# Highly Memory-Efficient Recursion

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M R 2021 年 1 月 24 日
コメント済み: Stephen23 2021 年 1 月 26 日
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.
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Rik 2021 年 1 月 26 日
Is it really impossible to clarify your question with edits and comments? I doubt it.
Stephen23 2021 年 1 月 26 日

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### 回答 (1 件)

Sindar 2021 年 1 月 24 日
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
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Sindar 2021 年 1 月 24 日
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 年 1 月 25 日

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.

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