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Master pyraminx God's number bounds

xyzzy

Member
Joined
Dec 24, 2015
Messages
1,614
tl;dr: Ignoring the tips, GN is between 14 and 26 moves (inclusive) in OBTM.

We will completely ignore the tips in this analysis, because they're trivial and boring. If you care about the tips a lot, you can just add 4 to the numbers here as appropriate. We split solving the master pyraminx into two steps: (i) reduction to Jing's pyraminx and (ii) solving Jing's pyraminx.

Excluding the tips, there are four types of pieces on the master pyraminx, with exactly these restrictions:
- 4 corners (same as pyraminx); three orientations each (no permutation)
- 6 midges (same as pyraminx); even permutation, two orientations each, even number of flipped edges
- 12 wings; even permutation (no orientation)
- 4 centres; even permutation (no visible orientation)

A quick computation shows that the total number of states is \( (3^4)\cdot(\tfrac126!\cdot2^5)\cdot(\tfrac1212!)\cdot(\tfrac124!) = 2\,681\,795\,837\,952\,000 \). (If you see a figure on another website that's 81 times of this, that's because whoever wrote that included the tips.)

Phase 1: reduction

Reduction here has two components: edge pairing, and ensuring that the centres "line up" with the corners. The latter is how we ensure that we don't end up with a 3-cycle of centres after reduction, which cannot be solved with just Jing's pyraminx moves (Uw, Lw, Rw, Bw).

The alternating group \( A_4 \) has a copy of the Klein four-group \( V \) as a normal subgroup (which is also the unique Sylow 2-subgroup), and the quotient group \( A_4/V \) is isomorphic to the order-3 cyclic group. Considering the action of \( A_4 \) on the centres of the master pyraminx, the three cosets of \( V \) in \( A_4 \) can be labelled as \( V \) itself, the clockwise coset if it consists of the four clockwise 3-cycles of centres (as viewed from a tip), and the anticlockwise coset if it consists of the four anticlockwise 3-cycles of centres.

(It's easy to check that these subsets of \( A_4 \) really are cosets of \( V \). Conjugating a clockwise 3-cycle by any element of \( A_4 \) (which is equivalent to a proper rotation of the tetrahedron) will obviously result in a clockwise 3-cycle, possibly in a different location. Specifically, this also applies to conjugating a 3-cycle by any element of \( V \). Thus the coset of \( V \) containing any of the clockwise 3-cycles has all four of them, and since each coset of \( V \) has exactly four elements, it follows that the set of clockwise 3-cycles is a coset of \( V \). The same applies for the anticlockwise 3-cycles.)

As an arbitrary convention, take an anticlockwise twist to have the value 1 and a clockwise twist to have the value 2. Oriented centres and \( V \) get the value 0. Working in \( \mathbb Z/3 \), we now have an easy-to-verify invariant on the Jing's pyraminx: the sum of corner orientations is equal to the "value" of the \( V \)-coset the centre permutation is in. This invariant will always hold when doing only Uw, Lw, Rw, Bw moves on the master pyraminx; a clockwise turn will change the value of the sum of corner orientations by +2, but it'll also change the value of the \( V \)-coset by +2 since it's a clockwise 3-cycle of the centres; likewise for anticlockwise turns.

As for edge pairing, we do not care about how the midges are permuted, nor do we care about how the wings are permuted; what we care about is the relative permutation (a term I made up, but it seems to be a useful concept). I've written about this before in the context of 5×5×5. The rough idea is that we "split" each midge into two halves, so we're actually looking at a permutation of 12 midge halves and a permutation of 12 wings. Let the former be \( M \) and the latter be \( W \); then \( M^{-1}W \) can take on any value in \( A_{12} \). This gives us a right group action on \( A_{12} \), given by how \( M^{-1}W \) changes as we apply moves.

In total, this phase consists of \( \tfrac1212!\cdot3=718\,502\,400 \) cosets of the Jing's pyraminx states subgroup.

(is it just me or does the TeX stuff not work inside spoiler boxes, hmm)

This step takes at most 14 moves, and has the following distribution:

Depth

# states

0

1

1

8

2

72

3

744

4

6576

5

59210

6

479280

7

3579324

8

22094560

9

105140230

10

291360283

11

268127727

12

27628904

13

25480

14

1



The unique 14-move antipode is given by a superflip-like position, where all the wings are on the wrong side of their matching midges and the centre coset matches the corner orientation, e.g. U Lw L' B Rw Lw U L' R B Lw Uw L' B.

Phase 2: Jing's pyraminx

Jaap Scherphuis has done an exhaustive search for Jing's pyraminx; the results are on Jaap's Puzzle Page. This phase takes at most 12 moves to solve.

Adding the numbers from both phases, this leads to an upper bound of 14+12 = 26 moves. A lower bound of 14 moves follows from phase 1's GN being 14. (A careful analysis of the number of canonical move sequences might be able to bring up the lower bound to 15 moves.)
 

Ben Whitmore

Member
Joined
Sep 28, 2019
Messages
4
@xyzzy I found a few positions that take 20 moves to solve two days before this post, so 20 is a lower bound, and I conjecture that god's number is exactly 20. In november 2017 I found that superflip takes 18 moves: (Uw Rw Lw' U' R' L)3

I solved the superflip*H coset where H is the subgroup of centres, corners, and middle edge permutation (|H| = 7664025600 cosets, |G/H| = 349920 positions), and superflip = flip the 6 middle edges. The distribution is:

Depth

Positions

12

4

13

40

14

550

15

8210

16

66930

17

193475

18

77928

19

2426

20

357



One example of a position that takes 20 is superflip + twist one corner: Uw Rw Lw Bw' L R' Rw Bw U' Rw Uw' B' L' Uw' R' U' Rw' B' L' Bw
 
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