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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。 8 i Y$ K% z# R
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。" G/ I c" X, o2 X
* @# v, j4 T7 J5 C+ D1 O" O好吧，那我们开始，首先是Python实现：; J4 N( ~, U+ g' a4 c% G" o" T) X
( K1 Z* L, _$ \+ b. g* Y+ v; LN维Kuramoto模型的数学描述如下：1 e1 E4 a" r! B0 [
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没错，是讨厌的数学公式，没事，它可以改写成这样：" b7 H0 e- F9 F( D
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( P8 C0 N; b7 g+ V9 I1 }好像还是有点长，那我们在改写一下：
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+ ^' ]& t! C& r+ M% }; H+ b7 g; W! I8 k% Z. ]
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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) z! N1 r" c7 b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？
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2 Z2 g# C5 @. s7 `. r6 D- T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？1 L3 e: c1 J1 W( m9 g* q
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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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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。) v4 V5 P4 {# p9 k9 i2 M" m, S% T
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组间同步能力与时间t的关系出现了！
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" N7 W, E# s2 C8 n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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! n* v. x! k1 p栗子是这样的栗子：
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* ~4 \0 i3 O% q8 F# P2 v `% A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4 T( s! M- f5 t! t$ N& M8 y8 g0 U
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! {, Y0 B) R: b# M3 y! {
完全分散编组 `" R9 L, r h. V% r6 n) S n
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! K( e( h9 s- J2 x, |0 p% `
9 p4 d5 C% v! K7 R: l0 Z( \& d% {; k不完全分散编组
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! T1 k% `* f% y1 n1 c参数数据# F' A/ T. x! K8 }/ r! J5 O( i8 |
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参数确定一下有没有未知量：
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! M" a) F: j+ \$ ~7 l首先N，数据数目已知，这个有了。: S& t6 h+ y; I9 z( `7 l
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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。- s% L( w7 m% U5 D" I- v
: O' O V4 F% ~5 ?& |0 n\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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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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& @; `+ ?& y( T8 ~7 n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！9 g; q2 W* k) k$ x
$ e( O3 i2 l+ k# W+ |% Q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：0 w4 |+ b$ b% B# Z
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#codingutf-8
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4 @' _2 F# f" d##ScriptName:KuramotoSimulation.py
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import matplotlib.pyplot as plt
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from pylab import 6 c8 b) `9 E( R3 K
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from sympy import ) }' ] }$ M" G; |1 [
. ~/ o% m0 k" X4 F" Ffrom matplotlib.ticker import MultipleLocator, FormatStRFormatter
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import math 6 K9 c/ ?% Z5 z3 _: C6 i3 m
8 q! }$ F- H6 p$ [import numpy as np & q2 \: A# _& c4 M5 E6 _
8 _; c# J+ V6 a$ z( \0 n8 Q; j* sN = 31 #总节点数 . d' G/ v% D; R& o- f
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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] 3 }- M$ ^* I2 h
# Q P, c4 f ^! i C1 jw = [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]
6 U8 z& H7 L) @
1 E2 @( Y& y* ?( J% F: Wk1 = [
9 [) r! \: Y: B: X, Q: e, x4 K' Y" x6 e% ]9 |' Q
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
; W9 A Z5 I* I" |1 g6 H9 _
, o+ V, O7 z5 G }1 j[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],
, k4 j/ ?/ E2 D: ?) t+ Z3 p! d) s. Z7 _
[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],
. i1 r3 J. p1 I5 t9 S3 {$ Z' n* L( u/ Q+ R: X7 ~& J
[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],
. q7 \6 v) b9 ?+ Y3 Q+ E
* K' x0 h& P. H9 S* r5 F9 Z[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], - H( W- B* H9 P7 B7 c" }- f0 h+ E
2 `' a5 |+ Z+ M, |1 L5 N[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], / O% S! B2 }: V, d4 n) T& G
" r; x! l3 ~" H& i! ][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],
+ _3 L+ q4 X" W" |) q2 z3 T. ?8 `' i- E% _. r, t
[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],
( f8 |8 r- V) G. S' J& M T: l, D5 z) W Z
[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],
% {* N P2 F" X/ x' E3 p
! t& y: l5 U+ L0 g[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],
2 |- a5 U0 ~! Z+ y% b2 H- H/ s$ O
[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], * H2 n& Z+ \0 @: Z; h- Y. r
6 ^1 [ F* S: ^/ t C: F5 C[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],
0 N; f, e, t* F4 J. i/ b o% O
! K- y% B4 O4 g[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], + T* z8 U+ v! t% E* O
3 Z# l; [9 q3 K: o$ @: R) C, i
[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],
' J" e. G$ x, x1 C; Y3 S
Z5 _4 G- l1 m4 z: B0 X8 N" a2 S[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 @$ a j. m# E( T/ ^: Q
! ?9 S1 w+ N$ u. [9 q[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],
+ Y" K4 |0 j" g* ?
