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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。
! D% w) c- A0 s6 l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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! j) l; d1 C! ]9 R好吧，那我们开始，首先是Python实现：8 H [2 L* L7 x' D1 V! t( U
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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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: s5 X2 `1 V7 M" _: L1 n- ]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 d4 K7 ~, N" {) j; f( d
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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？
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5 j) O/ C1 t: z/ t7 {8 o h; W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：, z" \% |+ C( z$ j$ ^$ t
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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。3 s2 F, Q. K6 G0 |9 _. K, Y
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组间同步能力与时间t的关系出现了！- ]6 m( Z: a5 b( o3 e/ V; n9 G# _
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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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* @$ x$ C/ ~0 }& h, v K% R2 P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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独立编组模式
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" u+ t) ~) v' k6 a9 h完全分散编组
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( m: I* d+ D6 x6 q不完全分散编组
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7 W, v. A9 z4 C) Q! v参数数据
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+ x+ u: U x/ _" T5 `参数确定一下有没有未知量：, G6 g% _, A7 h: `! N
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首先N，数据数目已知，这个有了。
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, s: X5 n( } z2 S' t; WK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。7 d# q+ P& b! L! g* {- @
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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。1 c3 y2 C- D+ x
; J7 v8 u/ y, `$ C\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：2 A2 H+ g& {7 g) j" r' l8 _& b- x$ N
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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。2 u* _7 i5 r- y) R/ ~; R
4 J2 I" L* i6 y( K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！/ u# Q0 l, |! n; `2 ]
* C8 N/ V3 D6 W! x) k) x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! `4 @$ D% c$ K6 h
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##ScriptName:KuramotoSimulation.py
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; H2 X2 {9 z6 W" s! L0 Oimport matplotlib.pyplot as plt 5 j( K1 E5 X% h! X) L
) |2 j6 N$ z) e% E6 d6 J* xfrom pylab import
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) S2 h5 l/ @: v ~' Rfrom sympy import
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/ F5 D+ S! _( U6 L4 t7 \from matplotlib.ticker import MultipleLocator, FormatStRFormatter
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import math
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import numpy as np
# o! Z3 |' `5 T
& J( C9 T4 U8 _. S3 O; {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]
0 U9 U: d/ H& X# T2 m4 Q0 R6 V k- q
+ O+ \* a( o0 m' b; {! ?& C0 z3 Hw = [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]
& |( [. \& B/ w) S; G& Z0 O; L0 M& W3 M7 j( A9 R& V
k1 = [
" }$ j0 e7 ~* D* H- i& a( k6 D. v1 m& q, V( t6 ?
[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],
* j7 v* c/ @% {9 L4 ]6 i# K: ~# e& H' Z
