Kuramoto模型在Python和MATLAB中的简单实现

haidaowang

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。
8 |, W3 P* j1 J6 u  t2 X: D* X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。
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好吧，那我们开始，首先是Python实现：
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N维Kuramoto模型的数学描述如下：
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没错，是讨厌的数学公式，没事，它可以改写成这样：
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好像还是有点长，那我们在改写一下：
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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。

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？

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？

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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- c$ `' W% r2 F6 x4 z$ F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。
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% S  G$ s8 c: c组间同步能力与时间t的关系出现了！

1 |+ g" \8 G( Z" s' l% I) l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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# Z7 @- b' G8 A8 Q% h5 I5 O0 y栗子是这样的栗子：

( U9 l3 i% x! u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：

独立编组模式
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3 B' h6 d6 @+ @2 b完全分散编组
% }8 X# U( U. ?# z3 `

& P/ [  F1 l- d7 ^/ j
不完全分散编组
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1 `$ `" |( H6 `! s
. d# a0 |" i; s7 |; e. }参数数据
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: ]. A  q; R" O) b9 z( J3 w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。

K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。
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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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  T! m& L$ y) c. |$ _' V\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：

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。

" @7 O) V/ q8 g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！

  L3 C$ S: s# j; A* ^, M. F& I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：

- z: t3 k' Y  L9 ]8 P6 i& x6 D5 Y& U#codingutf-8

##ScriptName:KuramotoSimulation.py

import matplotlib.pyplot as plt

# ?' _$ Z) K1 ?$ ?5 y1 Rfrom pylab import
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4 C" ?+ `8 A" ~1 F3 Bfrom sympy import
/ O, a; [! n: z0 k
( f* ?5 q% m) g% T1 O! A$ vfrom matplotlib.ticker import MultipleLocator, FormatStRFormatter
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import math
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import numpy as np
3 f% j$ w5 C1 Q# u4 c% J+ a
6 X/ K$ Z" d" @: VN = 31      #总节点数

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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/ F. |9 v8 d" k) p4 w3 Q' d; K9 Z, q0 cw = [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]

k1 = [
- C8 ]1 p/ g+ R2 z- p/ u' M$ C1 ?% _
[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],

) Q6 O* w* c: J+ q7 s! F[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],

! _% ]5 d  z/ q( o! q# W) C- |5 b[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],
0 U8 m& `* u4 \! s3 g
3 p1 H6 K0 W6 H- I7 A/ p: ^[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],
; o6 R# H- X# T1 ^6 x6 S, P
( l6 r& i/ V4 B: c[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],
: ~+ |3 d5 K3 s9 Y
[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],

[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],

& V, R% @6 R/ k. e! O& _[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],

5 |: ?8 O6 D& w% G$ J! N; _[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 \, @7 `# _/ @1 j4 v
. J8 m( `3 b2 E0 A, W- w5 F5 J[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],

[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],

2 T7 h1 U6 K0 l+ L, R, B6 \9 p% t[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],

2 v3 G  Q9 L8 z6 D: x, \# F[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],
8 w* ?8 E: _; T) k
0 @" [2 ]% q; D) p+ X( n" O- l9 ]+ t[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],
2 [0 k2 k% j# Q0 R5 b# t
[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],
5 F0 b& u3 }% h$ q. y1 G# V
[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],

[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],

[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],
' m( k/ n3 x% ^$ B
) X9 e  x( x3 J3 b# H[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],

/ S; y8 x. L7 @9 r" x. P[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],

[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],

  a. x1 m2 w$ u5 i" w$ r, t' |6 ?[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],
/ q9 Z" J, M* u, l$ U- q; j6 s" O
[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],

[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,0,0,1,0,0,0,0,0,0,0,0],

[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],
  O, m/ \, j  Q9 s
! e  R' ]; |1 Y# n* F* _7 g& t[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],
, W* u. N' [2 a( g4 o2 b+ e2 f
[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],
0 M' Q! t+ H3 }( N: w
. s7 y) s* l/ i' Q+ J  f[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],

[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],

[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],
4 S0 \! z3 _" n- A2 u, L
[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],

[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]]   

7 c4 U$ w7 \: a0 i4 w( A0 Y8 x4 Ln = [i + 1 for i in range(22)]                #目标划分,24个值,1-24

; Z9 ?; [5 N; O7 L3 ^+ c- Tt = [j  for j in range(1000)]

5 ~$ y) E+ k% X# h/ Z0 Aci = 0
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C = []
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C.append(c)

for d in range(1100)
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8 k( i# _% r" V9 ^$ D& l    for i in n
+ B! s- q4 m8 _
        for j in range(22)

