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1 L/ ~( D2 z% S! \: T6 j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。
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% j; w2 D0 t4 P6 ?5 {1 R- k7 a3 v 假设大路开这样的路:% C4 h! A9 Z2 Y) g. a. ]8 \; {7 u7 l
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12121212121212121212121212121212。
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按2珠路,就是BPBPBPBPBP。
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如果我们去掉第一口,就会出现完全相反的结果:3 X- D" q* O% v( i9 W9 K* Y/ U
" A" G0 Q4 H6 N+ ]& r6 ? 21212121212121212121212121212121。+ o$ T) ^5 F$ K
2 @" S) l& Q+ Z% f 变成了PBPBPBPBPB。( W( d6 \6 ]9 P
' s, {" P- m3 f; d 如果我们再去掉一口,又返回第一种情况了。* U! R$ ?& R% m! C
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所以每一条大路,按2珠路排列,有2种不同的路数。
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再举一个列子:
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大路:122122122122122122。8 w" M, v1 {+ F V+ j3 `% z
8 M4 ]) W# C( K! y2 ` 按三珠路排列:
' L1 r% W# R& c N/ h
( a+ }) @& L9 @- o) L 122,122,122,122,122,122。
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去掉第一口,变成:
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221,221,221,221,221,去掉前2口,变成:* F% }* O5 W3 ]
0 ]# y% T6 l0 T' J' G8 \) 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。5 T$ f$ I/ u- z0 M( }2 |7 [ }
- _; ^, V- U: y% S! b 同理:按N珠路排列,有N种不同的路数。" B! k; K2 D0 [/ |# D `
6 e1 G0 H/ C+ s4 t+ p+ R* t 我提出这个的意义在于:/ S! j( n; Y; q9 `1 Z
5 ^3 `1 P& z1 M! H3 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。
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2、为三多理论提供了下注的多面性奠定基础。
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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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