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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。
0 n7 `2 H9 Y; K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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8 m/ d2 ~9 m( K4 s& e; Z K好吧，那我们开始，首先是Python实现：; b6 c1 q( Q( D& ?4 q
: g/ D( L) y' S+ n; t9 K3 u9 K L3 NN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：: f+ l& L6 S( o, D, t9 s) D
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% ]# w0 Z& }; P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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: g/ p+ p$ T+ f A& R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？7 u- W M2 r! p8 M$ t. U, D* \
6 C( F! W! e# a, C# b g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？& ?0 n U6 @7 D" J# c6 c
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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：# I- z& i* Z: M$ p0 @2 |3 y
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7 u* S; I5 {9 d- s \- c* 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。
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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。4 u# F0 V6 v) o/ E: n0 b
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栗子是这样的栗子：
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% Q$ {7 z' |$ y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：) O$ t) F. G6 F! h$ G+ x6 ^
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独立编组模式
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/ |2 l4 d+ O; N完全分散编组7 W @0 l o# L
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6 K2 o4 u2 n6 H( O# A3 H& U不完全分散编组
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3 P) p( L& o7 g/ A) E ~: d: o参数数据
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参数确定一下有没有未知量：0 j# i% F- j! g6 Q7 v, ^3 {
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首先N，数据数目已知，这个有了。
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) v3 [# S. w9 l3 \4 K9 R* `" P3 TK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。4 A g8 W1 I4 i
u7 _5 p# V8 }0 `2 [; E\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。0 o5 _7 q' |) X1 Q* R
" c1 A. K( T4 G M\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：* p7 [6 o' s, L2 x# U
( L: P d( s/ X' g# ]1 {# {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。% O5 Z" l- M' ~' S- Q
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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！& t9 H+ y& g5 {6 R4 l7 P* N4 D! t
7 {6 A, q# T1 A5 {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：: t0 J# x& h& b0 T& ? O& t
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#codingutf-8% H. `* q, Z( d2 U
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##ScriptName:KuramotoSimulation.py
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; ^0 L! E: h2 H. Y& Uimport matplotlib.pyplot as plt ! k$ I/ J% N8 r
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from pylab import 8 ~% a% _( D' P" C L0 _
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from sympy import
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+ `& `8 s' f' H' R( Xfrom matplotlib.ticker import MultipleLocator, FormatStRFormatter
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import math , K# H3 S& o. f" l" ?3 s# D
+ _8 p$ X3 o8 k5 \' Fimport numpy as np # G) \$ a9 Y; {* k- w, n: N8 p1 z7 W
7 f- u6 s: o) G: @ z0 G; ^
N = 31 #总节点数 2 N8 R) ?) P8 a+ b9 y( A
5 {' S% L6 y" \1 Ic=[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] / X9 B [6 ]8 z7 [% B& |
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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]
) C( M9 }- F' B) C2 O4 h- {5 e6 H2 m5 R+ n# g
k1 = [ % G: F, a, F9 X7 S% {0 }
& V/ V3 ~, T+ {2 r( t[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
0 v, `( v A& m: H; a" d# c7 E* j! G. z* J' x. l
[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, Y. A+ i+ g: e3 _9 u/ l! R2 Z' ?4 Z
[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],
4 d1 z \: N; v" a$ R8 K7 _! ]% V. |$ r
[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],
% T+ f+ _1 {5 e I/ I9 P2 m0 S4 _
[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], . n1 ~/ m, r2 x. ?
