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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。$ Z; w9 a* P2 n6 `; k
# k0 x0 e9 w1 R0 D; ?( M4 _
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）( x( u( ~, s+ [3 Z
: g% k' s5 X- P. d0 w" \: v
第一行：OXOXOXOXOXOXOXOXOXOXOXOXOX（第一行永远是单跳）
+ h/ ]! e, h! k1 Y6 r6 R: r% x# w8 S& Z k4 I
第二行：O X X O O OX O X （第二行开始出现洞）) J' r8 t6 \/ }8 f) u3 X) m" F
% T! B, r- c3 p0 V( u. a4 x5 v
第三行：O X O X X
a7 J' N# T4 @$ \! W, k( v: {; H$ t0 W( [
第四行：O X X
5 f% a' M0 q7 b- [
& U6 D, N. n6 S 第五行： X$ _. i# ]. T0 K* y/ I
. y8 m* g9 h3 {4 A7 B! B3 ^6 |% C 那么新写出来的大路是怎么样的呢？
* d- k2 R: f1 ]: x7 k0 c
* q" q/ U6 E9 r) X- ~ 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。& r1 ?8 ?' T! s
2 h4 g- C" ]' b 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。）1 i5 x3 P7 Q2 A8 h
7 B# r4 W( N! u; ?& _ 第一行：OXXOXOXO OXOOX（第一行不再是单跳了）
8 p$ D/ R0 a# w! z0 M$ o6 r" v1 O( h7 i& e( e' I: X$ w" X* b
第二行：OOXXXOOX XXXXX （第二行也是全满的）5 D) x9 M5 ^+ t3 [" O/ u: Z" b& E
6 s2 u) [: c, |; I8 P$ ]
第三行：OX OX XO XOOX （第三行才出现洞）
. ?, ^4 q& Y# u8 W" D+ L9 g4 t
6 m. o& I5 ?/ @3 D# ~ 第四行：O XO O XXX 0 ^5 ~7 W! s& J
3 Q$ b7 B& s% p% J7 k 第五行： O
6 D) F5 C0 W* L+ I2 g8 i; Z! t* V0 L1 ^) a# q1 w- \4 A8 d
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）。9 i& _# W. X" y- i% i4 |# K
2 m' n s6 I. l, z 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。1 Z- h: T. A) F; u; Q; _/ l
) ?/ G0 d4 {' U- W8 w 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。
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