#### meskelto

##### Member

The reasons behind this:

- There is a high chance for slice move insertions to solve bad edges.
- Many bad edges can be solved for free due to a large amount of move cancellations.
- Not too difficult to solve intuitively, meaning multiple HTORs can be checked for good finishes in FMC.

**Steps:**1: First HTO Block (FHB) with Corner Permutation (CP) <U D R2 L2 F2 B2> [3 moves]

CP can be difficult so just try a FHB and test for CP. The CP can be tested on a corner pair within a potential FHB (with or without a good edge in between) by trying to “solve” all other corner pairs 2-gen (“solve” = all 4 corner pairs have the same pattern; it’s easier to see if both corners belong in either U or D).

2: Second HTO Block (SHB) with LL Corners (C) (i.e. AUF); mostly 2-gen <U Uw R2 Rw2> [7 moves]

Just form a corner pair around the correctly coloured edge and insert. You want to reduce the number of bad edges in the LL by forming corner pairs around the corresponding coloured edge. Then maybe there are only 2 or 4 bad edges in total to solve with one slice insertion.

3: HTO LSE (HLSE) <U Uw M2> [0 moves due to cancellations; 10 moves total]

Now the entire HTOR solution so far can be considered a skeleton, and the remaining 2 or 4 edges can be converted from bad to good using slice insertions (maybe with set-up moves if needed).

There could be an insertion opportunity if two adjacent coloured bad edges (e.g. red, blue) are diagonally opposite in the same slice. Then an M2 would swap those two edges into their correct slice (by the end of the skeleton).

However, the other two edges in the M slice must also be considered. If they are both the same/opposite colour (e.g. orange/red) (so M2 = nothing changes) or adjacent coloured

There could be an insertion opportunity if two adjacent coloured bad edges (e.g. red, blue) are diagonally opposite in the same slice. Then an M2 would swap those two edges into their correct slice (by the end of the skeleton).

However, the other two edges in the M slice must also be considered. If they are both the same/opposite colour (e.g. orange/red) (so M2 = nothing changes) or adjacent coloured

**bad**edges then an M2 can be inserted to make progress converting edges from bad to good.In FMC checking for CP is fine, just try to solve 2-gen, and HLSE can be solved via slice insertions. Step 1 is definitely the hardest step, but once you overcome this then the rest of the solve can be easy (see LOBE).

**LOBE:**One variant of this method that could be considered the go-to method is where you form a Line Of Bad Edges (LOBE) in DF, DB and then solving to HTOR is simple. This is because there is only one edge of the colour of the SHB, meaning that after solving the SHB the rest of the edges “sort themselves out” since they are all effectively the same.

By solving in this way the final bad edge case is always the same. In fact, 50% of the time the insertion to solve them cancels directly at the end, giving an easy finish. This makes this my go-to method, since one could find a decent HTOR quickly, without having to go back and look for insertions.

There may be a problem for the other 50%, as then you would have to go back and hope to find enough insertion opportunities (2 insertions are probably needed; 1 if you are lucky). This should be avoided by solving the SHB differently though.

HTORoux gives a reasonable opportunity for slice insertions since bad edges will likely appear in the M slice at some point, when any bad edges in DF, DB can be swapped. This is because the bad edges will spend most of the time in U.

Additionally, due to the entire second step being mostly 2-gen, there are 2 move cancellations between M2 and R2 meaning every M2 only slice insertion is free!

Finally, by solving 2-gen and especially by forming a LOBE, it can be very easy to finish the HTOR so a decent solution can be found quickly.

Additionally, due to the entire second step being mostly 2-gen, there are 2 move cancellations between M2 and R2 meaning every M2 only slice insertion is free!

Finally, by solving 2-gen and especially by forming a LOBE, it can be very easy to finish the HTOR so a decent solution can be found quickly.

CP is tricky.

Solving 2-gen can sometimes lead to a larger move count than you would want, for some bad cases.

If there are more bad edges in the top layer than in the bottom layer then it is difficult to set up slice insertions for all bad edges. This can be resolved by moving some bad edges to DF, DB with Rw2 (instead of R2; or an M2 cancelling with R2), but sometimes a good opportunity doesn't arise.

It may also be difficult to influence the final HTO step without adding too many extra moves, but I don’t know too much about this currently. These can mostly be resolved by building the FHB differently, picking a different FHB, maybe switching to the inverse, or even trying a different DR.

Solving 2-gen can sometimes lead to a larger move count than you would want, for some bad cases.

