadahmadi]

Because it's easier to recognise. I wouldn't bother with a non-adjacent case unless there were no adjacent cases or all the ones available had only slow solutions.

but i think we should learn fast cases however they may be hard.
anyway this algs are provided with myself program so if you have any ideas to delete additional algs say me. i should write one " if " in my program so say me what should I write?

 

[mark49152]

anyway this algs are provided with myself program so if you have any ideas to delete additional algs say me. i should write one " if " in my program so say me what should I write?

I'd like to see a list of optimal solutions where DBL and DFL are initially solved but I can search through your list and find those cases. What's the difference between 3v1 and 2v2?

 

[Mohammadahmadi]

What's the difference between 3v1 and 2v2?

2v2 mean 2 white on first layer and 2 on the second
3v1 mean 3 on first and 1 on second

I'd like to see a list of optimal solutions where DBL and DFL are initially solved but I can search through your list and find those cases.

I'll provide that for you soon.

 

[Mohammadahmadi]

I'd like to see a list of optimal solutions where DBL and DFL are initially solved but I can search through your list and find those cases.

your algs are ready
please see here

 

[mark49152]

your algs are ready
please see here

That's great, thanks. I've already found some new tricks.

How do you determine the cases? I'm not sure I understand why the third set has only 6 cases. What about the case where the top sticker faces up, and the bottom sticker is on the side?

 

[Mohammadahmadi]

That's great, thanks. I've already found some new tricks.

How do you determine the cases? I'm not sure I understand why the third set has only 6 cases.

it's too easy. in all of them DBL and DFL are fix. so we have two another white.

one case is that both of them are in first layer. in this case you can make a 'case1' alg with do R2
one case is one in first layer and one in second. and we have all of them in 'case3'
one case is that both on second layer. and they are in 'case1' and 'case2'

in 'case1' we have two white side by side. each white may be in 3 sides so we have 3x3=9 cases.
in 'case2' we have two white face to face. each white may be in 3 sides but some cases can be repeat with doing U2! for example, FLU on top and BRU on right face = FLU on left and BRU on top. you can make them from each other with do a U2. so we don't have 9 cases! 3 of them are repeated so we have just 6 cases.
in 'case3' just like 'case2' we have 6 cases again. and they can make with R2

What about the case where the top sticker faces up, and the bottom sticker is on the side?

do a R2 so we have third case in 'case3'

if you have any other questions, please feel free to ask me. tnx