As I stated in the Alternating Cube thread
, I have found a way of mapping the moves and states of the Alternating Skewb to a group, which proves the states of the Alternating Skewb do form a group. I can now easily complete the table:
0 - 1
1 - 4
2 - 12
3 - 36
4 - 108
5 - 324
6 - 966
7 - 2712
8 - 7296
9 - 18908
10 - 43739
11 - 81580
12 - 102224
13 - 69860
14 - 20535
15 - 1536
16 - 79
As you can see, the values of the even distances match my previous table, so my earlier analysis, while incomplete, was accurate. As I stated in the Alternating Cube thread, I can give any of these states (such as the hardest positions to solve) explicitly if someone can suggest a format.
Looking at # of skewb states of # of alternating skewb states:
3149280 / 349920 = 9
Anybody know where the 9 comes from? That seems to imply that it's from the corner orientations, which would make that part of the solution a bit interesting.
I looked into this and the answer is not what you may think... Let me demonstrate what I mean:
Perform this sequence, which is valid for the Alternating Skewb, on a solved Skewb:
UBL DFL' UBL DFL' UFR DBR' UBL DBR' UBL DFL' UBL DBR'
Here's another, again valid for the Alternating Skewb:
UFR DFL' UFR DBR' UBL DFL' UBL UFR' DFL UFR' UBL DBR' DFL DBR'
Clearly something sneaky is going on here... Where does the factor of 9 come from?! Don't worry, I found the answer
I want to compare to a normal Skewb, so bear with me for just a second here... On a normal Skewb, the corners come in two orbits. If we leave the physical core of the puzzle fixed, this also fixes the permutation of one orbit of the corners. If we calculate the number of configurations starting here, then accounting for the centers, then the other orbit of corners we reach the following numbers:
a) Each fixed corner can be in 1 of 3 orientations (3^4 = 81)
b) All even permutations of the centers are reachable (6!/2 = 360)
c) The permutation of the remaining corners is isomorphic to the Klein-4 group and can only be in 4 states (4)
d) The sum of the rotations of the remaining corners must be 0 mod 3 (3^4/3 = 27)
This leads to (3^4)(6!/2)(4)(3^3) = 3149280 configurations for the Skewb.
The algorithms for the Alternating Skewb I gave above prove that the factors from the corresponding c) and d) of the Alternating Skewb are the same as the Skewb - a fact which in my opinion is quite surprising! This also means the factor of 9 must come from parts a) and b) alone, and so it does.
a) For the Alternating Skewb, even ignoring the REST of the puzzle, not all 3^4 orientations of the 4 fixed corners are possible! By alternating CW and CCW, the sum of the rotations of these corners can be 0 mod 3. Furthermore, by adding a single CW move on at the end, we can also achieve a rotational sum of 1 mod 3 (say positive is CW). However, it is not possible to achieve a rotational sum of 2 mod 3! Isolating these four corners from the rest of the puzzle, this is actually fairly obvious, but I admit I missed it, and I bet many other great minds on the forum missed it too. The orientation of the last fixed corner depends on the orientation of the previous 3 fixed corners. Of the 3 possibilities for this last corner, only 2 are possible. (3*3*3*2 = 3^3 *2 = 54)
b) Although it is possible on a Skewb, it is impossible to cycle 3 faces on the Alternating Skewb independent of the orientation of the fixed corners. What exactly IS possible? That question isn't easy to answer... Here are some facts:
-The first center can be in any of the 6 faces independent of the orientation of the fixed corners.
-The second center can be in any of the remaining 5 faces.
-The third center has four faces to choose from, but it actually can only be in 2 of those.
-At this point, the fourth, fifth, and sixth centers have no choice - the centers' permutation has been fully determined
Now, you may be inclined to ask: for the third center, which two spots? That depends on both the selected orientations of the fixed corners and the locations of the previous two centers. Suppose the orientations of all four fixed corners are correct and two adjacent centers are solved. Then the third center could obviously be in the correct spot as well, implying that in fact all 6 centers are solved, or the 4 remaining centers could be like this:
UBL UFR' UBL DBR' DFL UBL' DFL DBR' UFR DFL' DBR UBL' DBR DFL'
If instead, the two solved centers are on opposite faces, then the remaining centers could either be solved or like this:
UBL UFR' UBL UFR' DFL UBL' DBR UBL' UFR UBL' DBR DFL' UFR DBR' UBL DBR'
Now it may seem like the first of these two configurations should be able to generate another 2, but that actually invokes a mirror symmetry which would require a switch between CW and CCW, forcing two CW turns in a row. If we use an odd number of moves to attempt to create the equivalent center configuration except with a CCW move next, we can only form the mirror image of the center permutation seen before (ignoring the messed up corners that are a required side-affect of the extra CW move) as you can see here:
UBL UFR' UBL DBR' DFL UBL' DFL DBR' UFR DFL' DBR UBL' DBR (For the experts, this sequence is LITERALLY the one I just gave above with the last move removed. Haha, gotcha!)
Any way you look at it, only 6*5*2 = 60 of the expected 6!/2 = 360 permutations of centers are actually reachable after the orientations of the fixed corners have been set
In summary, for the Alternating Cube:
a) Three fixed corners can be in any of the three orientations, but the final corner can only be in two orientations (3*3*3*2 = 54) (2/3 the possibilities of a Skewb)
b) Only 60 permutations of the centers are reachable (6*5*2 = 30) (1/6 the possibilities of a Skewb)
c) The permutation of the remaining corners is isomorphic to the Klein-4 group and can only be in 4 states (4) (same as Skewb)
d) The sum of the rotations of the remaining corners must be 0 mod 3 (3^4/3 = 27) (same as Skewb)
2/3 * 1/6 = 1/9 and indeed the Alternating Skewb has (3*3*3*2)(6*5*2)(4)(3^3) = 349920 configurations, or 1/9 that of a standard Skewb.
These are fun! Gimme another please!
PS: As most people have probably guessed, every Alternating Skewb algorithm in this post is optimal, so yes, one of the positions I addressed is a hardest position, as far from solved as possible!