+ O: y2 ]8 E6 }9 e8 F            cj = c[j + 1] - c
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, N$ F+ p- w, P+ T" R1 n& C% a            ci += k1[j + 1]  math.sin(cj)

        cii = ci + w
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        h = [0.01  cii + z for z in c]
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        C.append(h)

        c = h
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/ V8 ?& B, w) x; Ldef r_function(u)
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    y1 = 0
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    y2 = 0

    y12 = 0
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' J2 N0 m3 y( f: q    y22 = 0

    r = []

    for x in range(1000)

1 i( }# g, c6 {$ A        for ul in u

2 b) U8 a/ n3 C; ^! N& ~: d2 F) O. F            y1 += math.cos(C[x][ul])

7 R. K( y3 Z. z# @            y2 += math.sin(C[x][ul])
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        y12 = y1  2

7 ^+ n) L+ N3 S9 y0 ~        y22 = y2  2
$ u! |" X# P0 x3 B5 Y/ U0 f
        r.append((float(1)  N )  ((y12 + y22)  0.5))
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    return r
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r1 = r_function([1,2,3,4,5,6])

r2 = r_function([7,8,9,10,11,12])
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r3 = r_function([13,14,15,16,17,18])
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r4 = r_function([19,20,21,22,23,24])
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r5 = r_function([25,26,27,28,29,30])

: f" H: `% \4 K5 C#r6 = r_function([31])


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ax = subplot(111) #注意一般都在ax中设置,不再plot中设置
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ymajorLocator  = MultipleLocator(0.1) #将y轴主刻度标签设置为0.5的倍数
& m" q6 g* H* J' {
  F0 X8 A8 n, Y, Kax.yaxis.set_major_locator(ymajorLocator)

2 p( @" h# b; g, j8 G# gplt.plot(t, r1, marker='o', color='green', label='1')

plt.plot(t, r2, marker='D',color='red', label='2')
" k* V  X7 x9 u; w  |( O& v/ r
. T6 T% E# ~. Vplt.plot(t, r3, marker='+',color='skyblue', label='3')

plt.plot(t, r4, marker='h', color='blue', label='4')
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plt.plot(t, r5, marker='_',color='yellow', label='5')

#plt.plot(t, r6, color='red', label='6')
# M4 Z6 i' i+ k3 C( P


8 @! u' M" U3 s: R8 ]: ]/ pplt.legend() # 显示图例

plt.xlabel('iteration times')

plt.ylabel('r')
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plt.show()


独立编组结果如图：

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, \8 T8 d1 h# K. y3 T- Y  ck1独立编组
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：

  X2 r8 \* q* d6 m8 pk2 = [
; @+ }7 w0 s' o1 {# A# t1 B) d; @
8 d4 l$ P4 O& J. L( U[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],
  {, o( S3 w- e" R! r& \
[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],

3 k1 Q( ]; C; Z8 R9 A9 K5 W[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],

1 u6 Z9 m0 {* k; b, K[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],
) Q  _+ `- u' S8 U0 O
# @5 ^7 P) J& C. ?. W: Q4 D, F[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],
) ~$ j' F8 h# g8 a! k
[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],
% v2 O8 f7 v! H1 n3 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],

- [5 X* B) {/ g[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],

$ v6 B* s0 Y  Y- N3 }# W$ E[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],
* j$ s4 s. r' ]5 Y# \
5 o( A4 W8 O$ ?' z[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],

" A0 J/ c$ N; I( V8 q! l" r[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],

[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],

[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],
' b% c9 C8 ~1 G' |* c# S
[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],
1 D$ |$ V' v, J# _
. W9 Z& O/ L$ e' t  d) p  t[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 B( P) V7 H. }+ s0 T% z[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],
  w, V; b8 ?1 z0 z! |
+ ^; z& ?$ R, j- v3 s; E* j[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],
7 @  }4 d% J' [* f) t& W8 w6 A& u4 W
8 }, ?2 c, W8 a1 n( C! ]3 ~[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],
0 }+ S( F+ r3 ?! G8 [
[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],
0 g% Q! g% Y- }, s6 v$ ?
0 j. c- y0 w" q$ |& s[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],

) M& ]" d, G- r6 J+ S[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],
) n1 E9 i8 j% h& B) M6 p8 |) I
; @+ I8 k) N' T' K[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],
  J% e( J# C" }4 Z
[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],
) }6 M( W0 m& i% V, U; _
  P! j) g7 ]9 a6 F* e3 ]! w# s[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],
1 Y% o3 b0 C# D" S7 [
[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],
8 X; C* W6 r5 _8 j& P' I
[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]
* w: l* _2 X! X
+ y1 q, c1 o9 L# z2 n, A$ `/ E

k2独立编组
( m) T# G# M- k/ M
/ ~3 J# H$ `8 Q; ~! l

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？
9 N6 m/ ?0 {9 \/ Q: q
7 Z! {) a8 M& B# T这里还有一个公式，来解决这个问题，编组同步能力的量化：


