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事情是这样的，我最近在研究团队编组及内部模式对发挥团队能力的影响，以及如何正确编组让团队能力发挥实现最大化，别问我为什么研究这个，反正稀里糊涂就研究上了。我发现在描述团队编组间及内部同步能力的时候，人们对Kuramoto模型（藏本模型）作了大量的研究,其中包括模型达到完全相位同步的充分条件、耦合强度对于同步的影响、一定条件下振子的收敛速率等。但具体实现一般都在MATLAB中，且网上代码过于复杂（我运行了一遍一堆报错），这里我使用Python和MATLAB对Kuramoto模型简单模拟。
" y7 @7 ^& _/ Y8 b4 w1 m模拟的话还是一遍举个栗子，边分析边测试效果最好，百度学术上有一篇关于Kuramoto模型的简单论文，我们就用它来实现模拟。4 N) A0 b H8 m
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好吧，那我们开始，首先是Python实现：0 p) g/ k4 d/ I+ P" A1 K! ]" N
: C+ d2 l. I8 G9 R3 m, o3 Z% W$ [3 g) W( IN维Kuramoto模型的数学描述如下：
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6 g4 l# K8 t- U! s3 y; w( }没错，是讨厌的数学公式，没事，它可以改写成这样：( v$ n _2 D9 W- R, w/ G6 ~ r6 d
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; j& S- g% p/ S好像还是有点长，那我们在改写一下：
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4 H! B* y( s9 g; p8 A" B看着好多了，那我就来说说式子中参数的意义，Kij为耦合矩阵，是为了便于描述不同振子间耦合程度不同的情形。最下面那个式子的r就是我们的目标，反应振子间的相关性，这个相关性就可以描述我们想要的编组内部同步能力。
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哎呦，这个式子看起来好简单，这里要补充一下知识点：同步能力可不是一下子各组该怎么同步直接确定的了，它是一个从开始到稳定的阶段，也就是说随时间变化，最终反映在各组的同步能力才会确定，那么最后图像是什么样子才算同步能力好呢？5 n0 {$ x# q9 d0 `8 w8 N
. l& M4 D3 I4 [( Y; }同步能力好，是指随着时间的推移，各组的同步能力r逐渐稳定，波动现象消失或固定在某一个小范围内。需要注意的是这和各组r值之间的差距没有关系，我们要的是一个平稳的状态。那怎么办找r和t的关系呢？ A E8 {1 h0 X& A1 Q k- X- Q9 m
8 g0 k% q7 h- t3 N! K# Q注意看最上面那两个式子，相位（第一项，等号左边那个）上面有个点，这样他可就不简简单单是个相位了，它代表的是相位的变化值，是一个微小的微分值，好吧具体意思就是，那个式子左边展开之后是这样的：
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$ o" u; I$ p6 y' ^' \( ^哎呀，t出现了，其实\theta 与t有关，这里你可能有点绕，因为它们之间的关系是一一对应的，就是说每个时间的t对应了一个\theta ，我下面带入具体数值的时候你就知道了。& `: X2 x' C% o6 ~+ @
, K0 F7 U: B `% e. p" c4 n组间同步能力与时间t的关系出现了！
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也就是说我先用上面的那个公式4计算出来\theta 的值，在带入到公式5，那么t-r关系就可以明确下来了，那现在我们再回过头来看看文章中已经给的例子，看看还有没有未知量。
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栗子是这样的栗子：
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假设某机构内部有 4 个编组，每个编组包括 5 个节点(其中 1 个节点为领导节点) 。另外，将上级领导作为一个独立的编组，且只包含一个节点。假设在领导机关增加4名信息传递人员。当以独立编组模式编组时，指定1名信息传递人员为指挥者，其指挥关系与其他编组一致; 当分散编组时，信息传递力量节点的关系与所在编组其他节点指挥关系一 致。其中，完全分散编组模式时，各信息传递力量节点之间无信息共享通道; 不完全分散编组时，在各信息传递人员节点之间建立一条信息共享通道。各编组模式及其拓扑结构如下图所示：
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% m i4 I% n0 u, ~$ w独立编组模式0 o7 {8 l& L% w" B# u8 k; a
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+ x2 t4 d, s# m9 o3 w8 X3 E9 L完全分散编组
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不完全分散编组
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参数数据
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参数确定一下有没有未知量：
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首先N，数据数目已知，这个有了。6 i6 ?8 }6 d& p( G, y$ U% u! Q3 D
0 y1 w* X( T; H/ j! z# e7 y) uK值是分组内的连接强度，这个是看实际情况，由甲方提供或者自己看着给的，这里就是甲方给的编组图，i与j点的链接强度一目了然，这个有了。) o |# H$ m" ~' j, f# R- M& ~
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\omega _{i} 是振子i的固有频率，也称自然频率，甲方会给，没法自己估计，这个有了。
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\theta ，\theta 怎么办，初始的\theta 会给，自己也能测的出来，但那么多\theta {j} -\theta{i} 得多少不知道啊，这里通过翻看文章，我发现其实文章是有一个特殊条件的，不然的话是需要研究耦合因子求三种约束条件解情况的，特殊条件就在这：
