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3 I' W- P) ~( W2 C: w" E: K: k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。9 X! B9 a/ l8 ?! A
( O- B8 I( R& o. n7 U. q6 x/ T7 S 假设大路开这样的路:
. l4 u6 @: I- X* b) o z
- F+ Y& @8 E/ J9 x5 c" m 12121212121212121212121212121212。4 u1 g v8 t; s1 B. Y2 ~
8 d4 u+ \9 s1 w/ \7 a5 i 按2珠路,就是BPBPBPBPBP。, n( c+ E4 z' a0 P7 x, B
% X6 i( }) G0 s; B! D/ u# g 如果我们去掉第一口,就会出现完全相反的结果:1 t6 w- q- R$ q; L$ L% B$ G
; Y+ Z4 H0 N/ M
21212121212121212121212121212121。( t( Q! [: J) I2 M
F' l9 C- \! D) ~, d 变成了PBPBPBPBPB。
4 \* b3 }3 }% W2 D' \! C9 s- s& E5 ?% J$ }) v) r1 X8 D. |
如果我们再去掉一口,又返回第一种情况了。
4 h+ @( J" S0 u
; z5 N3 }3 k0 r8 K/ ~* R4 b7 ^9 L 所以每一条大路,按2珠路排列,有2种不同的路数。
9 Q4 L) v8 o, m9 b# y# X
1 f6 \- s' X! {3 k# L t( s9 r 再举一个列子:
6 T* ^$ i- P( k+ X0 L
2 c; O3 s- ~* U" q0 Z8 K: u9 x 大路:122122122122122122。
7 b! Y( O! x% X4 ~; H# G: }) M6 h% n9 ]& [" l
按三珠路排列:
& B# H+ @; c: x
8 Z! w7 _" q% f3 U" L! H( L 122,122,122,122,122,122。
: B, s6 {4 o9 `& s
/ Y$ @. @$ ]* C& Q7 F; L! n; i7 H" R 去掉第一口,变成:
; t# g1 z0 R& V8 A* K7 X
3 k3 N' W+ j0 A4 E5 c, ~0 f0 F) n9 ^ 221,221,221,221,221,去掉前2口,变成:
; f: W9 h a5 [9 {" g3 i- V" K. {- g/ c8 J; t& m1 j
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。& ]' o" |- b0 y5 t& H2 C2 B6 D
5 t" X; _0 ]9 g6 k
同理:按N珠路排列,有N种不同的路数。9 x$ ^+ x: Q$ I8 \! j& q ]( a) `* ]
) v1 U" W0 b h, q 我提出这个的意义在于:
9 r# \) p) O+ R
; }. E: l$ \1 e- l& e, F- |" X 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 [4 y ]2 N+ x* l! Y6 _, _1 ^ n7 ^; b; Q& H. I
2、为三多理论提供了下注的多面性奠定基础。" E% `* m# `& v. w' _4 d" o. s) Z! l
" x: V3 h% K2 H5 j t 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。
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