[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],
; B& \; ]( T& t3 y( n
9 f& C6 {, E1 r) b( e" A/ ][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], 6 Z% B! e! C3 a/ I+ u
# [% e& \- M$ z2 Z! S: U[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],
a- e" x \' e$ `+ R
5 D- M* w/ r) T9 D[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], & l! { N+ g- |3 i
% k( C& n# A, `$ l
[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], 5 r: K, P o% i2 r% i: Y
# [" O! j4 \( p' Y! [[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 u; \8 Y b% {" \9 x7 }# T% i4 J+ x0 ]! I3 {# J$ K
[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], & t5 ?9 A0 a# l- Y$ u" O
* a! J; H/ ]/ s" P5 S8 J
[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],
/ A7 X4 i; k9 L4 U
$ A; _8 ~& H G) t9 ?[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],
- v3 i6 m0 }7 {3 ~% m! A0 x
- ]% E e# |' X: W, h[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], 6 i5 Q+ E/ G3 `* ]
+ ]/ U( F4 _7 e, W[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], 9 H# |7 M7 p6 S; x2 A! i/ M
3 ?& s* m7 J$ N
[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], # M+ i% q4 A( P) J( X A) Z
5 P0 H) T, f) Q7 [[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],
) q. I+ U) Z/ H& l! m1 R; f' c$ X& ]+ ^- ~
[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],
! F; Y: `( J, h W1 V- Y9 |' Z
" ]+ F$ {. A2 V1 B[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], 8 N- q) |2 m' [
0 N3 ~$ ^: ?0 C2 U8 [2 H
[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],
! a$ X6 f% Z, Q
9 V* X" z& d# `9 [* h& y$ V" z[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],
8 e# o1 P5 R, Q$ d7 F. ^' ~ B! x7 P6 d; f, J7 Y" W! H: L
[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],
5 B) G) Y" X; i; |5 w- T% v+ ?' B4 _. o( K
[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],
- X% `% P& ~8 y+ t, d @! r, L
/ d5 y9 K. F9 C7 x[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],
5 p' ~% _5 I2 o" Z. e; K% C
1 O4 m3 v! A( y" n; `2 d! T" b[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],
0 m9 b4 c3 r* u7 b8 b8 ~, ]+ w% i" Y: a; v% E
[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], 2 y5 Q7 m' e% w0 G& d9 k8 G4 K6 S
( d& |8 t) k/ E) b; X* W9 u& \[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],
( }( M# z% v2 e. }5 Y) ^( {
6 U6 o! W3 {' N[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], - h8 Y6 E7 s3 d$ ~, v; m
4 |$ U( J& E3 f5 @- a0 ?9 Q( \[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],
" J* n. u8 v- V
% l) h" W% Y& }1 r w) n2 C5 U[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,1,0,0,0,0,0], 9 h! Y2 y' |5 L" H. {7 z
' o: `' e2 n3 @7 X4 U[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,1,0,0,0,0],
4 D8 _4 z/ r1 j" J. n$ F8 [3 d; F# o0 Y: p3 d4 z
[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],
2 T! z& }' E1 G& {& x
& W/ l9 p$ [# u v: I7 E2 B4 v# Q[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],
2 d8 X( t6 {% J$ f x" [% d/ t$ P4 O; S. @( \" m
[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],
! r9 d, W" A& c3 ~1 K& w
2 P' h; O5 H6 i2 [* U8 ? m( b[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]] 9 T: ^: O4 z! A: |, C$ i& @$ d3 g
! I v' v7 f5 F' A% rn = [i + 1 for i in range(22)] #目标划分,24个值,1-24
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t = [j for j in range(1000)]
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1 W/ L3 F+ ^# lci = 0 1 ]0 V/ L( N# E' ]0 B1 W
% R, \7 ~. a# U+ \% d' r3 y7 zC = []
+ v" m( a0 ?! {/ g9 b4 \+ O
: [( g: j' E0 Z x3 S" @C.append(c)
