i've been cubing for 10 years. gotta ask few things

[mencarikebenaran]

1. how sune OLL is possible ? because the possible case is only 2 or 4 corners twisted.
2. why ZBLL has only 72 cases for each ZBLL (except H-ZBLL) ? if i imagine ZBLL like PLL it will be 21 with 4 posibble position, so that should be 84
3. and why H-ZBLL has only 40 cases? the reason i asking this is same like question number 2

 

[coraleap]

1. You can get a case with three corners twisted because twisting a corner twice doesn't necessarily make it untwisted. For instance, twisting corners A and B clockwise and counterclockwise respectively, then twisting B and C counterclockwise and clockwise respectively, would lead to all three of them being twisted.
2 and 3. The reasons for this have to do with U face symmetry (or basically the ability to do AUFs before doing an algorithm.) The number of ways to permute four corners and four edges to create a solvable position is (4!) * (4!) / 2 = 288; adjusting for post-AUF gives you 72. For any non-H case ZBLL, executing any algorithm from one angle must lead to a different position than executing the algorithm from a different angle, since the corners aren't symmetric down to AUF. However, doing an H perm from any angle leads to the same position because the position is symmetric.

All of means that it isn't sufficient to just divide the number of U-ZBLLs by 4 to get the number of PLLs, since ones like the H perm, Z perm, E perm, and no-swap "PLL" would map to one or two of the U-ZBLLs, not 4. To count the number of positions, you can use this thing called Burnside's Lemma. When applied to cubing terms, it means that if you want to count the number of distinct permutations down to some symmetry (like the pre-AUF symmetry), you can look at each symmetry (for the pre-AUF it would be "the same as itself", "the same as it would be after a U", etc.), figure out the total number of permutations that satisfy that symmetry, then average them all.

If we use this to count the number of PLLs, we would get (72 + 4 (H, solved, Na, Nb) + 8 (H, solved, Na, Nb, Z perm from two angles, E perm from two angles) + 4) / 4 = 22 (if we include the solved position, we can see it's correct.)

For H-ZBLLs, we only consider the U2 symmetry. This gives us (72 + 8 (H, solved, Na, Nb, and the U' offset of all of them)) / 2 = 40.

Hope this helps!

 

[Tito_H]

1. You can get a case with three corners twisted because twisting a corner twice doesn't necessarily make it untwisted. For instance, twisting corners A and B clockwise and counterclockwise respectively, then twisting B and C counterclockwise and clockwise respectively, would lead to all three of them being twisted.
2 and 3. The reasons for this have to do with U face symmetry (or basically the ability to do AUFs before doing an algorithm.) The number of ways to permute four corners and four edges to create a solvable position is (4!) * (4!) / 2 = 288; adjusting for post-AUF gives you 72. For any non-H case ZBLL, executing any algorithm from one angle must lead to a different position than executing the algorithm from a different angle, since the corners aren't symmetric down to AUF. However, doing an H perm from any angle leads to the same position because the position is symmetric.

All of means that it isn't sufficient to just divide the number of U-ZBLLs by 4 to get the number of PLLs, since ones like the H perm, Z perm, E perm, and no-swap "PLL" would map to one or two of the U-ZBLLs, not 4. To count the number of positions, you can use this thing called Burnside's Lemma. When applied to cubing terms, it means that if you want to count the number of distinct permutations down to some symmetry (like the pre-AUF symmetry), you can look at each symmetry (for the pre-AUF it would be "the same as itself", "the same as it would be after a U", etc.), figure out the total number of permutations that satisfy that symmetry, then average them all.

If we use this to count the number of PLLs, we would get (72 + 4 (H, solved, Na, Nb) + 8 (H, solved, Na, Nb, Z perm from two angles, E perm from two angles) + 4) / 4 = 22 (if we include the solved position, we can see it's correct.)

For H-ZBLLs, we only consider the U2 symmetry. This gives us (72 + 8 (H, solved, Na, Nb, and the U' offset of all of them)) / 2 = 40.

Hope this helps!