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$ Q7 X8 }( s& Z/ Y. h假设编组内节点的初始相位差为π/2，且编号最小者为0，随编号增大而增大。
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: T8 A1 O* R+ p) R% J! O0 ?4 P哦，初始相位差知道了，你还告诉了我各个初始相位，那么\theta _{j} -\theta _{i} 的值就在一个范围内的几个固定值里面啊！
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好的，没有未知量了，就是找K的时候麻烦点，没办法，这个决定了编组的不同，写脚本算一下吧：
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#codingutf-8
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5 Z2 @6 ?/ ~- Z##ScriptName:KuramotoSimulation.py 8 @/ Y: X( ?% G9 y# f, H. c) g
( |2 d& `. E) h# M& N7 wimport matplotlib.pyplot as plt
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) M9 G1 O0 `5 R5 u# s9 q- p! o0 zfrom pylab import " \2 n9 a. p; ?
$ h% s' O u% i8 u, l! ]4 t9 A& `from sympy import @5 U C6 W0 y# l
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from matplotlib.ticker import MultipleLocator, FormatStRFormatter # ^ O0 \) N/ N" M
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import math
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import numpy as np , a% n, i8 y+ w& u0 _* c m
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N = 31 #总节点数
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c=[0,0,math.pi 2,math.pi,3 math.pi 2,0,0,0,math.pi 2,math.pi,3 math.pi 2,0,0,0,math.pi 2,math.pi,3 math.pi 2,0,0,0,math.pi 2,math.pi,3 math.pi 2,0,0,0,math.pi 2,math.pi,3 math.pi 2,0,0]
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w = [4,3,3,2,2,1,4,3,3,2,2,1,4,3,3,2,2,1,4,3,3,2,2,1,4,3,3,2,2,1,5]
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k1 = [
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. P' C* O; C# E* _& n[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
3 O6 |* v# I+ |8 J+ @6 ~' E1 v5 j$ I8 u! D" @5 R
[1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2],
1 g; S" B+ o5 J& a! v& G: O" i* j
* z [4 `0 \) k* G# R, K[1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], 3 Z& w6 @3 G! z& T; ^5 C$ }
8 b0 J8 F7 [8 t4 H+ m[1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
0 b7 L5 n: {/ g7 R$ ?" J
# x+ ~$ e0 @; y0 s9 k[1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
# z7 t' u+ ?3 U: A2 p# k& r3 t3 b" Y- N. I1 h- i: Z
[1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], $ a$ W* j& |9 l; @
) H4 p! q& P3 C1 s0 T2 w; x6 K: j
[1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], 5 ?$ p' m5 @' G G z$ N# N4 G
" N1 e. _* q1 O4 X+ }[0,0,0,0,0,0,1,0.9,0.9,0.9,0.9,0.9,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2], " k" d$ q: C9 F9 M# K- D
/ m) w* }, I$ J. |5 _1 o( D. @
[0,0,0,0,0,0,0.9,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], 7 c! u6 K. I, v& o
4 a2 ?7 P+ b7 }5 h8 w; Y[0,0,0,0,0,0,0.9,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], + G) Z+ n1 D& o% {( W
5 i( v5 a1 n9 _, @, p: V[0,0,0,0,0,0,0.9,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
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[0,0,0,0,0,0,0.9,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], y6 d& o# S# B7 `$ x _& Z e