: r! P7 l5 W% n) k% E* Z7 r) {[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],
1 y( b* U3 ^. f3 ?2 M" r
/ x1 ]1 m& k, Z: m[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], " O% S) y4 v, _8 O! j" a
* D9 ?* {0 z) M4 ~: m* 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], . r4 @/ y$ t s2 @
& G- p- r. v2 c
[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],
/ n% H% z4 z% V7 G, {8 ~* H# K2 Y6 h3 z
[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],
# C+ Q' V$ W" r1 U& F8 w; r) g4 g% O8 I4 W5 p' k
[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], 1 g$ L3 L' w+ o2 [8 ?# o9 V
0 S6 Q+ K F# u
[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], * I6 W7 m7 e) k
) r: c0 L j* E6 A) z[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], " W" p" J) V8 u, x
7 ]- G1 P9 X$ U[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], : `- p" r+ }9 ]" Q
5 z! h& I6 L0 ]* D9 D, @- o
[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],
* `; }$ l6 x- ~- X# v
4 A# D5 o+ D$ _; C& D. Z: f5 a[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],
. J- |0 J( D* i/ E8 [
f4 M2 t( y$ w t[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,1,0,0,0,0],
, T8 F0 `) _7 ^. B( G6 U2 X6 ]! @ y/ u9 t
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,1,0,0,0],
& o. c! d" G5 ~- { c- ?( e
2 ^$ d0 q# F7 l$ i[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,0,1,0,0], ; Z* o$ g4 x% i5 N4 h
$ `( B9 @9 x: {3 ?6 G" k4 i[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,0,0,1,0], * n# a8 ?& C4 ~/ [
4 i$ ^2 D% o/ G! x$ q
[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]]
, \* L8 N& v+ V5 C; e
5 N$ \+ J6 E7 Qn = [i + 1 for i in range(22)] #目标划分,24个值,1-24 . f q( ]" Z: m
! u0 ~7 L5 U& V/ F7 St = [j for j in range(1000)]
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2 N8 _2 O. T2 B* ~4 Eci = 0 ! k! W6 O. _% Q* o$ H
3 @; R# ]; i K) ~3 eC = [] * u$ D0 d0 K0 `9 _ u9 k6 Y
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C.append(c)
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for d in range(1100)
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for i in n
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for j in range(22) , K9 S4 R6 G+ J
0 ]' L7 i _$ u/ R cj = c[j + 1] - c
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0 a/ r. @' H g ci += k1[j + 1] math.sin(cj) P- l' j( P* O8 i& r
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cii = ci + w
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h = [0.01 cii + z for z in c]
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1 v1 g4 w+ V0 L+ F! @* N C.append(h) ! M$ X. [* z& F8 K! r5 {& E
; |' j& |# m F c = h
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def r_function(u)
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) G* ]; i" ^0 S; o y1 = 0 , s$ M7 i7 Y! d( ] |9 r
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y2 = 0