If there are more bad edges in the top layer than in the bottom layer then it is difficult to set up slice insertions for all bad edges. This can be resolved by moving some bad edges to DF, DB with Rw2 (instead of R2; or an M2 cancelling with R2), but sometimes a good opportunity doesn't arise.

It may also be difficult to influence the final HTO step without adding too many extra moves, but I don’t know too much about this currently. These can mostly be resolved by building the FHB differently, picking a different FHB, maybe switching to the inverse, or even trying a different DR.

**FHB+CP:**- One way to solve CP is by creating a new FHB/corner pair 2-gen after a trial FHB/corner pair, creating a 3:1 situation between corner pairs. However this always leads to bad 2-gen cases. Alternatively, create random FHBs and hope for the best…
- I think a lot depends on the 2-gen case you end up with; it’s better to have a 5 move FHB+CP and 3-5 move finish vs 3 move FHB+CP and 10 move finish. Unfortunately I have no idea how to influence this currently.
- You can try corner pairs that belong to different layers, so one U corner and one D corner (advanced).
- FHB+CP can be combined with LOBE.

**Bad Edges:**- Could try 1 bad edge in D, and hope you can find a solution with only 1 other bad edge and not 3.
- No bad edges in the bottom could be tricky, but doable if the solution is obvious. In this case I would transition to one of the other two cases by moving bad edges to D with Rw2.
- Could have a bad edge in one of FHB/SHB to solve with an S slice insertion. These don’t usually cancel though, but it could be useful in some cases. There has to be 0/1 bad edges in D for this to work.

**HTO:**- Try to form solved CE pairs/blocks when creating FHB/SHB to hopefully get a better HTO state.
- Try to insert slices in a different way (U instead of U’) to also influence the HTO state. This will probably add a few extra moves to HTOR but it could be worth it.
- I wouldn’t worry too much about E edges in HTO. These can sometimes be solved with inserted E slice moves after DR is complete, possibly in 0 moves due to cancellations.
- I would try to avoid having more than one 2e2e/3e alg to insert, otherwise the HTO probably isn’t worth it.

**Example Solves:**__Scramble 1:__U’ R2 B2 D’ F2 D2 F2 L2 U’ F2 D’ F2 R2 U

FHB+CP: F2 U F2 U x2 y [4 moves]

LOBE: R2 U’ Rw2 [3 moves]

SHB+C: U R2 * U [3 moves]

HLSE *: M2 [0 moves after cancelling]

Note: M2 = R2 Rw2

So really the finish is: U Rw2 U

__Solution 1:__F2 U F2 U F2 D’ B2 U B2 D [10 moves]

__Scramble 2:__R2 U B2 D L2 R2 D2 R2 D’ U2 B2 F2

FHB+CP: Skip

LOBE: U2 R2 U’ Rw2 [4 moves]

Finish: U’ Rw2 U’ [3 moves]

__Solution 2:__U2 R2 U’ L2 D’ L2 U’ [7 moves]

I also found this:

SHB+C: * U’ R2 U R2 U’ [5 moves]

HLSE *: U2 M2 U2 [3 moves after cancelling]

__Another Solution 2:__U2 L2 R2 D R2 D R2 D’ [8 moves]

__Scramble 3:__D’ B2 D2 L2 U’ F2 D’ B2 U L2 U’ F2 D2

FHB+CP: U’ B2 U B2 x2 [4 moves]

LOBE: Rw2 [1 move]

SHB+C: U’ R2 U2 R2 U R2 * U [7 moves]

HLSE *: M2 [0 moves after cancelling]

__Solution 3:__U’ B2 U B2 L2 U’ R2 U2 R2 U L2 D [12 moves]

I also found these:

() – Inverse

FHB+CP: (R2 U’ R2 x2 y’) [3 moves]

SHB+C: (* R2 U’ R2 U R2 U’ R2 U) [8 moves]

HLSE *: M2 [0 moves after cancelling]

__Another Solution 3:__(R2 U’ R2 F2 U’ B2 U B2 U’ B2 U) [11 moves]

FHB+CP+LOBE: (Rw2 U’ R2 x2 y’) [3 moves]

SHB+C: (R2 U’ R2 U’ R2 U2 R2 U R2 * U) [10 moves]

HLSE *: M2 [0 moves after cancelling]

__Another Solution 3:__(L2 D’ R2 F2 U’ F2 U’ F2 U2 F2 U B2 D) [13 moves]

__I think HTORoux is a solid way of completing HTOR in FMC, though the CP still needs work. I think consistent ~10 move HTORs could be found with this, and about 8 moves may be achievable by trying different DR solutions in FMC.__

**Conclusion:**