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[0,0,0,0,0,0,0.9,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], * J* X N, H( w. k' \
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[0,0,0,0,0,0,0,0,0,0,0,0,1,0.8,0.8,0.8,0.8,0.8,0,0,0,0,0,0,0,0,0,0,0,0,2], # z+ D1 C( h0 M# e
1 u$ p; y4 e! L[0,0,0,0,0,0,0,0,0,0,0,0,0.8,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], 0 J( o' Q! R; p6 {; x# o
7 X2 J6 F8 C& c& t& E% g[0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
# g7 t- e6 i3 ^# N4 q4 j& e8 i0 O( G1 x7 \- a
[0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
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[0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
: C. o4 i, Q6 M: T
4 C! t+ m2 G: Z3 ^6 d[0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0],
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[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0.7,0.7,0.7,0.7,0.7,0,0,0,0,0,0,2],
& M( ^5 w/ c& G3 W- y; m& l! ?* g9 G- |* ^
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,1,0,0,0,0,0,0,0,0,0,0,0], ( I( f) h3 }$ E8 G# _. J
* Q4 p/ P$ X( f' k7 y[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,1,0,0,0,0,0,0,0,0,0,0],
/ T m8 b; Z* [+ f
8 K* [+ V( e: G+ P7 b8 w[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,0,1,0,0,0,0,0,0,0,0,0], 0 i/ S' w2 d5 C9 E
1 @* f& [- S& T. d, p2 @[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,0,0,1,0,0,0,0,0,0,0,0],
' s o* X5 b$ v- ^% Z
5 b8 v+ I7 e0 z[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,0,0,0,1,0,0,0,0,0,0,0],
$ R- y/ M! T3 {# S2 x' d0 D2 ^1 R0 @7 K4 J, |
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0.6,0.6,0.6,0.6,0.6,2],
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5 n" W1 P' n c* u1 p! M7 c9 i[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,1,0,0,0,0,0], $ L& ~+ _" a' m3 J
0 y" c" d8 Y7 y& ]
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,1,0,0,0,0],
5 k7 m) g3 e% l% k0 u8 S# E6 @. c) W* W* e2 z0 v& u% s, A# x; n
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,1,0,0,0], 6 `5 Q( H& z* p9 }
6 P3 g/ h) n# d0 U2 u# q! |# M- N[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,0,1,0,0], 1 V& U& I1 g) g& {3 F- F
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[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,0,0,1,0], ( _9 W5 b0 I* g- L, J( Y
! L% T3 }9 N! z, s Z( u[2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,1]]
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) G3 R E' V- an = [i + 1 for i in range(22)] #目标划分,24个值,1-24 3 {: L8 J# W. q; U
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t = [j for j in range(1000)]
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ci = 0
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C = [] 3 F1 l1 F& N: d4 q" F& G
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C.append(c)
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, C! @' E* |" Sfor d in range(1100)
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0 F. @3 M2 k; l for i in n
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for j in range(22)
