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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。7 H7 B9 s4 C: A
; R1 S) x% u) z% J 假设大路开这样的路:
( ]( \7 |& x/ i. F0 I; b8 k, f0 Q/ N! N% d6 J
12121212121212121212121212121212。5 k" k4 [4 i) F) l6 C
! g8 j7 C" x6 A5 _+ ^, H7 [
按2珠路,就是BPBPBPBPBP。
# c) E* v( q+ K* D) A6 N j4 E+ }- u% s6 v) U* B! \) P: P
如果我们去掉第一口,就会出现完全相反的结果:0 C n1 a$ N# o
! S* h' H' F d, X0 L7 d6 W
21212121212121212121212121212121。
- V; x$ _$ N2 G/ R) c5 d+ C
9 f& r0 t8 K4 d- U( G- M 变成了PBPBPBPBPB。0 F7 O( \7 J) |! R& \9 {
5 a1 y. L9 _* F! M; S1 C
如果我们再去掉一口,又返回第一种情况了。
6 b2 j% O5 y+ u4 E0 I# Y7 m
6 S- x1 ~! C( @. u1 ? Q9 e 所以每一条大路,按2珠路排列,有2种不同的路数。
5 j2 D0 E* O5 C, m5 [; L
9 o/ O, S9 e9 n1 v- T, c& j. U 再举一个列子:6 R! ^$ K( f* K
3 H- i. {3 E ^: j2 v
大路:122122122122122122。
+ c# ~4 T) g9 Q% K" T" N
9 N* U* B b9 f$ U% w3 P0 _% v* L& W/ _ 按三珠路排列:
* q+ m$ @. V& g, U2 C9 W3 _$ I( m" i0 K; w
122,122,122,122,122,122。
" s& }7 n1 }5 H& M- @
+ w. r; H) b: H3 k 去掉第一口,变成:1 b- L0 N8 n! W
" W/ ^7 b3 p; q# b$ O
221,221,221,221,221,去掉前2口,变成:
J* S$ i" J, j- E3 j; N3 ^( F8 D* C0 T6 I. g/ M
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。. E: l& V' T- O9 ]
G7 S3 G- X3 j8 f 同理:按N珠路排列,有N种不同的路数。, d# b" W, {6 L
x; B- q& ]* ?
我提出这个的意义在于:* j, _% h, L( _6 p
H# q2 w8 w3 ^1 a4 T
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。
& e' d; Q: z1 z$ L2 y7 o# {3 f# O
. A9 M! Z) y9 q" ~2 q 2、为三多理论提供了下注的多面性奠定基础。
8 T1 i8 R7 j1 ~ N( `+ D9 W6 Q& ~# }- E" {/ x
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 T) s" G$ b% ^9 z6 s: `8 b! j: e$ v0 O* E8 c6 }) M2 k1 ?6 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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