, Yin = 
(Originally published in Winter 2008 issue of JOAC.)
Abstract: A new scheme for representing emblematic lines is introduced, based on the topological principles behind the Cantor Set. This scheme is applicable not only to the classical group of 4 emblematic lines and the recent finite extensions of the "moving line" concept, but also to transfinite lines.
In a recent discussion on the way emblematic lines (of both the static and moving kind) are represented calligraphically, the pictures of the solid yang line and the broken yin line were drawn on the board. In a moment of insight, I mused about the yin line, "Hmmm, looks like someone removed the middle third of the yang line..."
Yang = , Yin =
Being a topologist, it didn't take long to start thinking about the classical Cantor Set or K, a well known example of a "pathological" topological space. Starting with a closed line segment [0, 1], the first step is to remove the middle third of this segment, the open interval (1/3, 2/3). This leaves 2 smaller segments, namely [0, 1/3] and [2/3, 1], separated by a big gap. In the next step, the middle thirds of these segments are removed, leaving the segments [0, 1/9], [2/9, 1/3], [2/3, 7/9] and [8/9, 1]. This process of removing the middle thirds can be interated a countably infinite number of times, producing in the limit the Cantor Set. This space has a number of curious properties. First of all, unlike the solid, continuous unit interval we started with (it's not without reason the real number line is called the continuum), the Cantor Set resembles more a dust of distinct points all separated from each other by gaps: this space is totally disconnected. It appears to be only an infinitesimal remnant of the unit interval, having an area of exactly zero. However, in a very rigorous sense, this dust has exactly as many points as does the entire unit interval.
In more precise mathematical terms, let U equal [0, 1]. For each non-negative integer n, let T(n) be [0, 1] with all open intervals removed of the form:
( (3*m+1)/3**(n+1), (3*m+2)/3**(n+1) ),
where m runs from 0 to (3**n) -1 inclusive.
The Cantor Set K is the intersection of U and all the T(n) sets. We are guaranteed that K is non-empty because U is a compact space and all the T(n) are closed sets satisfying the Finite Intersection Property.
So how does K help us draw emblematic lines? Recall that the 4 classical lines (static/moving, yin/yang lines) can be represented as digrams of basic lines. As pointed out in the various ground breaking articles by McFnordland (listed in the bibliography), there are actually several distinct ways of mapping the emblematic lines to digrams. We'll be less concerned about the details here (especially the issue of whether "raw" digrams are meant or else the "journey" version, as defined in the McFnordland papers, an issue he called the interpretation problem) and focus on the broad concepts involved. It's become the modern tradition to equate yin = 0 and yang = 1, so the set {0, 1} is called the set of Basic Lines or BL. A digram is simply 2 basic lines stacked on top of each other, which we will represent as a function f(n) with domain {0, 1} and function values in BL. The values of n are "times" for a moving line, whether the initial value of the line (n=0) or the final value (n=1) after it has moved, or equivalently, the position of the basic line within the digram (top or bottom). There are 4 such functions.
Similarly, trigrams are functions f(n) with a domain of {0, 1, 2} since there are 3 lines in such a figure, while hexagrams have domain of {0, 1, 2, 3, 4, 5} for 6 lines. In fact, the number of lines is arbitrary and can even be infinite. These "transfinite lines" are simply functions from the non-negative integers to the basic lines. Our calligraphy problem reduces to how one would map these functions into some space of geometrical drawings.
The algorithm for making these drawings is really quite simple. Given a function f(n), assemble the collection of sets U and all the T(n) sets for which n is in the domain of f and f(n) = 0. The required drawing is just the intersection of all the sets in this collection. In other words, the choice of whether or not to remove the middle thirds at any given interation is conditional on the value of f at that step.
For a finite figure such as a digram, we typically describe the function by listing the sequence of values in order. For instance, f(0) = 1 and f(1) = 0 would be abbreviated (10). The drawing for yang earlier in this article corresponds to the sequence (11) under this drawing algorithm, while the yin line is generated by (01). The drawings for (10) and (11) are below. The reader should be able to generalize this to more complicated drawings. (Take note: these drawing sequences differ from the more traditional digrams for the emblematic lines.)
(10) = 
, (11) = 
Unlike the technique of using circles or X's to represent movement, the usage of Cantor Lines generalizes easily to more complicated cases than the digrams, including the transfinite case. In the later situation, it's important to consider certain mathematical details given that we are potentially taking the intersection of an infinite number of sets. We are guaranteed that such an intersection is in fact non-empty (that is, we actually draw something on the parchment) because K (which we know is non-empty) is a subset of any such intersection. Uniqueness of the mapping, that is, distinct functions map to distinct drawings, is also easy to prove.
Granted, for larger domains, these drawings become unwieldy, if only because the ink would run badly on the parchment, obliterating the finer details, but in an ideal geometric sense, they are a remarkable way to draw arbitrary moving lines.