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6 E& d3 y% f# g8 @+ F6 D cj = c[j + 1] - c
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ci += k1[j + 1] math.sin(cj)
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7 J; \8 C0 x. B _" X- H8 } cii = ci + w # K. I; W5 u5 t& i: }+ E/ D
7 i1 c" E% ]& r) s h = [0.01 cii + z for z in c] 9 p* @9 p' D* e/ I% |5 ~
7 i& B9 f1 b9 a" W9 U5 g C.append(h) ! Y/ g% N5 q _; _9 S9 l' S
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c = h 3 a2 L; J% {* ^0 Z" @) J
8 H4 k" u9 \ L5 J _8 {def r_function(u)
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y1 = 0
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v+ s2 A& a- {* i/ I y2 = 0
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y12 = 0
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0 |+ p: I" V8 u& m1 q3 q% p e y22 = 0
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r = []
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for x in range(1000)
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for ul in u
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0 g- f) L$ C. u y1 += math.cos(C[x][ul])
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3 ]6 `$ V" [' L: W3 w y2 += math.sin(C[x][ul])
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' L" `" ^: h1 @6 N) u y12 = y1 2
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6 T. R0 ], a2 x' E y22 = y2 2
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r.append((float(1) N ) ((y12 + y22) 0.5)) + c' n) E( r1 q1 L) r
" M" ~2 f0 W% b, X o7 E return r
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r1 = r_function([1,2,3,4,5,6]) / Q4 y4 u4 q, f1 t \) {' E
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r2 = r_function([7,8,9,10,11,12])
) H$ C9 i' {- o0 d2 S. k3 ]9 P) l/ P' y- }- Q* K N- B4 }
r3 = r_function([13,14,15,16,17,18]) ' d/ H2 u T6 e$ O5 u
3 F& \$ T) l1 g3 }3 ~" C
r4 = r_function([19,20,21,22,23,24]) 9 a& ~( B# V; m3 H) X T
1 C1 B! x, G( t+ i/ ] {! Z2 {
r5 = r_function([25,26,27,28,29,30]) 3 Y G! b* n, C+ P7 K( l
: l5 h5 M+ c8 Y2 M4 y( ]9 U
#r6 = r_function([31])
; E' k, K1 M) ~ Y& {) K1 O! w* u; G& @6 g3 Q2 }5 X
# {+ h; a' {' k$ V$ I& Y9 l0 e C% }* c6 `2 o* c
ax = subplot(111) #注意一般都在ax中设置,不再plot中设置 , P. O/ e/ t9 Z
9 e! S. r1 @4 j( y7 EymajorLocator = MultipleLocator(0.1) #将y轴主刻度标签设置为0.5的倍数 6 F6 I7 d8 {$ \0 ~( ^! t
: k& H+ M6 ^: ] U1 f# n$ R
ax.yaxis.set_major_locator(ymajorLocator) 6 Q3 t$ h" ?0 R, h$ r, u
) N, C+ v* `# @5 j7 G' Z& Yplt.plot(t, r1, marker='o', color='green', label='1')
4 R2 K* h0 @) t+ K1 d8 y" T1 ^/ t, d& H# C5 n5 Q' D R# D+ d
plt.plot(t, r2, marker='D',color='red', label='2')
( L2 k, u8 b# f3 h1 m |- ?- _3 V) @( f3 E/ P/ q7 ~* Y
plt.plot(t, r3, marker='+',color='skyblue', label='3')
7 L- O ^0 O0 k" Y$ D0 H) ~% N9 C7 F4 \
plt.plot(t, r4, marker='h', color='blue', label='4') ) h5 k7 O3 }% l8 q0 Y* @
* A1 f C6 S3 ^- Y5 K. Y
plt.plot(t, r5, marker='_',color='yellow', label='5') 1 U% `# M, R9 b4 z0 q$ T
. a7 |' j. z/ W1 x V#plt.plot(t, r6, color='red', label='6')
& v% i n7 W. J) y$ S$ z" k) Q( Y" U X* S
/ h. @; Z7 S+ r9 ^# ?