2 h$ n8 y- r+ g* `
[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],
4 X0 i5 F- ]* i# y/ t- V! ?( ^* P9 r& [# Y6 F- N
[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], ! L: |. Q# t/ o Q% K H0 P1 K
: z7 ]3 i, A4 {[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],
3 I& q6 a( i+ G
" N% S" }9 f/ Y' V[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% ] J9 F( v3 b3 z2 Y5 ^8 o- t) @7 S/ W- x
[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],
' h/ ~% e# k4 E9 D+ ~* r6 Q% J( }9 K N3 D5 V+ J) `1 E/ Z
[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],
; d! E3 `! Z* L" O. I L J
8 }4 C( x! R" x# i0 q1 e[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],
3 ]1 g5 _6 ? z% [% W! ?8 v) Y3 s3 o ~# i4 h" K
[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], 2 `. I5 A, P# k! j
, R- Y+ P8 C- G7 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], 2 w6 y0 S# k& Q. b! K
5 d i' Z8 }5 E1 F5 X1 j$ n' K[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],
# w2 C) Y2 P* C9 B) x$ R) o# ]. h9 R% @2 ^' z
[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],
/ F H7 Z* |% a; T3 `
" ^4 l1 y" o( \[0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], % \$ S, h: O- ]2 y4 [- N& z( D
) y* e* O: L/ r; d# j7 |! }
[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], 4 K2 N& g4 D; @ a$ W# \# Y3 O$ C
' _+ c" d4 \3 x/ P6 h7 [) \& I
[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], . O) a: k8 d1 X8 G3 E
- a7 o0 O0 ^) f6 A+ w# z[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], 4 v2 n* M# q- A; l5 F* }( w8 W6 S" I
% Z& @( k" r$ g$ k: D) 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],
* r& d0 t8 S* g4 X+ s9 U' b. L' O2 P# F9 \
[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],
3 C' e3 ?" | i$ ^5 E5 B! h" {1 i1 Q8 S# I2 C( 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], . ?6 d7 L! d# H2 w
9 R! _' u, }4 z. T, C4 ?+ @[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], / D6 J: |" L$ C& R3 C& s7 {" \
/ r' U2 \# v* d% p; F% d& S# i
[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], 7 C. U8 e1 _: L1 Q5 U6 F
& b) |2 r. x7 ^2 z$ s$ ]/ Y
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0.6,0.6,0.6,0.6,0.6,2],
5 d) J) |7 ]6 G4 ], j! W# ~- ]$ X) C7 o8 c/ e, k7 p4 n
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,1,0,0,0,0,0],
4 O# Q; |; C. H5 e
. u' v+ C$ l0 @% M[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,1,0,0,0,0], 7 m- b; _7 E4 M- \9 N0 g
) E8 O" A0 L, d/ m! U; n/ i[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,1,0,0,0],
2 _6 o$ Y* {0 m/ g E0 c* ]' X7 E( o. S, k3 t, {" R) G j
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,0,1,0,0], . J. P m2 u+ k" T2 l" _0 D
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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], F# t& Q: G( k* P+ X
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[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 K( b! p; t2 @" K8 t
" N K1 ^& f7 D& }2 Y ~/ l: Bn = [i + 1 for i in range(22)] #目标划分,24个值,1-24 % A. l+ ?1 m" \' b+ U
' h6 _, K ?0 ` I8 ?. r1 Lt = [j for j in range(1000)] $ m6 t) t9 j2 y7 B& x7 `2 D
9 e( n; i5 l# m! }ci = 0 1 A5 N- j4 D& A& B
