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6 A. V: E9 O9 F
, W) J, Y# ]$ C! g# U 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。$ E1 h" O3 e6 R8 B# c+ W
r5 h9 f$ i8 M! a* f( m" W% f
假设大路开这样的路:
: \4 w) ]3 A* V) Q0 G( |
* _( @% c8 l% Z N 12121212121212121212121212121212。
# t2 }* V8 V8 X4 K5 O5 F3 Y, l
* o: d" z7 O2 d 按2珠路,就是BPBPBPBPBP。
* p; c! i+ ?, W* R' q2 \3 n
! [! X7 i/ r/ i% s 如果我们去掉第一口,就会出现完全相反的结果:9 c! W; p' [+ V2 V+ @3 E
( `" d: {( E, R1 ?5 C, @# x
21212121212121212121212121212121。# X. {) w6 H1 D( }+ I( n6 O
. i& |3 v8 Q: F2 o- w3 r
变成了PBPBPBPBPB。
/ a5 \* P9 e7 f2 e/ s, m3 L" S& ~, E- W* }5 X" O
如果我们再去掉一口,又返回第一种情况了。( B, t: w2 j. H; u- P
$ |! T3 l3 k) B o
所以每一条大路,按2珠路排列,有2种不同的路数。3 ?4 r9 x$ f! _/ M8 ?0 v0 Y2 p+ R
9 t3 K& t s/ a/ }
再举一个列子:0 N' l! N, T! z" _5 h5 n0 H
0 @1 ]2 R( H. |6 x+ U 大路:122122122122122122。! q& J! P* s0 d, n, |. X# I
$ A0 W# P9 u7 S, @. {) n, E 按三珠路排列:
# \2 |. J" z- i$ |: T' M' `* X0 p
" Z% Y* i2 Z2 o- v 122,122,122,122,122,122。
# s2 {9 R# }; g3 Q
, Z$ a0 Y b! s+ y/ |1 N6 e/ T 去掉第一口,变成:
* ], z: [8 t* j
. e6 R0 N0 R, x, x* s! g% V$ A9 \ 221,221,221,221,221,去掉前2口,变成:0 F+ t: b& t) i, Y2 v( _
) \2 n9 u; r2 S5 o8 S 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。( R( J: `& y7 x' q! W- M1 m; W
N* r9 ]+ _% P; u. ^8 K! I 同理:按N珠路排列,有N种不同的路数。' a4 H. g% ]) Y$ N
# _& C+ {3 Q9 H; W, j! i* x0 g+ Q
我提出这个的意义在于:
# r% B$ x B6 j, g' G5 l) T0 U" v. W3 G1 |' V. P
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。& O+ M7 U$ b/ T. s
, l6 K% `8 t3 r+ a! r
2、为三多理论提供了下注的多面性奠定基础。
5 ~1 y/ Z" O2 T# |% z- y8 n. W6 q
4 N1 s" J/ O# y' o9 ?' a* m 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。# l2 K4 g6 w; P0 C: [8 H; U/ f
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+ C5 a2 e$ a9 V( z" Z7 P0 c% D* e+ r) Z4 W# F J8 d
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