Of some new acyclic canons

Talk at MaMuX seminar Ircam, Paris, january the 24th

The new canons where obtained by some of the following actions :
• completion (given R, find S)
• affine orbite (use the affine maps x -> k×x mod n to get other tiles from a given one)
• augmentations/diminutions. On a tile, it works in one of two ways :
	﷿ Replace each note (or silence) by p consecutive notes (or silences)
	﷿ Change the tempo by the factor p.

Solutions are given in intervalic form, reduced up to cyclic permutation.

The relevant functions are written with Mathematica in canonCrawler.nb. There names are respectively :
• complete (or randomCanonFromList, for the quicker but randomized version)
• orbite
• zoomIn, zoomOut (and obviously tile -> p×tile)
• basicForm (i.e. first element of orbite)

In the following, a few words are given to explain the howTo. The solutions are listed next.

144=6×24 :

•I cannot complete a 6 notes motive on period 144. The reverse, using any element (24 long) of  (Subscript[R, 144  bis]), gives back only the 3 known elements of (Subscript[S, 144  bis]).
• Something weird :  doubling (Subscript[S, 72]) gives 2  (Subscript[S, 72]) = (Subscript[S, 144  bis])already given by Vuza's algorithm: nothing new there. Thus the 2-augmentations of (Subscript[R, 72])should be in the complements of (Subscript[S, 144  bis]). They are not, in the 2808 given by Vuza's algorithm. So I computed them and their orbits of 48 elements each.

orbite[zoomIn[R72[[1]], 2]]

orbite[zoomIn[R72[[2]], 2]]

orbite[zoomIn[R72[[4]], 2]]

Nothing new in the complements of these complements.

144 = 12×12

• Completing ((Subscript[R, 144])^1)(35 tiles of size 12 and period 144) gives 162 solutions, including the 6 given by Vuza's algorithm. Here they are (only the acyclic ones are given):

Select[%,  aperQ]

32 of them are palindroms.

• Doubling (Subscript[R, 72])(the dual augmentation from the preceding case) one gets 6 tiles of length 12,  and period 144. They have 324 complements, all of which are acyclic, all new, for 6×324=1944 new canons. Here comes:

complete[{2, 6, 6, 12, 22, 8, 18, 12, 10, 2, 6, 40}]

Complementation ot these gives  2 (Subscript[S, 72]) =  ((Subscript[S, 144])^bis), as expected, and nothing more.

• I cannot use complete on my computer for the 3 elements of  but with zoomOut and orbite I found a few— 432 — of the colmplements, that is to say 3×432 = 1296 more.

Pour 108 :

En complétant, tout simplement, le (Subscript[S, 108])de l'algorithme, on trouve 72 nouvelles façons en plus des 1780 données par l'algorithme (soit 216 nouveaux canons). Il resterait à tenter de compléter des motifs acycliques cousins des Vuza canons connus, obtenus comme compléments des compléments de (Subscript[S, 108]), comme {3,3,21,3,3,75} (mais c'est long !). Ceci dit cela fait déjà 45 motifs non dans (Subscript[S, 108]), qui complètent les 252 trouvés soit 11340 nouveaux canons de plus !

Pour 120 :

En quintuplant (Subscript[R, 72]) et divisant par 2, on obtient trois motifs de période 120 et de longueur 6. Une recherche aléatoire donne moult compléments, mais je n'ai pas la puissance nécessaire pour les trouver tous; remarquablement, tous ces compléments sont acycliques ! Pour l'instant cela donne 3×239 = 717 nouveaux canons,  mais on sait que c'est incomplet.

120

Completion of the Vuza canons ((Subscript[r, 120])) gives nothing new. One strange thing : a lone complement has inner periodicity, all others are acyclic. The funny one is {10,30,10,30,10,30}, the others form two orbits of four. I am not able to complete (Subscript[S, 120])for lack of computer power.
So I found another track : compute S72*5/2 which gives length 6, period 120 tiles. The random method gives a few of the complements ; members of their orbits are

%

This is known to be incomplete, as some elements of (Subscript[r, 120 ])are missing.

108

Completing any element of the one orbit that forms (Subscript[S, 108]) gives 672 solutions, 252 of which are acyclic. Here are the 72 new ones (12 orbits of 6 each) among them (no palindroms):

Complement[%,  R108]

Now, I tried completing one of the 672-252 periodic complements, for instance {1,8,1,7,1,1,7,9,19,1,8,1,7,1,1,7,9,19}, This gives some 48 aperiodic solutions. Strangely not all of them are complements of an acyclic tile, indeed we find back only the known 3 elements of  (Subscript[S, 108]) :

%

Select[someAcyclicTiles,  
      canonQ[#,  
{1, 6, 4, 7, 2, 9, 7, 1, 6, 11, 11, 7, 
            1, 1, 5, 4, 7, 18}
] &]

This puts to light two interesting (or maybe depressing) facts :
• by double complementing of a periodic (not acyclic) tile one may sometimes find acyclic canons,
• some tile may have as complements a subset of the complements of some other (more regular ?) tile. For instance, the complements of {1,7,10,1,7,10,…,1,7,10} are 1296 (!) but they include the aforementioned someAcyclicTiles which did include already (Subscript[S, 108]).