# 4x4x4 Fastest Center Solves For All 16 UF cases



## unsolved (Apr 1, 2014)

I ran my brute force 4x4x4 solver on an entirely-finished cube except for a single center on the U and F faces (mostly for optimization and speed testing). Every center configuration can be solved in exactly 8 moves. Here is the data if anyone is interested.


```
Center scenario U1:F1 solved after 722,657,630,017 nodes by: u L' u' l u L u' l' 

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|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
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Center scenario U1:F2 solved after 25,060,299,743 nodes by: U r U' l' U r' U' l 

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|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
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Center scenario U1:F3 solved after 34,111,149,775 nodes by: U b d' b' U' b d b' 

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|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
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Center scenario U1:F4 solved after 36,601,594,190 nodes by: U l' d2 l U' l' d2 l 

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|####|####|####|####|   |OOOO|OOOO|####|OOOO|   |XXXX|XXXX|XXXX|XXXX|
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|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
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Center scenario U2:F1 solved after 33,786,950,785 nodes by: U b u b' U' b u' b' 

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Center scenario U2:F2 solved after 723,853,410,535 nodes by: u L u' r' u L' u' r 

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|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
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|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
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Center scenario U2:F3 solved after 25,973,341,490 nodes by: U r d2 r' U' r d2 r' 

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Center scenario U2:F4 solved after 32,943,679,535 nodes by: U b' d b U' b' d' b 

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Center scenario U3:F1 solved after 22,089,356,201 nodes by: U f u' f' U' f u f' 

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Center scenario U3:F2 solved after 36,277,395,200 nodes by: U l' u2 l U' l' u2 l 

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Center scenario U3:F3 solved after 765,444,600,500 nodes by: u' f u F u' f' u F' 

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Center scenario U3:F4 solved after 23,482,897,075 nodes by: U f' d' f U' f' d f 

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Center scenario U4:F1 solved after 25,747,270,856 nodes by: U r u2 r' U' r u2 r' 

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Center scenario U4:F2 solved after 23,256,826,441 nodes by: U f' u f U' f' u' f 

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Center scenario U4:F3 solved after 22,315,426,835 nodes by: U f d f' U' f d' f' 

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|####|####|OOOO|####|   |OOOO|####|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
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Center scenario U4:F4 solved after 727,561,803,916 nodes by: u f' u' F' u f u' F 

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|####|####|OOOO|####|   |OOOO|OOOO|####|OOOO|   |XXXX|XXXX|XXXX|XXXX|
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```


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## ryanj92 (Apr 1, 2014)

It's nice to see your solver is coming along - obviously these are all very standard comms (just ask a 4BLD solver), but it's cool that your solver can recover them


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## cmhardw (Apr 1, 2014)

unsolved said:


> I ran my brute force 4x4x4 solver on an entirely-finished cube except for a single center on the U and F faces (mostly for optimization and speed testing). Every center configuration can be solved in exactly 8 moves. Here is the data if anyone is interested.



Great job on your solver! I'd be interested to see what your solver does with the 4x4x4 center 3-cycle: (Ufr Fur Ruf)
I've always wondered whether that case can be done in fewer than 10 moves, but I've never discovered one.

A couple of observations:


Spoiler






unsolved said:


> ```
> Center scenario U1:F1 solved after 722,657,630,017 nodes by: u L' u' l u L u' l'
> 
> TOP                     FRONT                   RIGHT
> ...



I've never thought to do a center commutator like that, that's cool! The interchange slice is the *l* face, and you're inserting Fur to Ful via the L face. Cool!





Spoiler






unsolved said:


> ```
> Center scenario U4:F3 solved after 22,315,426,835 nodes by: U f d f' U' f d' f'
> 
> TOP                     FRONT                   RIGHT
> ...



There are a couple cases like this in your list, and I'm curious about how your solver works. This case can also be solved with U' f d f' U f d' f'. Does your solver search the moves in a certain order? For example it would search for algorithms that start with U and then all possible continuations before it would start with U' and all possible continuations?



Thanks for this post. It's neat to see your solver working, plus I learned a new type of commutator reading it!


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## unsolved (Apr 1, 2014)

ryanj92 said:


> It's nice to see your solver is coming along - obviously these are all very standard comms (just ask a 4BLD solver), but it's cool that your solver can recover them



Yes, I wanted to start with something simple. Plus, I was curious about something. My own center-solving algo that I used, over and over, required an "alignment" before applying. Sometimes this would add 1 or 2 moves (trivial, I know). But I was wondering if every center-pair configuration could be solved, without needing this pre-alignment. The answer is now: "yes."

So all of these algos were previously known? I thought maybe there was just one/two known, then everything else would be the alignment + algo.


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## cmhardw (Apr 1, 2014)

unsolved said:


> So all of these algos were previously known? I thought maybe there was just one/two known, then everything else would be the alignment + algo.



The algorithms your solver generated are all commutators, and the process of forming commutator algorithms is very well understood.

I asked this in my post above, but it was buried in a spoiler. Can you let your solver examine the 3-cycle (Ufr Fur Ruf) ? I know of a way to solve this case in 10 turns, and I have always wondered it can be done in 9 turns (Block turn metric).


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## Lucas Garron (Apr 1, 2014)

We can has animations? :-D

I'm definitely interested in checking out these algs, but I'd rather be able to just see them instead of having to figure out your notation conventions and very large diagrams.


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## unsolved (Apr 1, 2014)

cmhardw said:


> Great job on your solver!



Thank you kindly.



cmhardw said:


> I'd be interested to see what your solver does with the 4x4x4 center 3-cycle: (Ufr Fur Ruf)
> I've always wondered whether that case can be done in fewer than 10 moves, but I've never discovered one.



If you can copy/paste my cube ASCII drawing and give me a configuration, I can run a test on it.
It looks like a non-commutor solution, which takes longer to solve (I'll explain when I get to your question about my solver).



cmhardw said:


> I've never thought to do a center commutator like that, that's cool!



That's the elegance of brute force!



cmhardw said:


> There are a couple cases like this in your list, and I'm curious about how your solver works. This case can also be solved with U' f d f' U f d' f'. Does your solver search the moves in a certain order?



I generate all moves at each level of the search in the same order, and that order is:


