Generator-offset property: Difference between revisions

Line 31:

# ''S'' is unconditionally MV3 (i.e. MV3 regardless of tuning).

# ''S'' is of the form ''ax by bz'' for some permutation (''x'', ''y'', ''z'') of (L, M, s).

# The ~~cardinality (size)~~ of ''S'' is either odd, or 4 (and ''S'' is of the form ''xyxz'').

# The length of ''S'' is either odd, or 4 (and ''S'' is of the form ''xyxz'').

# S = aX bY bZ is obtained from the (single-period) mos aX 2bW by replacing all the W's successively with alternating Y's and Z's (or alternating Z's and Y's for the other chirality).

# The two alternants differ by replacing one Y with a Z.

# ''S'' is ''pairwise-mos'' (PMOS). That is, the result of identifying any two step sizes of ''S'' is always a mos.

# ''S'' is ''monotone-mos'' (MMOS). That is, each of the following operations results in a mos: setting L = M, setting M = s, and setting s = 0.

# ''S'' is a [[billiard scale]]. (?)

# If len(''S'') ''S'' is a [[billiard scale]]. (?)

In particular, odd GO scales always satisfy these properties (see Proposition 2 below).

Line 43:

==== Proof ====

Assuming SGA, we have two chains of generator ''g''0 (going right). The two cases are:

CASE 1: EVEN ~~CARDINALITY~~

CASE 1: EVEN LENGTH

O-O-...-O (n/2 notes)

and

CASE 2: ODD ~~CARDINALITY~~

CASE 2: ODD LENGTH

O-O-O-...-O ((n+1)/2 notes)

O-O-...-O ((n-1)/2 notes).

Line 67:

# ''a''4 − ''a''2 = ''g''1 − 2 ''g''2 + ''g''3 = (''g''3 − ''g''2) + (''g''1 − ''g''2) = (chroma ± ε) != 0 by choice of tuning.

By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus ''g''1 and ''g''2 must themselves be step sizes. Thus we see that an even-~~cardinality~~, unconditionally MV3, AG scale must be of the form ''xy...xyxz''. But this pattern is not unconditionally MV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and ''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''. This proves (3).

By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus ''g''1 and ''g''2 must themselves be step sizes. Thus we see that an even-length, unconditionally MV3, AG scale must be of the form ''xy...xyxz''. But this pattern is not unconditionally MV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and ''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''. This proves (3).

In case 2, let (2, 1) − (1, 1) = ''g''1, (1, 2) − (2, 1) = ''g''2 be the two alternants. Let ''g''3 be the leftover generator after stacking alternating ''g''1 and ''g''2. Then the generator circle looks like ''g''1 ''g''2 ''g''1 ''g''2 ... ''g''1 ''g''2 ''g''3. Then the combinations of alternants corresponding to a step are:

Line 77:

(The above holds for any odd ''n'' ≥ 3.)

For (1), we now only need to see that SGA + odd ~~cardinality~~ => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.

For (1), we now only need to see that SGA + odd length => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.

For (4), assume S is aX bY bZ, a odd. If b = 1, there’s nothing to prove. So assume b > 1. If Y’s and Z’s don't alternate perfectly, then (ignoring X's) you have two consecutive Y's somewhere and two consecutive Z's somewhere else. Assume that g_pf = iX + jW with j >=2. (If this is not true, invert the generator, since b > 1.)

@@ Line 31: / Line 31: @@
 # ''S'' is unconditionally MV3 (i.e. MV3 regardless of tuning).
 # ''S'' is of the form ''ax by bz'' for some permutation (''x'', ''y'', ''z'') of (L, M, s).
-# The cardinality (size) of ''S'' is either odd, or 4 (and ''S'' is of the form ''xyxz'').
+# The length of ''S'' is either odd, or 4 (and ''S'' is of the form ''xyxz'').
 # S = aX bY bZ is obtained from the (single-period) mos aX 2bW by replacing all the W's successively with alternating Y's and Z's (or alternating Z's and Y's for the other chirality).
 # The two alternants differ by replacing one Y with a Z.
 # ''S'' is ''pairwise-mos'' (PMOS). That is, the result of identifying any two step sizes of ''S'' is always a mos.
 # ''S'' is ''monotone-mos'' (MMOS). That is, each of the following operations results in a mos: setting L = M, setting M = s, and setting s = 0.
-# ''S'' is a [[billiard scale]]. (?)
+# If len(''S'') ''S'' is a [[billiard scale]]. (?)
 In particular, odd GO scales always satisfy these properties (see Proposition 2 below).
@@ Line 43: / Line 43: @@
 ==== Proof ====
 Assuming SGA, we have two chains of generator ''g''<sub>0</sub> (going right). The two cases are:
-  CASE 1: EVEN CARDINALITY
+  CASE 1: EVEN LENGTH
   O-O-...-O (n/2 notes)
   O-O-...-O (n/2 notes)
 and
-  CASE 2: ODD CARDINALITY
+  CASE 2: ODD LENGTH
   O-O-O-...-O ((n+1)/2 notes)
   O-O-...-O ((n-1)/2 notes).
@@ Line 67: / Line 67: @@
 # ''a''<sub>4</sub> &minus; ''a''<sub>2</sub> = ''g''<sub>1</sub> &minus; 2 ''g''<sub>2</sub> + ''g''<sub>3</sub> = (''g''<sub>3</sub> &minus; ''g''<sub>2</sub>) + (''g''<sub>1</sub> &minus; ''g''<sub>2</sub>) = (chroma ± ε) != 0 by choice of tuning.
-By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus ''g''<sub>1</sub> and ''g''<sub>2</sub> must themselves be step sizes. Thus we see that an even-cardinality, unconditionally MV3, AG scale must be of the form ''xy...xyxz''. But this pattern is not unconditionally MV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and ''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''. This proves (3).
+By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus ''g''<sub>1</sub> and ''g''<sub>2</sub> must themselves be step sizes. Thus we see that an even-length, unconditionally MV3, AG scale must be of the form ''xy...xyxz''. But this pattern is not unconditionally MV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and ''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''. This proves (3).
 In case 2, let (2, 1) &minus; (1, 1) = ''g''<sub>1</sub>, (1, 2) &minus; (2, 1) = ''g''<sub>2</sub> be the two alternants. Let ''g''<sub>3</sub> be the leftover generator after stacking alternating ''g''<sub>1</sub> and ''g''<sub>2</sub>. Then the generator circle looks like ''g''<sub>1</sub> ''g''<sub>2</sub> ''g''<sub>1</sub> ''g''<sub>2</sub> ... ''g''<sub>1</sub> ''g''<sub>2</sub> ''g''<sub>3</sub>. Then the combinations of alternants corresponding to a step are:
@@ Line 77: / Line 77: @@
 (The above holds for any odd ''n'' ≥ 3.)
-For (1), we now only need to see that SGA + odd cardinality => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.
+For (1), we now only need to see that SGA + odd length => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.
 For (4), assume S is aX bY bZ, a odd. If b = 1, there’s nothing to prove. So assume b > 1. If Y’s and Z’s don't alternate perfectly, then (ignoring X's) you have two consecutive Y's somewhere and two consecutive Z's somewhere else. Assume that g_pf = iX + jW with j >=2. (If this is not true, invert the generator, since b > 1.)