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眼里没有庄闲,只有连和跳,因为庄闲是一对一的,而连跳都是一对二的。理解也好不理解也好,这个讲起来费劲,就不过多讲解了,你只要记住百家le只有连跳没有庄闲就可以了,通常大家所看见的大路,第一行是满的,从第二行开始才有洞出现。
8 S6 R; z+ P$ R; B. o8 n6 Y
) H( X9 R, i- P: }% F' Q; L. ^: q- H! \ 用什么办法把第二行也变满,让洞出现到第三行去呢?这就需要重新排列大路。一种全新的方式去排列大路(又是新用什么排列,用连,和,跳连就可以比如大路是(随机的)" n1 d1 {' A1 r, Q* |! Y w+ n- U
# |" g5 u {( n/ N4 b: M 第一行:OXOXOXOXOXOXOXOXOXOXOXOXOX(第一行永远是单跳)
3 e/ s, w# T8 m8 H! s5 w) f! G: k- |' M [8 X
第二行:O X X O O OX O X (第二行开始出现洞), J7 t2 d/ G+ G, w7 R& s. X$ y
- R: j d" y! q* O" Q( _5 r 第三行:O X O X X
2 I4 C; a' ]% ?; }9 [4 h" Y
2 M4 P5 L# I# [1 h1 k; @ 第四行:O X X
) t% o% k4 P- B6 a; U$ G6 V" E/ I1 a4 b$ ~) r
第五行: X
1 ?& D$ Z" a8 {0 M1 G/ Y, T4 t) [ e4 ^
; j. n5 s) @: V4 C3 _1 B* \' h 那么新写出来的大路是怎么样的呢?+ ?8 J" h5 J2 h0 U3 H# V @
* s& e) L3 q& [: Z* p/ }& G 记住一个简单的口诀就可以写出新的大路:1.连是连;2.跳连也是连。: H$ ~( l8 `" ~/ |, z
+ C0 \1 w+ b) H7 Z% y# L2 G
如下(大家凑合着看吧,实在不行就自己拿笔抄一遍,和上面的路一一对应。)
9 M7 ? B2 N. g, i$ K! b" C/ A8 E1 U; M( {, h* [7 T2 ?& k
第一行:OXXOXOXO OXOOX(第一行不再是单跳了)3 p5 `9 d* l/ E* Y& u% u
1 B# }. S' `; B6 ^
第二行:OOXXXOOX XXXXX (第二行也是全满的)
% d" T+ L' b9 L. l( N7 K |( j* {1 G& d K, t7 E* Q8 @4 u8 a/ C
第三行:OX OX XO XOOX (第三行才出现洞)
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2 y; [6 m( `5 J* l! m# O 第四行:O XO O XXX
, z* h6 m9 S1 @- o
- W: h# d: G2 i1 \ 第五行: O: `+ G! T5 Z; H! l
# C6 ?0 | L4 z 通过以上排列,大家看见了,变换了组合方式以后,大路不再显的那么杂乱无章了。每一列要么是连,要么就是跳连。这就似乎是最初期的关系形态变化组合排列,通过这样的排列,从形态上看,整个路只有12个形态上的跳,其他全是连贯的(可能已经有人开窍了)。
2 ^: u9 b; w# G- Y! G' ]8 o1 ~1 m* N' z5 M
大家看第一行不再是完全的单跳,第二行没有洞的出现,第三行才开始出现洞,每一列最少是2个符号,通过这样的形态组合,实际上是减少了原本大路上的跳40%-60%,比如原来一靴牌60把,基本上是30个跳。30个左右连,而且是间隔无规律,杂乱无章的,按常规跳连的次数是基本一致的,周期越大,数字越接近。
0 O- @% E2 W) @' W; z2 b7 m2 ?/ M( S( A
通过配列 可以把原本大路的跳精简到10-15个左右,相反增加了40-60%的连,以上是对大路形态转换的最初级形态,相信已经能对大家有帮助。至于怎么样去下单 ,一目了然了。% y2 e0 o, i6 ^+ h% U' i ~' ?7 A8 A4 S
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