```
#define U___PLUS____ 0
#define U___MINUS___ 1
#define U___TWICE___ 2

#define F___PLUS____ 3
#define F___MINUS___ 4
#define F___TWICE___ 5

#define R___PLUS____ 6
#define R___MINUS___ 7
#define R___TWICE___ 8

#define D___PLUS____ 9
#define D___MINUS___ 10
#define D___TWICE___ 11

#define B___PLUS____ 12
#define B___MINUS___ 13
#define B___TWICE___ 14

#define L___PLUS____ 15
#define L___MINUS___ 16
#define L___TWICE___ 17

#define u___PLUS____ 18
#define u___MINUS___ 19
#define u___TWICE___ 20

#define f___PLUS____ 21
#define f___MINUS___ 22
#define f___TWICE___ 23

#define r___PLUS____ 24
#define r___MINUS___ 25
#define r___TWICE___ 26

#define d___PLUS____ 27
#define d___MINUS___ 28
#define d___TWICE___ 29

#define b___PLUS____ 30
#define b___MINUS___ 31
#define b___TWICE___ 32

#define l___PLUS____ 33
#define l___MINUS___ 34
#define l___TWICE___ 35
```


So at depth 1, U,U', and U2 start things off. But here is a technical detail that is worth a little explanation.
I use an array of functions pointers as a move generator. Each function has the same name, the only difference is,
you pass in a number to the array, and it magically makes the correct move.


