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[lastpost] => 2763058 【星宝】9月5日投注流水334998元 1757246556 舞出精彩
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1 A- y+ i. \5 J3 v& j
2 E. Z( ~/ }- t, m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。' S: n9 \$ o3 x, z, }
' ^! h1 N1 t9 s4 h 假设大路开这样的路:7 o' {# W2 L9 c4 ~5 @
T6 Y8 T% V- h6 t' Q: U7 ^( e 12121212121212121212121212121212。% ?: r. `- R' e3 m6 A
* }! W/ ^, s( E0 k& ~+ X A 按2珠路,就是BPBPBPBPBP。
$ g% X1 S: k) |* }0 d0 m0 s1 r' t% J P: C3 f3 I- p1 n+ I
如果我们去掉第一口,就会出现完全相反的结果:# ] u' D' ?: s( S4 O, H
/ T% Z! f" G' _, ?0 x; w5 ?7 x 21212121212121212121212121212121。
+ x6 a- |) |2 P# j
1 b8 N9 r3 p9 |5 Y, y2 }. i% G 变成了PBPBPBPBPB。
4 X0 e! z% O: s4 X" v }$ l9 Z8 J! Z+ Y1 F
如果我们再去掉一口,又返回第一种情况了。/ o# b5 i) W! a: R7 T
. y6 W6 t6 n6 j O
所以每一条大路,按2珠路排列,有2种不同的路数。" w$ T! C$ x: ]1 A$ t
% q/ O6 j" Z7 p( m
再举一个列子:$ X P: p4 O: v6 {, N" B
, M$ q0 v+ d/ }8 l( T9 K2 q 大路:122122122122122122。
! s& C6 U3 q+ Y$ P$ H& L% U# D' C+ o9 M& q+ s( k* l7 b8 j d) {. R
按三珠路排列:
6 N2 |+ ?7 i" I) ]2 o8 {$ a: T+ ~4 [; h+ O
122,122,122,122,122,122。) q" A1 D0 K; v8 ~ P
* Z5 V# q% _7 N6 ?# c
去掉第一口,变成:% ]$ J# ~. d( Q# x
* T/ F6 P8 ]* N
221,221,221,221,221,去掉前2口,变成:: [! E! [7 d6 W
5 `" Q# f0 O3 ]5 q. y) _4 q, E: ` 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。: I, S6 s$ i2 F7 I1 k7 [( I- y
9 N6 |3 E. O; @* y 同理:按N珠路排列,有N种不同的路数。1 j; F- A, V+ `+ R3 L- O. s3 _. C" [
1 B. K b9 ]; h5 A 我提出这个的意义在于:. p! F! [ i* ~$ G$ q/ f4 y
, @; |! x. I9 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。5 l- W1 i [6 {9 X- q' G9 C
H/ T$ M! \8 T4 m 2、为三多理论提供了下注的多面性奠定基础。
" ?! l) i s n9 f6 J9 s9 X% I0 q' ?9 \) p; C
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。5 A/ a- U" v% `/ M9 y _% I
* `, A: Z. Y0 J- y. v H* h
$ S& ~, U9 s9 j) A" I/ `. X2 j
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