, j v( D+ [- m' r
C = []
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C.append(c)
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for d in range(1100) - V. \. K8 O- E! a' [
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for i in n & I7 O5 \! Z, g# D6 G
( B% H1 j4 V' j: Y5 `5 c" e for j in range(22) 6 X* i8 Y6 s+ W( N
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cj = c[j + 1] - c
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ci += k1[j + 1] math.sin(cj)
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2 C$ C& v1 A" z5 F) t cii = ci + w * E9 x$ q! s* S: I8 q
+ P& {, r, d; `! R! z; [ h = [0.01 cii + z for z in c] 8 Z5 V! e5 A# F
5 U- x: D% v' E6 E) N% @) }4 {) B C.append(h) 9 {" f: T% {- |3 Y, J4 `
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c = h
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def r_function(u) 7 S# E p$ n9 P3 r4 n
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y1 = 0 ) G+ V+ L) R( ]" V _: i' E( d
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y2 = 0
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) F& e3 d+ I7 j4 q& e9 ~ y12 = 0
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y22 = 0 " w6 N# |5 @; _6 l% H0 o9 d. U# L
9 y1 M. { v' l; C/ j r = [] & P6 K6 l6 Y# v# {8 |8 d
: Y- Z, [# q6 ?( a for x in range(1000)
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6 x, l W# T) e$ [3 F' K Q for ul in u " Y6 W* u) n. \- k" A
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y1 += math.cos(C[x][ul])
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6 ?+ }! k- t& q" @ y2 += math.sin(C[x][ul]) |) A) u; J3 N5 z) L1 K
/ w4 R9 p+ D# k7 | y12 = y1 2 ; J. {1 \8 }4 V7 y9 b9 T
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y22 = y2 2
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% P2 M* Z6 U, X9 x: `+ s+ I! ^ r.append((float(1) N ) ((y12 + y22) 0.5))
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, w2 t0 s; z+ ?) b return r . M2 D4 Y1 C6 E6 }9 J1 e# `" p9 o
' S: X$ L# G/ R: B; e$ H7 kr1 = r_function([1,2,3,4,5,6]) 9 z; `4 l1 {7 ?: k/ w2 @2 {) {& Z' X$ I% M
, Z9 L( _4 n0 v7 Z" Hr2 = r_function([7,8,9,10,11,12])
T8 | I) X# ?% _( v( g) O# |: s- {/ P- Q9 w0 E
r3 = r_function([13,14,15,16,17,18]) : k+ h3 ]" O' j7 @! Q
8 }0 s, z+ o( H/ y( n( H" w
r4 = r_function([19,20,21,22,23,24])
$ {7 I( b5 ~9 B$ B1 E( s4 j7 n6 \1 c7 ]/ T3 }+ [% l& |- U
r5 = r_function([25,26,27,28,29,30])
' O$ s5 i; ^0 h
) l9 A/ }1 B; v- |! x1 Z# y$ L#r6 = r_function([31])
4 a( {% i) b! X# X4 S
4 g) f- Q8 Z& b7 ?4 }. \( {8 J3 z9 g. n5 Q; @) W
% v. G5 n! D. Q% S9 c( A/ m
ax = subplot(111) #注意一般都在ax中设置,不再plot中设置 7 F. `% F: [' u; g) E
: t! F w9 q- B0 _9 F
ymajorLocator = MultipleLocator(0.1) #将y轴主刻度标签设置为0.5的倍数
$ E+ H: X8 O9 C7 _: J! W; N) e. R
ax.yaxis.set_major_locator(ymajorLocator) - i/ n6 L; O. u w) i$ e! m, @
! f: l C8 `; z1 uplt.plot(t, r1, marker='o', color='green', label='1')
8 U' R" r$ d y$ }6 v, ?9 {! z7 o: t: G( D. z3 j
plt.plot(t, r2, marker='D',color='red', label='2') " B: Q& a/ g G5 j
d( w/ q( ~- S4 Y; Hplt.plot(t, r3, marker='+',color='skyblue', label='3') " Z+ D; k% F# A1 z* A
) G' ^3 r! M3 @- t' F3 y
plt.plot(t, r4, marker='h', color='blue', label='4')
