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通过我几个月来对珠路理论的研究,发现了一个有趣的现象:每一条大路,按2珠路排列,有2种不同的路数;按3主路排列,有3种不同的路数;按4珠路排列,有4种不同的路数,按N珠路排列,有N种不同的路数何解?我举一个例子,假设 B= 1 , P=2,T暂且忽略吧。2 C$ K. [. E4 ~. G
1 B# u% f2 X+ i' E$ Z2 X* C 假设大路开这样的路:9 d& Y' n/ \, q, N" c4 `
9 a# i y! ?- |7 x, `
12121212121212121212121212121212。
% s. |6 F' y6 ?- ~1 e v; n- ~! n$ b7 [; S* m& `
按2珠路,就是BPBPBPBPBP。% I4 `6 R, n/ x& u; _! w, l
1 q' \% h# `6 B* B6 m* S5 X 如果我们去掉第一口,就会出现完全相反的结果:; K* a5 l8 X& x3 h, v
8 E( }4 T- D7 ?, } 21212121212121212121212121212121。- Z& N; p- Q: s
' }" N! c# g0 h& O5 ` 变成了PBPBPBPBPB。
; l- p3 t2 Y0 e' B6 Q; e* J h+ l3 T' r- z( N
如果我们再去掉一口,又返回第一种情况了。* k/ p8 F) L z1 }& S1 i/ ^
9 i& g! i, m1 g( F( p 所以每一条大路,按2珠路排列,有2种不同的路数。
6 R7 M) H) X7 c1 y+ t3 K
, j3 H1 \% ?& C8 M, k% Z: G/ i; T* ` 再举一个列子:7 i" \( V* }& w' f Y" m! q+ ~+ P `
' y- ]& {3 Q7 t4 S% s7 W5 { 大路:122122122122122122。& @9 a g2 K5 s: \
4 {1 @4 p2 B& K. j) I 按三珠路排列:6 l/ u$ [0 v3 c' j3 T+ B
+ z* S) V, @4 n2 s6 A k; n( t3 V( | 122,122,122,122,122,122。! U6 A9 B/ ^, y' t1 R0 ^
2 w& k% j8 m7 q* Z# g0 M" G
去掉第一口,变成:" d& O* \! h5 J" X
5 M4 J+ l% P3 d2 f. e+ q: a# T
221,221,221,221,221,去掉前2口,变成:
) v$ D: k7 u: Z' y
2 K+ i/ [8 `" V- [/ T7 I# e# V 212,212,212,212,212,去掉3口,又返回122,122,122,了所以每一条大路,按3珠路排列,有3种不同的路数。
* e5 J' D1 g/ j3 ^) D5 g9 }5 p5 d0 W) v g2 L8 z
同理:按N珠路排列,有N种不同的路数。4 `1 n$ a4 p2 g1 x7 M
: s9 n! e I1 @1 P* }) E 我提出这个的意义在于:
& ~5 P, Y8 u: L4 \. F0 n! n6 a1 }9 U! q' P; x
1、字串81、每靴的第一口为起点来编排二三珠路,与第2口,第3口开始的排列是不同的结果。; M( R" ]8 J/ \* G$ h7 q
! x7 R0 y' G) j+ b4 |( \
2、为三多理论提供了下注的多面性奠定基础。
+ e& J9 ~& Z$ U7 q4 J9 n! o3 J, |
0 Q9 c0 {5 \7 W @ 3、某一靴牌,按第一口开始的珠路可能是 烂路,按第2口,第3口开始的珠路可能是上上路。% _3 y9 I7 [7 @0 u" U0 K d' [
, w u4 b9 _/ ]! n* i8 R |
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不用花时间去研究这个思路,n年前就见过。结果都一样,下三路就是个例子。 |
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稳稳的发现没问题了的啊。- a: m: }% x/ \/ k9 M2 \
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楼主的理论打法不错的呀,赢钱了应该是打法对你有用哦~来收藏下了解下了 |
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