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[lastpost] => 2766041 【吉祥坊】9月16日投注流水1000元 1758212484 中大奖
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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。
! W: J- n5 w. g' z
: h9 u, e5 S8 j( x" S$ {& t 假设大路开这样的路:
) L0 l$ F$ m5 @9 I0 m
1 g- ]0 \6 {% ^1 _: \9 g# G3 s' r 12121212121212121212121212121212。
! T3 ?; j% K+ s, ^' Y6 b: a( y+ y/ u/ I7 m5 _* j
按2珠路,就是BPBPBPBPBP。
. u( `& V0 ^# q* z- P6 U& d7 o% H9 C- K: d! _
如果我们去掉第一口,就会出现完全相反的结果:9 `/ }. n: e! i. r% u
W# u7 F/ }' o: X, p1 U
21212121212121212121212121212121。
4 m$ U6 X: m0 x6 W7 v+ T% j5 Y( t
; J6 C H8 _* ] A" | 变成了PBPBPBPBPB。4 y8 A: s- ^8 t6 u: m/ E |; j
+ E, J4 `- m- a9 N
如果我们再去掉一口,又返回第一种情况了。
2 W8 Z d3 _3 n2 l
# a" c; O/ D' L. `" n7 D) O& C 所以每一条大路,按2珠路排列,有2种不同的路数。
' t+ p9 W8 j: r; ~1 I$ h' n3 [# ~6 g2 J1 M& f6 n7 T
再举一个列子:
2 S3 ^, s' S9 d7 m+ E9 A$ R) i. @# ~' f# @' m* y
大路:122122122122122122。* S6 J$ m# Z! ^) Q3 P& `7 h
4 _9 h [" f# O$ v( v% }* I/ m
按三珠路排列:+ ^$ U5 J7 I& e# Y3 m& g6 {4 Z
; R) f9 O' ^3 k5 u* I( \
122,122,122,122,122,122。! S" v4 W% t5 Z& H* g9 w! i
/ o/ k1 _0 V: c- X& d; I. i1 f
去掉第一口,变成:) @! ~% ]/ a! S% y- C8 {9 t
0 M ~" K1 D" L- A# ^6 d( Z 221,221,221,221,221,去掉前2口,变成:2 }4 v- \% p1 b6 \& L3 [0 [. }
2 `: I$ Q, w" E) H: K
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。
, R8 e+ Y4 d9 g# ^: @( X# ~, V% Y! _/ C- }
同理:按N珠路排列,有N种不同的路数。! }" f3 q6 o- N: P2 J# v
z2 V% }0 ]4 D4 m0 N 我提出这个的意义在于:( q' |7 V/ F3 Q4 s, V- z) b4 ]
* w/ ?( y. g. x9 M 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 r$ O/ ~0 E1 s
3 l% b) R7 S! L( z 2、为三多理论提供了下注的多面性奠定基础。( f4 a k& ^# F7 f8 F; f# R
5 w2 a& Q5 s# l0 P+ q! u 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。' t5 A& n# N. s, {( e$ p
3 [' E( G+ Z9 h- b7 z4 m0 G
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我现在还是研究一下理论打法,感谢你的分享,我也来学习 |
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