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？

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：

0 F( k/ [; A) p5 g9 I  b  x
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。

" s. T4 s/ |/ R, M! i8 H  ]/ M: h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：

clc;

9 `8 t% K/ W1 l) H) J/ Kclear all;

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;

8 Y) w1 ^, z# ?; a0 H; n  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;   
, }9 f7 [: u8 L; z* B  B( g
# |  C7 K5 T! g, 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;   
; w  u: o4 F. ^% J) K
& c" m' g! W  V! B  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;   

  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  0  1  1  1  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5;
$ v$ i! o( v) l$ R4 c. a' q
* M: {: ?  X4 f9 z  t. {8 Z0 H  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;   

  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;   
  [. E8 D7 G, K
  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;   

$ S( `4 U4 A, i4 Q/ b  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;   
( j4 N* Z4 G" \6 x6 B; r
  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;

  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;   
  H% V5 u: y3 L( a  E
  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;   
# j+ {/ W% T: g* N/ P' r7 ~5 a
  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;   

  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;   
+ `1 @3 r9 ~2 p# |7 Z
& E0 m9 F6 n$ b, h; ]% o  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  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0;   

* L, ?0 \, e8 P8 ?5 @  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;   
8 ?$ ]: K4 A" |  e+ m
& a# a. F9 {0 `5 ?& @" F' b& q  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;   

# J7 U8 |# h/ W8 r  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;   
) h- L3 s; r" c. f7 @6 u" p
  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;
8 J2 |* v) @' c1 s5 F2 d7 N
" Q. k; @$ q9 C0 W3 ~  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;   

  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;   

! O: V  ?. d2 d. a+ c  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;   
, `/ c: @" i  X5 w5 e4 t
  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;   
" X+ L! h; A9 {+ }" R
  ]     

N = 25;
( q+ G: I& F" w5 ^/ N" v- c, s3 N
Step= 10000;
8 C- |+ \- O$ r, k* N* {
* ]( x* Q7 \" |  S4 U0 x. u. C# uTheta=zeros(Step,N);

Omega =zeros(Step,N);
7 h+ A  x7 G' O( Y
DeltaT=0.01

Theta(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];

: X% K; a6 `/ h+ \Omega(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];

2 _0 m, P! O* X* u8 a0 Z# i%  求解微分方程
4 ^% s7 S' a( u+ }- t3 @
for n = 1:Step
% i4 V$ ^  d) L$ v
    for i = 1:N
( m3 x8 f* M) e0 v1 G, I
& c8 H6 E' M1 M* e& _- Q) b0 W, t            Sigma = 0;

        for j = 1:N

            %%判断k对同步效果的影响

4 U/ V" _1 X: h5 ]5 Q3 a= Sigma + k(i,j) * sin( Theta(n,i) - Theta(n,j) );

0 j. I& B# p; J& g6 ^        end
9 e' o! r0 f4 G1 m' J$ a
; R5 F6 j+ R" |            Omega(n+1,i) = Omega(n,i) - Sigma;
, u* M4 }5 Z4 j3 b3 s
            Theta(n+1,i) = Omega(n+1,i)*DeltaT + Theta(n,i);

# t: m% T3 l5 k" h/ H3 W8 @. K/ e  ]    end
/ t8 ^# `; A. t0 v$ I1 }
+ _- J7 p! Q5 x! A) b2 L) ^end

. O  L! _: T2 L7 A# q( E0 \; _%  求解序参量r(t)
! E7 x# Z) J: [6 v) y
, l) x4 c0 D- i- jgroupN = 5;

) a, ~- c$ W+ t$ dStep = 1000;

6 N( ]6 b& {* s, ]  ?2 r6 ?for i = 1:Step

    sigma1 = 0;
" Y$ o" G$ d5 U1 Y; ~: A! v! n
    sigma2 = 0;

5 S& Q) N) x7 Q/ B$ D    sumTheta = 0;

    for j =6:10  %表示相应的群组

6 z, Q" n$ ^: U; ^- _# b9 @1 k* [        sigma1=sigma1+cos(Theta(i,j));

! u8 T" X9 K+ L, q* w        sigma2=sigma2+sin(Theta(i,j));
4 R7 I: x! J- o
  u8 L' `+ T3 [  V; h; }; K    end
3 d5 b) m% X, H# }, q' ~
    r(i)= sqrt( sigma1^2+sigma2^2 )/groupN;

- s8 \$ e0 W# t" d9 w% P! hend
( S0 Z5 j) Q: d
% Q7 x* [$ b8 Q4 k; Ix= 0.01：DeltaT：DeltaT*Step;

plot(x,r);
7 G3 m+ Q( {3 L/ I4 N  Z- Y# NMATLAB版不完全分散编组结果如下图：
  ?) h5 S, F+ C9 U
* c- o! Z9 O% A# U1 y7 d) T

regngfpcb

Kuramoto模型在Python和MATLAB中的简单实现