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4 \ b( \" z& g% u! T9 `8 e0 O" S: A6 Z" Z
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。6 l) c/ b, I8 u( L( c& V
1 F3 i5 [/ \7 _ q 假设大路开这样的路:1 B& G% v+ [) `: R8 H4 ~
/ S; k( C( H, c( Z: s6 J 12121212121212121212121212121212。
) I4 f( t7 O9 O" W( `7 [ U, z
按2珠路,就是BPBPBPBPBP。
) W1 K2 k& _$ t& Q; }* Z( z1 k) x: X% S1 V& Y
如果我们去掉第一口,就会出现完全相反的结果:/ s- w# P( Q8 @# \& S L
2 }- g6 M$ y$ A( l, H) y 21212121212121212121212121212121。! g5 `! i" s$ N( k0 S T5 O2 w( N
% A6 D, U, w# X( u+ z6 X 变成了PBPBPBPBPB。1 W" F3 e$ l4 S# H# u+ T: {
* z6 p: D' _* X5 g7 g! z: a7 i4 Q
如果我们再去掉一口,又返回第一种情况了。5 |* h. b; ?9 H5 `4 J+ C
# z5 c" p* I3 D, ?" K, L- _; t, [ 所以每一条大路,按2珠路排列,有2种不同的路数。: p- g% b* ` t/ F: t- ]- X2 B
6 v3 ?/ a- p( l. O9 J2 ^- o 再举一个列子:4 b# K! m l4 m$ e' B
$ z% g, l* t) f, |$ F# O 大路:122122122122122122。
* \! q0 G, f6 ] V+ Y6 B2 o$ ~
2 U, u" I, I- N' u- W 按三珠路排列:% l: \$ M: n, J1 u' S/ S
+ l# h Y9 i% c3 ~" \
122,122,122,122,122,122。) w& X# ?4 {1 k5 E1 p5 n5 j) O
* d% Z3 |2 E) u" i
去掉第一口,变成:* ~3 B1 ?0 ?0 Z1 k
; \- x" ^* i/ V: H
221,221,221,221,221,去掉前2口,变成:# j: ?; B: z& _3 C
5 E* u6 A. W. A, ^+ J! H1 K1 W
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。
4 j, u7 D& Z; w, x: P6 i* s, T" Y# v2 Q7 s
同理:按N珠路排列,有N种不同的路数。
/ E8 s$ Q+ t" O% T- N
: R5 W+ Y# y( d9 [4 ], Y H8 B0 g 我提出这个的意义在于:
8 h8 |/ Q* d- [- t. C1 \# _
8 p5 [ T0 C) v$ \% a 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。# _! V: D* Y: q g4 G3 q. v" t8 Y- k
4 g7 j: C4 ]! e 2、为三多理论提供了下注的多面性奠定基础。' f y, A ~7 s( Y6 S
7 R9 ?/ \: m. D 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。
; f" r* }4 _1 G L. {9 |* q
4 A; E$ E- s" b8 {; L3 A6 p* [3 g+ O; V2 @6 e
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