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通过我几个月来对珠路理论的研究，5 R+ y* d$ ? k6 I7 G/ O1 }! e5 r- C _$ ?; g% i6 F
5 M, J; s) O* q! s, O
发现了一个有趣的现象，
, i# |. K/ W' v" _' U @' p每一条大路，
$ g; r0 K, o D0 E按2珠路排列，有2种不同的路数1 _% l2 j Y8 @" u) {# f8 a2 _' A) t5 P+ C- g) {! B
按3主路排列，有3种不同的路数& n( |2 F1 f; ?, q' L3 A: Z4 z% q$ N& n7 \% i( X6 }+ {$ L
按4珠路排列，有4种不同的路数
$ E0 V0 `3 g0 f ~5 A& I按N珠路排列，有N种不同的路数。: n1 f3 R9 @ p% ]* C
( G& {3 H; A& C% C- P p- ~什么意思呢，可能大家一时不清楚，
; l" B" K" a0 J: M我举一个例子，假设 B= 1 ， P=2，T暂且忽略吧。7 f# f3 c1 ~2 l
9 `2 q3 Y" k) I& v# S* M假设大路开这样的路：! z ~0 G3 v. K' o5 A4 ]" }5 w# u
3 J8 f/ j1 _" F6 o! K$ M12121212121212121212121212121212, j0 j; e6 p* q3 Q5 h @+ g5 c" W* `1 T; z2 h& y3 ^' }) n2 `
够极端了吧
, q0 \& J7 i0 b% I按2珠路，就是BPBPBPBPBP....., m8 f) O! x+ S9 L2 x) F0 b. o4 W7 J5 o3 H6 v+ l
如果我们去掉第一口，就会出现完全相反的结果：, c4 Y9 {1 N) I5 v
21212121212121212121212121212121 ]+ S8 U! X# U4 c
变成了PBPBPBPBPB.......1 u5 g6 c6 F/ L4 g! f7 Z# E+ C' }4 b4 r' t+ I l: H3 I
如果我们再去掉一口，又返回第一种情况了。! [( g, L- u! O3 @# G4 D) O$ Z: y9 y' n w1 O1 D8 a: ?( e
所以每一条大路，! R6 V# q$ h. c; z% O- l$ i7 A& s" j: ^; u8 B
按2珠路排列，有2种不同的路数。
% ^% T9 V+ o y1 f' [2 a再举一个列子：
$ P9 h; U* Y6 O% H N大路：122122122122122122。。。
9 T T$ `/ b& e% R# L7 b- ?按三珠路排列：' v3 j! O7 S7 G$ u4 ?. y
122，122，122，122，122，122。。。$ R+ F: q" g2 E0 {5 F
去掉第一口，变成：6 G2 V! \" L' c" k: d; X) G6 K- A- h: R: G/ I1 t
221，221，221，221，221，+ R$ P2 d/ f7 Z, P5 M
1 I* F$ |3 h+ Z4 P/ v9 T' J去掉前2口，变成：3 v1 q) z( ?2 f! }) U/ W
1 T) R' e0 K$ H2 s2 Y, @212，212，212，212，212，: y& l, B; I6 e- m. J# c) `
6 q+ I- T. M! [: E6 j2 ?去掉3口，又返回122，122，122，了
T# y% P- ]: Y5 Y/ P所以每一条大路，( j7 U B* W/ d @
按3珠路排列，有3种不同的路数。, \2 t& p) ]1 t; `
同理：按N珠路排列，有N种不同的路数。& e* q! _& ?3 c' S# t8 v
我提出这个的意义在于：" v) X/ h3 }* D1 {5 J! a1 s. O* q1 ]* ~$ l1 U1 I7 T/ T/ t
字串83 L: R) G1 V- s2 Y+ Y
1 }7 D% y4 s, l" }" m A* F2 [* n- p, d3 [6 U. s5 J' [0 {1 A" m' k( @
$ S2 e5 F" B* V, {, O1、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。2 g: X/ W# h7 N" o7 |9 l* |
$ {$ W8 `# u ]& E) `2、为三多理论提供了下注的多面性奠定基础。
. M0 C4 K8 w- \$ L0 }- p# Q) r3、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.; T3 T+ k y( [. w2 H$ b
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