递归最小二乘（RLS）算法的复杂性

Question

哪个操作使recursive least squares (RLS)算法的复杂度等于O（n^2），为什么？

% Filter Parameters 
p  = 4;    % filter order 
lambda = 1.0;    % forgetting factor 
laminv = 1/lambda; 
delta = 1.0;    % initialization parameter 
w  = zeros(p,1);  % filter coefficients 
P  = delta*eye(p);  % inverse correlation matrix 
e  = x*0;    % error signal 
for m = p:length(x) 
    % Acquire chunk of data 
    y = n(m:-1:m-p+1); 
    % Error signal equation 
    e(m) = x(m)-w'*y; 
    Pi = P*y; % Parameters for efficiency 
    % Filter gain vector update 
    k = (Pi)/(lambda+y'*Pi); 
    P = (P - k*y'*P)*laminv; % Inverse correlation matrix update 
    w = w + k*e(m); % Filter coefficients adaption 
end 

full code

Answer 1

This纸有一个很好的解释，这里是一个图像中的相关章节：