Highly Memory-Efficient Recursion

Question

M R 2021-1-24

1
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/725787-highly-memory-efficient-recursion

评论： Stephen23 2021-1-26

Let us assume we observe data over a 6-month horizon on a monthly basis (k = 6). Let us also assume the following recursive structure (example below).

I wonder, how this structure can be programmed most efficiently considering the following aspects (Efficiency will be absolutely necessary):

The period of 6 months is for simplicity and is not generally fixed to this value. Accordingly, this value needs to be redefined;
The monthly calculations have to be stored because I need it for further calculations.

MWE:

beta = sym("beta", [1 2]).';
phi = sym("phi", [2 2]);

In k1 we calculate a matrix of the following form:

K1 = beta * beta.'

In k2 we have a similar expression, but now it also depends on phi times the previous period.

K2 = beta * beta.' + phi * K1

In period k3 it is the same as in k2, only that K1 changes to K2. So from here on it is recursive and repeats itself. So for the last observation (k6) it is the same formula with K5.

2 个评论
显示无隐藏无

Rik 2021-1-26

Is it really impossible to clarify your question with edits and comments? I doubt it.

Stephen23 2021-1-26

http://web.archive.org/web/20210126191006/https://www.mathworks.com/matlabcentral/answers/725787-highly-memory-efficient-recursion

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Sindar 2021-1-24

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/725787-highly-memory-efficient-recursion#answer_605282

在 MATLAB Online 中打开

Do they need to be symbolic? That probably adds a decent cost

Similarly, if you have many zeros, sparse matrices will store more efficiently

Are both beta and phi measured monthly?

If k remains small, it may be more efficient to store phi_i and beta_i, then manually unwrap the recursion and directly calculate the Ki you want, like so:

% K6 = beta6*beta6.' + phi6 * K5
% K6 = beta6*beta6.' + phi6 * (beta5*beta5.' + phi5 * K4)
% K6 = beta6*beta6.' + phi6 * (beta5*beta5.' + ...
%     phi5 * beta4*beta4.' + phi4 * (beta3*beta3.' + phi3 * K2))
K6 = beta6*beta6.' + phi6 * (beta5*beta5.' + ...
    phi5 * beta4*beta4.' + phi4 * (beta3*beta3.' + phi3 * (beta1 * beta1.' + phi1 * K1)));

This will save memory, but there's also good chance it will run faster than recursion when you need a single Ki

5 个评论
显示 3更早的评论隐藏 3更早的评论

Sindar 2021-1-24

Ah, I'd assumed that beta and phi were much larger. So, is each individual K 3x3?

Can you explain why symbolic expressions are necessary? This is probably the main reason for large memory requirements, and may interact poorly with recursion

How long do iterations take?

M R 2021-1-25

编辑：M R 2021-1-25

Yes, each in dividual K in this case is 3 by 3.

Symbolic Expressions are necessary because I want to differentiate these expressions afterwards. Moreover, I need these expressions for an optimization.

请先登录，再进行评论。