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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。: E/ X0 k" W1 T' o. y1 O$ d3 `6 K! e
. ~* }; D A! }# H( s) ]+ @ 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)
* }* a) i" [6 q9 b; d/ e' }% @
7 M1 D1 J1 _! u$ a3 g( ] 第一行:OXOXOXOXOXOXOXOXOXOXOXOXOX(第一行永远是单跳)' k! j' R8 c; L( w0 i% _# i
$ `) N9 K: Z2 X1 s, ]* O 第二行:O X X O O OX O X (第二行开始出现洞)7 N: G! r, n* J( w$ h# R
% }& e: w; ^4 w" Y& D$ Z0 G
第三行:O X O X X
/ N( @9 A% K6 Z8 n z: e9 o) j3 Q I3 t" Z1 f' C
第四行:O X X
0 n) k( F, R+ w2 I; k; K; `9 k# B$ L( E5 q N5 F2 z
第五行: X# m0 |5 \8 {& R! s: I- l8 c9 p
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那么新写出来的大路是怎么样的呢?- y: p' p+ k$ ^# x1 O( c" |' v
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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。7 i& J( K) g8 G! j
' c- l2 ~4 r0 r. a0 B% g7 C 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。)
" e/ D8 |$ q K5 [
! i6 J% k% u% W2 \) H2 N 第一行:OXXOXOXO OXOOX(第一行不再是单跳了) z: Z" [/ c$ G, L x
$ `) T3 \7 |- b2 A 第二行:OOXXXOOX XXXXX (第二行也是全满的)
) r2 H) H% q# N; ^% E. B) W8 ^/ B& {/ W* |
第三行:OX OX XO XOOX (第三行才出现洞)
5 h4 ^2 A" n1 E1 j. ?1 r! G5 v; j% x @7 N* s4 u8 M8 x
第四行:O XO O XXX $ R+ p1 b( Q! r; k: T7 H
1 f# \ R$ @' U& H9 o) W0 [2 |3 U
第五行: O
. m5 v) \% \0 v4 j" m
: m* m7 u6 u* v: F! \( ^+ o2 m 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)。
9 p/ G0 T% V' d* i5 _: @- g7 `) j( c3 b1 d2 y& I: b2 `* J2 ^ `
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。" H9 x' H% v; q q E& y2 ~
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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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