; ^' v; w& D5 ^, } _: N c
( N" _, c: U( O7 v7 ]- p+ z( c1 qplt.plot(t, r5, marker='_',color='yellow', label='5') * g% w& e( i; U0 \4 A
# n, L& s5 Y5 X) c7 ^) Q
#plt.plot(t, r6, color='red', label='6') ) G) W+ x2 I. W H8 ?7 D
, f, d4 u( |* T
9 H4 _) b7 f, F7 v, a9 _- R6 U) C$ I& ?0 `! ^
plt.legend() # 显示图例 5 `7 l4 I# P/ h4 C: F& e8 P0 \ e
) i$ ` W1 q$ e; w, M. [plt.xlabel('iteration times') 8 E* ?, x; r+ D' D4 B+ ]$ K2 e
, H+ h3 g9 ?2 r( b
plt.ylabel('r') + i0 o+ Q9 k* [2 q
) n @) V! R0 C {plt.show() " d8 Q1 B( J% {; v; N# t
! n& D' ^: |- M% R/ v
# ]9 t, |3 B. k+ A# z独立编组结果如图：
7 e! {1 I- M, b I; u
; \8 q; u/ p! P4 q" p. @$ _( u+ [
6 f! G. U+ T* z6 d# n+ s6 K
7 j' x, r" [0 F& Zk1独立编组- t8 i4 n5 Y, D: B
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：+ Q6 L( X+ L% A* s. A v- \
/ w1 E/ H; K( U" [0 ~2 e0 U) p; bk2 = [
) g( E1 C. ]4 W/ m5 O$ x# {1 b2 q! d4 d3 q. X: d
[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 e! J5 u$ c# V% l" [' J i- g( b J7 g* @% p: `6 l) P
[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], ; t+ d+ \7 J/ F7 K8 d4 \- u
; r( t6 g) d* J. ][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],
7 V) p F( n' Z/ `- x# X+ K
. u8 K$ } {/ M, [" W4 @5 h[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], 9 f* w2 M" I) u
" I: C# v8 \ y: u
[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$ h% H; p- d- u3 _# A% `
5 I/ O9 ] q5 o) G0 }6 v[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],
" n ]$ K* A0 a1 q6 s% d7 o: W) R( ]9 O; Z
[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], , ~: i! o e" \, D
' N$ r# v' }& f9 [5 C8 k% g7 E[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],
: c: [3 z1 ?# F" P: g. V ?" [3 [' Z" { g/ V; m* `* W
[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],
9 I: |$ e h4 ]* X6 u; b7 u$ e9 ~4 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], ) P0 Y4 l. P0 b: ]7 D3 s* ~1 Y
) e1 K, L( @- F" F[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], / b* i9 D" p, f& J9 r
; G1 c5 y1 ~! N7 u7 W! a9 l7 F
[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],
2 q6 E. U3 n9 G5 T$ L. c! C7 w* ?' n6 E$ g; n: R( a9 [
[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],
4 @9 R P0 c) c7 l* A' E% L; e
2 C1 x( J$ `, H* J7 N7 ^% k[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], & p8 Q/ Z0 ~& t* J1 M3 o/ @
: u9 `) B& `2 \3 ?* ~+ \[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], % B4 `' T4 d$ j. f* P2 A v
$ x( f! @3 n5 V0 D+ V
[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], + k- z0 ~* v3 e" H0 E' d
: {% U% y4 ]7 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],
6 Q% r! C7 j- _+ t8 J r5 }3 N; i5 v+ k/ C- s5 d" y
[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], , x; Y! m: G8 M5 }
! L1 {. [- _8 u# W
[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], 1 `7 h! I+ I4 H7 d" p# e$ x% d
; }5 t# ^( y. A/ N$ C# R* F
[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 j' [7 c* f5 l2 \, A- D. Z- a; n, P4 k4 V
[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 R. K' @! E: q. w. c3 {. p& o5 }' {- O, k, `: r2 f3 G
[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],
3 }+ g7 W: Z' m5 h+ S3 _9 T$ L2 x9 Q" W