```
cube_move_array[U___PLUS____] = plus_top_face_move;
cube_move_array[U___MINUS___] = minus_top_face_move;
cube_move_array[U___TWICE___] = double_top_face_move;
```

What I wanted to avoid was code like: if(time to move U') then (call make_U_prime_move) because that introduces branch
point to the code, and "if" statements will slow the code down in the long run.

So now all I do it pass in indices from for loops, and the code looks like this:


```
for(i = FIRST_MOVE; i <= LAST_MOVE; i++)
	for(j = FIRST_MOVE; j<= LAST_MOVE; j++)
	{
	if((i/3) == (j/3)) continue;
	for(k = FIRST_MOVE; k <= LAST_MOVE; k++)
	{
		
		if((k/3) == (j/3)) continue;
		if(prune_parallel_rotations(i,j,k,PASS_NO_MOVE_AS_ARGUMENT)) continue;
		if(evade_evaluations_3(i,j,k)) continue;

			modified_cube = cube_move_array[i](starting_cube);
			modified_cube = cube_move_array[j](modified_cube);
			modified_cube = cube_move_array[k](modified_cube);

			g_total_nodes++;

			solved_info = solved_distance(modified_cube);

			if(solved_info == ALREADY_SOLVED) 
			{
				depth_3_solution(i,j,k);
				if(quit_on_first_solution) return;
			}
			else
			if(solved_info == SOLVED_IN_1_MORE_MOVE)
			{
				depth_4_solution(i,j,k);
				if(quit_on_first_solution) return;
			}
	}//k loop
	}//j loop
```


A few notes to help explain things.

FIRST_MOVE = U___PLUS____ = 0
LAST_MOVE = l___TWICE___ = 35

So the for loops just range over 0 to 35.

The code if((i/3) == (j/3)) continue basically means

if(i is 0,1,2) and (j is 0,1,2) skip to j = 3.
if(i is 3,4,5) and (j is 3,4,5) skip to j = 6, since we already did j = 0,1,2 at this level also.

That's why I code the move generator indices as triplets.
If ever I moved U,U' or U2 in a prior level, I won't make any of these moves at the next level.
It's a very simple pruning mechanism that requires no "if" statements.

if(prune_parallel_rotations(i,j,k,PASS_NO_MOVE_AS_ARGUMENT)) 

I implemented the idea posted by Jakube about how to filter out U D U moves, (since U2 D was done at a prior level)
but I made a 3-ply and 4-ply version of it. When at level 3 in the game tree, I use PASS_NO_MOVE_AS_ARGUMENT
so that the routine will only test 3-ply scenarios since we are not at a 4th ply yet.

if(evade_evaluations_3(i,j,k)) continue;

This is a heuristic I developed and tested on several trillion positions at various levels of search.
It is a statistical estimation of the likilood the cube can be solved by moves i,j,k given the starting cube.
If there is a "snowball's chance in hell," I don't bother with the attempt.

It is a very fast, very simple function that executes 40 times faster than the move generator,
so it's worth applying this test then skipping the attempt at a solution. It speeds things
up tremendously.

if(solved_info == SOLVED_IN_1_MORE_MOVE)

This probes the 1-turn database in RAM after the moves to the starting cube have been made.
If it wasn't solved, maybe it is 1 move away from being solved, and we can avoid taking 33 times as long
to get through the next level of search.

I have this for 2- through 5-turns as well, and use them on deeper searches.

if(quit_on_first_solution) return;

This may be the answer to your final question. Since I had this enabled,
it did not try any other solutions (since I was solving 16 scenarios at once).

I can go re-run it with this turned off and see if your solution pops up. I think it will.


----------



## unsolved (Apr 1, 2014)

Lucas Garron said:


> We can has animations? :-D
> 
> I'm definitely interested in checking out these algs, but I'd rather be able to just see them instead of having to figure out your notation conventions and very large diagrams.



It was easy enough to change my string array to output U instead of my T, etc., and I have done that now.

As for the ASCII notations for the centers I solved:


```
TOP                     FRONT                   RIGHT
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####| U1 | U2 |####|   |OOOO| F1 | F2 |OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####| U3 | U4 |####|   |OOOO| F3 | F4 |OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
```

There are 4 centers, so I just numbered them 1-4 with the face designation in front of the number.

So U1:F1 is one scenario, U1:F2, etc., through U4:F4. That is every possible combination of U and F centers.


----------

