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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。 n# t* z6 | v k5 K% a2 d2 Y
1 ^/ }; a, Q% a }3 y3 R 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)
* @) G% Q7 t9 J' C. S6 g8 B9 X. ?3 p# A( H; x4 @( n* C6 K
第一行:OXOXOXOXOXOXOXOXOXOXOXOXOX(第一行永远是单跳) L* L: |# d1 _7 S9 `1 x# a' x
" I( q2 y) a6 P- e: ` O 第二行:O X X O O OX O X (第二行开始出现洞)
) f& K f. Z. C8 S) E
, N. {9 b" A' p6 j, Z! }/ B 第三行:O X O X X5 g6 x8 o( F9 L% h3 t
/ k( I$ Q: c! L. h4 o7 g* K4 u 第四行:O X X q# V! C2 n0 ^1 r' @/ q
3 Q1 Z5 q+ e$ S3 G 第五行: X0 |# W3 V3 c7 E% [6 h
% W! W+ ~6 ]) {: I% \ 那么新写出来的大路是怎么样的呢?
& n, t* Y0 R1 I6 z7 v( N' \! G4 C4 C, V+ 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。
$ c* x: l1 b5 z$ v. v
3 v& |) ? r6 y0 D% G9 M 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。)( E4 m( a4 B$ g+ s' ^
5 f- W4 V/ ?# U) n; f! t g* } h7 w
第一行:OXXOXOXO OXOOX(第一行不再是单跳了)
) v2 t( b" D4 \# s7 } C& b
$ `/ r9 E* q# l4 i* |) p 第二行:OOXXXOOX XXXXX (第二行也是全满的)
, R$ ~) t" a! r c; T
0 l. d1 `7 |, l0 a0 u! @: @7 n 第三行:OX OX XO XOOX (第三行才出现洞)2 m5 G7 O% { K6 K
/ r5 X7 T V9 Y) \# D2 b0 l 第四行:O XO O XXX * b* s; e9 ~) x: o
6 M; j* t: q3 b k( S9 S 第五行: O' J7 m' s% S# Z9 R ^% d7 P2 q
& A/ ?: ]# o" }7 Y: ` ]: c. N' F
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)。, {. `, a7 y. P, g1 B" T
# F- T$ C$ y4 }' S 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 ]$ G+ l& W
0 L6 U* I- r/ h/ W) S1 q$ X& o 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。' ?/ f4 F8 U) o
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