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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。
% G7 r8 u; h1 |) ^* H8 p: a. W6 p( `7 u9 A6 t) p
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)
' j! ]5 E3 c, J* Q. w9 Q# K7 \$ U: a+ [5 [
第一行:OXOXOXOXOXOXOXOXOXOXOXOXOX(第一行永远是单跳)* _! W4 c; g- }8 C/ x- A
* P7 j( K, I9 Q; a
第二行:O X X O O OX O X (第二行开始出现洞)
Q! I3 h. ^: R. t% Y. O4 ]. K9 e# |3 g
第三行:O X O X X
7 T9 k2 n, I: S+ f- h& e) o5 y
' U& {! O( z: v7 H: \( ? |$ b 第四行:O X X+ w( l4 j7 V8 {& D4 C8 E" \ Y
, E7 p; s. i8 V 第五行: X- ~) Q1 I5 h' t, g: }
9 h2 E5 J x2 X( J" j, w, j7 L
那么新写出来的大路是怎么样的呢?
# k9 O, m ^! B( ?2 F
@1 Y; B! v. j9 o( 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。
+ [1 i0 `7 q8 y3 U
! W1 ~+ m+ W, w- C# s/ ~2 o 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。)
( s! n; p$ P1 b4 g. b' p0 U6 L' e! u0 Y) S5 M( `
第一行:OXXOXOXO OXOOX(第一行不再是单跳了)1 e+ z# v+ V/ d
$ j {7 j) H5 s: V9 c d
第二行:OOXXXOOX XXXXX (第二行也是全满的)' k' v8 R) A* k/ }; W! y( {6 t' {
4 | D$ Q" W$ n$ f# a. L. |) e
第三行:OX OX XO XOOX (第三行才出现洞)
4 Y1 B& {! K# W/ L% R' E) q6 U
3 A4 ^; o$ Q( R% v: w7 P' e 第四行:O XO O XXX 2 @* Y* E6 P/ a: A
: r3 N! ]+ |% u' t' c 第五行: O; s' e- P$ f1 `: S
) n, q3 R6 n4 K/ ]% 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)。* M4 {( V; X3 o8 {! r8 e
7 V7 q% P7 Y8 q+ d0 s0 y4 U: Q7 X
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。
% }3 B% U X" j( B
# u2 c( l o3 e5 q5 l$ ` 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。! u" L6 m2 S: u" m5 W
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