Generator-offset property: Difference between revisions
→Theorem 4 (PWF implies SV3 and either GO or abacaba): italicizing latin variables that don't refer to interval sizes |
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=== Theorem 4 (PWF implies SV3 and either GO or abacaba) === | === Theorem 4 (PWF implies SV3 and either GO or abacaba) === | ||
Let S(a,b,c) be a scale word in three '''Z'''-linearly independent step sizes a, b, c. Suppose S is PWF. Then S is SV3 and has an odd number of notes. Moreover, S is either GO or equivalent to the scale word abacaba. | Let ''S''(a, b, c) be a scale word in three '''Z'''-linearly independent step sizes a, b, c. Suppose S is PWF. Then S is SV3 and has an odd number of notes. Moreover, ''S'' is either GO or equivalent to the scale word ''abacaba''. | ||
==== Proof ==== | ==== Proof ==== | ||
===== If the generator of a projection of S is a k-step, the word of stacked k-steps in S is PWF ===== | ===== If the generator of a projection of ''S'' is a ''k''-step, the word of stacked ''k''-steps in ''S'' is PWF ===== | ||
Suppose S has n notes (after dealing with small cases, we may assume n ≥ 7) and S projects to single-period mosses S<sub>1</sub> (via identifying b with c), S<sub>2</sub> (via identifying a with c) and S<sub>3</sub> (via identifying a with b). Suppose S<sub>1</sub>'s generator is a k-step, which comes in two sizes: P, the perfect k-step, and I, the imperfect k-step. By stacking k-steps, we get two words of length n of k-steps of S<sub>2</sub> and S<sub>3</sub>, respectively. These two-"step-size" words, which we call Σ<sub>2</sub> and Σ<sub>3</sub>, must be mosses, since m-steps in the new words correspond to mk-steps in the mos words S<sub>1</sub> and S<sub>2</sub>, which come in at most two sizes. Since S<sub>1</sub> is a single-period mos, gcd(k, n) = 1. Hence when 0 < m < n, | Suppose ''S'' has ''n'' notes (after dealing with small cases, we may assume ''n'' ≥ 7) and ''S'' projects to single-period mosses ''S''<sub>1</sub> (via identifying b with c), ''S''<sub>2</sub> (via identifying a with c) and ''S''<sub>3</sub> (via identifying a with b). Suppose ''S''<sub>1</sub>'s generator is a ''k''-step, which comes in two sizes: P, the perfect ''k''-step, and I, the imperfect ''k''-step. By stacking ''k''-steps, we get two words of length ''n'' of ''k''-steps of ''S''<sub>2</sub> and ''S''<sub>3</sub>, respectively. These two-"step-size" words, which we call Σ<sub>2</sub> and Σ<sub>3</sub>, must be mosses, since ''m''-steps in the new words correspond to ''mk''-steps in the mos words S<sub>1</sub> and S<sub>2</sub>, which come in at most two sizes. Since ''S''<sub>1</sub> is a single-period mos, gcd(''k'', ''n'') = 1. Hence when 0 < ''m'' < ''n'', ''mk'' is ''not'' divisible by ''n'' and ''mk''-steps come in ''exactly'' two sizes; hence both Σ<sub>2</sub> and Σ<sub>3</sub> are single-period mosses. | ||
index: 1 2 3 4 ... n | index: 1 2 3 4 ... n | ||
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Σ<sub>3</sub>: [some mos] | Σ<sub>3</sub>: [some mos] | ||
===== Two sizes of k-steps in S project to S<sub>1</sub>'s perfect generator ===== | ===== Two sizes of ''k''-steps in ''S'' project to ''S''<sub>1</sub>'s perfect generator ===== | ||
We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c). | We can write sizes of intervals in ''S'' as vectors (''p'', ''q'', ''r'') using the basis (a, b, c). | ||
