4x4x4 Superflip of 11 out of 12 edges

[unsolved]

The fastest Superflip I've seen for 11 out of 12 dedges for the 4x4x4 is 28 moves. Is there a solution with fewer moves?

 

[Christopher Mowla]

The fastest Superflip I've seen for 11 out of 12 dedges for the 4x4x4 is 28 moves. Is there a solution with fewer moves?

Are you referring to the BTM metric or single slice turn (ply) metric?

If you are referring to the BTM metric, then I assume you are talking about cuBerBruce's algorithm. If so, then here's a 26 BTM I just found in under an hour.
2R U' F2 U 2L F2 2R U2 2R U' B2 U 2R D2 L' R' B2 L2 B' R2 B D2 B e F m y z

Of course, cuBerBruce's is 35 ply, and this one (which I made from one of his parity algs) is 28 ply.

If you are somehow speaking in terms of the ply (single slice turn) metric, I am curious to see another solution besides the above. However, I don't have any idea of how to find a shorter solution than 28 ply.

 

[unsolved]

Are you referring to the BTM metric or single slice turn (ply) metric?

I use SST almost exclusively, so by this count your solution is 26 in length.

I'm trying to solve the width of the longest dedges-only-displaced position. I've computed a 9-turn dedge database but it's not deep enough to solve the 11-dedge-flip.

 

[Herbert Kociemba]

I use SST almost exclusively, so by this count your solution is 26 in length.

No, in SST his solution is 28, the e and m move count 2 moves each.
Christopher wanted to see some other solution. I still have not implemented the phase 3 of my solver (lack of motivation in the moment). So I just took an arbitrary 14 move reduction (there are lots of 14 move reductions) and cube explorer found a 14 move solution, which seems quite lucky. So 28 moves too in SST in total without much effort.

3F' R2 3u 3f2 U' 2R' 3f2 3u' R2 3u 3R 2B2 2L2 2B2 U2 L2 U2 D B2 U F B' U2 F2 B2 R2 U' D x' y'

 

[unsolved]

No, in SST his solution is 28, the e and m move count 2 moves each.

You are correct. I missed that.

At present I am experimenting with a 3-phase solver that tackles edges first, then performs a table-lookup for the corners-with-edges-solved database, then executes a pre-computed list of cascade center cage algs to finish.

Phase 3 is the most fun to watch.
Phase 2 is pretty fast.
Phase 1 takes 3 weeks to produce decent results.

 

[Christopher Mowla]

Christopher wanted to see some other solution.

Thanks for posting. The fact that your alg is also just 25 btm is icing on the cake!

 

[Herbert Kociemba]

3R' F2 3u2 3R' U2 3R 3f2 2R2 U2 3R' 3u2 3R' y z U2 2F2 R L' U D R L' U2 R2 D2 B2 L2 D2 y

is another one. It is 26 sstm (12+2+12) and 24 in btm. I just fed about 10 more reduction solutions into cube explorer which in the moment is quite time consuming because I have to color the facelets manually. The success should motivate me at least to write some code to import the results with a text file into cube explorer.

 

[unsolved]

It is 26 sstm (12+2+12) and 24 in btm.

Excellent result Herbert!

26 is the magic number I was hoping for (or smaller, of course). I have the computing resources for a 10-turn dedges-only database and a depth-16 nominal search. This means I have a 26-move horizon for dedges.

If 26 moves turns out to be the hardest dedge configuration to solve, my 3-stage, Dedge First approach will work fine.

This brings up another important question, which is worth a separate topic.

Nicely done!

 

[Christopher Mowla]

For what it's worth, I found two 23 btm algs by combining a 15 btm "pure" 3 flip alg from CE (which I posted in the wiki) with an 11 btm 8 middle edge flip from CE. Both are 27 sstm though.

2L B2 2R' U2 2R' B2 2L2 B2 D2 2R' U2 2R D2 2R2
R2 s' U2 B2 m' U2 R2 e' F2 U2 m' x2
=> 2R B2 2R' U2 2R' B2 2L2 B2 D2 2R' U2 2R D2 r2 s' U2 B2 m' U2 R2 e' F2 U2 x2

2L B2 2R' U2 2R' B2 2L2 B2 D2 2R' U2 2R D2 2R2
R2 s' U2 B2 m' D2 L2 e' B2 D2 m' x2
=> 2R B2 2R' U2 2R' B2 2L2 B2 D2 2R' U2 2R D2 r2 s' U2 B2 m' D2 L2 e' B2 D2 x2

Using the same process, we can find dozens of 24 btms, but the above were the only unique 23 btms through depth 11 of 8 middle edge flips (in slice turn moves).

