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Cube Explorer 5.00 released - with slice moves

xyzzy

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You have B, D and L moves restricted. I haven't checked the source code (which was actually publicly released somewhat recently!), but my guess is that the internal representation of move sequences consists of just the six single-layer moves, and the "slice" moves are just stored as L2 R2 or whatever, and only converted to M2 x2 for display. The restriction of B, D and L moves applies to the internal representation, not to whatever the alg looks like after the slice moves are written as slice moves. The checkboxes are greyed out when in slice moves mode, so you'll have to go back to the usual mode first, uncheck those boxes, then go back into slice mode.

Also, use ksolve++ or HARCS or something along those lines if you want to generate MUD algs and the like. Cube Explorer is entirely the wrong tool for this task.
 

Thom S.

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In addition on what WoowyBaby and xyzzy have said, I have two things
First, your best Algorithm always moves the centers. CE doesn't like that as it assumes you want your centers to be like this.
Second, CE won't care if the found algorithms are good, it just gives it to you
 

xyzzy

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CE is close to optimal but not actually optimal right? idk
It searches only for optimal algs when there are any "wildcards" (greyed pieces, pieces with only orientation specified, pieces with only the colours specified).

If you specify the whole cube state (which is not the case here), it'll search with the non-optimal two-phase algorithm by default, and you can check the "optimal" checkbox to make it find optimal solutions.
 
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I am not out of the world and if there are some bugs in CE I will try to fix them. I will not add any additional features to the program any more - just fix bugs or eventually improve the user interface if you can convince me that something is seriously confusing or impractical.

Before I upload a fixed version I would like to know if there are other annoying things which could probably be fixed easily.

Edit: In the fixed version you now can also use the exclude feature when using slice moves. If you exclude B moves for example and you solve a cube with the B face turned you get:

F' E2 S2 E2 S z (5s*)
F' M2 S2 M2 S z (5s*)
S M2 S2 M2 F' z (5s*)
S E2 S2 E2 F' z (5s*)

So you can replace a B move by 5 slice moves.

If you exclude B and F you get for example

L' E' M' E R U' S E S' x' z (9s*)

Edit:
The above is wrong. It's always dangerous to change code which you did not see for years.
S' F' z (2s*)

should be the right answer. I will have to revisit the code.
 
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I'm trying to work on this software, for the latest version. Just getting to grips with it and following some tutorials. When I run it, it doesn't return the most optimum solutions. See screenshot.

Sent from my CLT-L09 using Tapatalk
 

shadowslice e

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Screenshot? But for cube explorer, just blank out one of the pieces and it'll generate all solutions of depth n, then n+1 etc. You should probably also specify orientation for that piece since that allows ce to use bigger pruning tables.
 
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What it seems to be doing is not putting the double turns as R2 but as R R
Sent from my CLT-L09 using Tapatalk

Better image

66a657e53ffb0c83fa1ddee352d15855.jpg


I figured it out. Was using the qtm file instead of htm

Sent from my CLT-L09 using Tapatalk
 
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WoowyBaby

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You have downloaded the qtm version insetead of the htm version, or you changed some settings in it. You should try downloading the other one from Kociemba's site to get the optimal outputs you're looking for.
 

CubeExplorer

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I think I have misunderstood something of the basics of permutations.

FRUR'U'F' is a well known maneuver of the beginners method. I use it just as an example.
To undo it, I can use FURU'R'F'. This is quite clear. The Rotations are inverted and executed in reverse order.
It can be verified with Cube Explorer of Herbert Kociemba.
There I type in FRUR'U'F' FURU'R'F', apply it to a solved cube and it does not change. The expression is equal to Identity.

Now I thought, the inverse of FRUR'U'F' can be written as
(FRUR'U'F')' and this would be the same as FURU'R'F'.
and (FRUR'U'F') (FRUR'U'F')' should yield Identity
But when I test this expression in Cube Explorer, the result is different, CE shows as generator:
F U R U' R' U R U' R' F'
which is equvalent to (FRUR'U'F')4 == ((FURU'R'F')2

I had a search in the wonderful permutation-puzzle-book but could not find an answer.

So my question remains: what does the last apostrophe operator really do in (FRUR'U'F')' ?
 

xyzzy

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Cube Explorer supports only the basic notation; no brackets, no commutator/conjugate notation, no wide moves.

In effect, the last apostrophe does nothing because it's not recognised as being attached to any move.

