关于九宫数的一点研究

九宫旋转1,8,3,4,9,2,7,6,针对这个周期序列计算它的序列公式,结果非常复杂:

道家阴符派博客--关于九宫数的一点研究--九宫

1/8 cos(1/4 π (n - 8)) + 1/8 cos(1/2 π (n - 8)) + 1/8 cos(3/4 π (n - 8)) + 1/8 cos(π (n - 8)) + 1/8 cos(5/4 π (n - 8)) + 1/8 cos(3/2 π (n - 8)) + 1/8 cos(7/4 π (n - 8)) + 1/4 cos(1/4 π (n - 7)) + 1/4 cos(1/2 π (n - 7)) + 1/4 cos(3/4 π (n - 7)) + 1/4 cos(π (n - 7)) + 1/4 cos(5/4 π (n - 7)) + 1/4 cos(3/2 π (n - 7)) + 1/4 cos(7/4 π (n - 7)) - 3/8 cos(1/4 π (n - 6)) - 3/8 cos(1/2 π (n - 6)) - 3/8 cos(3/4 π (n - 6)) - 3/8 cos(π (n - 6)) - 3/8 cos(5/4 π (n - 6)) - 3/8 cos(3/2 π (n - 6)) - 3/8 cos(7/4 π (n - 6)) + 1/2 cos(1/4 π (n - 5)) + 1/2 cos(1/2 π (n - 5)) + 1/2 cos(3/4 π (n - 5)) + 1/2 cos(π (n - 5)) + 1/2 cos(5/4 π (n - 5)) + 1/2 cos(3/2 π (n - 5)) + 1/2 cos(7/4 π (n - 5)) - 1/8 cos(1/4 π (n - 4)) - 1/8 cos(1/2 π (n - 4)) - 1/8 cos(3/4 π (n - 4)) - 1/8 cos(π (n - 4)) - 1/8 cos(5/4 π (n - 4)) - 1/8 cos(3/2 π (n - 4)) - 1/8 cos(7/4 π (n - 4)) - 1/4 cos(1/4 π (n - 3)) - 1/4 cos(1/2 π (n - 3)) - 1/4 cos(3/4 π (n - 3)) - 1/4 cos(π (n - 3)) - 1/4 cos(5/4 π (n - 3)) - 1/4 cos(3/2 π (n - 3)) - 1/4 cos(7/4 π (n - 3)) + 3/8 cos(1/4 π (n - 2)) + 3/8 cos(1/2 π (n - 2)) + 3/8 cos(3/4 π (n - 2)) + 3/8 cos(π (n - 2)) + 3/8 cos(5/4 π (n - 2)) + 3/8 cos(3/2 π (n - 2)) + 3/8 cos(7/4 π (n - 2)) - 1/2 cos(1/4 π (n - 1)) - 1/2 cos(1/2 π (n - 1)) - 1/2 cos(3/4 π (n - 1)) - 1/2 cos(π (n - 1)) - 1/2 cos(5/4 π (n - 1)) - 1/2 cos(3/2 π (n - 1)) - 1/2 cos(7/4 π (n - 1)) 

化简下来,可得

1/2 Cos[n π] ((1+Sqrt[2]) Cos[(n π)/4]-(-1+Sqrt[2]) Cos[(3 n π)/4]+3 (1+Sqrt[2]) Sin[(n π)/4]+3 (-1+Sqrt[2]) Sin[(3 n π)/4])  

如果要在九宫上,试图直接计算先天八卦序列也是很复杂的,会变成如下样子:

