4x4x4 God's Number Position Hunt

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The subject of the original post asked for candidates for antipodal positions.

I hereby nominate the 10 positions presented in the 4x4x4 fewest moves challenge run earlier.

I doubt the sort of mostly-solved positions we've seen will be antipodal.

I also doubt any of the 10 positions I nominate are antipodal, but I think those positions are much
more challenging than the ones I've seen posted. And I think they serve as a good benchmark
for how effective solvers of any sort are getting.

And I'm interested in three metrics (I call them OBT = Outer Block Turn, SST = Single Slice Turn,
and BT for Block Turn).
 
The subject of the original post asked for candidates for antipodal positions.

I hereby nominate the 10 positions presented in the 4x4x4 fewest moves challenge run earlier.

I doubt the sort of mostly-solved positions we've seen will be antipodal.

I also doubt any of the 10 positions I nominate are antipodal, but I think those positions are much
more challenging than the ones I've seen posted. And I think they serve as a good benchmark
for how effective solvers of any sort are getting.

And I'm interested in three metrics (I call them OBT = Outer Block Turn, SST = Single Slice Turn,
and BT for Block Turn).

To record this hopefully permanently, here are the second set of 5 (from your website:
1) U L F D B2 F2 U2 R' F2 R2 B2 D2 R' D2 L B R D' U' R2 L' F L2 D F2 Uw2 B Uw2 F' R' Fw2 F' Rw2 R' B' R2 Fw2 L' Uw L2 D' B2 U Rw' Fw Uw' F' Uw' U' R B R U
2) F U R U B' D R U2 R B U2 L D' F' U2 L R B2 D2 R D2 R' B2 Uw2 Rw2 R' D L D2 L2 R Fw2 U L2 B' U Fw' L D2 F2 Rw' Fw2 Rw' Uw' Fw2 Rw D' F U F D
3) F D B D L' U2 F U2 D' B2 F L B F U2 L' F2 R' U2 F2 L B2 R2 B2 Fw2 Uw2 D' B F' Uw2 B R2 Fw2 Uw2 D' U2 Rw U L B U2 Uw' F' Rw2 R2 Fw F2 Uw' B' L B U L
4) R F D L D2 F D U B2 F2 L' B2 U F2 L D2 U2 L U2 R B2 L B2 Uw2 Rw2 F U2 R' B' Uw2 F2 R Uw2 F' Uw' L2 D2 L B2 R' Uw' D2 Fw' Rw R Uw' L2 B D F D
5) F U F U F2 R2 D' R2 B2 D B2 D2 B2 R2 U' F R F' B' D' R2 D' L' Uw2 F' Uw2 U2 Rw2 Fw2 B D F' B D' L2 B2 Rw' Fw2 R' U B U' Rw' Uw R' Fw B U L D R

I can't find the first set - do you have them somewhere?
 
To record this hopefully permanently, here are the second set of 5 (from your website:
1) U L F D B2 F2 U2 R' F2 R2 B2 D2 R' D2 L B R D' U' R2 L' F L2 D F2 Uw2 B Uw2 F' R' Fw2 F' Rw2 R' B' R2 Fw2 L' Uw L2 D' B2 U Rw' Fw Uw' F' Uw' U' R B R U
2) F U R U B' D R U2 R B U2 L D' F' U2 L R B2 D2 R D2 R' B2 Uw2 Rw2 R' D L D2 L2 R Fw2 U L2 B' U Fw' L D2 F2 Rw' Fw2 Rw' Uw' Fw2 Rw D' F U F D
3) F D B D L' U2 F U2 D' B2 F L B F U2 L' F2 R' U2 F2 L B2 R2 B2 Fw2 Uw2 D' B F' Uw2 B R2 Fw2 Uw2 D' U2 Rw U L B U2 Uw' F' Rw2 R2 Fw F2 Uw' B' L B U L
4) R F D L D2 F D U B2 F2 L' B2 U F2 L D2 U2 L U2 R B2 L B2 Uw2 Rw2 F U2 R' B' Uw2 F2 R Uw2 F' Uw' L2 D2 L B2 R' Uw' D2 Fw' Rw R Uw' L2 B D F D
5) F U F U F2 R2 D' R2 B2 D B2 D2 B2 R2 U' F R F' B' D' R2 D' L' Uw2 F' Uw2 U2 Rw2 Fw2 B D F' B D' L2 B2 Rw' Fw2 R' U B U' Rw' Uw R' Fw B U L D R

I can't find the first set - do you have them somewhere?

So we are trying to find faster solutions to the positions created from the scrambles above?
 
So we are trying to find faster solutions to the positions created from the scrambles above?

