Octave reduction

Octave reduction is the process of replacing an interval by the unique equivalent interval between the unison and the octave. In practice, this is done by adding or subtracting octaves from the starting interval as necessary.

Practical methods

An easy way to find the reduced form of an interval is to use a specialized calculator such as xen-calc. This is especially useful when working with very complex ratios.

There are also several methods that can be followed. The choice of an appropriate method depends on the interval size measure being used: linear measures (e.g. frequency ratios), or logarithmic measures (e.g. scale steps or cents).

Linear measures

Stacking intervals expressed as ratios corresponds to multiplying those ratios. For instance, going up an octave means multiplying by 2, while going down an octave means dividing by 2.

1. If the starting interval is greater or equal to the unison (1) and less than the octave (2), it is already in reduced form.
2. If the starting interval is less than 1, multiply it by 2. Repeat until the resulting interval is greater than 1.
3. If the starting interval is greater than 2, divide it by 2. Repeat until the resulting interval is less than 2.

Examples:

• 3/4 is less than 1, so multiply by 2 to get 3/2.
• 7/2 is greater than 2, so divide by 2 to get 7/4.
• 4/1 is greater than 2, so divide by 2 to get 2/1, which is equal to 2, so divide by 2 to get 1/1.
• Adding 4 just perfect fifths (3/2) corresponds to (3/2)⁴, thus 81/16 (or 5.0625), which is greater than 2 octaves (2² = 4), but less than 3 octaves (2³ = 8), so divide by 2 twice to get 81/64.
• Subtracting a just perfect fourth (4/3) from a classic minor third 6/5 corresponds to 6/5 divided by 4/3, thus 9/10 (or 0.9). This interval is less than a unison (2⁰ = 1) but greater than one octave down (2^-1 = 1/2), so multiply by 2 once to get 9/5.

Logarithmic measures

Stacking intervals expressed with logarithmic measures corresponds to adding those measures. For instance, when working in cents, going up an octave means adding 1200 ¢, while going down an octave means subtracting 1200 ¢.

1. Find the logarithmic measure of the octave in the same unit as the one used for your starting interval; e.g. 1200 ¢, 19 steps of 19edo, 1900 r¢, etc.
2. If the starting interval is positive and less than the octave (e.g. 1200 ¢), it is already in reduced form.
3. If the starting interval is negative, add the octave. Repeat until the result is positive.
4. If the starting interval is greater than the octave, subtract the octave. Repeat until the result is less than the octave.

Examples:

• 1442¢ is greater than 1200 ¢, so subtract 1200 ¢ to get 242 ¢.
• In 31edo, the octave is 31 steps and the patent val of the fifth harmonic is 72 (steps). This interval is greater than the octave, so subtract 31 to get 41, so subtract 31 again to get 10.

General formulas

Linear measures

For a starting interval [math]\displaystyle{ r }[/math] expressed as a ratio, the reduced form [math]\displaystyle{ \text{red}(r) }[/math] of that interval can be found using this formula: [math]\displaystyle{ \text{red}(r) = r \cdot 2^{-\left\lfloor{\log_2 r}\right\rfloor} }[/math]­.

Example:

• Octave-reducing 4900/243 can be done by using the formula with [math]\displaystyle{ r = 4900/243 }[/math]:
  [math]\displaystyle{ \begin{align}\text{red}(4900/243) &= 4900/243 \cdot 2^{-\left\lfloor{\log_2 4900/243}\right\rfloor} \\ &= 4900/243 \cdot 2^{-\left\lfloor{4.33375\ldots}\right\rfloor} \\ &= 4900/243 \cdot 2^{-4} \\ &= 4900/243 \cdot 1/16 \\ &= 1225/972\end{align} }[/math]

Logarithmic measures

For a starting interval [math]\displaystyle{ l }[/math] and octave [math]\displaystyle{ e }[/math] expressed in the same units, the reduced form [math]\displaystyle{ \text{red}(l, e) }[/math] of that interval can be found using this formula: [math]\displaystyle{ \text{red}(l, e) = r \bmod e }[/math]­, where [math]\displaystyle{ \bmod }[/math] is the modulo operation.

