Solving cubes and (mathematical) series

[kwan cake]

As I wrap up my low-level sequences & series class this term, I've wondered if/how 3x3s could be compared to a sequence or a series. I'll try to list out my thought process:

• A set is a collection of elements and a sequence is just an ordered collection; the elements of a set or a sequence can be basically anything
• A group is to a set what a series is to a sequence; groups take elements of a set and somehow combine them whereas series combine the terms of a sequence in order
• I know of the Rubik's Cube Group, where we have the set of all legal moves and the operation is "combining" them by turning the cube
• If we take a set of some legal moves---say the moves to solve some scramble---and give those moves a specific order to be executed in, then I assume that means we've turned our set into a sequence
• Is actually executing the sequence of moves an example of a series? We have our sequence of moves, and we "combine" or "add" then one after another

Mostly, I'm curious if comparing 3x3s to a series would have any point; maybe patterns we know about similar series could reveal patterns about 3x3s or other twisty puzzles...or maybe it's essentially just an over-complex way of saying "if we do moves in the right order, we solve the cube." I'm not sure how that would work or what kinds of patterns you could find, but I'd be surprised if there wasn't something there.

Please let me know if my understanding of sequences/series/sets/groups is completely bonkers lol, or if this post makes no sense. I haven't found an answer after scouring the internet (or anything I understood, at least). I've only recently gotten curious about that kind of math and am having lots of fun confusing myself

 

[Christopher Mowla]

or maybe it's essentially just an over-complex way of saying "if we do moves in the right order, we solve the cube."

I'm not sure about that (left-to-right moves in a twisty puzzle "algorithm" can be represented as a right-to-left product of permutation matrices perfectly . . . well, the piece permutation aspect—Corner and edge piece orientation is a different story).

However,

Mostly, I'm curious if comparing 3x3s to a series would have any point; maybe patterns we know about similar series could reveal patterns about 3x3s or other twisty puzzles..

I know of three "phenomena" of the nxnxn that pertains to series.

1. As I explained in this post, this is a generating function by @Herbert Kociemba (source post) which gives the number of sequences (algorithms) of length x for a cube size n (in the outer block turn / face turn move metric).
2. I used the a series to find a closed formula for the number of K4 "OLLs" for the nxnxn cube. (Open formula / algorithm, paper illustrating it, and a second paper which (on page 2, etc.) uses geometric series to find a closed formula.)
3. I used a series on PDF page 20 (actual page # is 19) of my supercubes and odd parity paper to help find a closed formula for the minimum number of (non-fixed nxnxn supercube) center pieces which must be in an odd permutation for a given number of orbits of wing edges in an odd permutation. (I gave concrete examples in this "recent" post on rCubers.)

Please let me know if my understanding of sequences/series/sets/groups is completely bonkers lol

In my humble opinion, I think the set of integers is an easier way to learn what a group is (compared to the cube).

 

[kwan cake]

In my humble opinion, I think the set of integers is an easier way to learn what a group is (compared to the cube).

Oh you're absolutely right about...it definitely took me multiple hours of reading over the past 5 days to just kind of understand the "phenomena" you mentioned. They're very fun to explore though.

As I explained in this post, this is a generating function by @Herbert Kociemba (source post) which gives the number of sequences (algorithms) of length x for a cube size n (in the outer block turn / face turn move metric).

This generating function in particular might be one of the coolest I've seen, and it's interesting how simple and (mostly) intuitive it is. I wonder what this function would look like for other non-nxn puzzles.

Thanks for taking the time to link so many things in your reply, there's a lot of great stuff to read through here!

 

[Herbert Kociemba]

From ChatGPT I got a really nice hint how to find the asymptotic branching factor for a NxNxN cube given my generating function for the OBTM which is

gf[n_, x_] := 3/(6 - 4 (3 x + 1)^(n - 1)) - 1/2

The asymptotic branching factor is the quotient of the number of algorithms of length k+1 divided by the number of algorithms of length k for k->infinity.
For the 3x3x3 it is for example about 13.3485. A trivial approach would give you 18, since there are 18 possible ways you can extend a sequence of length k to a sequence of length k+1. But we only count the canonical sequences where no trivial cancellations can be made between adjacent moves and hence the factor is smaller.
A trivial approach for NxNxN in the outer block turn metric gives 3*(3*(N-1)) =9*(N-1) since on each of the three axes you have N-1 outer blocks which you can turn 90°, 180° or -90°.
Now ChatGPT tells me that "For a generating function with a simple dominant pole, the coefficient ratio tends to the reciprocal of that singularity".

gf[n_, x_] has a simple dominant pole, namely where the denominator is 0. This means simply

6 - 4 (3 x + 1)^(n - 1)=0 with the solution x= (-1 + (3/2)^(1/(-1 + n)))/3

so that we have for the asymptotic branching factor 1/x which gives the formula

bf_OBTM[n_]:= 3/(-1+(3/2)^(1/(-1+n)))

Numerically, for n =3..10 this is
{13.3485, 20.7305, 28.121, 35.5148, 42.9104, 50.3069, 57.704, 65.1015}

What even is more interesting is the behaviour for large N. The asymptotic branching factors for N=10, 100, 1000, 10000, 100000 are
{65.1015, 730.993, 7390.01, 73980.2, 739882.}, and bf_OBTM[n]/n gives

{6.51015, 7.30993, 7.39001, 7.39802, 7.39882}

which seems to stabilise around 7.4 .
And indeed bf_OBTM[n]/n for n->infinity exists and gives

3/Log[3/2] which is about 7.39891.

So we have the nice result:

For NxNxN cubes, the asymptotic branching factor in OBTM grows asymtotically linearly with N, with proportionality constant 3/Log[3/2].

 

[Herbert Kociemba]

For a even better approximation it is possible to show that the absolute difference between the linear function

3/Log[3/2]*(N-1) -3/2 and the asymptotic branching factor goes to 0 (with order O(1/N)).

Here is a table of 3/Log[3/2]*(N-1) -3/2 for N= 3..10:

{13.2978, 20.6967, 28.0956, 35.4946, 42.8935, 50.2924, 57.6913, 65.0902}

Compare it to the exact asymptotic branching factors for N=3..10 above and you see what I mean.