Suppose for sake of contradiction that only one size (α, β, γ) in S projects to P in S<sub>1</sub>. Then projecting to S<sub>2</sub> shows that S<sub>2</sub>'s generator is the k-step (α + γ)*(a~c) + βb, and Σ<sub>2</sub>'s imperfect generator is located at index n, like Σ<sub>1</sub>'s imperfect generator is. Then S<sub>1</sub> and S<sub>2</sub> are the same mode of the same mos pattern (up to knowing which step size is the bigger one). Assume the L of S<sub>1</sub> (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S<sub>1</sub> come from a. Then the steps of S<sub>2</sub> corresponding to the L of S<sub>1</sub> must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S<sub>1</sub> and S<sub>2</sub> are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S has exactly three step sizes. | Suppose for sake of contradiction that only one size of ''k''-step (α, β, γ) in ''S'' projects to P in ''S''<sub>1</sub>. Then projecting to ''S''<sub>2</sub> shows that ''S''<sub>2</sub>'s generator is the ''k''-step (α + γ)*(a~c) + βb, and Σ<sub>2</sub>'s imperfect generator is located at index ''n'', like Σ<sub>1</sub>'s imperfect generator is. Then ''S''<sub>1</sub> and ''S''<sub>2</sub> are the same mode of the same mos pattern (up to knowing which step size is the bigger one). Assume the L of ''S''<sub>1</sub> (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in ''S''<sub>1</sub> come from a. Then the steps of ''S''<sub>2</sub> corresponding to the L of ''S''<sub>1</sub> must be either all b’s or all a~c’s, thus these steps are all b’s in ''S'' (otherwise they would be identified with the a, against the assumption that ''S''<sub>1</sub> and ''S''<sub>2</sub> are the same mos pattern and mode). So ''S'' has only two step sizes (a and b), contradicting the assumption that ''S'' has exactly three step sizes. | ||
Only two sizes of k-steps of S can project to P in S<sub>1</sub>, for if there are three sizes of k-steps (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S that project to P, then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S<sub>3</sub>, contradicting the mos property of S<sub>3</sub>. | Only two sizes of ''k''-steps of ''S'' can project to P in ''S''<sub>1</sub>, for if there are three sizes of ''k''-steps (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in ''S'' that project to P, then β, β’ and β’’ are three distinct values. Thus these would project to three different ''k''-steps in ''S''<sub>3</sub>, contradicting the mos property of ''S''<sub>3</sub>. | ||
===== Some implications of the above ===== | ===== Some implications of the above ===== | ||
Suppose Q = (α, β, γ) ≠ R = (α, β’, γ’) are the two k-steps in S that project to P. Then T = (α’, β’’, γ’’) projects to I. Here the values in each component differ by at most 1, and α ≠ α’. Then the cyclic word Λ<sub>1</sub> formed by the a-components of the k-steps in P is α...αα’. Since Σ<sub>2</sub> is a single-period mos pattern of βb + (n − β)(a~c) and β’a + (n − β’)(a~c), the cyclic word Λ<sub>2</sub> = the pattern of β and β’ must be a single-period mos. Similarly, Λ<sub>3</sub> = the pattern of γ and γ’ is a single-period mos. | Suppose Q = (α, β, γ) ≠ R = (α, β’, γ’) are the two ''k''-steps in ''S'' that project to P. Then T = (α’, β’’, γ’’) projects to I. Here the values in each component differ by at most 1, and α ≠ α’. Then the cyclic word Λ<sub>1</sub> formed by the a-components of the ''k''-steps in P is α...αα’. Since Σ<sub>2</sub> is a single-period mos pattern of βb + (''n'' − β)(a~c) and β’a + (''n'' − β’)(a~c), the cyclic word Λ<sub>2</sub> = the pattern of β and β’ must be a single-period mos. Similarly, Λ<sub>3</sub> = the pattern of γ and γ’ is a single-period mos. | ||
Suppose Λ<sub>2</sub> is the mos λβ μβ’. Then Λ<sub>3</sub> is the mos (λ ± 1)γ (μ ∓ 1)γ’. Since neither Λ<sub>2</sub> nor Λ<sub>3</sub> are multimosses, and at least one of μ and (μ ∓ 1) are even, it is now immediate that n is odd. | Suppose Λ<sub>2</sub> is the mos λβ μβ’. Then Λ<sub>3</sub> is the mos (λ ± 1)γ (μ ∓ 1)γ’. Since neither Λ<sub>2</sub> nor Λ<sub>3</sub> are multimosses, and at least one of μ and (μ ∓ 1) are even, it is now immediate that ''n'' is odd. | ||