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y12 = 0
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U3 W& ^$ M. \3 D9 n" u' ` y22 = 0
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0 C6 T* j6 v6 K r = []
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% h& \$ L+ H8 D3 W- q; k for x in range(1000)
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for ul in u / g8 K1 ~# Z1 y- ?6 _$ _
" ^0 L$ c6 t) \9 v& `! @ y1 += math.cos(C[x][ul]) 6 X; D0 \" R1 y/ v
/ n! j8 S, `, u. c y2 += math.sin(C[x][ul]) - h& n6 f, ~4 K4 X, d; U) u; Y7 D
# ^% [; \8 ?( }7 V# {9 ^! p y12 = y1 2
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4 z$ m# e5 m- S- O* X( N5 z y22 = y2 2 - [4 ~7 ?+ v N h8 F8 V) V
1 h% x* s, z! a r.append((float(1) N ) ((y12 + y22) 0.5)) 5 h; h' x, b: H5 O
4 }- ]3 I/ N# H K' ~
return r $ [! I* Y7 h. O+ G5 B# ~1 l
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r1 = r_function([1,2,3,4,5,6])
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r2 = r_function([7,8,9,10,11,12]) ! O' a3 E( v# `2 m; o* @
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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])
* l5 r- D8 ~% X/ z, M- u; w: U# A1 U8 S) `, y
r5 = r_function([25,26,27,28,29,30]) * j& d- G6 K6 [# E2 d; B
x" }. }% Q: g
#r6 = r_function([31])
# u& f8 _. [3 L4 c2 t$ V( U2 n G3 q" e8 \, P! x
$ \! J/ f4 j- T# n) p4 C
) X- A1 s0 S5 @ax = subplot(111) #注意一般都在ax中设置,不再plot中设置
$ I! l8 h9 h+ j7 ~# U4 @ {: R+ C- l$ W- u, d3 ?
ymajorLocator = MultipleLocator(0.1) #将y轴主刻度标签设置为0.5的倍数 & f1 ]4 F% r' ~9 K- ~+ n, ]5 L# i
2 J8 M( N6 ]+ i1 @7 K
ax.yaxis.set_major_locator(ymajorLocator)
# r! S) V# p8 t" _) I' _; q( j( \% x- Q- M
plt.plot(t, r1, marker='o', color='green', label='1')
* ~1 z5 G" j) ^( G1 z
9 h8 B* I- u- Iplt.plot(t, r2, marker='D',color='red', label='2') 5 d+ y) R+ i7 W, o, P
' u" e. j- }4 k6 j0 k% D& ]1 d5 V R
plt.plot(t, r3, marker='+',color='skyblue', label='3')
1 g$ Y2 {; t* `: Y: U6 o/ k1 c$ n5 I
- @1 S p/ Y- S; L* x- B% j2 m+ U N* s: Iplt.plot(t, r4, marker='h', color='blue', label='4')
5 @4 i" o# n t9 D3 A7 C( L [' ?: x3 [# }4 j6 q
plt.plot(t, r5, marker='_',color='yellow', label='5')
& L' w8 A3 d. R& \) B$ Z& ? ]8 ~) z( x+ y* l
#plt.plot(t, r6, color='red', label='6') 4 {. ^) P3 a4 l) o2 j) d
1 u% X" B/ m9 o. d+ l7 h
1 k" R. T" ^( Y3 b/ F( q* n* N% v+ P( H0 F) L, \$ s8 p8 A, g- W8 l3 Q, H) Y
plt.legend() # 显示图例 ' n# O3 m3 t* z- P0 u B: n- P
6 h" M a. H2 I9 X/ i6 z) I( \/ Y
plt.xlabel('iteration times') & n( e" q6 R# w2 t
. R$ P) B2 Q B. p/ g2 Rplt.ylabel('r')
9 x ?3 @3 }5 n( S
/ P7 B$ j) C9 ~; xplt.show()
- k9 z/ H2 O; H. M$ Q7 J* ^) ~; G8 K7 A0 z$ V% Y8 ]( h$ F
: v$ M% _& a) f0 j$ `. X2 Z
独立编组结果如图：
3 F2 r @) J( W: m( H6 M" X
y) X4 Y" ~4 N, a% c
- Q+ M/ ~2 {) j( s1 Q
5 |% y/ j4 s) d; U" g. T: |k1独立编组
+ u- q! q% a* 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：( I1 M& i5 g9 m# a( Z$ z# m
' n0 n, n( O, m: @1 g$ e+ k
k2 = [
