在7D空間爬山

Question

我在7D空間（意思是x =（x1，x2，x3，x4，x5，x6，x7））有以下函數，我想用matlab中的爬山算法找到這個函數的最小點。

我發現this link有用，但我不知道如何在Matlab中實現我的功能。

更新：

我實現下面的代碼，但我真的不知道這是否是正確的。

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
%%% Create a grid of states  %%% 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
clear all ,close all; 
n=7; 
range=[-32.768:0.1:32.768]; 
x=[0,0,0,0,0,1,1]; 
F=-20*exp(-0.2*sqrt(1/n*sum(x.^2)))-exp(1/n*sum(cos(2*pi*x)))+20 +exp(1); 
F1=zeros(7,2); 

best = -100000000; % Best value found so far. 

for (j=1:20) 
    % Pick a starting location at random, and try and find the maximum state by hill climbing. 
    % Repeat this a (to be precise, repeat it until j = 20). 
    s=floor(100*rand(7,1)) ; 
    % Generate successors, and compute the one with the maximum value. 
    % Only consider states to the N, S, E, W, and NoMove. 
    for (i=1:100) 
     % Find successors 
     S0=s; 
     F0=-20*exp(-0.2*sqrt(1/n*sum(S0.^2)))-exp(1/n*sum(cos(2*pi*S0)))+20 +exp(1); 
     for tt=1:7 
      arr=[0;0;0;0;0;0;0]; 
      arr(tt)=1; 
      S1=s+arr; 
      F1(tt,1)=-20*exp(-0.2*sqrt(1/n*sum(S1.^2)))-exp(1/n*sum(cos(2*pi*S1)))+20 +exp(1); 
      arr(tt)=-1; 
      S1=s+arr; 
      F1(tt,2)=-20*exp(-0.2*sqrt(1/n*sum(S1.^2)))-exp(1/n*sum(cos(2*pi*S1)))+20 +exp(1); 
     end 

     [v,vi] = max([F1(:,1)',F1(:,1)',F0]); 
     arr=[0;0;0;0;0;0;0]; 
     index=mod(vi,7); 
     if(index==0) 
      index=7; 
     end 
     if(vi<=7 && vi ~= 15) 
      arr(index)=1; 
      s=s+arr; 
     elseif(vi>7 && vi ~= 15) 
      arr(index)=-1; 
      s=s+arr; 
      else 
      s=s ; %% for better understanding 
     end  
    end 

end 

Answer 1

我在這裏實現它。我希望對有問題的另一位讀者有用。

clear all ,close all; 
clc; 
n=7; 
range=[-32.768:0.1:32.768]; 
%x=[0,0,0,0,0,1,1]; 
%F=-20*exp(-0.2*sqrt(1/n*sum(x.^2)))-exp(1/n*sum(cos(2*pi*x)))+20 +exp(1); 
F1=zeros(7,2); 
for (j=1:20) 
    s=floor(rand(7,1)*64-32) ; 
    i=0; 
    convergence=0; 
    while(convergence~=1 && i <10000) 
     % Find successors 
     S0=s; 
     F0=-20*exp(-0.2*sqrt(1/n*sum(S0.^2)))-exp(1/n*sum(cos(2*pi*S0)))+20 +exp(1); 
     %step=rand(); 
     step=0.005; % this is step of climbing 
     for tt=1:7 
      arr=[0;0;0;0;0;0;0]; 
      arr(tt)=step; 
      S1=s+arr; 
      F1(tt,1)=-20*exp(-0.2*sqrt(1/n*sum(S1.^2)))-exp(1/n*sum(cos(2*pi*S1)))+20 +exp(1); 
      arr(tt)=-step; 
      S1=s+arr; 
      F1(tt,2)=-20*exp(-0.2*sqrt(1/n*sum(S1.^2)))-exp(1/n*sum(cos(2*pi*S1)))+20 +exp(1); 
     end 

     [v,vi] = max([F1(:,1)',F1(:,1)',F0]); 
     arr=[0;0;0;0;0;0;0]; 
     index=mod(vi,7); 
     if(index==0) 
      index=7; 
     end 
     if(vi<=7 && vi ~= 15) 
      arr(index)=step; 
      s=s+arr; 
     elseif(vi>7 && vi ~= 15) 
      arr(index)=-step; 
      s=s+arr; 
      else 
      convergence=1; %this means no neighbor has better value than current a 
          %maybe this point be local optimom 
     end  
     i=i+1; 
    end 
    disp('*****************************'); 
    disp(sprintf('Step of convergence %i:',i)); 
    disp('coordination of optimum point :'); 
    disp(s'); 
    disp('*****************************'); 
end