Oh, absolutely! Ideally at some point our software will
reach the point where we can find near-optimal solutions
(I don't think we are quite there yet) and start finding
some actual candidates for antipodal positions.

Probably symmetric positions may be more fruitful, as
well as being easier to "prove hard" (less search
required).

It's early days yet. But let's see how good our solvers
can get.
 
Oh, absolutely! Ideally at some point our software will
reach the point where we can find near-optimal solutions
(I don't think we are quite there yet) and start finding
some actual candidates for antipodal positions.

Probably symmetric positions may be more fruitful, as
well as being easier to "prove hard" (less search
required).

It's early days yet. But let's see how good our solvers
can get.

My solver is just a brute-force, "zero-elegance" solver. It just generates moves, no goals in mind, no intermediate evaluation function, and it hunts for any-and-all solutions at the current depth. It does store every position 5 moves from the solved state (and fewer) into a RAM buffer, so it can trim up to 5 plies from the game tree. Here is what I have so far for the first attempt:

so_far.png


Obviously it will not carve too much further into this one.

It does have another specialized database of "centers only" which are solved through 7 turns. That is, if you have everything solved but the centers, it can make an announcement when you are 7 turns from the solved state.
 
Here is what I have so far for the first attempt:

Excellent, so now we have a position that we know is at least 13 moves from
the start (if I interpret your screenshot correctly). Do you agree? This is a
start.

Can we find positions that we know are at least 14 moves from start? 15?
(I think we have some from an earlier investigation, but it would be nice
to give them here.)

While these are clearly not proof of antipodal positions, the fact that we've
actually proved a concrete position to be at least a particular distance away
is progress.

If we find a symmetric position and use symmetry to reduce the search, we
can probably get another ply deeper (maybe even two plys).

Long term, in order to raise the lower bound, we need to prove a particular
position to be deeper than the simple pigeonhole argument (or else use
another technique). For instance, some of the subgroup DP approaches
(to calculate counts of canonical move sequences for cosets of subgroups)
may be able to increase the lower bound without finding a concrete position.

Some may argue we are so far from being able to solve a random position
optimally, that this is a fool's errand. But I think there is a lot to be
learned from the attempt.
 
Excellent, so now we have a position that we know is at least 13 moves from
the start (if I interpret your screenshot correctly). Do you agree? This is a
start.

The program is still searching the 13th ply. So we can say with certainty it is at least 12 moves from a solved state. You can see (08/36) right before the PV. That means it is on move number 8 of 36 in the current depth.

Can we find positions that we know are at least 14 moves from start? 15?
(I think we have some from an earlier investigation, but it would be nice
to give them here.)

Do you mean other positions, or "play into the line" one ply at a time and do 13-ply searches to functionally test for a depth of "13 + d" where d is the number of plies we advance manually?

For example, from the scrambled position, if I play the move F, then search 13-ply, we have functionally searched 1 out of the 36 moves to a depth of 14.
If we recreate the scramble, then play the move F', then repeat the 13-ply search, we have no searched 2 out of the 36 moves to a depth of 14. Etc., Etc.

While these are clearly not proof of antipodal positions, the fact that we've
actually proved a concrete position to be at least a particular distance away
is progress.

Yes but that branching factor of 36 is a real killer.

If we find a symmetric position and use symmetry to reduce the search, we
can probably get another ply deeper (maybe even two plys).

I am on a 16-core machine but I have parallel searching off. My earlier parallel implementations were very fast, but ultimately flawed, as the intrepid Bruce Norskog was able to demonstrate. I disabled all of the parallel code and stripped out a great deal of the dead wood. If someone wants to take a crack at parallelizing the code, I'm willing to donate the code.

By the way, my move generator only executes 6 lines of C-code to generate a legal 4x4x4 move. It is slowed down by probing the 16 GB hash tables and generating 64-bit hash numbers and resolving hash conflicts. Otherwise it's speed is in the 50 million moves per second range using a single CPU core.

Long term, in order to raise the lower bound, we need to prove a particular
position to be deeper than the simple pigeonhole argument (or else use
another technique). For instance, some of the subgroup DP approaches
(to calculate counts of canonical move sequences for cosets of subgroups)
may be able to increase the lower bound without finding a concrete position.

Some may argue we are so far from being able to solve a random position
optimally, that this is a fool's errand. But I think there is a lot to be
learned from the attempt.

I have a few ideas regarding optimal solutions as well. Are you familiar with so-called "Dropout Expansion" by Thomas Lincke? Programmers Martin Fierz and Edward Gilbert used this technique to develop automated opening books for their checkers programs using this technique. Their implementation was very effective, making their programs virtually bullet proof in the opening.

http://www.fierz.ch/strategy4.htm

I am convinced this approach would work well for scrolling into a solution-attack line, then backing up as needed. All that is required is an evaluation function, which can be as simple as an enumeration of correct colors per cube side, maybe with goal-directed bonuses for centers to be prefered over edges, and so on. Fixed depth searches with conspiracy numbers to guide the way will insure that positions with color counts improving get searched more deeply than those positions where there is little to no progress, or worse, lost progress.
 
