Roots of permutations (continued)

The method we described works well for small bases but as soon as we try to find roots of a dodecaphonic sequence, it reveals insufficient. I realised this working on the row “A0B281947356” which is what dodecaphonists called an “Alleintervallenreihe”. Its elementary partition is (“A510”, “B6932”, “874”) and we see immediately that no lower partition can produce by division such uplets, therefore the roots of “A0B281947356” have to share the same type.

Now as a partition containing an odd uplet is accompanied by a mirror of itself in the same cycle and as a mirror is a root for its reflection, we can suppose that the mirror of our row is one of its roots. It is easy to verify that the permutation corresponding to uplets (“15A0″,”396B2″,”784”) contains “A0B281947356” within its cycle (this is even easier to verify using the scripts provided  in the Code section).

We can vary the composition of uplets choosing one uplet in the original row and one in the mirror – respecting correspondence in subsets of digits – as we did above, mutatis mutandis, when we varied the order of figures in a root partition in order to discover new roots.

This way we find easily among the 480 millions permutations in base 12 a small subset of 8 partitions working as roots for our Alleintervallenreihe:

  1.  (“A510”, “B6932”, “874”) : “A0B281947356”
  2.  (“A510”, “B6932”, “784”) : “A0B271984356”
  3.  (“A510”, “396B2”, “874”) : “A03981B47652”
  4.  (“A510”, “396B2”, “784”) : “A03971B84652”
  5. (“15A0”, “B6932”, “874”) : “15B28A947306”
  6. (“15A0”, “B6932”, “784”) : “15B27A984306”
  7. (“15A0”, “396B2”, “874”) : “15398AB47602”
  8. (“15A0”, “396B2”, “784”) : “15397AB84602”

It is nice but will it suffice? Unfortunately not. As long as we don’t reach a deeper comprehension of permutations we risk to miss some source for our roots. Now we could simply rely on the idea of complete cycle (A3) and test each one of the 60 members of our suite, but suppose that for non-musical purpose we were searching roots of a much higher base, then this test itself would reveal prohibitive. As we’re about to find some way of doing what we want in base 12, we can still ask ourselves if we didn’t leave some other cases apart: being reluctant to dive into group theory has a cost.

However, let’s finish with this imperfect method.

What really are n-tuples?

We can see our n-tuples as subsets of binary permutations. This way they can be read in one direction or the other so as to describe what happens to our natural row (“0123456789AB”) in order to become our permutation (“A0B281947356”). In the elementary partition of this last row  (“A510”, “B6932”, “874”), we can read the first uplet from right to left as “0” takes the place of “1”, “1 takes the place of 5”, “5 takes the place of A”, “A takes the place of 0”.

Dancing a round

This  description of the relation between our rows reminds me of a nice illustration of algorithm with Hungarian folk dancers. We can also say that this is one manner for each sign in this uplet to be located at the right of another sign: “0 is at the right of 1”, “1 is at the right of 5”, etc.. And we have to consider these signs as dancing a round and it is indeed a useful addition that can be made to our method in order to get all the roots. For if we require from each sign in each uplet to be at least one time in one uplet placed at the right of each other sign from the same original subset, then we get all our roots.


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