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for d in range(1100) 7 c' I: p: F$ Y v+ H9 |/ v; }% e+ o
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for i in n
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for j in range(22)
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cj = c[j + 1] - c
u! y: y: C' k2 r. ]( e! V
8 q4 }3 u. S1 ^. ?% ^ ci += k1[j + 1] math.sin(cj) * S' h: m7 D* H% U& z) K3 y& J
- V3 t' I% g1 {3 u, l# s" o5 c cii = ci + w . e- s0 _+ y7 C+ S; w: Y, i$ Y
7 b! Q! M4 l7 o+ H9 K h = [0.01 cii + z for z in c]
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C.append(h)
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l% o0 \0 O1 f; d( n, B c = h 8 N2 Z/ S- s8 V$ t
. A0 \( a( X* X6 K. Bdef r_function(u)
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1 u" E/ b7 m9 f) {0 P0 S* R y1 = 0
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y2 = 0
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y12 = 0
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y22 = 0
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! Z& L3 H+ e2 h' L0 U" u+ S r = [] 6 w' H7 j6 b& Z/ W" t D6 w2 Q5 ]
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for x in range(1000)
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for ul in u ( s9 s4 h/ ^1 p8 V7 L/ u
' |- G# T5 k# w3 w, g y1 += math.cos(C[x][ul]) 0 k- \9 x* r3 e( ?5 a. Z s
8 n3 @5 E4 K. j7 j
y2 += math.sin(C[x][ul]) $ n0 a+ t' z$ Q3 s
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y12 = y1 2
+ q3 W+ Y, ^% t1 N3 Q* V; p9 _ Z) e" l" S
y22 = y2 2
# W4 c/ ^! D! `' C8 ~0 R; w$ R) f# }) Z( A: u) f
r.append((float(1) N ) ((y12 + y22) 0.5)) + Z, \# `0 ]1 T, c! t
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return r . c" [) L/ U- x$ v
5 T9 }* A$ O3 Q4 r. w9 b" }
r1 = r_function([1,2,3,4,5,6])
! W0 y; y/ i# D# V: r, U4 a% L+ G$ [. q) T0 P( w) n
r2 = r_function([7,8,9,10,11,12]) + H: i+ W% A7 m! M/ Q
0 C! C' W8 l; W% w, M9 Y$ C
r3 = r_function([13,14,15,16,17,18]) 3 \$ x! Y. ]* X0 C
0 u* d# w' a4 z% [+ v
r4 = r_function([19,20,21,22,23,24])
) p0 g* H' [; J) L0 v7 J. _
# A J# Z& T( I) I3 y) |0 qr5 = r_function([25,26,27,28,29,30]) $ m8 E8 N5 Y. _8 [& S, I, d+ O( c
+ P/ h. |7 l) T( d#r6 = r_function([31]) 8 j2 c' C- }/ t, `" X# I! R8 a
1 x/ U: \; I7 C, N
+ B' m* d/ M/ a! R. J0 m
+ ]; B' x" H& dax = subplot(111) #注意一般都在ax中设置,不再plot中设置 8 b1 f0 S, Z- Z. p+ T( R, V0 B
' P$ Y* ~% p! [. OymajorLocator = MultipleLocator(0.1) #将y轴主刻度标签设置为0.5的倍数 2 f" m/ {$ n. q$ y; j' }
8 k4 I! T# T" j, U. s8 Iax.yaxis.set_major_locator(ymajorLocator) ( v% a6 S7 n0 N% V0 e) j/ I
" V3 b; k* [: Nplt.plot(t, r1, marker='o', color='green', label='1')
. c4 D9 ]3 I2 M) w# B; l$ O) t6 E
plt.plot(t, r2, marker='D',color='red', label='2')
1 N% F- F( j+ d3 |" y( f% V
5 M* q6 x! H1 j+ H J( ]2 i, xplt.plot(t, r3, marker='+',color='skyblue', label='3') + p$ u/ i7 P# H1 w$ s2 W7 J
7 f+ C. [1 ^' i6 E
plt.plot(t, r4, marker='h', color='blue', label='4')
. b7 @+ Q! y* ]# `' `
0 r Q7 m, u/ B) R5 ]plt.plot(t, r5, marker='_',color='yellow', label='5')
7 v8 y2 O5 f9 n: n+ i$ w1 z
! _: S' S- U0 J. v2 h) a# W+ b#plt.plot(t, r6, color='red', label='6') / {/ l& K1 R; t1 \. X6 i
8 U$ B$ c" A' ?* Q+ K
+ p. M# o3 w2 a6 Z, Z, m
6 j! c _: x! K# Z1 nplt.legend() # 显示图例
% F% }1 c7 q% R! k. s* D8 |- w+ u6 g* I2 ?9 S/ I
plt.xlabel('iteration times') . G- a. b7 X! Z% i
. I% U: J, x6 n" l9 d, Pplt.ylabel('r') # Y, C, G7 T) ~/ _. H% A# V
+ c" F7 J& _! I' r6 R5 {) Hplt.show() 6 D1 @/ D! ^9 e- W: p3 n
/ H* m& f8 [- K9 w4 p/ c2 ~3 L
q* _: j; V, Q- M, ^3 ^+ ?
