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2 I& d4 t6 N& V1 P8 e
" V- O+ i9 f& P( e4 _2 c通过我几个月来对珠路理论的研究,发现了一个有趣的现象:每一条大路,按2珠路排列,有2种不同的路数;按3主路排列,有3种不同的路数;按4珠路排列,有4种不同的路数,按N珠路排列,有N种不同的路数何解?我举一个例子,假设 B= 1 , P=2,T暂且忽略吧。
. G; M& O" D& d* I8 }# s: m/ Y; L1 M! S! f+ L @% P9 V- A5 k
假设大路开这样的路:
2 j: {/ y& `) B9 f3 B& ~ ~3 e% \) f6 P$ z1 m
12121212121212121212121212121212。
4 f2 x7 z8 @2 o/ n5 E. U; Y7 v" l8 C3 H3 e& a( a% m
按2珠路,就是BPBPBPBPBP。
Y1 f! W# C! W3 Z! j* l: u( H, G ]
如果我们去掉第一口,就会出现完全相反的结果:0 U3 @ y7 Q/ i" F. j, {( a
2 W. X# u9 P% r& a: l1 k3 E5 V- U
21212121212121212121212121212121。
, e! m9 i4 W, J3 n) e6 s# s& N) Q3 b8 p, \/ [8 K
变成了PBPBPBPBPB。0 Y; \; @5 y. q2 [/ t* } c
6 B* q( B( C. S. X5 c 如果我们再去掉一口,又返回第一种情况了。
# @+ b" d6 d2 O
+ u; |: m F$ a) i3 G" U% d 所以每一条大路,按2珠路排列,有2种不同的路数。
9 R+ ~; o- a/ u5 I4 u o# F, F
6 V2 R/ d+ g& {2 s% y4 H: p 再举一个列子:
( I D( o, d% U# ]. v0 M. A+ `
' G6 @ z( x" X# s: p% n' D 大路:122122122122122122。
# _* B8 _& N4 y! r9 b7 d, T4 X$ a6 u8 _4 t9 n, ]
按三珠路排列:" V$ k* A! o8 I9 m& q+ Q
' [4 ?9 M% k' q8 t; |) ~ 122,122,122,122,122,122。
3 m" t3 Y; C1 w" H- @7 E0 Q4 P; C. W
去掉第一口,变成:
4 ~: A9 d# g) G+ d& d$ F8 d! c6 t$ W) o3 J1 C
221,221,221,221,221,去掉前2口,变成:( `$ F. D8 p6 Y& F7 c0 }' g! o9 L
0 L/ ^8 [# O/ |# C
212,212,212,212,212,去掉3口,又返回122,122,122,了所以每一条大路,按3珠路排列,有3种不同的路数。8 P* R+ h& P2 b0 m
9 Q1 d: w$ c% B' D- N
同理:按N珠路排列,有N种不同的路数。& k+ {6 w' M+ L
+ K1 {# L! R ~) d1 l# a
我提出这个的意义在于:7 ^( }0 E( a0 L- L4 ~
; O' {* F% r. ^5 Q( ~( j" }; Y- F( } 1、字串81、每靴的第一口为起点来编排二三珠路,与第2口,第3口开始的排列是不同的结果。/ g; B9 ], {( \/ b% Q1 u
6 o$ U6 I" _# a: `- n1 n6 j, g
2、为三多理论提供了下注的多面性奠定基础。9 W( K; R$ J( e+ K) N8 r: {
E, \8 i5 N: j: A8 e7 x2 G 3、某一靴牌,按第一口开始的珠路可能是 烂路,按第2口,第3口开始的珠路可能是上上路。0 |. h! g0 a; c6 V6 M
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