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[lastpost] => 2760735 【YZ】8月26日投注流水75095元 1756454320 rainwang
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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。5 _9 o+ _/ {8 P) W1 L
* B- A- a% y3 o% M3 q
假设大路开这样的路:( f/ k; v* p8 i% p8 i6 _
. Y) O! ?+ W ?% @ g2 ^
12121212121212121212121212121212。
+ j7 S9 H; \; k: R; w9 q# d' C2 I6 Z
按2珠路,就是BPBPBPBPBP。
2 F9 r1 s0 ^( y$ n8 |* Y% x/ }5 p& }0 v5 O- F4 y! n9 v5 Q
如果我们去掉第一口,就会出现完全相反的结果:& H& |% S/ B0 P3 M
* M5 J* G7 p+ Q5 ?; U) H 21212121212121212121212121212121。
8 z3 ?, x1 L/ W M. F% ?9 Q- \
7 e# r, `, c% a- g' K% W$ Z 变成了PBPBPBPBPB。( B0 j% S5 t4 H# Q
( \; x" n7 C! I- g
如果我们再去掉一口,又返回第一种情况了。
! Y' S) }' t$ ^
: A. T: h- Q! h1 f" b& q 所以每一条大路,按2珠路排列,有2种不同的路数。7 c$ e2 d* n. Z0 s2 X& K# ? A+ j7 a
3 O& D9 H' g& L
再举一个列子:
, x5 y9 _4 I2 u+ y4 D; |4 H" g8 b2 G4 B8 V3 I! Y' Q
大路:122122122122122122。
+ O: x# Z& e. y' E' O" V) a
4 b6 B0 w9 W0 c9 s/ W5 q) f+ p0 M+ ?5 ? 按三珠路排列:& N. b# Q+ `# D; P( G& T( ~+ k
9 g( k4 \6 [ O' b 122,122,122,122,122,122。# f: p& [5 y0 c& k9 q5 Y2 C, w" ^4 K
/ M! p3 }) n0 A. t3 z# ^7 u8 g
去掉第一口,变成:
. V. Z2 ]$ l j/ q7 v6 `/ g' P: t- @* I4 d+ i
221,221,221,221,221,去掉前2口,变成:
- }/ g+ o% g( m. \ s2 o: t8 a- Z6 s( M: [& T
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。' w5 U# ~: F. h( \; o2 I& m
5 j: r: W2 w, Z r2 | 同理:按N珠路排列,有N种不同的路数。
% c1 G" `" p# S7 t+ R
6 P" M/ d3 T2 n n. L p2 }' [& ^ 我提出这个的意义在于:3 X5 l# n/ f7 C; c) }9 G$ x
2 \. e7 D. Y/ C 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 v. i$ E. B# S! s# ?* f
% i0 j# H: n, `/ S ^ 2、为三多理论提供了下注的多面性奠定基础。- J5 ]/ o; i& `/ B$ k" Y
& |" K/ t, n( T4 g1 J, g
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。
; }+ P. `+ A& @" a& E n7 Y% H7 |8 ^- Q2 I
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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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