; y" W) J u) X( \4 b* r _( E( M. n* ^+ T1 ~
[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 t& d8 M0 y$ n3 H. O
2 X8 @0 [' [, N c6 ?6 w
[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], / W' T8 a' y/ |# a* M
1 W8 ]" [# p( E Y
[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], ) @% u4 y, W C
+ t0 A' y) S: Z3 l: b; g[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],
7 c/ F2 {1 W& ]& e3 ^+ C* s5 e- A- Z' E. q, 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], $ B+ v0 Z4 R& k& j- X, V' e
% q7 x; _9 L2 N* T2 C2 h
[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],
' t* q: y* x" _: u% I6 z! c/ C0 v+ j1 |
[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], 7 d) L: y1 E8 m0 M0 Z
! D [9 T* n; O2 M# D
[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],
# s1 Y. [ t A" Q% I8 }8 A- N3 {2 H: o
[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],
. a0 ?) A$ g/ A+ E) \8 _- i5 A
[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],
" N G- h& T' P, r" V" F9 f/ j& d1 V8 e1 B! f6 m# T+ 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],
. R* t( j Z, @3 `
+ E: r6 F6 n( M! a# H+ [3 ~[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], 5 D: Z1 _# m- U9 n
6 b( W: z7 u: ^2 ~% f- e
[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],
) w0 v$ y- N$ l5 [3 x0 X: |
& J$ u* c0 ]: F( U) u2 ]$ d[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 G# z- N ?7 O# w* r# T
# h0 I; g m, _1 }$ E8 [) N, B+ `[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],
% \5 p' [( a- G' r \% b- E& g
" ~* }& r$ o, \! p: s* W[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], * e' a' H$ z; Q0 N3 t
7 o% V" W2 m: J* B G" 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], # {& h% V; J m6 f) j2 X
' k, J- i2 }$ B$ r2 h[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],
. z- t9 E: d" c; X( _5 y9 g9 {2 X5 W- c. H3 j
[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],
. ~8 t6 F% p. E; s! |( q( f8 [+ P- j
[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], 3 w1 j0 E# i2 G% l
! {! a+ L& j. ?! a. B1 ~
[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], 7 H' N# E1 L- v/ J! o
* v2 R X0 G$ X% S" r
[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], . l' d" J% l+ e& k3 o& B8 N1 e4 f
6 s. m7 L1 s; b6 |[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; G7 }; X! M; B+ I! H k) Y9 h
- b3 @5 Q1 i: |2 ^' A, X[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], 4 V4 h* g8 [' ^4 w
4 _ B; M- `9 L4 q" q8 i$ n! D[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],
# K3 M1 Z. h$ [/ B/ X# f+ r! q) N! ~
[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] ; R" t+ Y) [4 F8 l
; l1 N Z0 Z+ ~ r' E6 p" X4 k C! e& X/ O; x( o# Q
; m: L9 t& T! J1 h( Y4 hk2独立编组( d" c5 I0 c. @0 S, {4 U- ~
" C+ Z9 Z4 q! G' V3 x- V% D2 J# j
, N, Y# u2 W9 _) F; ^
; g; m$ _* G6 ^( J
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？
