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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。+ W1 `1 u& W" N0 A/ e1 x1 s. t
8 Q! R5 j7 f' b: @' {6 j1 s 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）8 F4 y% {) P6 ~4 a' n E8 s. _
5 M# ^0 ^$ p" O/ N5 x) a+ ]! Z; }7 ^ 第一行：OXOXOXOXOXOXOXOXOXOXOXOXOX（第一行永远是单跳）
0 p7 s- s: y$ |" V3 t, `2 W0 F( p) d. `; g
第二行：O X X O O OX O X （第二行开始出现洞）
; D8 g/ q+ O% a2 j- s0 l0 n' I+ V4 m" f. j1 K. z
第三行：O X O X X
7 H: R k4 u6 a+ N6 m& ^
2 e q$ b/ K8 ]2 \, E/ c 第四行：O X X
% Q* M( z0 ^+ I5 A( @- C. u' n1 E2 R7 G0 q9 c8 W9 z
第五行： X! w) V& _+ \8 L4 K, ]7 R. N( {
& o' ?0 _+ u& A: I
那么新写出来的大路是怎么样的呢？) j/ ^. L# c$ z0 X) S" P
$ }1 W0 n1 h6 t5 [% d: k! r4 B4 H
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。
0 w X" r, C# x! q5 E* S4 u% Y6 P/ D, V7 z7 j; d0 K' W! m* ?1 p
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。）
) M' ~7 j/ S' r: D% {! L+ i) `, ?2 v5 x
第一行：OXXOXOXO OXOOX（第一行不再是单跳了）! n$ g1 @* m0 ]6 L: V$ F
; k k$ \. W5 @) g2 ~" W 第二行：OOXXXOOX XXXXX （第二行也是全满的）
- r% ]+ W" i. z* i2 G+ N( l6 t, c8 A5 O$ O2 b9 m' c
第三行：OX OX XO XOOX （第三行才出现洞）) Z- H" {2 [" D; D3 F
; f; M# b; ]) A5 _) A- B& B 第四行：O XO O XXX
/ C% T r4 ^+ p2 x) Q* T3 K( r+ i9 o' J2 `" D' W o3 k/ Q
第五行： O$ d+ E9 q h# s
/ L! A4 H% S( d4 [7 ], D- r. P/ ^1 F9 n 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）。) D( ]% `# \+ F5 T
" j' r* }9 Y4 A1 [4 B# U% i 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 \1 ]! Y0 x6 K3 [8 y& E! e0 G
% e1 I( d8 ^& `& s2 u9 l7 s 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。! s7 \$ p2 E. u' U
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