Can we find positions that we know are at least 14 moves from start? 15?
(I think we have some from an earlier investigation, but it would be nice
to give them here.)

Has the pure dedge flip been proven to require 15 BTM?

Of course, we know of over 86.5 quintillion positions that are at least 13 BTM from solved. In fact, we know a set of over 86.5 quintillion positions that are known to be at least 13 BTM from another set of positions of the same size (and that contains the solved state).

(By "from start" I assume you meant "from solved," since the 4x4x4 positions form a coset space, not a group. So the "start" position can not be any arbitrarily chosen position.)
 
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Has the pure dedge flip been proven to require 15 BTM?

Of course, we know of over 86.5 quintillion positions that are at least 13 BTM from solved. In fact, we know a set of over 86.5 quintillion positions that are known to be at least 13 BTM from another set of positions of the same size (and that contains the solved state).

(By "from start" I assume you meant "from solved," since the 4x4x4 positions form a coset space, not a group. So the "start" position can not be any arbitrarily chosen position.)

It requires 15 moves using STM I am sure of that.

I compiled a list of positions as a function of depth up to move 32. Here are the counts through depth 16, I have not added the other numbers yet:

http://lightningcloudcomputing.com/OO_4x4x4/info_05.shtml
 
It requires 15 moves using STM I am sure of that.
How are you sure?

I'm sure many thought that the adjacent double parity case (y' 2R2 F r' F') 2R2 U2 2R' U2 2R' U2 2R' U2 (F r F' 2R2 y) could be solved in no fewer than 18 BTM, but this (my) solution is 16 BTM. It uses two wide (double) turns, which each still count as a single block turn each.

Moreover, it probably wasn't imagined that the 2-cycle case r2 u' b 2R' b' u b r' u' 2R' u r 2R b' r2 could be solved in 15 BTM, but I found five 15 BTM length solutions (there are probably more). This algorithm, for example, is 27 turns in your solver's metric but is 15 BTM.
 
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Are we talking about the single dedge flip?
Yes. The examples I gave are illustrations of when the presence of wide (double) turns lessens the number of BTM an algorithm has. Who's to say that there isn't some single dedge flip algorithm of this category? Sure I researched this particular position for 9 months, but I can't say I'm positive I explored every possible path.
 
Yes. The examples I gave are illustrations of when the presence of wide (double) turns lessens the number of BTM an algorithm has. Who's to say that there isn't some single dedge flip algorithm of this category? Sure I researched this particular position for 9 months, but I can't say I'm positive I explored every possible path.

I have searched it to completion. Recall I have the 5-Turns-From-Solution database now, so I only need to explore every 10-ply node in order to see 15 plies distant.

r2 D2 r' F2 r F2 D2 r D2 l' F2 l F2 D2 r2 is one of the solutions.
 
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I have searched it to completion. Recall I have the 5-Turns-From-Solution database now, so I only need to explore every 10-ply node in order to see 15 plies distant.

r2 D2 r' F2 r F2 D2 r D2 l' F2 l F2 D2 r2 is one of the solutions.
BTM means "Block Turn Moves", not single slice turn (which you call ply) moves. Bruce specifically mentioned BTM.

Reread this portion of my post.
Moreover, it probably wasn't imagined that the 2-cycle case r2 u' b 2R' b' u b r' u' 2R' u r 2R b' r2 could be solved in 15 BTM, but I found five 15 BTM length solutions (there are probably more). This algorithm, for example, is 27 turns in your solver's metric but is 15 BTM.
 
Has the pure dedge flip been proven to require 15 BTM?

I've been reviewing the earlier runs of my OBTM optimal solver.

The position:

2L 2R U' 2R2 F2 2R2 U 2R2 U' F2 U 2L' 2R

(I'm using 2L here instead of l to mean a slice turn)

has been searched through 16 moves in the OBTM without a solution.

So here's a concrete position that requires at least 17 moves in the OBTM.

I've got a few more positions proved to require at least 17 moves.
(They were when I was trying to solve some top edge cases.)

Proving these positions took a great deal of CPU time. I should
probably start with some more-random positions.
 
I've got a few more positions proved to require at least 17 moves.
(They were when I was trying to solve some top edge cases.)

Proving these positions took a great deal of CPU time. I should
probably start with some more-random positions.

How much CPU time to search through depth 16?
 
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