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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。
+ L- k2 n4 \" v; l# {' f. V6 R' K1 S
k' N( N3 S* T 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)
b7 K# G. C( f. z. K' z
# f0 S+ W) D E7 r) t 第一行:OXOXOXOXOXOXOXOXOXOXOXOXOX(第一行永远是单跳)
) |4 n- v4 T8 i
8 f( M6 _3 s! o% E9 U% p! d. j 第二行:O X X O O OX O X (第二行开始出现洞)- @, ^# b. x. @ U& D8 Q/ v$ E
1 h D! b0 H# f2 m3 L$ K! i
第三行:O X O X X. w* _" U5 j4 b# u7 w8 `
@' v/ ~0 n+ Y: n) d 第四行:O X X3 \0 q+ y3 z" V) H0 }( F/ T
# q/ Y& E" l4 A
第五行: X
7 O% I: e, l7 Y
: |6 l: T7 @* q' Z- a 那么新写出来的大路是怎么样的呢?, O( T* t3 P$ _9 @2 X/ `( Z
8 J1 b5 t4 K! z: D' K9 z2 R
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。: V3 t( H5 E/ L/ ~9 E' O8 b; Y
; [4 F% v+ q) N5 d- X2 m- g1 Y3 b
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" j6 B4 i* p5 }
. V# p$ p, F9 M/ g* b9 f 第一行:OXXOXOXO OXOOX(第一行不再是单跳了)
) \3 \$ r2 q2 K8 H% a8 P' q4 C5 ]- }
第二行:OOXXXOOX XXXXX (第二行也是全满的)
. Q* l7 a4 K1 b6 |
" {% ]5 b4 ]3 h) R9 c5 E$ F) O8 [ 第三行:OX OX XO XOOX (第三行才出现洞)) f8 i5 W" p4 R. q! M" X2 ]; H; D b; e
, e8 L* _1 \3 E2 j/ }5 ~1 g; ? 第四行:O XO O XXX & l% u# w7 n* N0 R+ ^ R
1 z0 X- p) }: @- C! X J6 _' M: F 第五行: O
- _/ X/ k' a: @
" c& E. E+ c6 H& s: N& s4 t 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)。- L- n; Z9 C8 I8 H1 f8 ?
! X% u, C5 ?8 I& t% S5 }# W# E! O
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。
4 W. `0 V' l: q2 B7 I+ C. F; @$ x- E5 B& z8 z5 Q
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。0 @9 K' C/ p0 b8 C- I' _& y) s
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