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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。4 ?8 }4 i6 Z! w9 b9 Z
* n1 L0 g$ z& b$ i
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)
5 }! Q0 g( P; j6 G, {2 ?, v! w% i% h( I' J$ j/ Q
第一行:OXOXOXOXOXOXOXOXOXOXOXOXOX(第一行永远是单跳)
. I4 i7 {) K+ c6 z( H
1 W* Y) H( I: Q% e0 N 第二行:O X X O O OX O X (第二行开始出现洞)* A$ v9 b& g, \) K' q
) X* W* J( L' M- u' x
第三行:O X O X X
% G$ Z2 R* J, U! h$ }! l
! O' d) h" R7 C8 t8 s1 j 第四行:O X X$ f& P# I. @$ v# X
; o/ U# ~( [+ i; j
第五行: X+ m0 _8 o c; e$ C6 D$ S0 q
8 x [0 \5 Z/ T; V6 R% n 那么新写出来的大路是怎么样的呢?4 X. g/ ^: x9 J/ x8 W9 X' t1 ]
3 [! O6 I* ?% m+ D8 r" k 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。1 K) Q8 n& L8 E2 l- @
$ B3 z& O. B1 D: Q3 v
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。)
$ q/ o/ ]' r3 s, A
" A8 Y; p* X! V# ` 第一行:OXXOXOXO OXOOX(第一行不再是单跳了)
8 D, @2 b! y- f ^, U) k
; W- p/ ^) }& n 第二行:OOXXXOOX XXXXX (第二行也是全满的)
$ S6 I7 m- T! I# V4 d6 C- T1 n
* ]: a3 ^( ^9 M6 N y 第三行:OX OX XO XOOX (第三行才出现洞)+ X% p4 Z7 }( C4 b8 V
$ A; e7 b/ _5 i: \+ i
第四行:O XO O XXX
+ @& s" Q( a& }6 o3 ?& t9 f
/ a7 X" n& K! x3 X* { } 第五行: O
* X3 ]8 h/ N/ u, o% X% b8 q% Z
' R$ |0 c3 m) m9 A1 I+ G, l: B2 l 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)。" o, p* h% B: x( p; s, g* M; z
5 l% Y. N A, J n L0 q 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。9 Z" X7 B ~. I+ ^4 k# q. I* @
4 X) W7 d$ C0 h |
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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