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Suppose I have $n$ features as $Y$ = ($y_1 , y_2 ...., y_n$), and a matrix of $J$ of dimension $M$x$N$, one feature of $Y$ is selected randomly to be convolved with one random column of $J$ resulting a new vector called, for example, $X$.

The question, if I have the matrix $J$ and the vector $X$, can I detect the selected feature from $Y$ and selected column from $J$ using any deep learning algorithm? such that CNN or DNN ...... etc. What's about if I don't have the matrix $J$ am I still able to detect the selected feature $y$ and column of $J$ based on $X$ ?

For example, let's $Y = [1,2,3,4]$ and matrix $J$ is any random matrix of dimension $4$x$4$. Hence, one random number of $Y$ either 1 or 2 or 3 or 4 is going to be convolved with one column of matrix $J$, let's say element 2 of vector $Y$ is convolved with with second column of matrix $J$, so the resulted vector is $X$. So the question can we estimate the element selected from $Y$ based on matrix $J$ and resultant vector $X$ using machine or deep learning algorithm? I think yes we can, but what's the most appropriate algorithm to do that?

thank you

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  • $\begingroup$ I think your question is missing some information. Are all $y_i$ eventually multiplied? Can you give an example of such a system? $\endgroup$ – havakok Jul 17 at 5:21
  • $\begingroup$ @havakok .. thank you for you reply, I tried to write a basic example, I hope it's somehow clearer now. . No, only one element of $Y$ is multiplied and we need to detect it and detected with which column of $J$ was multiplied. $\endgroup$ – New_student Jul 17 at 8:54
  • $\begingroup$ Are your features non-negative? $\endgroup$ – jojek Jul 17 at 19:21
  • $\begingroup$ @jojek Yes, they are non-negative $\endgroup$ – New_student Jul 21 at 3:10
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Your question can be interpreted in many ways. First, convolution with a number is not a common operation. Convolution is defined by:

$$(f*g)=\int_{-\infty}^{\infty}f(\tau)g(t-\tau)d\tau$$

Assuming a constant signal $g(t)=a\in \mathbb{R}$ this is reduced to: $$(f*g)=\int_{-\infty}^{\infty}af(\tau)d\tau=a\int_{-\infty}^{\infty}f(\tau)d\tau$$

I am assuming, therefore, that you meant multiplication and not convolution. Correct me otherwise and I will edit the answer accordingly.

Secondly, assume for your example the given parameters:

$$Y=[1,2,3,4],J=\left[ \begin{matrix} 1 & 2 & 3 & 4\\ 1 & 2 & 3 & 4\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{matrix} \right], X=\left[\begin{matrix}4 \\ 4\\0\\0\end{matrix}\right]$$

It is easy to see that a couple of use-cases could give us this result. For example, it could be that the chosen $y$ was $y_4=1$ and the chosen column was the most right column $j_3$. On the other hand, the combinations $\{y_1,j_1\},\{y_3,j_0\}$ are also plausible. I have purposedly gave a very bad, linearly dependent, matrix $J$ to illustrate that the choice of $J$ in the design process or formulation of the problem is of major interest and is fundamental for answering you question. The following example illustrates the complete opposite:

$$Y=[1,2,3,4],J=\left[ \begin{matrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{matrix} \right], X=\left[\begin{matrix}x_0 \\ x_1\\x_2\\x_3\end{matrix}\right]$$

Obviously, $x_0=y*j_0$,$x_1=y*j_1$, and so on. Here, it is easy to deduce $y$ and only a small number of $X$ are even possible.

Generally, if we call $J$ a dictionary, you are looking at something called Orthogonal Matching Pursuit (OMP) problem, which is not a machine or deep learning problem. Otherwise, you can try and use a deep learning model to try and learn the dictionary, though this would be unrecommended as such a solution is not compatible with such a problem in my opinion.

Hope this answers your question. If not, or I misinterpreted your question, feel free to comment and I will try to supply more information if I can.

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