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Cube Explorer 5.00 released - with slice moves

How do you ignore pieces? For example, if I wanted to ignore the M-slice in CMLL.
 
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')' ?
 
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?
 
Last edited:
xyzzy said:
In effect, the last apostrophe does nothing because it's not recognised as being attached to any move.
Click to expand...
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.
 
Last edited:
CubeExplorer said:
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"
Click to expand...
Yes I think this matches with the definition of inverse.


CubeExplorer said:
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')
Click to expand...

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.
 
CubeExplorer said:
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')
Click to expand...
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.
 
Thank you, Lucas for your suggestion.

Alg does not work on this old windows xp machine wihout java, im operating here. So I can not say much. I will try it later on another machine.
 
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Reactions: qwr
xyzzy said:
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.
Click to expand...
oh yeah that makes a lot more sense.
 
xyzzy said:
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.
Click to expand...
What an interesting find! Glad @CubeExplorer posted about the confusion. :-D

Cube Explorer handles such algs in these two files, but I can't easily find the bug:

github.com

CubeExplorer/FaceCube.pas at b759a797655bb14e952555add0167ee541ce30d0 · hkociemba/CubeExplorer

A program about Rubik's Cube with a lot of features. - hkociemba/CubeExplorer
github.com github.com

This line (and a similar one on line 949) looks like an obvious culprit, except that it's guarded by a check that only applies it to moves.
It might be worth telling Kociemba: https://github.com/hkociemba/CubeExplorer/issues


CubeExplorer said:
Thank you, Lucas for your suggestion.

Alg does not work on this old windows xp machine wihout java, im operating here. So I can not say much. I will try it later on another machine.
Click to expand...
Hmm, neither alg.cubing.net nor Twizzle have ever used Java. But Twizzle especially (which I linked to) relies on newer browser features, so unfortunately breaking in Windows XP is pretty expected. :-/
Let me know if you're find issues on something new!
 
Lucas Garron said:
github.com

CubeExplorer/FaceCube.pas at b759a797655bb14e952555add0167ee541ce30d0 · hkociemba/CubeExplorer

A program about Rubik's Cube with a lot of features. - hkociemba/CubeExplorer
github.com github.com
https://github.com/hkociemba/CubeExplorer/blob/ee7772e2da38d6538dedd3b158b6277ae3eb144e/RubikMain.pas
Click to expand...

Should be this. Replaces apostrophe and hyphen with the digit '3', then later parses it as an integer. So e.g. (R U' F B2 D)2'1' gets preprocessed into (R U3 F B2 D)2313.
 
Lucas Garron said:
... neither alg.cubing.net nor Twizzle have ever used Java. But Twizzle especially (which I linked to) relies on newer browser features, so unfortunately breaking in Windows XP is pretty expected. :-/
Click to expand...
Thank you again, Lucas. Its interesting background info to me.

Anyway my original confusion concerning the inversion of a maneuver ist cleared now. The using of sage did help a lot.

Looking for a visual representation of the algebraic output of sage I put this into Cube Explorer Program and I found a surprising result again.
Code: 
print('(F*R*U*Ri*Ui*Fi)*U^2*(F*R*U*Ri*Ui*Fi)*U^2 ==',(F*R*U*Ri*Ui*Fi)*U^2*(F*R*U*Ri*Ui*Fi)*U^2)   
#   (F*R*U*Ri*Ui*Fi)*U^2*(F*R*U*Ri*Ui*Fi)*U^2 == (4,7,26)(5,10,18)
print(((F*R*U*Ri*Ui*Fi)*U^2*(F*R*U*Ri*Ui*Fi)*U^2) == (F*R*U*Ri*Ui*Fi *U^2* F*R*U*Ri*Ui*Fi *U^2))   
#   True

CE wrong:
(FRUR'U'F')U2(FRUR'U'F')U2
CE OK:
 FRUR'U'F' U2 FRUR'U'F' U2
Maybe the brackets have a different meaning in CE, but I could not find Info to this point.

I post it here, because I will notify Herbert Kociemba with a link to this thread.
 
CubeExplorer said:
Thank you again, Lucas. Its interesting background info to me.

Anyway my original confusion concerning the inversion of a maneuver ist cleared now. The using of sage did help a lot.

Looking for a visual representation of the algebraic output of sage I put this into Cube Explorer Program and I found a surprising result again.
Code: 
print('(F*R*U*Ri*Ui*Fi)*U^2*(F*R*U*Ri*Ui*Fi)*U^2 ==',(F*R*U*Ri*Ui*Fi)*U^2*(F*R*U*Ri*Ui*Fi)*U^2)  
#   (F*R*U*Ri*Ui*Fi)*U^2*(F*R*U*Ri*Ui*Fi)*U^2 == (4,7,26)(5,10,18)
print(((F*R*U*Ri*Ui*Fi)*U^2*(F*R*U*Ri*Ui*Fi)*U^2) == (F*R*U*Ri*Ui*Fi *U^2* F*R*U*Ri*Ui*Fi *U^2))  
#   True

CE wrong:
(FRUR'U'F')U2(FRUR'U'F')U2
CE OK:
 FRUR'U'F' U2 FRUR'U'F' U2
Maybe the brackets have a different meaning in CE, but I could not find Info to this point.

I post it here, because I will notify Herbert Kociemba with a link to this thread.
Click to expand...
Did you try putting spaces in between moves? I think it's customary to put a space in between every move.
 
qwr said:
Did you try putting spaces in between moves? I think it's customary to put a space in between every move.
Click to expand...
Thanks for the hint. Could not find anything like this in the help file.

The Program normally works correctly without spaces BUT after the closing Brackets there is a space needed.
Code: 
(FRUR'U'F') U2(FRUR'U'F') U2
FRUR'U'F'U2FRUR'U'F'U2
Both lines work well.

Anyway the issue with the apostrophe remains unsolved.
 
Last edited:
My program does not support bracketing at all. The only brackets it is intended not get confused with are the eventually existing brackets at the end of a maneuver which include the maneuver length.
 
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