# t) c% q& n, k
plt.legend() # 显示图例
6 ^8 s/ ?$ v' l W1 s4 V4 k5 |
! x. P. x1 s2 cplt.xlabel('iteration times')
2 M( R+ M- }9 B
* P' S# q( Z5 o4 |5 rplt.ylabel('r')
P, ~1 W# |, v2 W1 c: K% g" N! i. D$ S) W7 j# t$ c
plt.show()
& a+ c( Q5 ]1 o, H2 M- H9 o
0 h# F$ d3 j, y5 ^0 U% ^0 h8 y9 A; x7 Q
独立编组结果如图：
1 V; h( j: r" N
7 v! O0 }3 u% g0 J- a& ^& u& m5 R6 J; O
4 ]9 C1 `+ |1 b# e- b9 \$ `( Ak1独立编组
8 ]( z0 Y! n, B Y好的，从图像我们来看看Kuramoto模型在描述这个编组的时候，5组最终稳定，我们说这个团队编组还算科学，但我们改变一下K的值，换成分散编组：9 T' w2 { O U% K& c# z
% O3 j' D* ~. `& y% R1 tk2 = [
$ J6 l+ _: X, b! X# _+ P
k1 v7 i# n8 c {) _[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], q9 ?) V% S% D+ w2 D3 h' M
0 v. }; o7 L- p5 Q; p. n[1,1,0.9,0.8,0.7,0.6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2], ! p' z6 K ~2 \3 Y, E
# u3 _3 A- H: L1 z! n
[1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
3 R( P( `3 O8 F, v' D7 F9 {
! A8 t5 A7 J; ]3 O7 C! ~) U# \[0.9,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
8 `; w; ]4 { G4 A
" g# G6 w, L9 g% |/ x9 x[0.8,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], # B3 M# S" v4 N) X4 B0 d
9 h' L8 C+ b N# H) _) Q8 S, M[0.7,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
) J5 u. G# L! d7 ?( v* E. P( p( K. f
[0.6,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], " c) K1 x7 ^" G: L& r
0 x" t4 F) }% w2 `# Y0 {[0,0,0,0,0,0,1,1,0.9,0.8,0.7,0.6,0,0,0,0,0,0,0,0,0,0,0,2], 2 E& j6 \! ~4 A* [
$ E3 V1 ]! y( Q* `; N
[0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
, C, Z7 Z0 p0 A
/ S) p, a7 z5 K* T# L/ ^# A3 R. s[0,0,0,0,0,0,0.9,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], . C7 l! w* w, [' z0 Q: e/ [5 j, X
7 a/ E+ Q! F1 a2 M0 h3 z$ _[0,0,0,0,0,0,0.8,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
2 s2 V4 `# N; h+ ~& F
9 e9 G- U2 B! S5 G; j+ a+ b! L[0,0,0,0,0,0,0.7,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
# v( A+ w" B1 \ A
0 [( W5 n2 ]) c[0,0,0,0,0,0,0.6,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0], 0 Q+ X) E) b0 N8 d
3 h+ D. {; n i5 Y3 V! V[0,0,0,0,0,0,0,0,0,0,0,0,1,0.9,0.8,0.7,0.6,0,0,0,0,0,0,2], * i! e. e' M# M% V9 X e
8 s; ^6 I% D4 R
[0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0], 2 C& Z) w0 ^: A$ [1 T
# i) ?2 f* F( J
[0,0,0,0,0,0,0,0,0,0,0,0,0.9,0,1,0,0,0,0,0,0,0,0,0], # ]( Z% ]6 T) H& o
, A& j' O b& g q9 N. E$ O[0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,0,1,0,0,0,0,0,0,0,0,0],
! Q& d4 |2 ?* _9 p7 f7 \4 l' W1 r1 L( E. j) M. G% L. I0 h& l
[0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,0,0,1,0,0,0,0,0,0,0,0],
$ Q) X- P# H! |0 [* P" d3 W M7 Y4 r* |2 A0 d( X l7 }- j8 t: s4 n
[0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,0,0,1,0,0,0,0,0,0,0], 5 f2 R" [2 s1 H6 B; X* F- }
