The faces of the hexeract
Partial hexagrams and the f-vector of the 6-cube
Mark which lines you leave undetermined and watch the face of the 6-cube they stand for light up.
The faces of the 6-cube are the partial hexagrams: words of six symbols over yin, yang and undetermined. The vertices (nothing undetermined) are the 64 complete hexagrams; the edges (one undetermined line) are our 192 mutations; the whole solid (all six undetermined) is the fully open hexagram. A face of dimension k can be completed in 2ᵏ ways, and there are C(6,k)·2⁶⁻ᵏ of them.
tap a line to cycle yin → yang → undetermined
f_k = C(6,k)·2^(6−k), verified by formula and by direct enumeration of the 3⁶ words
The faces add up to 729 = 3⁶: three options per line, the expansion of (2 + x)⁶ at x = 1. It is the same 729 as the Kun/Qian ratio of the Markov chain, and the reason is structural in both cases: three states per line (there old yin, young, old yang; here yin, yang, undetermined), not a numerological coincidence. The Euler characteristic closes the picture: the alternating sum of the boundary faces gives 0 (a 5-sphere) and, adding the solid, 1 (a contractible ball).
The faces of the 6-cube are the partial hexagrams (words over yin, yang and undetermined): the number of faces of dimension k is f_k = C(6,k) 2^(6-k), verified by formula and by direct enumeration of {0,1,*}^6; the f-vector is 64, 192, 240, 160, 60, 12, 1.
verificar_hexeractoThe faces add up to 3^6 = 729 (three options per line, the generating function (2+x)^6), the same number as the Kun/Qian ratio of the Markov chain; and the Euler characteristic is 0 on the boundary (a 5-sphere) and 1 on the whole solid (contractible).
verificar_hexeractoThe types of claim and the full bibliography (APA) are in Foundations.