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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。0 T3 r/ r& P! b6 E: h1 x5 l3 S
" r! s$ R+ ?% @, P5 j- j 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)& t4 x. H8 A% |" R1 S) A
7 n4 q) g( r! @) v 第一行:OXOXOXOXOXOXOXOXOXOXOXOXOX(第一行永远是单跳)% I G# \5 r& Q# v# f- n9 W
) t2 F) u, s! R( b2 p1 p) u# s 第二行:O X X O O OX O X (第二行开始出现洞)
4 C* E3 N8 W$ p* o
2 n+ @9 p$ @/ d# E 第三行:O X O X X
; l. M/ o# b3 ^4 ^$ \+ f5 @; V3 u/ z) v
第四行:O X X3 x3 N# x3 U. s
- y1 ^* t4 ^; {5 U/ f: W
第五行: X, }' d6 [! w4 x9 A
, ^; b. ~# `. c, J; i1 k6 c 那么新写出来的大路是怎么样的呢?- S3 j0 `) U9 C& B' p
4 Y k+ i6 s% C
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。
2 ]$ m3 t3 R: ^7 S5 N* f9 S" c/ V% E3 W. X0 ]
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。)
& I- C8 ~" L o/ G: N4 }! X: `
$ U4 ^- @% }% A" P/ j- j0 |1 o 第一行:OXXOXOXO OXOOX(第一行不再是单跳了)$ m: e% j6 w2 A7 T
! e9 \" W" }3 A, J0 F9 \8 @4 _ z 第二行:OOXXXOOX XXXXX (第二行也是全满的)) L& Y$ N' O/ ?2 C
& z \/ W5 y4 F' L p 第三行:OX OX XO XOOX (第三行才出现洞)
$ C2 p! h! ?! G* B7 a: \1 S+ c! @( b
第四行:O XO O XXX 4 I3 Z3 o1 z# m& t$ |! k
" t6 A' }1 r! W. P; J, N 第五行: O R7 S: K- L9 k7 h. q8 s* f
) x" z1 T* c' h5 c
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)。1 V2 ?% P) z& v! n" }1 g9 L$ T9 [
6 t9 K: j! Y6 m: e, }+ ?& Y7 n 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。
; {* F6 `- j4 y, i: x( x" `/ E E% J2 [( `: U7 h9 E$ @9 R
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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