No solutions were less than 27 sstm.

 

[unsolved]

That's the beauty of brute force: give it enough time and it will explore everything.

But the thing is, I don't want to tie up my computer for a week if it won't be worth the depth-16 search. If it requires 17 moves, it would take months, and this technique won't work for every case.

 

[Herbert Kociemba]

The import to cube explorer now works, and it was worth the time to implement this.

2R U2 3f2 2R F2 3R F2 2R2 3u2 3R' F2 3R' y z R2 2F2 U2 F2 R L' D' U R' L F2 x2 y

This one is 23 sstm and 20 btm! I do not believe now that the 11 dedge flip has any importance with regard to an antipodal position, not even if we only look at the edges alone.

 

[unsolved]

This one is 23 sstm and 20 btm! I do not believe now that the 11 dedge flip has any importance with regard to an antipodal position, not even if we only look at the edges alone.

This is good news for the 11-flip test position but not the best news for the Dedge First solving method. I was hoping antipodes would not be beyond the mid 20s but based on your result, maybe there are.

I guess now the hunt is on for a "deeper than 23 SSTM" dedge-only antipodal position.

I hope we cannot find one!

I am glad you spent the time and effort on this Herbert. I hope you found it enjoyable and worthwhile.

 

[Herbert Kociemba]

If I understand you right, in your phase 1 you want to repair the edges, disregarding the corners and middle pieces. This is of course a totally different situation and surely this needs considerably less moves compared to regarding all pieces like in the pure 11-dedge flip. So the number 23 does not say anything about this situation anyway. But indeed it was fun to find such a short maneuver for the 11-flip. I only computed maneuvers up to 12 moves in phase 1, so maybe there are even shorter solutions if I also run depth 13 in phase 1. But this takes a lot of time, so I will not do this in the moment.

 

[Christopher Mowla]

The import to cube explorer now works, and it was worth the time to implement this.

2R U2 3f2 2R F2 3R F2 2R2 3u2 3R' F2 3R' y z R2 2F2 U2 F2 R L' D' U R' L F2 x2 y

This one is 23 sstm and 20 btm! I do not believe now that the 11 dedge flip has any importance with regard to an antipodal position, not even if we only look at the edges alone.

It only requires 5 more btm to flip 11 dedges versus flipping a single one? I didn't see that coming.

Your alg is two fewer OBTM than my (I should properly say your) 23s as well (because I used CE to construct them). Well done!

 

[unsolved]

If I understand you right, in your phase 1 you want to repair the edges, disregarding the corners and middle pieces.

Correct.

This is of course a totally different situation and surely this needs considerably less moves compared to regarding all pieces like in the pure 11-dedge flip.

Yes, in general, that is my hope. But it's always nice to know the longest solve length that might be needed. I am guessing 18 moves might be needed on average. I'll know for sure once my 10-turn database completes in October.

 

[tseitsei]

Yes, in general, that is my hope. But it's always nice to know the longest solve length that might be needed. I am guessing 18 moves might be needed on average. I'll know for sure once my 10-turn database completes in October.

But in this case you ask for a solution to 11 flipped edges that PRESERVES centers and corners.
And on the other hand in your solver (if I understood correctly) you first solve edge pieces while you IGNORE (don't need to preserve) centers and corners...

It is quite possible that the movecount for latter is (significantly) lower than for the former...

 

[unsolved]

It is quite possible that the movecount for latter is (significantly) lower than for the former...

I know

I was looking for a "far worse than likely" type of position.

But, curiously, on average, it still takes 16 moves to solve 20 of the 24 dedges. I'm tracking "best results per move count" for a bunch of random scrambles.

 

[cuBerBruce]

But, curiously, on average, it still takes 16 moves to solve 20 of the 24 dedges.

There are only 12 dedges, not 24, so you are not using the word correctly here. A dedge is two edge pieces with the same color stickers that have been paired up (so that they are adjacent). A dedge tends to be thought of as if it were an individual piece.

 

[unsolved]

There are only 12 dedges, not 24, so you are not using the word correctly here. A dedge is two edge pieces with the same color stickers that have been paired up (so that they are adjacent). A dedge tends to be thought of as if it were an individual piece.

Thank you Bruce.