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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。2 W2 z# h: Z' a
; k" z/ v- J8 w2 ?6 V+ F, U
假设大路开这样的路:
! J% N2 R$ G m6 _: d2 f. s5 v% z z* e$ `3 ^4 n9 m" U. t
12121212121212121212121212121212。4 W: E" ^# d- `% x- s
* a: l1 n0 `- E& W; B1 M' L K* T
按2珠路,就是BPBPBPBPBP。! H1 Z% M% o' G/ x$ L, y# W
, E4 t" a/ R$ R+ J- y4 h$ D) M 如果我们去掉第一口,就会出现完全相反的结果:+ n8 D4 }7 `- \+ l1 i* }
6 b5 \8 c2 ?* G! t
21212121212121212121212121212121。' O7 e7 j2 V. L6 k; x. B o
) ~) L/ G3 d3 @& @2 Z& A/ Y
变成了PBPBPBPBPB。
- ~" B+ e T6 p: I9 X6 X
; E1 v8 X! z) R' ]2 P 如果我们再去掉一口,又返回第一种情况了。
/ i8 T. B3 \4 R f5 `4 o5 R) s' v2 V# X; V5 |
所以每一条大路,按2珠路排列,有2种不同的路数。
6 V) `8 k- R7 ~6 [/ D; g* B0 {8 d
再举一个列子:7 f+ c5 |+ P! w; i: q
1 `% j; ]' A5 u7 x
大路:122122122122122122。9 O/ t6 {# R6 Y+ \! r
: v, Z" g* G5 A- W0 \ 按三珠路排列:
; p3 v: n- y- z% s. g% ~8 z( _ {( |
122,122,122,122,122,122。
- i0 P5 H- u) A" @4 ~' r) L$ {8 V$ h. D" \" V
去掉第一口,变成:
- X- o% [: r$ c( N- X* u$ ~/ M+ o& H- k' C5 e1 C( Z2 v: ?+ \
221,221,221,221,221,去掉前2口,变成:
; p$ q( F4 b' @' {% m% _" p5 R- x# F3 f( V
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。
1 {5 N V; }/ [8 ?' O7 Z, s$ z: R9 j4 W/ O y; k
同理:按N珠路排列,有N种不同的路数。# W$ @ { s- n) m7 \1 ~. \
' P9 c' w+ H+ f7 z9 k# ^2 ]8 K. F 我提出这个的意义在于:
! B/ q/ O, {5 C# c( ^3 [2 i0 e( y# A, C7 ]; t i6 v e! e/ L' C1 S3 I
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。0 }" [* {* e0 K* V% a5 T
- H' ?) i3 d# A7 {! H X9 L
2、为三多理论提供了下注的多面性奠定基础。* t I J& i/ |, d% @4 E1 f
! x( V" T$ L0 R9 p# j v& n/ @$ f' p 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。9 N0 B& i- Q" Y8 h- i C
W4 P2 Q$ Y, u8 i6 f: a% n
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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。' b) H" Z, T0 X: C. Z' y
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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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