[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], $ g5 L3 S8 f) K; o4 [) y# z7 G9 J" f3 M
7 f1 }1 G1 N& Z# K1 l[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], ; H9 X; h" h6 Z
( B# V8 l7 U; ?4 }[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 U2 A5 Y8 q' _6 @1 Z6 e
+ H6 L0 v6 Z" }6 [4 i6 g/ C, k4 B
[2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,1]
8 q( Q6 K7 v& F! g( M
' Y! \1 F& \/ N+ o/ k/ i& N6 j8 B; @
- Y N: ]2 [+ z* | m0 }k2独立编组+ D# w: D, {& q3 T. U
- h |" }, U/ q2 G" h2 T* e6 g8 T
7 m# @. f: \8 p h/ a; f- m- T: `, @8 s9 s, T& ?7 y" v
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？
# ?# a4 O9 v! ?1 A! X" N( {, n9 e
这里还有一个公式，来解决这个问题，编组同步能力的量化：
! n- W% l& q l9 \
. b1 W5 l# M4 x! R) O! a# M9 C- B4 P/ K
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？$ c, f5 W+ _- [3 p# i
& Q& |3 d1 t; }4 n( Y- @/ I1 n. F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：
+ Q8 ^/ d" I8 R Q% v: o6 g
5 m1 _6 Z8 d9 Y/ w
) V& |: h6 l4 G& a9 G$ P4 C+ B+ s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。
8 b1 x! z; @+ [4 c7 o: p5 h( T0 T! q. o
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：
! t8 i% B3 V% j6 V8 R6 W" p. V1 r6 K* o, T% r$ g& u
clc;
" h, R2 v; V6 S9 s, ~, O) r: Z& M6 H4 e* Z) e
clear all; * _ a5 D& w# e( s
& o! i+ Q& z' Y4 X+ P+ f& q
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 ], k5 r5 n
- h& q. Q3 Y, u; c 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;
" c) F7 O7 q+ q! [' u j) ]& V2 w9 {- E8 a4 x( m9 k+ m
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;
+ ?3 V& V; [/ j$ L& l% _4 d8 |
2 p! p8 t, F, v* J$ o; Y 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 w9 e! B$ Y3 ?4 R; C. P u0 a, o9 P( ^6 {6 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; ! {; i' [3 J0 x& ~
, p, h" J; N, W1 N8 R& C
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; * K: y, @' T0 |. A4 K& A
; m: {* V2 q) }5 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;
$ x+ o1 @" t' p/ Z- E; R$ K7 w; @7 D" D# n
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; ! {1 S( L. ]1 o$ p% y% c
/ P. t2 N! y7 P 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;
7 A! q! O; `5 m& P! W3 a% G1 b! O3 ^
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; $ q; c& t" L# A* ^! @
/ u" M" T, \& S( c, P 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; . y3 y4 N9 b- z
( a; \ ]' Z; W 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;
, X* s1 q; V! m2 ?( C( N5 i, ~' d4 Z3 h, m% l! o9 ^& D
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; ) [+ I" b' R4 n$ J& k% M) k
+ J1 L, a8 o) B& F3 B( ~4 | 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 X! T0 ?2 N# x5 ?# a# R
; N1 e+ U7 m F7 W
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 C5 [9 I$ S, Z3 y
1 R, a2 s1 r& R4 _; C! m+ M
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;
6 ^: W3 k2 u9 B/ K9 N+ J9 c& a
# {1 X% U! A6 A 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; " M+ Z: R; C1 s% r2 H0 E+ l
6 L! ]! v& z2 {9 S- 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; 5 \* A: w9 s i- g i) R9 M, K& M8 g
! z9 R* A& v0 @' q4 L8 N7 q3 ]6 E
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; 3 u0 J, P- r4 L- q7 N/ S