Example:

• Octave-reducing 412 steps of 97edo can be done by using the formula with [math]\displaystyle{ r = 412 }[/math], [math]\displaystyle{ e = 97 }[/math]:
  [math]\displaystyle{ \begin{align}\text{red}(412, 97) &= 412 \bmod 97 \\ &= 24\end{align} }[/math]

This formula can also be written without the modulo operation: [math]\displaystyle{ \text{red}(l, e) = r - e\left\lfloor{l/e}\right\rfloor }[/math]­.

Example:

• Octave-reducing 412 steps of 97edo again:
  [math]\displaystyle{ \begin{align}\text{red}(412, 97) &= 412 - 97\left\lfloor{412/97}\right\rfloor \\ &= 412 - 97\left\lfloor{4.24742\ldots}\right\rfloor \\ &= 412 - 97 \cdot 4 \\ &= 412 - 388 \\ &= 24\end{align} }[/math]

Generalizations

Other equaves

Octave reduction is mainly used in the context of octave-equivalent tunings (eg. 12edo), where equivalent notes are separated by octaves. However, this operation can be generalized to any periodic tuning by replacing the octave by the interval of equivalence or equave of that tuning.

For example, tritave reduction is the analog of octave reduction in a tritave-equivalent tuning (eg. Bohlen–Pierce), where the equave is the tritave. Therefore, a tritave-reduced interval is always obtained through transposition by tritaves, and the reduced interval lies between the unison (1/1) and the tritave (3/1).

The general formula for an interval [math]\displaystyle{ r }[/math] and an equave [math]\displaystyle{ e }[/math] is as follows: [math]\displaystyle{ \text{red}(r, e) = r \cdot e^{-\left\lfloor{\log_e r}\right\rfloor} }[/math]­. Note that [math]\displaystyle{ e }[/math] is a variable and not Euler's number.

Examples:

• Consider a tritave-equivalent tuning; 7/9 is less than 3, so multiply by 3 to get 7/3.
• Consider a just perfect fifth-equivalent tuning; [math]\displaystyle{ \text{red}(81/64, 3/2) = 81/64 \cdot (3/2)^{-\left\lfloor{\log_{3/2} 81/64}\right\rfloor} = 1 }[/math]­.
• In the equal-tempered Bohlen–Pierce tuning, a tritave can be expressed as 1300 hekts and a BP fifth down as -500 hekts. This interval is less than the unison, so add 1300 hekts to get 800 hekts.

Balanced reduction

Balanced reduction is an alternate operation where the values are equally distributed around the unison instead of being situated between the unison and the octave (or the equave).

Examples:

• Balanced octave-reduction with ratios will lead to values greater than or equal to [math]\displaystyle{ \frac{1}{\sqrt{2}} }[/math], but less than [math]\displaystyle{ \sqrt{2} }[/math].
• Balanced octave-reduction with cents will lead to values greater than or equal to -600 ¢, but less than 600 ¢.
• Balanced tritave-reduction with ratios will lead to values greater than or equal to [math]\displaystyle{ \frac{1}{\sqrt{3}} }[/math], but less than [math]\displaystyle{ \sqrt{3} }[/math].

Here are some formulas for balanced reduction:

• Balanced octave-reduction of an interval [math]\displaystyle{ r }[/math] expressed as a ratio: [math]\displaystyle{ \text{reb}(r)=\frac{1}{\sqrt{2}} \text{red}(\sqrt{2} \cdot \text{red}(r)) }[/math].
• Balanced reduction of an interval [math]\displaystyle{ r }[/math] and an equave [math]\displaystyle{ e }[/math] expressed as ratios: [math]\displaystyle{ \text{reb}(r, e)=\frac{1}{\sqrt{e}} \text{red}(\sqrt{e} \cdot \text{red}(r, e), e) }[/math].
• Balanced reduction of an interval [math]\displaystyle{ l }[/math] and an equave [math]\displaystyle{ e }[/math] expressed as logarithmic measures in the same units: [math]\displaystyle{ \text{reb}(l, e)= \text{red}(\text{red}(l, e) + e/2, e) - e/2 }[/math].

See also

• Octave complement

External links

• https://forum.sagittal.org/viewtopic.php?p=1296 (thread on mathematical functions)