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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。
* U/ l- e9 A" B0 S
/ P% V6 U: h4 [2 K+ Y% w3 n; G! Z 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）0 R8 l8 o9 m% k$ T
! c' ~+ `$ M; H" M* Q
第一行：OXOXOXOXOXOXOXOXOXOXOXOXOX（第一行永远是单跳）! ]6 F% \* \' c
. J1 q; c3 z5 W7 X+ X. J% p. D 第二行：O X X O O OX O X （第二行开始出现洞）7 ^/ ]5 h. v# J# k
1 t+ [" H; W H( R' _
第三行：O X O X X0 L5 U3 m* M* e
0 ]! r+ d: F" g. G* {4 d
第四行：O X X
' q: y7 n: V, E& R5 z$ N5 S
Y; y u( O2 z; |/ o5 k1 W1 x 第五行： X
% k; S. S" v/ U3 v2 d/ k5 _ h: P1 u# k" }8 S
那么新写出来的大路是怎么样的呢？
0 O2 |: ~: G1 A5 y: h( ?* b. [- g5 {9 e) Z
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。
" F$ A: I" w0 v8 w6 v# c; Q% a1 ]" @
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。）
8 p* v% }! k7 x8 }7 k+ `8 U5 C+ S+ s Q- o4 S; F1 z
第一行：OXXOXOXO OXOOX（第一行不再是单跳了） ^5 |. v: R8 s( |2 P! I& |" v; X( g# N
( m7 R! H. A* P 第二行：OOXXXOOX XXXXX （第二行也是全满的）: b" H6 e' P& I, {( m: x
( Q9 s* B9 P# B& ]; V' W# `
第三行：OX OX XO XOOX （第三行才出现洞）. i0 J7 n2 q. i/ M' g* c- r
; X3 D2 |7 L" A- l 第四行：O XO O XXX 5 }" u! c& d! q6 ? x% z+ l
2 t3 `! k: K8 D( i7 A" ~0 o2 w 第五行： O
( W" Z) y- R$ A9 R% J; w- o% D8 D* ?: N6 k( y
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）。
, C' j4 |9 ~: ~& @$ H/ U+ b2 H. v+ m: W' _5 H+ F2 a$ j
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。
# p9 E+ G; M( N1 _
7 W2 q8 N3 y7 m6 X$ B2 d. p 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。) ]4 ~8 Q- I" H5 \0 |
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