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Order of the Additive List

Submitted by cubex on Thu, 03/10/2005 - 02:39.
Conjecture: The order of the additive list always evenly divides the order of the generated group.

Time for some definitions.

First from the possible moves of the cube using Singmaster notation (U, D, F, B, L, R) pick any number of operators, this is the basis for the generated group.

The group generated by < U, F, D > is an example of the "generated group".

From these operators we can generate the "additive list" so the elements are

{ U, UF, UFD, UFDU, UFDUF, UFDUFD ... }

Now for some rules...

Pick only one of _each_ of the 6 operators UDFBLR. Any number of operators may be used from 1 to all 6. Using the inverse of an operator is allowed. For now we shall keep things in the quarter turn metric so there is no use of op^2 only op or -op.

Once we pick an op we don't use that op again until we repeat.

So in my example of < U, F, D > we can have { -U, -UF, -UFD, -UFD-U ... } but not { -U, -UF, -UFU ... } as this breaks the rule of only one use of a specific op before repeating begins.

The order of the additive list is the number of ops times the order N where N is { op1, op2, op3 ... }^N = I.

In my example using < U, F, D> the order of the additive list 3 * 90 = 270, since { U, F, D }^90 = I.

Sizeof(ufd) = 159993501696000.

Then the conjecture predicts these numbers divide evenly:

159993501696000 / 270 = 592568524800.

So let's try another additive list, this time we will pick -U instead of U.

Now we have {-U, F, D }^180 = I. Size of the additive list is 3 times 180 = 540.

Again the Sizeof(ufd) is 159993501696000. Then 159993501696000 divided by 540 is 296284262400.

There are a lot of additive lists but not a huge number so I think it is possible to try all examples. So far no one has found any counter-examples but we have not tried all the possible lists.
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It is well known that: The

Submitted by jaap on Thu, 03/10/2005 - 08:31.
It is well known that: The order of an element of a group divides the order of the group.

In your example, this means that the order of the element U'FD divides the order of the generated group <U,F,D>. This takes care of the factor 90.

Your conjecture claims that the length of the sequence (in the example it is 3, the length of U'FD) further divides this quotient (also called the index).

The fact that the only sequences you allow are those with all different faces ensures that this potential divisor is <=6. It is not surprising that this is an actual divisor because the group order will have an abundance of small prime factors.

Let's take any move sequence of length 5, involving 5 different faces. The group generated by the 5 faces will certainly have more than 5 corners and more than 5 edges that are permuted. The group order therefore has two factors 5. The element itself might have an order that is divisible by 5 if it permutes some of the pieces around in a 5-cycle. It cannot have an order divisible by 5^2=25, because there are no 25-cycles possible. Therefore your conjecture will certainly be true with regards to the prime 5.

The order of the group will have lots of factors 2 and 3, because of the orientations of the pieces. Your conjecture will therefore certainly hold for lengths 2,3,4 and 6 as well. For example a length 3 sequence will involve at least 7 corners, so the group has a factor 3^6 from the orientations alone. To get an element with an order high enough to be a counterexample, it would have to involve a 3^5-cycle at least.

I think this pretty much proves it.

Jaap's Puzzle Page: http://www.geocities.com/jaapsch/puzzles/

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More than a conjecture, but less than a theorem

Submitted by cubex on Sun, 03/13/2005 - 08:58.
I tend to agree, but I'm going to try to work out all the cases. I don't think it's a huge number.

Here's another problem but this time I'm not going to make any conjectures.

Can we reach all the positions of the cube if we only use processes formed by an additive list? In other words: Can all elements of the cube exist as elements of an additive list?
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No

Submitted by jaap on Tue, 03/15/2005 - 05:10.
No, that is not possible, and can be shown by a straightforward counting argument.

Let m1 .. mn be the moves of an additive list, and let P=m1 m2 ... mn be the whole move sequence that is repeated.

The elements on that list are then:
{m1, m1 m2, ..., P, P m1, P m1 m2, ..., P^2, P^2 m1, .... }

If I remember correctly, the maximal order of any element of the cube group is 1260. An additive list therefore contains at most 1260*n < 10^4 elements.

How many additive lists are there?
There are less than 18^6 of six moves, less than 18^5 of 5 moves, and so on. There are certainly less than 10^8 all together.

Therefore all additive lists together contain less than 10^12 positions, much less than the whole cube group.

Jaap's Puzzle Page: http://www.geocities.com/jaapsch/puzzles/

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