Emmanuel Amiot Modèles algébriques et algorithmes pour la ...

Online Supplementary I
7.3 Proof of Thm. 2.16
GGGE♭ 21
The theorem states that the maximum number of orbits (and hence of different notes in the melody) is
3n/4.
Proof Consider some autosimilar melody with period n generated by map f : x ↦→ a x + b. Obviously, for
the number of orbits to be maximal, their sizes must be minimal. Rejecting the case f = id, this means
that there are as many one note orbits (i.e. fixed points) as possible, the rest being arranged in two note
orbits. But from Prop. 2.14, the number of fixed points is the gcd of b, a − 1, n, and thus we can do no
better than a − 1 = b = n/2 (with n even). n/2 points remain; in order to arrange them by two note orbits
n/2 must be even also. So necessarily n = 4k, b = 2k, a = 2k + 1.
Now we just have to check the converse: for such n, a, b there are n/2 fixed points:
f(x) = x ⇐⇒ (2k + 1)x + 2k = x ⇐⇒ 2k(x + 1) = 0 (mod 4k) ⇐⇒ x is odd
and the other points (even values of x) come in pair orbits : x, f(x) = (2k + 1)x + 2k as then x = f(x) and
f(f(x)) = (2k + 1)[(2k + 1)x + 2k] + 2k = 4k((k + 1)x + 1) + x = x (mod n)
We have n/2 orbits with 1 element and n/4 orbits with two elements, which totals 3n/4 different notes.
7.4 Tesselations with autosimilarity
7.4.1 Proof of Thm. 4.2. The theorem characterized autosimilar melodies that tile by translation with
a 2-tile.
Proof It is easy to check that the given condition provides 2k orbits with two elements, as f : x ↦→
(2k + 1)x ± k has order 2:
f(f(x)) = (4k 2 + 4k + 1)x ± k(1 + 2k + 1) = x ± 2k(k + 1) = x as k s odd and 4k = 0
and f(x) − x = 2k x ± k is equal to ±k when x is even, and to 2k ± k = ∓k (mod 4k) when x is odd: so
all orbits, being pairs with equal diameter, are translates of one another.
Conversely, let us assume that f has order two: f 2 = id i.e. a2 = 1 and b(1 + a) = b + a b = 0.
If (as assumed) O0 = (0, b) is a translate of O1 = (1, a + b), then both pairs have same diameter:
either b − 0 = a + b − 1 or b − 0 = 1 − (a + b). The first case is forbidden as a = 1. So a + 2b = 1. We
must now consider O2 = {2, 2 a + b} (unless 2 is a member of O1 or O2, which case is easily excluded) :
either 2 a + b − 2 = b − 0 or 2 a + b − 2 = 0 − b, i.e. 2a + 2b = 2 which would imply again (substracting
a + 2b = 1) a = 1. Only the former case is possible: hence 2a = 2 which is allowed only if a = n
2 + 1. Then
2b = 1 − a = n
2 , implying that n must be a multiple of 4. Finally n = 4k, a = 2k + 1 and 2(b + k) = 0
(mod n) i.e. b = ±k. Moreover k must be odd, or else b(1 + a) = ±2k(k + 1) = 0 (mod n = 4k).
Going from one orbit to the next is like braiding a girdle, as the first note of one orbit is translated to
the last note of the next, and vice versa.
7.4.2 Tilings with augmentations. The main text has stated that
Any autosimilar melody whose orbits share the same length enables to build tilings with augmentation.
Under this condition, a tile is just any set with one note exactly in each orbit.