I have just noticed that this question, which I have found quite exciting (and so have at least 14 other users who have voted for it) has been withdrawn by the OP. Is there any way to figure out the reason for that?

## 1 Answer

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Mostafa's comment above:

This problem has imposed by someone to me for doing research on it. Today he told me that he did not allow me to present the problem to others. So I deleted the question and informed him. Sorry for this mistake, but I ask the moderators to agree with my decision.

especiallythose questions that have already received upvoted answers. Luckily it had not been answered yet. $\endgroup$andall coordinates of $v$ are in $\{0,1\}$. $\endgroup$Well, Zaimi is right that the problem 1.2 is for any vector $x$ (not just a 0-1 vector) see the definition in line 4 of Page 2. But you are right that since I quote Caro and Jacobson, then the special case where we require 0-1 is what they looked at in that paper. Also, my proof uses only 0-1 in the vector. In any case, as far as I know, even the less restrictive version (where you allow $x$ to be arbitrary) is open.$\endgroup$12more comments