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I am a newbie to the world of algorithms design and currently working on a algorithm which is related to cell-phone user associating with cell-phone tower. Let $\mathcal{K} = \{1,\dots,K\}$ be set of cell-phone users and $\mathcal{J} = \{1,\dots,J\}$ be set of cell-phone towers. Let $k$ and $j$ be index of cell-phone users and cell-phone towers respectively.

One of the step of this algorithm is to start with a random feasible initial activation vector $\beta$, where each element of $\beta$ is index of cell-phone tower to whom cell-phone user $k$ is associated.

I completely understood that length of vector $\beta$ will be $K$ and it will contain the various values of $j$.

But, my query is how should I initialize the vector $\beta$ in any software say MATLAB if we consider for e.g., $K = 20$ and $J =3$.

Any help in this regard will be highly appreciated.

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  • $\begingroup$ I can't understand what you're asking. If you're asking how to inititialize a variable in Matlab, that is off-topic here. If you're asking what a specific algorithm does, I can't tell what algorithm you have in mind; presumably you look at what it tells you to do. If you are asking for us to design an algorithm, it's unclear what the constraints are. $\endgroup$
    – D.W.
    Commented Aug 19, 2022 at 18:08
  • $\begingroup$ Thank u sir for the response...by this question I mean to ask the following: Say I have the case where there are 20 users and 3 towers, then I have to randomly initialize the vector $\beta$ that has index of the towers. So, since its random means we can choose any thing which is feasible. So, now if I choose that my first 10 users will connect to tower 1 , next 4 users to tower 2 and remaining 6 to tower 3 then can I say that my vector $\beta = \{1,1,1,1,1,1,1,1,1,1,2,2,2,2,3,3,3,3,3,3\}$. Is this representation correct sir? $\endgroup$
    – chaaru
    Commented Aug 20, 2022 at 5:34
  • $\begingroup$ Phones usually connect to the tower with the strongest signal that had available capacity and don’t care about any other phones. $\endgroup$
    – gnasher729
    Commented Aug 20, 2022 at 9:26
  • $\begingroup$ Yes sir....I agree with u...I am just considering initial random step as given in the paper... $\endgroup$
    – chaaru
    Commented Aug 20, 2022 at 10:17
  • $\begingroup$ @chaaru What paper are you referring to? $\endgroup$ Commented Aug 20, 2022 at 19:42

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