[superprismatic]

Let P be a permutation on 12 objects

which has cycle length 12. As an

example, suppose P were given by

 1  2  3  4  5  6  7  8  9 10 11 12

 2  8  6 10  4  9 11  7 12  3  5  1

[/code]
which tells us to move the object in

the 1st position to the 2nd position,

the object in the 2nd position to the

8th position, ..., and the object in

the 12th position to the 1st position.

We will use this to permute the 12

letters ABCDEFGHIJKL.  Applying P

to this yields LAJEKCHBFDGI. For this

we write P(ABCDEFGHIJKL)=LAJEKCHBFDGI.

We can continue to apply P so that

P[sup]2[/sup](ABCDEFGHIJKL)=P(LAJEKCHBFDGI) and

we can compute P(LAJEKCHBFDGI).  Since

P has cycle length 12, applying P

over and over again 12 times will

get us back to ABCDEFGHIJKL, written

P[sup]12[/sup](ABCDEFGHIJKL)=ABCDEFGHIJKL.

So, I can easily produce all 12

sequences of letters which powers

of P can make when applied to

ABCDEFGHIJKL.



Now, suppose I do this with another

permutation Q on 12 objects with

cycle length 12.  The 12 letter

sequences which were produced (in

alphabetical order -- [b]NOT[/b] the order

of their generation) is:

[code]

ABCDEFGHIJKL

BDGKAILFJECH

CGFLKAIEBDHJ

DKLCBJHIEAGF

EAKBJHCLFIDG

FIAJHCBKGLED

GLIHCBJADKFE

HFEILKADCGJB

IJBEFGDCLHAK

JEDAILKGHFBC

KCHGDEFJABLI

LHJFGDEBKCIA

Find all the possible permutations

which Q could be?

[bushindo]

Let P be a permutation on 12 objects

which has cycle length 12. As an

example, suppose P were given by

 1  2  3  4  5  6  7  8  9 10 11 12

 2  8  6 10  4  9 11  7 12  3  5  1

which tells us to move the object in the 1st position to the 2nd position, the object in the 2nd position to the 8th position, ..., and the object in the 12th position to the 1st position. We will use this to permute the 12 letters ABCDEFGHIJKL. Applying P to this yields LAJEKCHBFDGI. For this we write P(ABCDEFGHIJKL)=LAJEKCHBFDGI. We can continue to apply P so that P²(ABCDEFGHIJKL)=P(LAJEKCHBFDGI) and we can compute P(LAJEKCHBFDGI). Since P has cycle length 12, applying P over and over again 12 times will get us back to ABCDEFGHIJKL, written P¹²(ABCDEFGHIJKL)=ABCDEFGHIJKL. So, I can easily produce all 12 sequences of letters which powers of P can make when applied to ABCDEFGHIJKL. Now, suppose I do this with another permutation Q on 12 objects with cycle length 12. The 12 letter sequences which were produced (in alphabetical order -- NOT the order of their generation) is:

ABCDEFGHIJKL

BDGKAILFJECH

CGFLKAIEBDHJ

DKLCBJHIEAGF

EAKBJHCLFIDG

FIAJHCBKGLED

GLIHCBJADKFE

HFEILKADCGJB

IJBEFGDCLHAK

JEDAILKGHFBC

KCHGDEFJABLI

LHJFGDEBKCIA

Find all the possible permutations

which Q could be?

Nice problem

4 ways. There is a permutation Q that cycles through all 12 states given here, and the transformations Q⁵, Q⁷, and Q¹¹ also cycle through all 12 states.

Edited April 15, 2010 by bushindo

[Guest]

SP are you a bellringer by any chance? This is exactly the construction used (numerically) in the design of differentials and principles in 12 bell ringing but re-phrased in mathematical rather than ringing terminology.

A permutation that has cycle length 12, produces a 12 part (or section) principle. You can get further 12 part lengths by using co-prime multiples of that permutation as Bushindo suggests.

Multiples that are factors of 12 produce differentials that may be shorter or longer than 12 parts, depending on whether the sub-cycles are co-prime. On 12 they are all shorter.

5 1 11 2 10 8 3 12 6 9 4 7

2 4 7 11 1 9 12 6 10 5 3 8

8 6 5 9 12 11 1 4 3 7 10 2

7 12 9 8 3 2 10 1 4 11 6 5

Edited April 15, 2010 by Ringer

[bushindo]

Multiples that are factors of 12 produce differentials that may be shorter or longer than 12 parts, depending on whether the sub-cycles are co-prime. On 12 they are all shorter.

This part seems counter intuitive to me.

I assume you meant that the cycles produced by the permutation Qⁿ can be longer than L when Q has cycle length L. It seems that for any L, Qⁿ has to be member of the set [ Q¹, Q², ..., Q^L ]. We can't go from one element to another in that set more than L times without falling on a previously visited element, and thus all multiples of Q should have cycle length upper bound of L.

Edited April 15, 2010 by bushindo

[Guest]

Sorry, you are right, I was stretching the definition of the permutation P to not having cycle length 12. It is the definition of P rather than Q that can define principles and differential ringing-method designs. Even then I've over-simplified, as it is the way you move the bells into those positions at the end of the first section that is the interesting part rather than simply deciding how they rotate section by section.

...but I guess I'm starting to speak a different language now..!