独立编组结果如图：$ `2 k8 ^7 U4 r# L# r1 W8 R0 Y
% v6 P$ G w5 a: ^5 ]( r7 Y& h5 b
0 D% n. i3 d. W3 H6 r. {4 S- K
+ S' K6 Z+ f- [0 V/ N+ tk1独立编组$ G* O! i, p' f* y# x2 N; Y# F
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：3 m8 ?* M* D; F9 q4 n& R9 L3 ]8 r
7 l5 J7 q1 E3 \/ w1 a3 B4 N
k2 = [1 A$ d5 C& p ~1 m6 D
1 [/ A) y* `! t3 y1 N9 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], * V( i" B8 B7 h( x* u$ S3 V
) c* L. @5 I( y: a4 A- Q[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],
! ] G( k4 D! ?
$ ^3 r" d: P. |6 M1 p[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],
- e+ ~' d! Z$ F: \
5 F9 n- I3 W5 r* P[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], ' G9 D6 O0 P; w
& o- M$ [9 g8 }8 Y3 e8 C[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 k$ e& ?+ F U! v2 M
/ K5 u& U' I6 ^+ T7 u2 Q
[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],
7 q& E8 m" m" r7 \' n- P2 U
6 r) K8 a, X, r0 i% k- R[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],
6 {4 @3 r; a9 r" G
' N* A$ g$ p7 B6 J" w[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],
! a0 u0 w$ `. x$ ^6 Q" w* ^9 Z" {) k `( p2 a! V" G9 M: ?
[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], 8 H2 Y2 j& \4 O/ C! C# V2 h0 o
& Q1 P- Q+ L; Q- Q1 k2 Y* D[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], & T* A) d% f& U) y( O
5 N2 D( Y: C! c5 p1 s# x& a
[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 z: u: L! D0 K/ e+ H2 x3 m
7 l0 \) L3 @7 ~8 x$ W! A. f R- T1 l" t[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],
/ r) ^& C! ?# W% Z* O1 s
1 n% Q" Z) X3 M2 X4 T: f3 L[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], " h, H' F- k0 S- J* H* a
$ D. Q* T0 F+ L
[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 O+ ^0 ~, U& o1 a" {( W: U. i+ @% u
[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],
1 G7 W) E9 L) A. g. h3 J
% l* L h% X k# e& i: ^6 _[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], 8 j; w' J0 S4 n W7 ?$ R
: O, ^- l/ Y0 o- j) C[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],
& b( V7 q0 B2 m; T2 _: Z; z7 O$ e s$ k: k+ r
[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], " g, a- H2 S0 L( s+ l/ g/ C. i
3 O8 P+ D1 a, _6 D8 c8 a$ C% U& M[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], $ o4 a+ f9 f+ Z$ E# |) g; ~
$ |6 T6 X) B. v. V[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],
* A" o: P) m- g# B( `+ y
x% ^; D! E( u8 T2 c4 x[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],
a Q% I/ \+ \: m! ? a9 f: ?1 l# y
[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],
" ?- E. r, R5 i, R" ^# o' [
* J5 j, }3 Z, p[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 e& E- l& A7 d: |+ D' G
" X5 o9 L2 H7 e- k" e: u[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],
# N% F8 [: q) [
" p, o6 q4 |7 E; |4 m[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], 1 I- P x& l7 [2 `# e( S5 X8 a; V* Y
+ ~; h2 D; M/ X
[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] - N: s% @" b8 R* S x
0 C$ ~ u' N0 B1 n# m
# E# q3 n# d( d6 N* \4 u3 S
: R5 v; ^6 B l- c5 N. t5 Uk2独立编组! x" r4 p" i& [0 m# ]4 m
( \" s" K- G, s
$ u' a5 [4 ?7 r! G$ i
. \3 A; a+ ]( Y- U7 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？" k# U/ v1 E/ h/ Q& L* E1 Q2 L8 ?