# H& {8 T$ h- W5 ?
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0.9,0.8,0.7,0.6,2],
' H( P2 V) w; G2 q
U: D, l1 g4 b) |- D1 i6 y[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0], # {6 s5 n j( G; J
# h2 @, @! s6 q7 a
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.9,0,1,0,0,0,0],
8 _- x$ b5 b2 V7 i) g2 f* Q7 W/ g: h9 r
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,0,1,0,0,0],
( A. K: x( d0 T% `1 e
& l+ w7 P; P# C1 c1 I[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,0,0,1,0,0],
" b: ]; W8 Z1 F, ? \( H0 Z
; \; c1 b& r0 v- V2 w% o+ }[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,0,0,1,0], + ?6 C8 X+ P j$ _7 z( Z
6 a0 e+ y2 J( K0 M2 C+ x' k
[2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,1]
& [$ }1 x& _6 V. N1 n) O
6 H# J: h6 F* o8 E1 q: Y0 a/ c9 C [# a7 Y7 t+ d! o$ Z5 T
3 n8 W" I1 `, q) s
k2独立编组
6 A7 A! Y: g3 e$ Z
: T; v/ C3 o/ D0 Y) g
6 a* G0 D5 J' L$ x8 B# S
* n3 p0 }7 _# { |这两幅图像都在开始阶段大幅波动，而后在一定范围内趋于稳定，那么到底哪个分组模式最符合实际，最能突出编组能力呢？6 L5 o9 I1 z' h" h+ ]* I) S7 o
8 @; A( @ w" ]# }( [
这里还有一个公式，来解决这个问题，编组同步能力的量化：8 C$ @$ ?( D$ X5 h0 J
# n& f5 t" |# Z4 Z4 |
( |7 s. S8 @2 m5 Y
M_{s} 就可以描述某编组的同步效果，r_{s} 是达到稳定状态后序参量的均值，β∈(0,1)是调节因子。我们可以用M_{s} 比较编组内部的好坏。那编组间能力的好坏怎样比较呢？5 c' H4 D# n* G/ o/ o
4 y) n6 ^5 |2 k) p) c
这个Kuramoto模型同样有所考虑，它有一个描述整个系统编组能力的公式：
% Q, J/ Z/ n2 I7 ]# |0 s
: S3 G. y' G2 r* P' `2 P% u% c( Z! L
4 ^1 H/ L0 K( J5 V! N% ^其中，P是编组的数量，M_{i} 是第i个编组的同步能力，\omega _{i} 是编组在整个系统中的权重，\psi {e} 是各编组平均 相位的均值，\Delta \psi 是各编组平均相位的标准差。具体的计算不是这篇文章的重点，就不在计算M{s} 和M的值来比较上述例子独立编组和分散编组的好坏了，本篇文章主要是讲下Kuramoto模型的解决思路，尤其是上面解决\theta 值的方法可以套用在其他Kuramoto模型中，做一个目标估计绰绰有余的。
9 s9 N$ @( Q8 b/ X
/ { B/ K: {2 N% z4 @5 E% ]下面是解决Kuramoto模型常用的MATLAB编程方法，具体思路与上述基本一致，这里不再赘述，K的值我们给另一种编组模式：不完全分散编组模式，也是现在实际上最长用的编组方式，直接上代码：
( M5 v h+ H4 c3 J2 \& p) J h9 n' v5 `) F
clc;
9 W0 Q( ~" f+ ~% M; m, N8 q5 f4 u/ Q' V
clear all; ( E+ y& R+ v7 w
. W% w6 G% K+ @0 ~) p* p! \
k=[0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.5;
$ D) v r6 H5 s' r3 a/ r9 W3 q( F1 g
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; " p/ o4 w. l2 f9 t+ k* |
: M9 D0 W- z+ L8 {! M" v
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; 0 @4 G% i. {6 A6 B' q( y. n
