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. ~0 n, Y; m( B+ a5 c! X& ~% V
" F8 G/ P5 f% [5 n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。/ ^" f( N+ P7 u
+ h0 _* Q, t" {& R1 t6 d0 e
假设大路开这样的路:
! e. y2 n( D4 S1 Z/ N# ~
2 R M1 J! m- E6 x& Z, w7 k# @ | 12121212121212121212121212121212。7 H9 @2 o7 i$ E& i# z
/ g4 z# g2 X; h; y9 M5 T
按2珠路,就是BPBPBPBPBP。
! i$ M0 [. `1 U' K8 O
# Q3 B0 R( L3 f0 h3 R+ ^ 如果我们去掉第一口,就会出现完全相反的结果: b$ A6 c% @+ s5 ?& d0 O1 {
% T- k- e/ Y: D0 p( B. M8 P, T 21212121212121212121212121212121。# U6 n( ]; K( e$ p
/ \5 h6 e; J, {
变成了PBPBPBPBPB。
' C% J& K2 G; c! z, C5 u1 D4 @4 W, w- r6 Y: b) z! b
如果我们再去掉一口,又返回第一种情况了。
% ^; K4 H* J- W# z1 U8 S6 Y @- J
& V/ j+ K0 q, X3 X% d+ [( U5 f# S 所以每一条大路,按2珠路排列,有2种不同的路数。
' S4 s5 d9 ~; M9 T2 K |% P& D% k g0 x
再举一个列子:
# r h2 z. U/ t3 B: K# p/ ]( j, N) q: S5 _. l+ M0 F) q' a
大路:122122122122122122。5 d5 S" z6 H" v, N, j6 ]% E1 A0 ^
3 ~' D3 X8 t$ C- ]; q 按三珠路排列:
7 z; g1 A$ @* ?( O0 `1 o7 U/ n n* |1 l* K- ?% W1 u
122,122,122,122,122,122。
8 o, Y2 X x: K! U3 ^: u0 d
- x" k# D) I* T( b; \# x+ f: a( T 去掉第一口,变成:% e9 h- Y5 Q, v6 o- L9 H6 {0 a4 l
. M' |1 F5 c0 W# f) J 221,221,221,221,221,去掉前2口,变成:
. H8 ~- `5 `) R5 S" L, C, V. a6 ^2 F
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。
- \% u4 f+ R: K6 i; C8 E% R# x2 M% T( s7 @1 _0 Y$ ]6 X# j
同理:按N珠路排列,有N种不同的路数。/ A9 i* J# Q# k* Z; j; f
+ y8 I# k! v5 b
我提出这个的意义在于:
2 T2 l( N% d9 }3 @
% Z X) m/ J! 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。
. M' ^4 P! s5 p( Z L
3 W5 z. V! x# h% W* d' v 2、为三多理论提供了下注的多面性奠定基础。
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' C7 ?* ?( A" L: N4 |9 R; O 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。" W- O& R p0 q+ R! g9 A1 E9 y% n
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