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! @& U2 @ A8 p: ?: u1 `, T3 e8 L! {1 R4 I W
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。
9 H1 ~0 Y8 F" J" `2 u2 v3 d/ t- m! n$ q' V' x: ?% I
假设大路开这样的路:
/ Z) E+ t0 \7 n; Q8 X
& {0 m; |2 D" {/ i& z% Q! Z 12121212121212121212121212121212。
$ [: c& j6 A7 d; y; G+ B! s, x) }) X- G' \+ V7 u/ T% u
按2珠路,就是BPBPBPBPBP。
6 g- U0 u# [9 p% y- ~) n$ z+ A) z2 J, G+ K
如果我们去掉第一口,就会出现完全相反的结果:
, r( e R2 T9 e' B, r* c. M0 L, K9 z/ F: A; d- s4 r" [
21212121212121212121212121212121。
0 G* z1 ?7 {3 [) s5 o
4 R6 ?- c. u, [* W# ], K8 x/ ^ 变成了PBPBPBPBPB。 h4 ^( I3 ?5 z$ B& m1 j M) m/ e
+ Q( j' `! u6 W 如果我们再去掉一口,又返回第一种情况了。) T2 q7 C* Q8 O
}! l$ m( S3 B" K- { d" s, Y: L, [
所以每一条大路,按2珠路排列,有2种不同的路数。9 d* x2 k! ^* P, T/ @- n
' m! Y% Y- ?+ m6 i5 h 再举一个列子:4 ^- o8 M2 q0 D
- {" B. m4 l5 j8 z4 G h 大路:122122122122122122。
. T& I7 V0 Q+ _9 f" x5 D/ P+ G! l# W2 Y3 Y; D* `
按三珠路排列:
+ { m8 h2 t* }; P
; n' k% B( f& ^" Z+ j( B 122,122,122,122,122,122。8 K' I6 o e; V# W1 H) X
! ~9 A( X2 Z I# e) ^
去掉第一口,变成:4 }. x3 g( x/ O: V. y
6 g( A' @, x5 t' j5 ^
221,221,221,221,221,去掉前2口,变成:5 w) o# ?3 J6 ]+ J- r, y: H
r, Z* K+ C8 `% p/ K( m 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 y2 h+ E% V' \
1 r9 e' y1 g0 u/ ` 同理:按N珠路排列,有N种不同的路数。; Q \3 K0 l" `5 K- r# V3 G# G% w( _
/ _* R% B9 c0 w J
我提出这个的意义在于:( `, @4 }$ B. c% Z* g8 Y0 P
+ G9 {) p: C7 s! H: }4 z G 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。
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3 }! f$ r z. W& N# N8 `' h! f 2、为三多理论提供了下注的多面性奠定基础。
% [: C b6 Q) f' ]! K
[: ?( r" N/ o% o6 o 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。3 z, {$ a) k. F: \5 l# x5 {: ~6 n
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