(4!) * (4!) / 2:
I think some people do not understand why it is necessary to divide “(4!) * (4!)” by 2, if he/she does not know the concept “parity” of the cube.

 

[coraleap]

(4!) * (4!) / 2:
I think some people do not understand why it is necessary to divide “(4!) * (4!)” by 2, if he/she does not know the concept “parity” of the cube.

Hi! For anyone who doesn’t understand why we divide by 2, here’s a quick explanation.

There are these things called “even swaps” and “odd swaps”. Any even swap can be undone by swapping two elements an even number of times, and any odd swap can be undone by swapping the elements an odd number of times.

When you compose them (by applying two swaps sequentially), they work sort of like how adding evens and odds work. Composing two odd swaps leads to an even swap, composing two even swaps leads to an even swap, and composing one of each leads to an odd swap.

When you do a quarter turn, you are doing an odd permutation on both the corners and edges. This means that any algorithm you do has the same parity on the corners and edges. Because of this, any position where the corners and edges have different parity is unsolveable. When counting the number of permutations of four pieces, this is why we divide by 2.

Hopefully this is helpful to anyone!

 

[That0newe1rdo]

When you compose them (by applying two swaps sequentially), they work sort of like how adding evens and odds work. Composing two odd swaps leads to an even swap, composing two even swaps leads to an odd swap, and composing one of each leads to an odd swap.

Would 2 evens not equal an even ?

 

[coraleap]

Would 2 evens not equal an even ?

Yes. Thanks for catching the typo!

 

[That0newe1rdo]

Yes. Thanks for catching the typo!

Np !
Thanks for the guide thing on parity, i find it interesting.

Yes i only noticed the typo because you said its like adding evens and odds lol

 

[mencarikebenaran]

i tried to do all 72 ZBLL cases to make me understand why it has only 72 cases.
i tried for ZBLL-U. then i twisted the corner to know what PLL it is

i got :
Aa = occur 4 times
Ab = occur 4 times
F = occur 4 times
Ga = occur 4 times
Gb = occur 4 times
Gc = occur 4 times
Gd = occur 4 times
Ja = occur 4 times
Jb = occur 4 times
Ra = occur 4 times
Rb = occur 4 times
T = occur 4 times
Ua = occur 4 times
Ub = occur 4 times
V = occur 4 times
Y = occur 4 times
E = occur 2 times
Z = occur 2 times
H = occur 1 time
Na = occur 1 time
Nb = occur 1 time
Solved = occur 1 time

if you sum it all, it will be 72.
the reason H only occured once is because its only edge, no matter what position, the corners will always the same.

 

[Tito_H]

I also had the same experience.

After I started Rubik’s cubing, one day I was wondering why each COLL (except H set) has 72 cases, instes of 144.

There are 6 swap types (Front, Right, Back, Left, Diagonal swaps, and No swap) for the corners. And, there are 4 edge locations with 4 different-colored edges, so the there are (4!) = 4 x 3 x 2 x 1 = 24 patterns (see the following <4 Edges' locations>)

In the class of high school, we were taught by the math teacher that if there are 4 boxes in line and 4 color balls, then there are 4 x 3 x 2 x 1 = 24 possible combinations/patterns to put in.

So, I thought mathematically the combination of corners and edges should be 6 x 24 = 144 patterns.

But, Not in the Rubik’ cube!

<4 Edges' locations>
There are following 24 cases of permutation of the 4 edges. The 4 letters correspond to the 4 edge locations(UF, UR, UB, and UL).

BRGO, BROG, BGRO, BGOR, BORG, BOGR,
RBGO, RBOG, RGBO, RGOB, ROBG, ROGB,
GBRO, GBOR, GRBO, GROB, GOBR, GORB,
OBRG, OBGR, ORBG, ORGB, OGBR, OGRB
(B=Blue, R=Red, G=Green, O=Orange)

After having checked web sites, I now know that the half of them can not be combined with each COLL.

For example, This is a normal T25, which exists and is solvable.

But T25 with UB and UL interchanged, does not exist because this can not be solved.

After I understood the reason, I could start to sleep well at night.

 

[abunickabhi]

Anything else to ask?