Either β’’ = β or β’’ = β’. Assume β’’ = β’. Then γ’’ = γ, and Λ<sub>3</sub> is (λ + 1)γ (μ − 1)γ’. Also assume that the first k-step in Σ is Q. Then we have: | Either β’’ = β or β’’ = β’. Assume β’’ = β’. Then γ’’ = γ, and Λ<sub>3</sub> is (λ + 1)γ (μ − 1)γ’. Also assume that the first ''k''-step in Σ is Q. Then we have: | ||
1 … n | 1 … n | ||
Σ = Q W(Q, R) T | Σ = Q W(Q, R) T | ||
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Λ<sub>2</sub> = β W(β, β’) β’ | Λ<sub>2</sub> = β W(β, β’) β’ | ||
Λ<sub>3</sub> = γ W(γ, γ’) γ | Λ<sub>3</sub> = γ W(γ, γ’) γ | ||
where W = W(x, y) is a word in two | where W = W(''x'', ''y'') is a word in two variables ''x'' and ''y'', of length ''n'' − 2. | ||
===== Case analysis ===== | ===== Case analysis ===== | ||
Since, by our assumption, Λ<sub>3</sub> has two γ’s in a row, Λ<sub>3</sub> must have more γ than γ’, so μ − 1 < n/2. Since Λ<sub>3</sub> is a mos, μ − 1 ≥ 1. So we have 2 ≤ μ ≤ ceil(n/2). | Since, by our assumption, Λ<sub>3</sub> has two γ’s in a row, Λ<sub>3</sub> must have more γ than γ’, so μ − 1 < ''n''/2. Since Λ<sub>3</sub> is a mos, μ − 1 ≥ 1. So we have 2 ≤ μ ≤ ceil(''n''/2). | ||
We have three cases to consider: | We have three cases to consider: | ||
'''Case 1''': μ = 2, i.e. Λ<sub>2</sub> is the mos (n − 2)β 2β’. | '''Case 1''': μ = 2, i.e. Λ<sub>2</sub> is the mos (''n'' − 2)β 2β’. | ||
For Λ<sub>2</sub> to be a mos, the first, and only, occurrence of R must be at either f = floor(n/2) or ceil(n/2). We may assume that it is at f; otherwise flip the chain and reindex the words to start at | For Λ<sub>2</sub> to be a mos, the first, and only, occurrence of R must be at either ''f'' = floor(''n''/2) or ceil(''n''/2). We may assume that it is at ''f''; otherwise flip the chain and reindex the words to start at 2''f''. | ||
1 … f … 2f n | 1 … f … 2f n | ||
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Λ<sub>3</sub> = γ … γ γ’ γ … γ γ | Λ<sub>3</sub> = γ … γ γ’ γ … γ γ | ||
We need only consider stacks up to f-many k-steps. Either: | We need only consider stacks up to ''f''-many ''k''-steps. Either: | ||
# the stack has only Q | # the stack has only copies of Q and R; or | ||
# the stack has one T and does not contain any R (since it’s more than f − 1 generators away). | # the stack has one T and does not contain any R (since it’s more than ''f'' − 1 generators away). | ||
These give exactly three distinct sizes for every interval class. Hence S is SV3. | These give exactly three distinct sizes for every interval class. Hence ''S'' is SV3. | ||
In this case S has two chains of | In this case S has two chains of Q, one with floor(''n''/2) notes and and one with ceil(''n''/2) notes. Every instance of Q must be a ''k''-step, since by '''Z'''-linear independence Q = αa + βb + γc is the only way to write Q in the basis (a, b, c); so ''S'' is well-formed with respect to Q. Thus ''S'' also satisfies the generator-offset property with generator Q. | ||
'''Case 2:''' μ ≥ ceil(n/2), i.e. Λ<sub>2</sub> has fewer β than β’. | '''Case 2:''' μ ≥ ceil(''n''/2), i.e. Λ<sub>2</sub> has fewer β than β’. | ||
