What is the minimum code size for CDMA codes for synchronuous CDMA (not asynchronous CDMA!) with N number of participants?

Furthermore, suppose N parties want to uniquely transmit a message of length K0 each, where K0 is small (e.g. 16 bit) and N is very large (e.g. 1000). What is the minimum size of the total message (i.e., the message including all N different messages of size K0).

Is the minimum size of the transmission K0*N = 16000 bit or can it be made lower?

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    $\begingroup$ I will ask you the same question: do you have any reference for this? $\endgroup$ – BlackMath Nov 17 '19 at 4:00
  • $\begingroup$ @blackmath: No I have not of course, otherwise I would not need to ask this question. By the way, the answer “No, this is not possible, for N parties, the code length must be at least N, regardless of the size of N” would be a valid answer as well. $\endgroup$ – divB Nov 17 '19 at 5:46
  • $\begingroup$ Because N no of participants can addressed with logN bits ex:-8 people with 3 bits. $\endgroup$ – Ch.Siva Ram Kishore Nov 17 '19 at 6:52
  • $\begingroup$ @divB point is that we could infinitely many questions of the scheme "1=2, is this true?", but we don't honor these questions in general, because the effort of asking that question and negatively answering it oneself is way below the effort of answering it as a third party here. So, if you don't have a source, and you don't have a reason to believe this is true, there's not really a question to ask here, to be totally honest. $\endgroup$ – Marcus Müller Nov 17 '19 at 11:54
  • $\begingroup$ @divB I however doubt you don't have a source, because you even use quotation marks in "with a delicate design of the orthogonal codes", so that comes from some context that you're omitting here. Please try to add more context! The way it's asked now, you have answered your own question (no, orthogonality isn't possible with less than K dimensions), but I think you're confusing a specific method (orthogonal spreading codes) with the ability to send multiple messages at once. $\endgroup$ – Marcus Müller Nov 17 '19 at 11:57

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