递归最小二乘（RLS）算法的复杂性

Question

哪些操作使recursive least squares (RLS)算法的复杂度等于O（n ^ 2）以及为什么？

% 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论文有一个很好的解释，这是图像中的相关部分：