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标题: 请问对于以下的黄金分割搜索算法如何改动？ [打印本页]

作者: 勇攀高峰 时间: 2014-12-13 03:53
标题: 请问对于以下的黄金分割搜索算法如何改动？
我要求解f(x)=x^2+sin(x)在(0,1)上的最小值，然后精确度大概到0.000001位就行了，以下的代码该如何改？需要在前面加些什么？如何加呢？求大神！

function [S,E,G]=golden(f,a,b,delta,epsilon)

%Input  - f is the object function input as a string 'f'
%    - a and b are the endpoints of the interval
%    - delta is the tolerance for the abscissas
%    - epsilon is the tolerance for the ordinates
%Output - S=(p,yp) contains the abscissa p and
%       the ordinate yp of the minimum
%    - E=(dp,dy) contains the error bounds for p and yp
%    - G is an n x 4 matrix: the kth row contains [ak ck dk bk];
%       the values of a, c, d, and b at the kth iteration

%  NUMERICAL METHODS: Matlab Programs
% (c) 2004 by John H. Mathews and Kurtis D. Fink
%  Complementary Software to accompany the textbook:
%  NUMERICAL METHODS: Using Matlab, Fourth Edition
%  ISBN: 0-13-065248-2
%  Prentice-Hall Pub. Inc.
%  One Lake Street
%  Upper Saddle River, NJ 07458

r1=(sqrt(5)-1)/2;
r2=r1^2;
h=b-a;
ya=feval(f,a);
yb=feval(f,b);
c=a+r2*h;
d=a+r1*h;
yc=feval(f,c);
yd=feval(f,d);
k=1;
A(k)=a; B(k)=b; C(k)=c; D(k)=d;

while (abs(yb-ya)>epsilon)|(h>delta)
k=k+1;
if (yc<yd)
   b=d;
   yb=yd;
   d=c;
   yd=yc;
   h=b-a;
   c=a+r2*h;
   yc=feval(f,c);
else
   a=c;
   ya=yc;
   c=d;
   yc=yd;
   h=b-a;
   d=a+r1*h;
   yd=feval(f,d);
end
A(k)=a; B(k)=b; C(k)=c; D(k)=d;
end

dp=abs(b-a);
dy=abs(yb-ya);
p=a;
yp=ya;

if (yb<ya)
p=b;
yp=yb;
end

G=[A' C' D' B'];
S=[p yp];
E=[dp dy];

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