1x1x1 2-simul | How to get into any position into needed position in 2 rotations

[acrystal]

So I keep seeing 3 rotations to get to a position sometimes. My honest opinion, its very easy to find a 2-rotation solution only! Here is how it works

Step 1: Check how many rotations it takes
Let's use poles to help us find out this case. A pole are 2 opposite centers placed in a specific position, that doesn't cahnge when a specific rotation is done. Poles are named after the rotation that doesnt change its place. There is the x-pole, y-pole, and z-pole.
For example, x-poles are named that because the position doesn't change when you do a x move.
Now compare the needed state to the current state. If:
3 poles matched: 0 rotations needed
1 pole matched: 1 rotation needed
0 poles matched: 2 rotations needed
For example, in the case of white-green as the needed state:
white-top and green front: x-pole, y-pole, and z-pole matches - 0 rotations needed
white-top and red-front: y-pole matches - 1 rotation needed
orange-top and blue front: no poles match - 2 rotations needed

Step 2: Solving 1 pole matched / 1 rotation needed
Let's make a blueprint: "0 poles ->1 pole-> 3 poles".
Now, you might skip the 0 poles part and get straight into 1 pole. Then the position can be done in 1 rotation, and it's very easy actually. Check which pole is solved, then find the optimal solution containing the rotation in the name of the solved pole.
For example, if you have white-top and red-front, you'll notice that only the y-pole is solved. Then, you would have to find an optimal solution with a y move, in this case, y'.

Step 3: Solving 0 poles matched / 2 rotations needed
Sometimes, you can not get any poles matching. Then the position can be done in 2 rotations. Think of it as a parity case, but instead of learning a parity alg, you'd make one! First, you could do a x/y/z move to solve parity, and solve one of the poles. Then, you solve all 3 poles by using the 1 pole trick.
For example, if you have orange-top and blue front, there is 3 solutions depending on what type of rotation the parity alg is made up of:
x: x2(solve parity/z-pole) z
y: y2(solve parity/z-pole) z'
z: z(solve parity/y-pole) y2

I hope this helps (even though this is kinda unintuitive).

 

[acrystal]

Bump
I realized 1x1 should be done without inspection, while this version uses inspection
So I guess this works for rotation purposes

 

[Mr.Ducky]

Bump
I realized 1x1 should be done without inspection, while this version uses inspection
So I guess this works for rotation purposes

Still works for non inspection if you think about it you memorize all the colors on 2×2 and instantly recognize your case to get the white top and green front