Idea for subseting ZBLL

[sz35]

As you can see from my signature I know almost half of ZBF2L and I want to learn ZBLL in some stage after I finish all the F2Ls,but I looked at some ZBLLs and they are really really hard to learn. I thought about :"how can I reduce them??" and suddenly this Idea came to me:
Step 1: making the cross+one ML edge( 0 algs/~8 moves)
Step 2: 2 F2L pairs (41 algs/~13 moves)
Step 3: ZBF2L (306 algs/~7.5 moves)
Step 4: Last corner+ any LL reduction
Step4a(My idea): permuting 2 opposite LL coeners while insertion of the corner of the LS. ( 15 algs/~8 moves)
Step 5: Reduced ZBLL (With my idea: 146 algs/~12 moves)
Total: moves:48.5(3 moves more then ZB) algs:508(333 less than ZB)

So what do you think?

 

[irontwig]

I still think Lars Petrus' idea is better, less algs, no extra LL preperation and less moves.

 

[Davepencilguin]

I still think Lars Petrus' idea is better, less algs, no extra LL preperation and less moves.

I've never seen this before.....
To my understanding, it's similar to the 1-look, 2-algorithm form of OLL, right?


I think this has potential...

 

[trying-to-speedcube...]

You will need some extra algs for ZBF2L when the corner of the last slot is stuck under the solved middle edge.

 

[Davepencilguin]

It sounds very similar to MGLS, in my opinion.

It'd be interesting to incorporate ZBF2L with CLS from MGLS though....

Cross + a middle edge
2 F2L slots
ZBF2L with extra algorithms for the cases when the ZBF2L corner is under the solved single edge
CLS
PLL

6 steps rather than CFOP's 7 steps.... hmm......
I'll look into it.