Chord complexity: Difference between revisions

ArrowHead294 (talk | contribs)
ArrowHead294 (talk | contribs)
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$$
$$
\displaystyle T_s\left(x_1, x_2, \ldots, x_N\right) = \log B_s\left(x_1, x_2, \ldots, x_N\right) = \frac{1}{N} \log\left(x_1 \cdot x_2 \cdot \ldots \cdot x_N\right) - \frac{1}{s}\log N
\displaystyle T_s\left(x_1, x_2, \ldots, x_N\right) = \log B_s\left(x_1, x_2, \ldots, x_N\right) = \frac{\log\left(x_1 \cdot x_2 \cdot \ldots \cdot x_N\right)}{N} - \frac{\log N}{s}
$$
$$


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$$
$$
\displaystyle \log W_s\left(x_1, x_2, \ldots, x_N\right) = \log \max\left(x_1, x_2, \ldots, x_N\right) - \frac{1}{s} \log N
\displaystyle \log W_s\left(x_1, x_2, \ldots, x_N\right) = \log \max\left(x_1, x_2, \ldots, x_N\right) - \frac{\log N}{s}
$$
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$$
$$
\log \max\left(n, d\right) = \frac{1}{2}\log\left(n\cdot d\right) + \frac{1}{2} \left|\log\left(\frac{n}{d}\right)\right|
\log \max\left(n, d\right) = \frac{\log\left(n\cdot d\right) + \left|\log\left(\frac{n}{d}\right)\right|}{2}
$$
$$


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$$
$$
\displaystyle \log W_s\left(x_1, x_2, \ldots, x_N\right) = \log B_s\left(x_1, x_2, \ldots, x_N\right) + \frac{1}{N}\log x_N/x_1 + \frac{1}{N}\log x_N/x_2 + \ldots + \frac{1}{N}\log x_N/x_{N-1}
\displaystyle \log W_s\left(x_1, x_2, \ldots, x_N\right) = \log B_s\left(x_1, x_2, \ldots, x_N\right) + \frac{\log\frac{x_N}{x_1} + \log\frac{x_N}{x_2} + \ldots + \log\frac{x_N}{x_{N-1}}}{N}
$$
$$


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Now, it is interesting to look at how the above expression scales for intervals that are fairly small. It is relatively easy to see that this denominator will be maximized when {{nowrap|''b''/''a'' {{=}} 1}}, meaning the span is zero, so that the cosh term equals 1 and thus the denominator is 2, leaving only the numerator of {{nowrap|(''ab'')<sup>''s''/2</sup>}}. For relatively small intervals, we'll get something close to this, meaning span is irrelevant&mdash;perhaps what we want.
Now, it is interesting to look at how the above expression scales for intervals that are fairly small. It is relatively easy to see that this denominator will be maximized when {{nowrap|''b''/''a'' {{=}} 1}}, meaning the span is zero, so that the cosh term equals 1 and thus the denominator is 2, leaving only the numerator of {{nowrap|(''ab'')<sup>''s''/2</sup>}}. For relatively small intervals, we'll get something close to this, meaning span is irrelevant&mdash;perhaps what we want.


For relatively large intervals, on the other hand, the entire thing simply tends to {{nowrap|min(''a'', ''b'')<sup>''s''</sup>}}. Since we have the identity <math>\log \min(a, b) = \frac{\log(ab) - \left|\log\left(\frac{b}{a}\right)\right|}{2}</math>, this means meaning we are subtracting the span from the Tenney height, this is definitely not what we want.
For relatively large intervals, on the other hand, the entire thing simply tends to {{nowrap|min(''a'', ''b'')<sup>''s''</sup>}}. Since we have the identity
 
$$
\log \min(a, b) = \frac{\log(ab) - \left|\log\left(\frac{b}{a}\right)\right|}{2}
$$
 
this means meaning we are subtracting the span from the Tenney height, which is definitely not what we want.


So what we will do is simply modify our formula so that the behavior for relatively small intervals is preserved across the entire interval spectrum, thus "span-correcting" our original formula. Doing so, we simply keep the numerator (the "complexity" part) the same, while pretending that we have always plugged 1/1 into the denominator. Thus, we simply get
So what we will do is simply modify our formula so that the behavior for relatively small intervals is preserved across the entire interval spectrum, thus "span-correcting" our original formula. Doing so, we simply keep the numerator (the "complexity" part) the same, while pretending that we have always plugged 1/1 into the denominator. Thus, we simply get