6 c. y9 ?8 M1 l这里还有一个公式，来解决这个问题，编组同步能力的量化：0 M0 K/ t( O; ^8 ^( v% s( \. _0 B
' Q9 T+ h$ h$ H3 O& ~2 K: v
% Y+ L3 x# {/ D# P- m
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？
! O4 V# {9 S+ I3 ]! t) d6 t& q; @* S9 K4 V" n/ P" E" _
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：1 _4 f: p0 h t; u) z& a0 [- _5 T
. V4 h) q& [& N
/ i5 f7 b5 X) ]. 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。6 W$ n5 Q. R) C& J9 T
, d' t; {- H4 t q- m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：) v) g1 O. r9 a: R" s1 h4 n
9 t0 W5 p5 X8 C% ]& sclc;
, V) Z& N# \& f. ]- h* {$ Y6 z3 [: q$ l" t$ _5 x3 u' m/ L% C
clear all; + J" O6 \6 |3 R# I" w e. t4 Z5 c
6 M. {$ K, \8 i2 M- }" p- Z# Ok=[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; ; h% q/ r: I) C) G _$ |
, `& |: a- d/ z+ 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;
- {- _" q* e; c
6 Y4 e* k! c4 f' W) |+ ` 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; . j+ N) Y6 B3 b
, O8 M6 m2 ?0 A' p2 [ 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;
. [& ], x+ F3 u
' H4 N# h. b5 D- y! U. ?# j 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;
4 g# C- a4 x$ E/ `: [3 {# M
8 ]& @5 z8 ?7 U ? 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;
" x, `6 ^) N1 ?% a! N0 w8 d
W* O' J9 Y, G8 E6 x! 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;
) ]9 H+ z& w2 I7 w! Y7 V5 o$ Z1 t4 k1 S0 V0 C8 o. ?0 t2 k: 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;
3 e* N& `2 u2 J2 E% ?2 P' [4 j
2 M% I: B1 ^8 M" W 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;
+ D' b0 o2 I5 R5 [: F; R" e
; e w P4 I5 t5 _3 c( S. o | 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;
) }5 X2 ~* @0 g( U) W& Y
" d5 N; [; ^, c 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 J6 F5 H$ Y" U8 m+ K3 q3 _# O9 n" y4 Z s# I3 a) Y
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; ( H1 T1 F6 m" {" c6 {& \
2 x# u- S* g! \. d/ k+ o3 p, ^( W2 o
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;
2 L3 C) b" U& F, }3 p2 s# P4 r9 l& n5 S) I
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;
) p; X0 } L; i. [! U/ F5 n3 k/ o- U1 a6 B4 C1 O! Y; ]8 P
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; 7 T* {1 A6 ^, Q0 U0 U. q0 z
, i7 D x2 ~) u9 x3 f! g/ F; p 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; ! T* k. h, T5 X6 W
$ `) w* w; I. x 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( S! g" ~, q+ c0 f9 S& }, Z) g1 i/ Z) f: d6 f: g7 B2 y
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;
, k. S! { i; T5 d4 N; w" V' s' N4 Y. ?% w* 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; * `2 n6 N/ l; E' }+ m6 A j
* b" Z. x2 Q& I& u 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; % T0 |* d& q( M; G5 x; e* n
U+ j0 J) ?5 R8 O+ b, M C3 Q 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;
4 s; ^6 Z4 @/ F1 t8 z7 r# Q! }1 k/ Y) l# ~8 Q6 \5 U6 J
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;
9 ?6 N5 ?8 \7 {. \, { M: h o) U! B) C" Q$ ^% X) p6 z0 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;
, `3 _6 \# }, j: R; T+ v: Y# ]7 `4 Q$ H; _4 m
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;
7 w$ r- W1 u8 r6 \
. _8 P! I% m4 T. ^% C 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; # I# V9 K% P0 |" H* k
4 V* u! ?4 ^2 _+ K* s% r o' A ] 1 \9 Z% }6 @( f6 s8 L
9 ^: ^9 m- |+ l
N = 25; 9 P' F9 [5 {/ w5 }5 [
6 m% m7 \% Y4 X/ Y* Y9 ^
Step= 10000;
* J9 o* v' c! K4 e$ g+ b
3 f; W0 Q! N' O8 I/ h8 s- qTheta=zeros(Step,N); ) E1 f1 f% B8 _ l/ \# K
; c# O; w9 o1 Q, O8 u) qOmega =zeros(Step,N); p2 N' C8 |. |5 g, J \' a
/ s7 }& h- e# x; Y8 K! wDeltaT=0.01 % e7 o8 \- ?5 }9 ^
/ M; \8 h9 s& T0 Y" [8 B# _( [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]; . w) t; K' O+ L1 _+ z
( u+ v9 @/ ~1 {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];
7 ]( M3 G, F1 D* A' n0 ~7 K4 l. b; P! L& p" g- O
% 求解微分方程
+ K; m# U0 e/ s
8 ^, o% R& ~7 S9 a; X5 Hfor n = 1:Step
4 U3 J% I3 y1 J% Z. S
7 c7 V+ @7 k8 y$ N! V# \% G8 o for i = 1:N 9 E9 ` k/ p! S9 J* B* x
4 ` T4 H1 s0 A' D Sigma = 0; . y' v0 r3 @$ v2 ]! W9 _) h0 Q) D
: \2 H5 q* L h) m+ G for j = 1:N / ?0 ]7 d" h2 W% {, ^$ T! U+ A& U6 ^
, b/ z5 [9 a9 t" u: N
%%判断k对同步效果的影响 % _( @- `& s- v. N' ?