" `+ N- S* j i$ a$ Q( T1 J5 I- _; N6 P9 A, d 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：
+ ~ s0 E/ P; I, w/ ?8 g( ]
o$ M' U( _1 F& r
3 Y- N! q5 ^" ?0 S0 nM_{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？
- U) p8 x4 }8 p/ R. q @: Q \. c* j7 F: T( ?! Q
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：
& ]: f0 N6 \6 D: _. l8 q/ P8 R
9 g" O8 e& b1 v" k9 G2 \% m; L1 c5 H
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。
, `7 r7 N% h* C( H2 E8 x$ H2 r+ \+ M
/ d1 p$ R! C+ J; K2 L! N& L P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：
' J* t* x, ?& z4 @2 O0 a8 N( ^: c7 f' T: a& ~2 z" e
clc;
# u1 E9 Y- j3 b5 ^) r# y7 N o
" f" `: M: N4 Q0 O: U- `7 Sclear all;
% [4 @4 t6 o+ N- r! }. r5 w( i
" s" f4 ~! u3 w v5 f. P2 G+ J4 d8 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; 4 R4 j9 Q* y. w3 |8 a& r6 N
$ S* t- H, R9 x! m- m, }. r+ G* 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; * H& L4 j3 r4 m9 `" O6 k! j& W
; I5 N4 d4 q, _( E8 Q 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# O" W% l+ s( |5 v; ?" S
) b6 H, M9 Y; c( f- F$ _8 \( 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;
`( E5 h% }, e# ^3 P$ J2 o( ?9 @# l" l" }7 ]2 _# R
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; 6 B3 w+ j' B3 J2 W d' P) ]4 y8 ~8 M
2 U, T6 W4 H' I4 W
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;
8 d- U) i) Q7 Q$ c' M
& c- U1 S; x% C 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;
$ k# T9 k4 m/ i9 a8 ~ e
: A$ l* u3 }8 @7 J! 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; 8 M1 { ^8 ~! B: C! b) q
$ i. d/ a: 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;
2 e, g1 @) D+ v# {* m
' d# O' k2 w7 X8 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;
+ V& C8 s3 g+ u C
2 f) h5 a7 C; |7 K$ ^ 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;
& w* Y9 Y( }0 [, I3 o! {, r
4 J2 h7 O7 I0 M/ k3 Z 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; ! T# Q, z/ @8 f, |* c1 M' `
6 T4 P; |/ p3 T6 K3 D8 z 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;
5 T* r6 }. g7 R# t5 Z' ^
~& |# p0 J0 O9 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; 3 a9 `2 r# ~6 x' e" O
) [, q# I( j/ r5 z7 g: F 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; / L- I, B* T# e, o5 \; p3 Q$ q. _, M; N' n
6 O. M) E: o& k% v* u& [ v 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; ' `* N& E% H0 N* H2 j3 V% a: G
0 u) |7 m- T5 _; Y8 @ 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 ~& t y7 K4 `+ g4 C; q' {2 y; @( _
7 |! a1 ?7 c- X% b, F 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;
- s8 n" m. ^5 ?% L- }$ W4 Y% u! ]! o7 E' p5 N. z
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;
7 u- A" f3 f2 G# F, |/ ]; w T6 S: l7 F$ y# B$ ]' t# 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; ( w" t2 {8 W/ i2 Y: X1 B
- W. D! e- X# S6 p0 R& N9 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;
- c# V3 J3 f5 z, o3 A5 T; ^6 s
