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' x8 B. T7 t; w, k& g h7 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。' t& Z# h$ a3 S& l! t: H
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假设大路开这样的路:( v, G, p* ~8 M4 p/ J' {
7 e T/ B- J5 n6 X# X$ e
12121212121212121212121212121212。
q$ \/ {/ u5 {9 g K
4 ~! u. B* V. T( e 按2珠路,就是BPBPBPBPBP。
2 {7 T; }* s& v, A6 H! J
, b, p4 f3 [- k1 f: r/ }( f8 K 如果我们去掉第一口,就会出现完全相反的结果:* g, Q. ]& s4 [- v8 F( C( \$ h! G
4 g- N, v! m- x2 r- W' Z 21212121212121212121212121212121。
: Z1 y+ B; W' G
, k$ ?1 Q( ~; ?+ w 变成了PBPBPBPBPB。
8 v! Z( ?, V, Z# g7 p6 j# k
* |! A, k* w* k& J9 x 如果我们再去掉一口,又返回第一种情况了。7 @' D# T5 [) C5 b# s9 h3 _# g! \
% p( n% k, T. ` 所以每一条大路,按2珠路排列,有2种不同的路数。9 P" r7 u, r. t y& }* ^& Z) }
S9 a/ b" a! f, F) J8 G z' W 再举一个列子:
- z5 P. H# \& a
( [' [* Q8 j# N8 [; U7 l 大路:122122122122122122。' v5 k$ m* E+ G5 M
9 v5 A. c6 J% @) R2 ~$ C3 S. @ b 按三珠路排列:
- ^; W) @+ X: M9 @& e* ^: P" h/ u4 i/ X, p2 ^) K( B2 \( X0 {0 {" R8 g
122,122,122,122,122,122。8 J! l! \% c3 x4 Z5 a
" z, `4 W: g# u( l7 r 去掉第一口,变成:- t, ?+ i+ g) }" ^
; D# L8 C2 c( O& D' N 221,221,221,221,221,去掉前2口,变成:( Z6 B! i2 p m; K2 \" l H/ A- M! t# ]
; D: {9 S4 a) n& q. l! ?; J) N
212,212,212,212,212,去掉3口,又返回122,122,122,了所以每一条大路,按3珠路排列,有3种不同的路数。$ h" p* y1 _; j" o) W( m
5 a( W. o/ a# E8 |& d. Z
同理:按N珠路排列,有N种不同的路数。+ l5 Q& P z! L8 ~: \: ]( c
\) T# a9 r# G9 Q
我提出这个的意义在于:
4 k6 L1 {6 g9 `4 j7 ]0 c% l( m+ K* E* k& l p5 o
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。8 M2 a; K* f" j! ~- `! l6 ^! w
! C8 c0 @4 O0 M: G3 c4 Y0 D, h0 ~# a' L
2、为三多理论提供了下注的多面性奠定基础。8 V: t, \! z; h$ N! t
& ^; K+ B) V# ?" C& F 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。, n( g& k% O" h
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