(-3 Cos[((-8 + n) Pi)/4])/8 - (3 Cos[((-8 + n) Pi)/2])/8 - (3 Cos[(3 (-8 + n) Pi)/4])/8 - (3 Cos[(-8 + n) Pi])/8 - (3 Cos[(5 (-8 + n) Pi)/4])/8 - (3 Cos[(3 (-8 + n) Pi)/2])/8 - (3 Cos[(7 (-8 + n) Pi)/4])/8 + (3 Cos[((-7 + n) Pi)/4])/8 + (3 Cos[((-7 + n) Pi)/2])/8 + (3 Cos[(3 (-7 + n) Pi)/4])/8 + (3 Cos[(-7 + n) Pi])/8 + (3 Cos[(5 (-7 + n) Pi)/4])/8 + (3 Cos[(3 (-7 + n) Pi)/2])/8 + (3 Cos[(7 (-7 + n) Pi)/4])/8 - Cos[((-6 + n) Pi)/4]/2 - Cos[((-6 + n) Pi)/2]/2 - Cos[(3 (-6 + n) Pi)/4]/2 - Cos[(-6 + n) Pi]/2 - Cos[(5 (-6 + n) Pi)/4]/2 - Cos[(3 (-6 + n) Pi)/2]/2 - Cos[(7 (-6 + n) Pi)/4]/2 - Cos[((-5 + n) Pi)/4]/8 - Cos[((-5 + n) Pi)/2]/8 - Cos[(3 (-5 + n) Pi)/4]/8 - Cos[(-5 + n) Pi]/8 - Cos[(5 (-5 + n) Pi)/4]/8 - Cos[(3 (-5 + n) Pi)/2]/8 - Cos[(7 (-5 + n) Pi)/4]/8 - Cos[((-4 + n) Pi)/4]/4 - Cos[((-4 + n) Pi)/2]/4 - Cos[(3 (-4 + n) Pi)/4]/4 - Cos[(-4 + n) Pi]/4 - Cos[(5 (-4 + n) Pi)/4]/4 - Cos[(3 (-4 + n) Pi)/2]/4 - Cos[(7 (-4 + n) Pi)/4]/4 + Cos[((-3 + n) Pi)/4]/2 + Cos[((-3 + n) Pi)/2]/2 + Cos[(3 (-3 + n) Pi)/4]/2 + Cos[(-3 + n) Pi]/2 + Cos[(5 (-3 + n) Pi)/4]/2 + Cos[(3 (-3 + n) Pi)/2]/2 + Cos[(7 (-3 + n) Pi)/4]/2 + Cos[((-2 + n) Pi)/4]/4 + Cos[((-2 + n) Pi)/2]/4 + Cos[(3 (-2 + n) Pi)/4]/4 + Cos[(-2 + n) Pi]/4 + Cos[(5 (-2 + n) Pi)/4]/4 + Cos[(3 (-2 + n) Pi)/2]/4 + Cos[(7 (-2 + n) Pi)/4]/4 + Cos[((-1 + n) Pi)/4]/8 + Cos[((-1 + n) Pi)/2]/8 + Cos[(3 (-1 + n) Pi)/4]/8 + Cos[(-1 + n) Pi]/8 + Cos[(5 (-1 + n) Pi)/4]/8 + Cos[(3 (-1 + n) Pi)/2]/8 + Cos[(7 (-1 + n) Pi)/4]/8

这实际在用三角函数来体现这个周期函数,进行了变换,总共有54项,如果仅考虑描述一半图形的话,它们仅是180度颠倒,所以考虑以四组卦为序,并设:

A=Cos[1/2π(-4+x)]
B=Cos[π(-4+x)]
C=Cos[3/2π(-4+x)]
D=Cos[1/2π(-3+x)]
E=Cos[π(-3+x)]
F=Cos[3/2π(-3+x)]
G=Cos[1/2π(-2+x)]
H=Cos[π(-2+x)]
I=Cos[3/2π(-2+x)]
J=Cos[1/2π(-1+x)]
K=Cos[π(-1+x)]
L=Cos[3/2π(-1+x)]

显然从A~L,初步可以认为这采用12个维度的描述,是从x-1、x-2、x-3、x-4,分别取乘1/2、1、3/2,然后得到的角度值,由于最终结果从-4至4,而3/2倍数的最大值为1.5,所以相当于被分解成了多个0~1.5间的小片断进行相加,所以两至三个为1.5的值便可以构成3~4.5之间的波动,然后再有两个修正值应当足够表达,所以化简可以到四到五个项,然而化简不利于观察内部结构,此处不化简。