EDIT: it actually seems like bracketing is supported?
 
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CubeExplorer

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In effect, the last apostrophe does nothing because it's not recognised as being attached to any move.
Thank you for answering.

The last apostrophe does NOT nothing in CE.
(FRUR'U'F')'
does the same as (FRUR'U'F')3
which is the same as (FRUR'U'F')(FRUR'U'F')(FRUR'U'F')

And the other part of my question is still unanswered, what does the last apostrophe do according to permutations theory?
Are my assumptions correct, when I wrote the following?
"Now I thought, the inverse of FRUR'U'F' can be written as
(FRUR'U'F')' and this would be the same as FURU'R'F'.
and (FRUR'U'F') (FRUR'U'F')' should yield Identity"

Code:
S48 = SymmetricGroup (48)
R=S48(" (25 ,27 ,32 ,30)(26 ,29 ,31 ,28)(3 ,38 ,43 ,19)(5 ,36 ,45 ,21)(8 ,33 ,48 ,24) ")
L=S48(" (9 ,11 ,16 ,14)(10 ,13 ,15 ,12)(1 ,17 ,41 ,40)(4 ,20 ,44 ,37)(6 ,22 ,46 ,35) ")
U=S48(" (1 ,3 ,8 ,6)(2 ,5 ,7 ,4)(9 ,33 ,25 ,17)(10 ,34 ,26 ,18)(11 ,35 ,27 ,19) ")
D=S48(" (41 ,43 ,48 ,46)(42 ,45 ,47 ,44)(14 ,22 ,30 ,38)(15 ,23 ,31 ,39)(16 ,24 ,32 ,40) ")
F=S48(" (17 ,19 ,24 ,22)(18 ,21 ,23 ,20)(6 ,25 ,43 ,16)(7 ,28 ,42 ,13)(8 ,30 ,41 ,11) ")
B=S48(" (33 ,35 ,40 ,38)(34 ,37 ,39 ,36)(3 ,9 ,46 ,32)(2 ,12 ,47 ,29)(1 ,14 ,48 ,27) ")
print(R*U)   #(1,3,38,43,11,35,27,32,30,17,9,33,48,24,6)(2,5,36,45,21,7,4)(8,25,19)(10,34,26,29,31,28,18)
print(R*U*U^-1*R^-1)   #()
#(FRUR'U'F') (FRUR'U'F')'
(F*R*U*R^-1*U^-1*F^-1)*(F*R*U*R^-1*U^-1*F^-1)^-1   #()
#(FRUR'U'F')'
print((F*R*U*R^-1*U^-1*F^-1)^-1)   #(1,3,9,33,35,27)(2,5,18)(6,8,11,19,17,25)(7,34,26)
#(FURU'R'F')
print((F*U*R*U^-1*R^-1*F^-1))      #(1,3,9,33,35,27)(2,5,18)(6,8,11,19,17,25)(7,34,26)
print(((F*R*U*R^-1*U^-1*F^-1)^-1)==(F*U*R*U^-1*R^-1*F^-1))  #True

OK, now I found a way to verify my assumptions. Being new to this topic I need do check, wether my understanding of the newly learned things is correct. The sage code above contains the results too and should be self explaining.

EDIT: Thanks to qwr for the second confirmation down there in the next posting.
 
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qwr

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Are my assumptions correct, when I wrote the following?
"Now I thought, the inverse of FRUR'U'F' can be written as
(FRUR'U'F')' and this would be the same as FURU'R'F'.
and (FRUR'U'F') (FRUR'U'F')' should yield Identity"
Yes I think this matches with the definition of inverse.


The last apostrophe does NOT nothing in CE.
(FRUR'U'F')'
does the same as (FRUR'U'F')3
which is the same as (FRUR'U'F')(FRUR'U'F')(FRUR'U'F')

Maybe this is a bug. F R U R' U' F' has order 6 so this is like a "square root" or order 3 version of that.
 

xyzzy

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The last apostrophe does NOT nothing in CE.
(FRUR'U'F')'
does the same as (FRUR'U'F')3
which is the same as (FRUR'U'F')(FRUR'U'F')(FRUR'U'F')
Ah, interesting. I stand corrected.

CE does seem to support bracketing (so e.g. "(R y)1260" works as you'd expect), but the apostrophe is being parsed as equivalent to 3. Probably because for normal moves inverting and power-3 are the same thing, an assumption that breaks down for arbitrary sequences.
 
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