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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。
# k( W3 V7 f0 w9 X. J6 [ M, Q& i4 V3 h5 i7 R$ e( r4 ?, z
假设大路开这样的路:4 T+ w/ X+ h" n, z% F7 |5 ?' M8 L
" n E" E- N3 V5 D( R* ^# U
12121212121212121212121212121212。
7 I/ ^6 q5 p1 q& E. H: U+ @$ h7 j& W# c- S0 y( Z
按2珠路,就是BPBPBPBPBP。5 w. B+ x7 N% e) |8 m& _) O
; [3 W! f. P% }- z9 n; M% } 如果我们去掉第一口,就会出现完全相反的结果:- R* p0 K+ ?6 `) k, L/ y) E1 W
9 u3 _8 B9 }# b/ k# k# I 21212121212121212121212121212121。2 M E" \3 t. i% Q1 }2 S+ V. L3 S8 l
" I1 _1 R$ s% [% o2 F6 E+ I 变成了PBPBPBPBPB。
3 A9 e9 V5 [# I! G- A, Y& G8 e7 O/ N) k
如果我们再去掉一口,又返回第一种情况了。4 y. Y# O: J% z# O' z1 Q
" Q) Q e1 r; T! ~! X5 y7 A 所以每一条大路,按2珠路排列,有2种不同的路数。
* ]! W* e% J: q* P* c. e+ p! K* r2 D% B9 P% S3 b1 M
再举一个列子:
% {) q* b$ p2 i* i; s+ @7 {
P) k4 L% Z- }6 R. h" i) _ 大路:122122122122122122。& @* ^$ F1 R* |* V7 W& ^$ w7 d
- E+ e8 s' }7 E6 T, W. P
按三珠路排列: d) {; P1 C1 _/ }4 v
0 S* I8 A, ]" c; N
122,122,122,122,122,122。
! q) I9 l6 T; ? Q4 h4 u
! Y* O9 E0 L. P. ]# m* T( r' v 去掉第一口,变成:
2 B* D( {8 n( s2 m E% T! r/ D
+ P# p4 i& r1 Q3 N( J7 g5 e; L% o 221,221,221,221,221,去掉前2口,变成:
+ ]0 j& x& h8 V; b! X! i9 e/ ], F! n
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。( L# p& ^, i5 G2 _4 y, H* |
* T" r n, t9 h! t0 t0 G
同理:按N珠路排列,有N种不同的路数。
% a- V7 {, O( _" R4 i7 R# p
' p! Q, C) u, e& A# N! Z 我提出这个的意义在于:/ X1 C+ J2 e- Z. s
1 @) n: u: X, m! S: b
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。
, G" ]! r7 e) I4 g! B( Z$ @8 A# E' i' n+ ?4 u& h
2、为三多理论提供了下注的多面性奠定基础。
Z7 k; p# ?( R1 k: @+ e1 K
9 d% O0 B7 p# O2 F) Q 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。
' q1 E( [( e2 }+ y+ v4 P! v9 [* E
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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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