4 z2 J$ Q. b5 x) d9 o/ U$ Z 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;
. y, ~5 ]1 u- S: A6 X0 D# k' I& X8 P8 u! l1 s5 W9 B
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; : `( w. n) Y9 _# D* Y% V7 k
! [% }' N! ]' ^( k 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;
% M* v. @- T6 R; n
1 e( R( X! K- Y" X 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; 9 k8 u) s+ q% q
/ B4 W8 Z. _! `) } |$ ] ] 4 U2 r# E; R2 L$ c& ^# Q
# \0 y. o7 b4 I9 i" rN = 25;
3 A3 a! h# z9 x
* s' W! Y* o! ^/ N v+ h* E( UStep= 10000; 0 _" A3 ^. w5 Q A: k
! \% i+ b( Z. DTheta=zeros(Step,N);
9 w4 z8 X5 h# Z7 F( ^
' ^; w8 K8 P! `+ E$ WOmega =zeros(Step,N); - R2 |% V* `4 H! q
: b9 ~# W* z# q' HDeltaT=0.01 ' [- P: T8 ^! [- R/ o; e
) A1 v& L/ Y( u! a; H4 ~: t
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];
4 j# ~( T: E6 k3 Q# p9 w* }/ B
1 `3 T+ |/ b ^. D6 r/ cOmega(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]; ( |% z% c" g' j T) _+ L
* ^: b* ` @4 l- S' I$ J- K# i# N
% 求解微分方程 / a- L5 m! M; n
% j1 N- h. I% Bfor n = 1:Step
/ L6 e2 I, X; u( B" l: J! _4 _. J( m
for i = 1:N 9 O& A. s3 T0 f% `
6 b" t, x% z' p' H ? c. N
Sigma = 0; 8 R. L$ p) r7 n3 C8 |. {3 r
+ o0 e0 S. l# a8 A2 H8 y1 D
for j = 1:N . c: }! J( v/ P' B1 x
5 S9 ^& J# f% o' E %%判断k对同步效果的影响 ; q2 p3 E: J, `- ]9 V+ j
2 \: s' B6 Q/ S. G= Sigma + k(i,j) * sin( Theta(n,i) - Theta(n,j) );
: Q2 N( g+ G. \; R. w" b! [: I$ _5 a; W$ T& b
end ; K) @* v" J8 |! @* a: h
/ }5 H+ ]3 W- H) l } Omega(n+1,i) = Omega(n,i) - Sigma; ' u" P" _2 F# r
, k* Q; q* h+ C% o; R
Theta(n+1,i) = Omega(n+1,i)*DeltaT + Theta(n,i); # w, H" \$ E; w+ j* e
/ {, v3 k' ^% m# O' N/ O, d
end
1 R6 R0 E5 B3 e, f3 o7 Z, s4 m
. A2 Z, r& \) Z3 F7 A/ bend 0 M$ V5 I `0 [4 k& J/ a
5 \& ^$ [2 J/ e" i0 h$ ] z
% 求解序参量r(t)
, l+ _" V- E. w* x* a. m1 x1 {. c! d5 |4 j1 O$ a' T% D; f
groupN = 5;
6 o) R* D9 p/ ?9 R5 e" C/ G) p' z0 t; v7 k( m" {+ M
Step = 1000;
1 V9 r9 g5 B# \( o0 ]
* ~8 ^# P1 ~! I) _4 F6 R8 C1 B: r) Rfor i = 1:Step / T. d* F; U1 D: _ t, J! R
/ I6 H3 Q4 r9 R) L) ?, W sigma1 = 0;
! o) o/ |8 t: y& k8 k8 @' x" x$ r
2 U! I. e3 U `/ f sigma2 = 0; ; A4 q% ^; a3 T* A* o
5 i- @, i, o: b9 d; v# ` sumTheta = 0; & f" x1 ?: K( M8 F
$ G- R6 z+ o& P4 v0 K
for j =6:10 %表示相应的群组 % O: @! u; ^( {. u
9 ]7 h: B, G8 f sigma1=sigma1+cos(Theta(i,j));
" O- H5 a" Z% z ~
% e! I3 ? L; j8 D% q: |3 c3 Q# @5 Q& L sigma2=sigma2+sin(Theta(i,j)); ( G/ J7 H, {' |+ U1 o
) c+ @% V" a7 B end ; X, @3 z' K/ e" |
: G" B+ w. k- u r(i)= sqrt( sigma1^2+sigma2^2 )/groupN; H+ ^$ h2 H7 {0 S
' s4 \1 _$ R0 t0 r: Y: |6 oend 3 O3 i, Q! j& C5 e
* m" U& g" H8 }2 a- f
x= 0.01：DeltaT：DeltaT*Step;
6 d# d6 R2 Q4 S3 K2 J! I' f2 O; u+ W/ E, Z: |
plot(x,r); ( t: [% R# h* y: }; o6 ]
MATLAB版不完全分散编组结果如下图：! u8 E' R& m, g5 U5 L/ m' W
' L; s% J8 u( `. N) Z6 Z6 k: C8 w
|
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发表于 2021-2-22 10:03
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