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$ {+ e# y* L8 W5 i
7 B) j% Y& q* M" {9 e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 r9 \3 ?3 i( x/ `9 k9 B! ?/ w; p$ v8 u3 y* D* g. ? k) U
假设大路开这样的路:
: Z I0 i% Q$ @7 \, r2 @: w4 V! F; a* W+ q* `2 c$ T
12121212121212121212121212121212。
+ n0 u. B4 i6 f% l+ @
9 D5 y/ g6 l/ J9 C0 H$ I 按2珠路,就是BPBPBPBPBP。
$ y7 T0 |+ H2 P- R: E( A$ O5 I( V9 p# l/ q0 Q& \1 F
如果我们去掉第一口,就会出现完全相反的结果:! a( i+ u: n: C7 @
* X% q! V8 J# W# R9 { 21212121212121212121212121212121。
" _, n7 z# C G9 [; W+ C
( I9 s) ]7 K7 t3 Q, s- U3 z 变成了PBPBPBPBPB。
. X( e, L" k; d0 z+ i6 h, b
x2 S% Q' k2 y1 ^3 o 如果我们再去掉一口,又返回第一种情况了。
5 Q( V4 S! I2 |' S( p' U! J& h2 L- D8 H/ R f7 A
所以每一条大路,按2珠路排列,有2种不同的路数。
5 r; p ^6 ~5 O9 W$ x' e6 o- a
" a% P* m4 h) m7 C 再举一个列子:: B1 p' a# ^" S' p: v
( B# ?0 N4 z7 B+ o2 Z# d8 D- z
大路:122122122122122122。
- a1 B* Y# ^$ ~) Q' \
) B7 y5 _( |5 h4 B) K0 x 按三珠路排列:
9 r7 F8 i% ], E: k8 m
: Z# P% d9 ^$ F' Q6 M. h- H6 i% V 122,122,122,122,122,122。5 v* @. A% \2 Z' T- Z$ k0 m
, T9 v Q: F# O7 \ y/ ]
去掉第一口,变成:
8 d: _# [1 R( v
/ u" Z, |5 @, @: v( \+ \( w3 p 221,221,221,221,221,去掉前2口,变成:
/ o: B" b5 |- n- L* G# e0 A( w; e. u: Z% z
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。+ W3 P% o+ W6 r6 }( Z/ v
& x5 [, e7 k# c0 O& x
同理:按N珠路排列,有N种不同的路数。4 v3 E$ S9 X o/ y% o# O. r& ~
4 `+ k4 Z0 Y- i* l; n- V! Q$ j
我提出这个的意义在于:: O; _. z4 Y+ m7 o
9 l/ Y3 E# N1 |) \0 \9 a 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。
k4 m( M9 G8 l0 n) n5 I- K
0 f0 I5 f- ], @$ w 2、为三多理论提供了下注的多面性奠定基础。: D+ \! x4 [# V2 G
8 t' ~8 ]6 f4 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。
3 V0 N& a# V+ e% e$ S0 F
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$ s8 I9 z! q/ r d1 m$ E1 B |
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