, x0 U, V! ]9 \( i 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 I, d, ?% X$ N+ U. x9 z3 M
2 X% {% Z+ j: C6 i& z! }; ^& j
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;
3 q% @8 e' ^1 e" y: k0 z7 K1 O# M, E# i" E& h9 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;
( P. Q B+ }- A, P7 m# D. x" h( a
) a2 p9 J3 x- Q! t 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; 8 X7 C( j; A$ Z6 K Y; k; n6 i
h; Q) }- f, U; L) \! {- E
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% T* c# r1 z
0 E5 t( ?$ `. W0 J; m& V 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; 1 [7 L9 v( W$ I3 K# F- Y V* a
" E' e, G ?0 [; r; V
]
9 W% O3 }) L% c, R3 S5 y8 ]3 Q
+ T8 a/ t3 I% w. V: QN = 25;
1 U' c' B/ X0 K" U( M$ m) E4 b: L0 @! R- ]7 e8 E' h' V
Step= 10000;
* [. i# g* d2 |& m! d8 U: u5 G0 k0 }% O ]4 t, V; a z z2 A; T# l
Theta=zeros(Step,N); 0 q0 o8 R" C$ ~+ W q `/ q
( E+ V2 h* t/ r; p- G# AOmega =zeros(Step,N); / W n6 l4 [0 \9 [) R, I
" D& v' u1 i! H# I+ _
DeltaT=0.01
- |2 {" F5 g8 z2 t/ x) d& ^7 x
) i3 @: }0 e. PTheta(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 F. J" ^6 O" d, C1 L6 c, A1 B7 l$ o: f u- I
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]; 4 V4 m, c# A! k2 R6 }
6 m! Z6 |7 J0 G% \/ S2 w0 T% 求解微分方程 % ?) b3 j* `- W$ p" }: v* v m4 ~; J5 B
8 f# v+ S4 y9 C5 E& q' A* Pfor n = 1:Step
5 R! Y& Q$ Q p- u& l% r6 @" q* c$ f) E
for i = 1:N
: x5 H& w1 u% |1 O
" G* W) e. U( G6 C: L* x- q( [ Sigma = 0; % j$ \! S( x3 T' c Q8 |
5 d3 }: t% W4 ^1 }2 b
for j = 1:N
& I! p9 l" `' g5 `9 z9 G
* C/ \2 e! z# h K4 |6 h) j %%判断k对同步效果的影响
7 X$ p; i# z; r% z/ u; e+ v7 [( j* Q# [! Y( J
= Sigma + k(i,j) * sin( Theta(n,i) - Theta(n,j) );
, ~. D. x( [ [% z
# N3 _( ]) b; q v end
/ s% z" }- n: G% m! y1 q4 J
# d, C0 b5 q0 x+ j Omega(n+1,i) = Omega(n,i) - Sigma; 0 |/ t; P$ x. T) N0 v8 M8 G
1 o) R: d0 ?' s2 q5 f Theta(n+1,i) = Omega(n+1,i)*DeltaT + Theta(n,i); 0 m' R- v: M4 O" U4 w. L: p6 d
1 N* d0 m( u6 y# Z9 N
end
: T% m- |/ D$ ~9 d. I% R" q/ B$ B; W# B/ l* U, g) i
end
X. V" U" |+ W1 x
. P% g( m& D5 ~0 \- ^% 求解序参量r(t) 1 ~6 ]8 I2 i: [4 l( W5 }
+ T; \9 `5 W: H2 M" g! j' o
groupN = 5; 7 N2 b1 \6 O9 z9 d
4 W' S8 F# F5 d5 O' J
Step = 1000; 7 y7 a% u! r0 x- S
) F J. W c/ v6 Y! K7 t! Jfor i = 1:Step
$ O7 c) v/ z, \3 M; e9 Q8 K, u" _6 }
sigma1 = 0; 7 ]+ h# J7 K- S, | Y1 X# Y
: u/ x% v. r; R( C5 a2 H/ N
sigma2 = 0; 1 B* N# r4 _4 L* f1 y. F6 i
; _! F/ y6 [3 j2 c sumTheta = 0;
5 s9 ~- L& U% ^8 r- x; A- u! T# Z; O" X5 _! a8 h
for j =6:10 %表示相应的群组 ' s' o+ Z- @ j% z& B! m
7 Q6 n( g; Q3 C1 N1 U2 w' c
sigma1=sigma1+cos(Theta(i,j)); * F1 d4 L# s% F" W$ I
* ^( }/ N" ` Y4 s4 R: D
sigma2=sigma2+sin(Theta(i,j));
. k1 i0 B1 \' J- k2 h
. {. l4 D' E* O7 T0 R1 n8 ]% u end
6 ^* i. P/ w6 q( r S& h
5 k; v' k, D+ Z5 _! | r(i)= sqrt( sigma1^2+sigma2^2 )/groupN; 2 B! e: X' s% r% F4 k+ Y
' l' [9 u( `* s' m: r0 A
end
8 U, r2 L2 h% |5 t% {" b) y3 ?. n# W
x= 0.01：DeltaT：DeltaT*Step;
& c$ h& A' c% e+ \
+ X& N9 T% V9 U/ eplot(x,r); ) z6 O; `, Z: w5 ^( a$ w
MATLAB版不完全分散编组结果如下图：
7 K3 l$ Z. K, s d/ ?
8 I7 n* h' O Z) c
|
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发表于 2021-2-22 10:03
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