# `. s2 Y, S$ k8 L= Sigma + k(i,j) * sin( Theta(n,i) - Theta(n,j) ); 7 \3 o& m& y& C$ Q; A7 q! V+ p2 K' R
6 X4 H6 F' l' q. Z7 Z9 r& R H
end
5 Q' p) y% K. }: o: r! o' C7 N
- @" [4 L; E' [2 {; X Omega(n+1,i) = Omega(n,i) - Sigma; + H; t- I% O$ ~2 Y6 W& y! h
% p/ z6 V$ ] H
Theta(n+1,i) = Omega(n+1,i)*DeltaT + Theta(n,i); ! t% e( I) B+ I8 ]9 [, H
( W5 M( Q! z: E/ j! d1 G
end 4 l* `7 _8 N, [- r" V
7 S$ U) }# }, U+ b. C: L
end 8 [- V4 j2 ~/ j) d3 M9 Q
( t. V \' w* I- v4 c% 求解序参量r(t)
# W$ P: O# x! g, \% ]
* A% n( p1 H1 ~( w cgroupN = 5;
5 X! m; S7 A( y2 Q. M
K. O$ C' c3 C4 e+ MStep = 1000;
, A/ ?* n9 n! r4 n( ]/ Z6 m* Y: K, a
for i = 1:Step & U8 F$ v* r \
$ `+ r+ {& L+ G$ O* B9 }' h$ }
sigma1 = 0; 6 F2 g6 m7 n' R
, L% ?6 N* L+ ]1 U. G
sigma2 = 0;
7 o" r! T! x" T& z4 \, a
% ^& l- A1 O e2 J0 f* o4 \0 P sumTheta = 0; 3 A2 I1 ^/ r; n9 T
3 h! d5 p4 V$ m
for j =6:10 %表示相应的群组
; x" b8 ?* L5 Z% Z7 |& U5 j. D
! G: N! F7 |9 A) e2 a sigma1=sigma1+cos(Theta(i,j)); % r [" p. Q5 j2 r4 v. K
# H, j5 C# D' f" ]6 h+ I5 ?: k sigma2=sigma2+sin(Theta(i,j));
) c5 W! P+ b5 C
) ]5 F$ r2 B2 l% y# }! { end 8 U) ]5 _1 M$ k1 c1 i
1 ~6 i/ ?: M( n& t6 [5 P
r(i)= sqrt( sigma1^2+sigma2^2 )/groupN;
5 N' C! {7 o' P$ j# j* C! N5 I% p6 x+ u3 \0 _# x9 |& ]0 {3 g
end
- ~- R* R) H$ h. N9 c. I$ a3 O
. e5 o) k4 ?) C: F9 k lx= 0.01：DeltaT：DeltaT*Step;
# R) M9 {! c: o- M+ H
2 i# H6 ]5 _- T7 y* s3 \4 p, }plot(x,r);
( A: V9 a- f) g) s, n; KMATLAB版不完全分散编组结果如下图：
1 m0 j0 m8 H" v' U: K
; P$ ^1 D3 y. z8 s! P* l
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发表于 2021-2-22 10:03
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