Since Λ<sub>3</sub> has more β than β’, Λ<sub>2</sub> is floor(n/2)β ceil(n/2)β’, and Λ<sub>3</sub> is ceil(n/2)γ floor(n/2)γ’. There is a unique mode of ceil(n/2)γ floor(n/2)γ’ that both begins and ends with γ, namely γγ’γγ’…γγ’γ. Thus Λ<sub>2</sub> is ββ’ββ’…ββ’β’. It is now easy to see that if the number of k-steps stacked is odd, then there are two sizes that do not contain T and one size that contains T; if the number of k-steps stacked is even, then there is one size that does not contain T and two sizes that contain T. Hence S is SV3. | Since Λ<sub>3</sub> has more β than β’, Λ<sub>2</sub> is floor(''n''/2)β ceil(''n''/2)β’, and Λ<sub>3</sub> is ceil(''n''/2)γ floor(''n''/2)γ’. There is a unique mode of ceil(''n''/2)γ floor(''n''/2)γ’ that both begins and ends with γ, namely γγ’γγ’…γγ’γ. Thus Λ<sub>2</sub> is ββ’ββ’…ββ’β’. It is now easy to see that if the number of ''k''-steps stacked is odd, then there are two sizes that do not contain T and one size that contains T; if the number of ''k''-steps stacked is even, then there is one size that does not contain T and two sizes that contain T. Hence ''S'' is SV3. | ||
In this case we have Σ = QRQR…QRT, and S is well-formed with respect to the generator Q + R, thus S satisfies the generator-offset property. | In this case we have Σ = QRQR…QRT, and S is well-formed with respect to the generator Q + R, thus ''S'' satisfies the generator-offset property. | ||
'''Case 3:''' 3 ≤ μ ≤ floor(n/2). | '''Case 3:''' 3 ≤ μ ≤ floor(''n''/2). | ||
Λ<sub>2</sub> has a chunk of βs (after the first β’) of size x = either floor(n/μ) (≥ floor(n/floor(n/2)) = 2) or ceil(n/μ) (= floor(n/μ) + 1). Hence Λ<sub>3</sub> has a chunk of γs of that same size. Λ<sub>3</sub> also has a chunk that | Λ<sub>2</sub> has a chunk of βs (after the first β’) of size ''x'' = either floor(''n''/μ) (≥ floor(''n''/floor(''n''/2)) = 2) or ceil(''n''/μ) (= floor(''n''/μ) + 1). Hence Λ<sub>3</sub> has a chunk of γs of that same size. Λ<sub>3</sub> also has a chunk that crosses index ''n'', which must be of size ''y'' = at least 2*(floor(''n''/μ) − 1) + 1 (Λ<sub>3</sub> might have chunks of size floor(''n''/μ) − 1 and floor(''n''/μ) instead) = 2*floor(''n''/μ) − 1, and at most 2*(floor(''n''/μ) + 1) + 1 = 2*floor(''n''/μ) + 3 (if Λ<sub>3</sub> has chunks of size floor(''n''/μ) and ceil(''n''/μ)). The difference between the chunk sizes is ''y'' − ''x'', which must be 0 or 1, since Λ<sub>3</sub> is a mos. We thus have the following subcases: (In the following, chunk of Λ<sub>2</sub> means chunk of β, and chunk of Λ<sub>3</sub> means chunk of γ.) | ||
'''Case 3.1:''' (x, y) = (floor(n/μ), 2*floor(n/μ) − 1). | '''Case 3.1:''' (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) − 1). | ||
Since y − x = floor(n/μ) − 1 and floor(n/μ) ≥ 2, we have: x = floor(n/μ) = 2 and y − x = 1; hence y = 2*floor(n/μ) − 1 = 3. The chunk in Λ<sub>3</sub> whose size is 3 is made from two chunks in Λ<sub>2</sub> of size 1. (So Λ<sub>2</sub> has chunks of size 1 and 2, and Λ<sub>3</sub> has chunks of size 2 and 3.) | Since ''y'' − ''x'' = floor(''n''/μ) − 1 and floor(''n''/μ) ≥ 2, we have: ''x'' = floor(''n''/μ) = 2 and ''y'' − ''x'' = 1; hence ''y'' = 2*floor(''n''/μ) − 1 = 3. The chunk in Λ<sub>3</sub> whose size is 3 is made from two chunks in Λ<sub>2</sub> of size 1. (So Λ<sub>2</sub> has chunks of size 1 and 2, and Λ<sub>3</sub> has chunks of size 2 and 3.) | ||
Λ<sub>2</sub> has two consecutive chunks of size 1. Since chunk sizes form a mos, Λ<sub>2</sub> has more chunks of size 1 than it has chunks of size 2. | Λ<sub>2</sub> has two consecutive chunks of size 1. Since chunk sizes form a mos, Λ<sub>2</sub> has more chunks of size 1 than it has chunks of size 2. | ||
