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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/ O
; l$ Y2 Z# x7 B$ w) O [) l7 l8 {3 s- A
发现了一个有趣的现象，
6 ~' M1 E3 k. r3 b$ _每一条大路，! u8 M0 f! y+ u. l* |* o+ J
按2珠路排列，有2种不同的路数1 _% l2 j Y8 @" u) {# f
# h" o& C& X/ u4 g) Z1 K; K按3主路排列，有3种不同的路数& n( |2 F1 f; ?, q' L3 A: Z4 z
' M8 o; E+ t. y8 U按4珠路排列，有4种不同的路数8 n: d$ P4 w5 u- K$ ~
按N珠路排列，有N种不同的路数。: n1 f3 R9 @ p% ]* C
2 C% a% Z1 g2 E" Y" \什么意思呢，可能大家一时不清楚，
5 E+ [) O# o% e7 N2 }7 Y我举一个例子，假设 B= 1 ， P=2，T暂且忽略吧。7 f# f3 c1 ~2 l+ G8 @/ a/ L9 S7 h* X/ E3 {
假设大路开这样的路：! z ~0 G3 v. K' o5 A4 ]" }5 w# u
. q6 x- | j4 g12121212121212121212121212121212, j0 j; e6 p* q3 Q5 h: @$ ]/ C2 L* b: b9 h+ M5 T
够极端了吧
/ v$ O) H+ u8 U按2珠路，就是BPBPBPBPBP....., m8 f) O! x+ S9 L2 x) F
- O8 Z; b3 r8 D6 G# J如果我们去掉第一口，就会出现完全相反的结果：
3 v; ^- D( \9 ]' H4 k! x21212121212121212121212121212121
, F d" F7 R/ \" d2 q0 |变成了PBPBPBPBPB.......1 u5 g6 c6 F/ L4 g! f7 Z# E+ C8 r" F. A! G9 G# S4 P; B
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' J1 p3 P$ i4 ?+ o
所以每一条大路，! R6 V# q$ h. c; z% O
* B3 b5 z8 j4 e" e5 d按2珠路排列，有2种不同的路数。% G( J+ [; M+ O' P
再举一个列子：4 ?# P O0 K( ]; S
大路：122122122122122122。。。
' K) ?( @! J5 A" X n, {: h按三珠路排列：& M. U7 h/ k8 ^; Q! Q: O
122，122，122，122，122，122。。。. p# b1 G: x4 |$ x2 e+ E
去掉第一口，变成：6 G2 V! \" L' c" k: d5 z9 B8 \7 Y! n, I' T8 d
221，221，221，221，221，+ R$ P2 d/ f7 Z, P5 M) `8 |* d5 J8 q4 M' s4 i
去掉前2口，变成：3 v1 q) z( ?2 f! }) U/ W
# Q2 T4 ^/ q- D8 I& ^$ Z212，212，212，212，212，: y& l, B; I6 e- m. J# c) `
% ?. Q& P8 Y. G. v2 c7 G5 d% t3 Q去掉3口，又返回122，122，122，了/ } z5 ?& x) _+ s' X* n
所以每一条大路，
# }! U% Y. E, ^2 ?! O! z' r: x按3珠路排列，有3种不同的路数。
2 Y- ?/ V1 A6 [3 H7 P3 g同理：按N珠路排列，有N种不同的路数。
, Z4 Q. |. n# k* j$ Z/ s1 @9 _5 p; D我提出这个的意义在于：" v) X/ h3 }* D1 {5 J! a1 s. O
( H) j7 [5 J0 d0 Y" f( Y字串83 L: R) G1 V- s2 Y+ Y
' j. d" Q4 d {, m5 [* a, i7 D7 Z: Z3 }- p, d3 [6 U. s5 J' [0 {1 A" m' k( @
/ y/ c$ M# g1 a: m' S; S2 T0 t1、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* |3 v2 y- i7 e% l" k# d
2、为三多理论提供了下注的多面性奠定基础。+ r+ }, O' B# g) J! a
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.
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