Generator-offset property: Difference between revisions

Line 29:

=== Proposition 1 (Properties of SGA scales) ===

Let ''S'' be a 3-step-size scale word in L, M, and s, and suppose ''S'' is SGA. Then:

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

# ''S'' is abstractly SV3 (i.e. SV3 for almost all tunings).

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

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

Line 66:

# ''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-length, unconditionally SV3, GO scale must be of the form ''xy...xyxz''. But this pattern is not ~~unconditionally~~ SV3 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 SV3, GO scale must be of the form ''xy...xyxz''. But this pattern is not abstractly SV3 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 come in exactly 3 sizes:

Line 76:

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

For (1), we now only need to see that SGA + odd length => unconditionally SV3. But the argument in case 2 above works for any interval class (unconditional SV3 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 SV3. But the argument in case 2 above works for any interval class (unconditional SV3 wasn't used), hence any interval class comes in (abstractly) exactly 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 29: / Line 29: @@
 === Proposition 1 (Properties of SGA scales) ===
 Let ''S'' be a 3-step-size scale word in L, M, and s, and suppose ''S'' is SGA. Then:
-# ''S'' is unconditionally MV3 (i.e. MV3 regardless of tuning).
+# ''S'' is abstractly SV3 (i.e. SV3 for almost all tunings).
 # ''S'' is of the form ''ax by bz'' for some permutation (''x'', ''y'', ''z'') of (L, M, s).
 # The length of ''S'' is either odd, or 4 (and ''S'' is of the form ''xyxz'').
@@ Line 66: / Line 66: @@
 # ''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-length, unconditionally SV3, GO scale must be of the form ''xy...xyxz''. But this pattern is not unconditionally SV3 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 SV3, GO scale must be of the form ''xy...xyxz''. But this pattern is not abstractly SV3 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 come in exactly 3 sizes:
@@ Line 76: / Line 76: @@
 (The above holds for any odd ''n'' ≥ 3.)
-For (1), we now only need to see that SGA + odd length => unconditionally SV3. But the argument in case 2 above works for any interval class (unconditional SV3 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 SV3. But the argument in case 2 above works for any interval class (unconditional SV3 wasn't used), hence any interval class comes in (abstractly) exactly 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.)