Use w[i : j] to denote the slice of the cyclic word w that includes both endpoints, i.e. the word w[i] w[i + 1] ... w[j] where indices are taken to be elements of '''Z'''/n'''Z'''. Λ<sub>2</sub> has only two chunks of size 1, Λ<sub>2</sub>[(n − 1) : (n − 1)] and Λ<sub>2</sub>[1 : 1], since otherwise Λ<sub>3</sub> would have a chunk of size 1 within Λ<sub>3</sub>[1 : (n − 1)]. Thus Λ<sub>2</sub> has exactly one chunk of size 2. Thus Λ<sub>2</sub> = ββ’βββ’ββ’ and Λ<sub>3</sub> = γγ’γγγ’γγ. Thus we have: | Use ''w''[''i'' : ''j''] to denote the slice of the cyclic word ''w'' that includes both endpoints, i.e. the word ''w''[''i''] ''w''[''i'' + 1] ... ''w''[''j''] where indices are taken to be elements of '''Z'''/''n'''''Z'''. Λ<sub>2</sub> has only two chunks of size 1, Λ<sub>2</sub>[(''n'' − 1) : (''n'' − 1)] and Λ<sub>2</sub>[1 : 1], since otherwise Λ<sub>3</sub> would have a chunk of size 1 within Λ<sub>3</sub>[1 : (''n'' − 1)]. Thus Λ<sub>2</sub> has exactly one chunk of size 2. Thus Λ<sub>2</sub> = ββ’βββ’ββ’ and Λ<sub>3</sub> = γγ’γγγ’γγ. Thus we have: | ||
1 2 3 4 5 6 7 | 1 2 3 4 5 6 7 | ||
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Λ<sub>3</sub> = γ γ’ γ γ γ’ γ γ | Λ<sub>3</sub> = γ γ’ γ γ γ’ γ γ | ||
Suppose a step of S is reached by stacking t-many k-steps. We have three cases after accounting for equave complements: | Suppose a step of ''S'' is reached by stacking ''t''-many ''k''-steps. We have three cases after accounting for equave complements: | ||
# t = 1: S is equivalent to abacaba. | # ''t'' = 1: ''S'' is equivalent to abacaba. | ||
# t = 2: S is QR QQ RQ TQ RQ QR QT => S is equivalent to abacaba. | # ''t'' = 2: ''S'' is QR QQ RQ TQ RQ QR QT => ''S'' is equivalent to abacaba. | ||
# t = 3: S is QRQ QRQ TQR QQR QTQ RQQ RQT => S is equivalent to abacaba. | # ''t'' = 3: ''S'' is QRQ QRQ TQR QQR QTQ RQQ RQT => ''S'' is equivalent to abacaba. | ||
(This also implies S is SV3.) | (This also implies ''S'' is SV3.) | ||
'''Case 3.2''': (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) − 1) is impossible because | '''Case 3.2''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ) − 1) is impossible because ''y'' = 2*floor(''n''/μ) − 1 can only occur if Λ<sub>3</sub> has chunks of size floor(''n''/μ) − 1 and floor(''n''/μ), which contradicts the size of ''x''. | ||
'''Case 3.3''': (x, y) = (floor(n/μ) + 1, 2*floor(n/μ)) is similarly impossible; y = 2*floor(n/μ) can only occur if Λ<sub>3</sub> has chunks of size floor(n/μ) − 1 and floor(n/μ), which contradicts the size of x. | '''Case 3.3''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ)) is similarly impossible; ''y'' = 2*floor(''n''/μ) can only occur if Λ<sub>3</sub> has chunks of size floor(''n''/μ) − 1 and floor(''n''/μ), which contradicts the size of ''x''. | ||
The remaining cases are all impossible because they imply y − x ≥ 2: | The remaining cases are all impossible because they imply ''y'' − ''x'' ≥ 2: | ||
* '''Case 3.4''': (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) + 1) | * '''Case 3.4''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ) + 1) | ||
* '''Case 3.5''': (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) + 2) | * '''Case 3.5''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ) + 2) | ||
* '''Case 3.6''': (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) + 3) | * '''Case 3.6''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ) + 3) | ||
* '''Case 3.7''': (x, y) = (floor(n/μ), 2*floor(n/μ)) | * '''Case 3.7''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ)) | ||
* '''Case 3.8''': (x, y) = (floor(n/μ), 2*floor(n/μ) + 1) | * '''Case 3.8''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 1) | ||
* '''Case 3.9''': (x, y) = (floor(n/μ), 2*floor(n/μ) + 2) | * '''Case 3.9''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 2) | ||
* '''Case 3.10''': (x, y) = (floor(n/μ), 2*floor(n/μ) + 3) | * '''Case 3.10''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 3) | ||
== Open conjectures == | == Open conjectures == | ||