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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。
* h9 R9 B5 n7 h3 e, _3 Y/ l( g( g! W' I& {6 g8 u2 M
假设大路开这样的路:% \( d" O, A/ m0 G4 T* P
2 S- m1 ~/ A/ N- o- H 12121212121212121212121212121212。0 I, A$ M" \3 H8 N9 c
" g ? S. \3 K- s 按2珠路,就是BPBPBPBPBP。
0 V5 E1 a( p/ p& N& A" a, i3 n
如果我们去掉第一口,就会出现完全相反的结果:) w" w m& J @, m& N6 w
7 S9 K7 A, G: o( s+ z; w2 H
21212121212121212121212121212121。( b1 L4 v% g" d2 H
4 j) J( H& Q$ R0 `8 r9 t
变成了PBPBPBPBPB。
, C6 F( j2 L) e2 R B8 }& V/ v
- A9 r4 n* I* Z0 T+ E6 w; b 如果我们再去掉一口,又返回第一种情况了。
6 w* Q8 m$ e+ G. j% F8 N! I" E0 [# |& o9 Z$ }. E
所以每一条大路,按2珠路排列,有2种不同的路数。) L, W+ r( ^: g" @+ p% A# R
" T8 t; d4 P5 r6 t: U
再举一个列子:3 T9 c# J+ O D3 F9 N
: l' P& j i# k* g
大路:122122122122122122。
! D0 h, S( R* T+ c
7 @% x. ~2 e0 p) h( m F' i 按三珠路排列:- W8 K+ x1 R/ f/ m7 P
9 w* S) |! W! b) i. h& _ 122,122,122,122,122,122。! _3 d- _9 v* \$ W
# L9 e! |, N8 V& M' E0 ?! D0 t 去掉第一口,变成:
( p% { U- j P& C$ G& W4 U+ b* z( }/ v9 Z. @: }4 m
221,221,221,221,221,去掉前2口,变成:
/ Y. z3 N m5 ]5 u5 y8 e
' [. g6 g; 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。
! @; y) [+ J1 S, f% w/ C/ {1 c+ T% B0 n" F( C- F2 J5 T6 m
同理:按N珠路排列,有N种不同的路数。
* D+ W/ G& [ X$ b p: F- L) @ o+ h
我提出这个的意义在于:$ X8 w L% H3 ?) v; r8 K
" i3 ]% ]: i& o- u# _3 B
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。$ w2 B: Q( u; h0 w {. X
! \1 o# d- l( c5 t2 _ 2、为三多理论提供了下注的多面性奠定基础。: K9 a2 e( {2 g0 p5 N3 Z2 j
0 \' b# Z4 ]% t! w7 A& Z$ 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。
8 @1 @# X) f( |$ g9 ]2 d2 }' a4 { t5 Y. U) D
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不用花时间去研究这个思路,n年前就见过。结果都一样,下三路就是个例子。 |
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稳稳的发现没问题了的啊。* e7 M# y" t% K' O) ?8 I
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楼主的理论打法不错的呀,赢钱了应该是打法对你有用哦~来收藏下了解下了 |
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