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0 P [2 P1 r. |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。, v' E$ X1 V, q" A
* m5 D! |6 O9 r* j2 K 假设大路开这样的路:( l% b8 j- P6 _! j! Y& `$ v2 ?
* [( P0 W6 C6 @2 V7 y1 J
12121212121212121212121212121212。6 `, f! q' r; V' v% B% R# |$ W( X
" a* N$ { R9 \2 Z b* m& i' B 按2珠路,就是BPBPBPBPBP。& a) M5 f x$ F/ D% S
! U( J& z( ?2 R9 g' q; R7 Z+ ?
如果我们去掉第一口,就会出现完全相反的结果:+ v; y& L* Q( b# ^' F8 S
9 l# [1 V* Z+ p5 N 21212121212121212121212121212121。
" P, u7 B2 R! ~0 e; l" f* f
P& T6 F# ~" D- t% X+ r 变成了PBPBPBPBPB。5 ^/ V9 }( P( v- d" J8 u
* M q3 y1 U# L, k+ {6 s
如果我们再去掉一口,又返回第一种情况了。
0 [; ~# Y" a8 x6 Q* D% o" Z& W) [
8 }! m) a/ q2 l2 o3 Q4 O; x* L 所以每一条大路,按2珠路排列,有2种不同的路数。
: J% c5 _- w8 w' [0 @1 X% E7 v, B& i6 ~- z# c4 Z# O
再举一个列子:8 b/ q! B7 h1 j8 T8 X' B v& }
& H& s: g1 o {5 P$ Z
大路:122122122122122122。" Q- N* B! M+ @4 S; U+ C% y1 w5 a
; d% F& f' y( a0 ?7 | \& m
按三珠路排列: R" n6 V$ Y3 q* Y
3 f( u% L0 ?8 A- U 122,122,122,122,122,122。
2 Z' x0 g0 I" B9 \8 L% g) D; F. {+ r) t0 A7 W& S" b! p: g4 t
去掉第一口,变成:! s! ]- X. J; k
+ w0 G: ^1 t5 S" @ 221,221,221,221,221,去掉前2口,变成:
: K% Q8 T2 r+ v% m7 z+ o8 P0 k" x9 ~, C5 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。
3 {! L) |0 W, w3 C: I. z3 h9 Y. X. Q$ O* Y, o$ ^
同理:按N珠路排列,有N种不同的路数。
# G4 T, o# m; b: c- d& X( U9 e6 d/ T. q% T( D! y9 l
我提出这个的意义在于:) D1 l% j- q8 S R+ V& _+ H# A
9 j* h) G' a- w, E. _ 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。/ k% @+ Z9 z, t8 Z A) n! M% U% p
R9 p$ ~# p2 f3 M! V6 }8 |+ ~ 2、为三多理论提供了下注的多面性奠定基础。" {, r$ b. U1 x
# p3 S+ r Y* f+ {& E 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。! w9 Q" ]1 h# ~% v9 K7 S1 O
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