# X9 e; g! C# Z- | 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; 7 _! P# ~' M" k
! r+ a1 G. A& ~. ~+ l 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
! l, t' K; y# A! Q# }9 V6 C) w& v* Q1 r
0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.5; 0 {+ w$ f; i# h }
3 \8 W5 ]" ^0 D' @5 I% ~
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; . {# t3 a# p) `8 `+ q6 b& W& M
Z$ @5 \- A& y. G 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
; I3 }- `; M, q. q7 b0 ^* l' e
6 G5 j$ L8 q( g 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
+ z0 h+ N' E# B: m5 b& L+ }$ U J y/ q
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; 9 c* I" Y J: a+ ~, Y
F* f- ~1 l: Q
0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0.5;
9 s2 S/ d; M! ~' e/ J, |$ t ]$ M& F3 n7 }
0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
* J8 h8 N8 P. H3 l* ?3 J5 [8 o% r2 ]' A2 ~0 Z2 X
0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
: p7 D$ F7 [6 M: m# [
# J7 | w! y7 O+ J 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
% v# W' m0 o# n1 h; ^" E' [/ c
' q9 P0 ]0 N7 S! [ 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0; / D d6 h- t2 {9 O+ p6 G0 u
( X: O, f! O* H) z0 q* g$ \
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0.5;
0 ?3 y. g" r% U7 J- [6 M4 k! k
4 s5 ^4 Y( m& Q8 j2 c 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0; 0 l+ J# H/ Y) l; g6 K
9 d, G* M1 M0 w 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0; T) l# u2 B" d6 j$ K- l5 A
8 C! @ H0 T/ B$ U$ s* | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0; , f' g9 b R: o( S* Y* O
, O. a3 w' q: c1 n# P" d 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0;
& p r ]2 s2 ^9 u- i J* G9 x" {+ a( b7 M1 D. Y/ L0 D$ K6 c
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 0.5; + ~( z; X' ^! `! P, K) g
% z, L/ l$ a0 i, g ^3 ~ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0;
6 I, ~6 n; g7 K) u, w- E: }0 L }, f8 v. M) c' I
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0; ; a W4 r3 Z4 n) x% d
& f: U% ?0 X* D
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0;
# Z; U9 P- ~ z1 N5 P1 D
+ s$ a% O+ T) `# Y6 C' K3 \2 H- Y 0.5 0 0 0 0 0.5 0 0 0 0 0.5 0 0 0 0 0.5 0 0 0 0 0.5 0 0 0 0;
7 | j# T. q( u+ S7 }7 D& w: }9 U
] , Y7 s5 J# s- w, [+ j3 o% u
- P3 s6 _ k8 k9 j( }
N = 25;
7 `% |" B$ ]/ P+ M8 T! Z. Y! K' R ~' ?
