Intuitive Square 1?

ChickenWrap

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Is there any way to do a square 1 without learning algs? I can get all the way to a square shape with 4 corners solved. I don't care about speed, I just want to be able to solve the puzzle without memorizing any algs, since I suck at that :)
 
Probably not. I guess if you are really good, you could figure out commutators, but you would still need the parity alg.
 
I haven't seen an intuitive method for Square-1 as of now.

Sorry, I meant commutators, or even a couple simple algs. I know I still have to memorize the parity alg and that is fine.

My problem is that I can't figure out how to do a "2 look pll" type thing on SQ1 (ie. only needing a couple algorithms that you repeat many times).
 
It's doable. Assuming you can "play around" and find an alg like J/J, you can use that with commutators to solve any non-parity case. You can "intuitively" determine a parity algorithm too: you need to get the cube into a state where you can swap an odd number of pairs of pieces with a single twist. The simplest way to do this is to get to the scallop/scallop shape (3 twists), swap 3 sets of corners with a / move, then undo.
 
I found the most intuitive part of square 1 to be cubeshape.
You can trace every single cube shape into a tree and make optimal cubeshapes this way. This is why I enjoy solving cube shape the most.
Of course, you could also be lazy and look up this awesome tree by Jaasp.
You can use this tree to verify optimal path.
 
The normal Lars method, just less algs:
- You need to learn a cube shape method, which requires one mini alg.
- An other "alg" you should know is the "M2", which is just ( 1,0 / -1,-1 / 0,1 ) done on a Square-1 in cube shape.
- To seperate (orient) the yellow from the white method, just do what you would do on a 4x4 at the last two centers.
- If you do R2 U R2 U M2 U' R2 U' R2 (always "missaligning" the top layer by an edge so you stay in cube shape), you swap the bottom back edge with the top right edge (for edge seperation).
- For PLL on both sides, if you have two normal PLLs or two unnormal PLLs, do M2. If one is normal and one is not, do something like / (3,3) / (1,0) / (4,-2) / (-4, 2) / (-1,0) / (-3, -3) /
(parity) and check again.
For normal PLL cases you can use Stefan's method.
 
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