坎艮震巽= -1    -1/4A-1/4B-1/4C-1/2D-1/2E-1/2F+3/4G+3/4H+3/4I-J-K-L
离坤兑乾=  1    +1/4A+1/4B+1/4C+1/2D+1/2E+1/2F-3/4G-3/4H-3/4I+J+K+L
震巽离坤=-(1/2) -3/4A-3/4B-3/4C+D+E+F-1/4G-1/4H-1/4I-1/2J-1/2K-1/2L
兑乾坎艮= 1/2   +3/4A+3/4B+3/4C-D-E-F+1/4G+1/4H+1/4I+1/2J+1/2K+1/2L
巽离坤兑=1/2    +1/2A+1/2B+1/2C-3/4D-3/4E-3/4F+G+H+I-1/4J-1/4K-1/4L
乾坎艮震=-(1/2) -1/2A-1/2B-1/2C+3/4D+3/4E+3/4F-G-H-I+1/4J+1/4K+1/4L
艮震巽离=1      +A+B+C-1/4D-1/4E-1/4F-1/2G-1/2H-1/2I+3/4J+3/4K+3/4L
坤兑乾坎=-1     -A-B-C+1/4D+1/4E+1/4F+1/2G+1/2H+1/2I-3/4J-3/4K-3/4L

在格式上有意将常数项独立了出来,这样便可以看到后面的项是如何匹配的,同样将先天往后天转化也排出来:

离乾艮震= 3/2   -1/2A-1/2B-1/2C+3/4D+3/4E+3/4F+1/4G+1/4H+1/4I+J+K+L
艮震离乾= 3/2   +1/4A+1/4B+1/4C+D+E+F-1/2G-1/2H-1/2I+3/4J+3/4K+3/4L
震离乾艮= 3/2   +3/4A+3/4B+3/4C+1/4D+1/4E+1/4F+G+H+I-1/2J-1/2K-1/2L
乾艮震离= 3/2   +A+B+C-1/2D-1/2E-1/2F+3/4G+3/4H+3/4I+1/4J+1/4K+1/4L

比较下来会得到有趣的结果:

坎艮震巽= -1    -1/4A-1/4B-1/4C-1/2D-1/2E-1/2F+3/4G+3/4H+3/4I-J-K-L
离坤兑乾= 1     +1/4A+1/4B+1/4C+1/2D+1/2E+1/2F-3/4G-3/4H-3/4I+J+K+L
艮震离乾= 3/2   +1/4A+1/4B+1/4C+D+E+F-1/2G-1/2H-1/2I+3/4J+3/4K+3/4L

震巽离坤=-(1/2) -3/4A-3/4B-3/4C+D+E+F-1/4G-1/4H-1/4I-1/2J-1/2K-1/2L
兑乾坎艮= 1/2   +3/4A+3/4B+3/4C-D-E-F+1/4G+1/4H+1/4I+1/2J+1/2K+1/2L
震离乾艮= 3/2   +3/4A+3/4B+3/4C+1/4D+1/4E+1/4F+G+H+I-1/2J-1/2K-1/2L

巽离坤兑=1/2    +1/2A+1/2B+1/2C-3/4D-3/4E-3/4F+G+H+I-1/4J-1/4K-1/4L
乾坎艮震=-(1/2) -1/2A-1/2B-1/2C+3/4D+3/4E+3/4F-G-H-I+1/4J+1/4K+1/4L
离乾艮震= 3/2   -1/2A-1/2B-1/2C+3/4D+3/4E+3/4F+1/4G+1/4H+1/4I+J+K+L

艮震巽离=1      +A+B+C-1/4D-1/4E-1/4F-1/2G-1/2H-1/2I+3/4J+3/4K+3/4L
坤兑乾坎=-1     -A-B-C+1/4D+1/4E+1/4F+1/2G+1/2H+1/2I-3/4J-3/4K-3/4L
乾艮震离=3/2    +A+B+C-1/2D-1/2E-1/2F+3/4G+3/4H+3/4I+1/4J+1/4K+1/4L

其中的轨迹表现得便很明显了。

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