$ b% K& } _3 l. C8 KStep= 10000; 7 F& V* b8 z1 O
3 B! f3 v! A) ^, s1 ]9 bTheta=zeros(Step,N);
/ j* f% t: R3 I2 d3 ?& G+ S: F- c- v# @
Omega =zeros(Step,N); , {( Q' d) E: F# T: n( r
2 B. Q0 |1 F6 K2 a8 nDeltaT=0.01 " r$ Z3 v( B( `( i) H+ |4 Q R
N: e* x0 U$ M2 W* v$ }% x: nTheta(1,：）=[0 pi/2 pi 3*pi/2 0 pi/2 pi 3*pi/2 0 pi/2 pi 3*pi/2 0 pi/2 pi 3*pi/2 0 pi/2 pi 3*pi/2 0 pi/2 pi 3*pi/2 0];
5 S P- R4 ~# O n5 q% v
5 X7 q4 M5 w U9 p" J- XOmega(1,：D=2*pi*[2 3 3 4 4 2 3 3 4 4 2 3 3 4 4 2 3 3 4 4 2 2 3 4 1]; . O9 X' |4 P, D' g4 t- c$ Y
0 U; v. C# b/ L7 [! w, x( f- c% 求解微分方程 2 ^( n& D+ y! u1 D( d
9 }2 M- \2 w( N# L' N, ^. Q! a
for n = 1:Step 0 l1 } f/ @$ @3 b7 }1 P
) W. {+ b) u0 C4 x; M+ w0 u
for i = 1:N 0 P+ C$ L6 x6 Y2 W2 y
3 X5 Y8 ^& B3 V
Sigma = 0;
3 Q) M( b2 ~' x+ _: g1 V4 n: X, q6 I6 T+ y
for j = 1:N
# d/ |( M s9 C; e* n" k: ?* y" C
" f4 U- e% ]% l% V %%判断k对同步效果的影响
( v3 B; P7 E' l# u, W; H; S
) I% q& `. l S! z= Sigma + k(i,j) * sin( Theta(n,i) - Theta(n,j) );
6 c- h4 t7 g% w& a) a# R, `! S! ]! g( T
end
3 B" {& Z! S8 ?: i4 F) M& f5 F# G4 v$ k: @
Omega(n+1,i) = Omega(n,i) - Sigma;
; T9 R; g! E/ B. v2 I, {, J. G. N) e# Y9 ~ B8 t% b
Theta(n+1,i) = Omega(n+1,i)*DeltaT + Theta(n,i); 7 T4 v/ C. _5 z8 Z
P1 ]5 S, {2 P; T; _+ a
end L8 s9 L: D1 G; N1 T2 o
' M5 a. e/ V% ]0 M e
end
, X- {! i0 ~3 R. \( n5 z% Y, k$ ]7 E# W P4 H: {
% 求解序参量r(t)
- Y; U9 f/ n, G
; ~6 |4 f( T8 G! v3 UgroupN = 5; " e* }6 r) C e) A1 x9 y
3 g( T. Z% k: D# V' \Step = 1000; 1 g2 w5 i& T( `& F
- r. ~! C p, r+ H# x5 dfor i = 1:Step . |! v- q c5 y- @
* S3 h1 i7 w/ f sigma1 = 0; # p @. d* m3 z, B+ i' F
! K& S0 f; L( ~+ ~% {1 j& I- d2 u sigma2 = 0;
( u: Q& A2 n. g/ s) g: T; {" Y7 ^0 i* |# l( J* z3 n; i
sumTheta = 0;
i$ l u0 A6 r# p1 O0 N; g% K7 A* v8 D) N6 m/ a \' g# g( ^0 r
for j =6:10 %表示相应的群组
+ |$ C$ W. {+ _4 W. M6 v, h6 w# w# L8 u9 t! [7 M
sigma1=sigma1+cos(Theta(i,j)); % B# ]6 `2 _! u
9 m7 V# ^) Z( H4 j; m! c6 ~ sigma2=sigma2+sin(Theta(i,j)); ( E. G$ E3 u: n2 p: k* o$ X5 [. Y; l
3 U: e3 J5 U8 }& x+ K! Z
end , V. a6 s- E7 C* ^
' ^& o3 _% l& e8 c( `
r(i)= sqrt( sigma1^2+sigma2^2 )/groupN;
% o' I' }0 x1 q/ `
& D& G3 X& |' ]9 M. Y Xend ' n: ^" x) v$ f* L) I% f0 Y
, y* [, I( h5 \- J4 I: w1 N* p2 a* V
x= 0.01：DeltaT：DeltaT*Step;
" q4 H- S' p1 g+ q/ I) q. B$ x2 p m
plot(x,r); + R. ? z# D* {! w9 i! d1 b8 P0 A
MATLAB版不完全分散编组结果如下图：5 P* g, @7 q3 g. l
2 a' i8 {* r7 P |
/1 
发表于 2021-2-22 10:03
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