314 reputation
211
bio website sites.google.com/site/…
location France
age 27
visits member for 3 years, 11 months
seen Jun 2 at 21:38

I am a PhD Student at INRIA Rennes Bretagne Atlantique research center, France. My PhD topic is on privacy preservation in user-centric peer to peer search environments. The field includes cryptography, differential privacy, data mining, secure multi-party computation and large scale peer to peer systems. I am also a staff member at the Computer Science department, Helwan University, Egypt. I currently hold a masters degree in Computer Science from University Rennes I, France.

My interests include computational complexity and computability theory, linguistics and natural language processing, cryptographic protocols and quantum secure multi-party computation


Jul
2
awarded  Curious
Oct
1
accepted Can the measurement matrix used for compressive sensing be a sparse matrix?
Sep
30
accepted Starting with $\frac{-1}{1}=\frac{1}{-1}$ and taking square root: proves $1=-1$
Sep
30
asked Can the measurement matrix used for compressive sensing be a sparse matrix?
Sep
12
accepted Handshaking with no crossovers in minimum number of rounds: has this problem been studied?
Aug
10
comment Starting with $\frac{-1}{1}=\frac{1}{-1}$ and taking square root: proves $1=-1$
But in this example, $e$, the base of the natural logarithm, is positive: $e^{z} \overset{!}{=} (e^{2\pi i})^{z/2\pi i} = 1^{z/2\pi i} = 1$ for all $z$. Is this an exception to the rule?
Aug
8
comment Starting with $\frac{-1}{1}=\frac{1}{-1}$ and taking square root: proves $1=-1$
So is it always safe to distribute the root over products and quotients if ["the number insider the root is positive" $\vee$ "the root is odd"]? Is that the rule?
Aug
7
comment Starting with $\frac{-1}{1}=\frac{1}{-1}$ and taking square root: proves $1=-1$
@ZevChonoles : The first question yes. But not the second question.
Aug
7
comment Starting with $\frac{-1}{1}=\frac{1}{-1}$ and taking square root: proves $1=-1$
Can you elaborate on why is this a problem?
Aug
7
comment Starting with $\frac{-1}{1}=\frac{1}{-1}$ and taking square root: proves $1=-1$
Related: math.stackexchange.com/questions/445585/… .
Aug
7
asked Starting with $\frac{-1}{1}=\frac{1}{-1}$ and taking square root: proves $1=-1$
Jun
30
accepted Proving that $\Pr[|X| > T\sqrt{n}/2] \geq \Pr[|X-\mathbb EX| < T\sqrt{n}/2]$
Jun
30
comment Proving that $\Pr[|X| > T\sqrt{n}/2] \geq \Pr[|X-\mathbb EX| < T\sqrt{n}/2]$
Yes. But isn't by using the triangle inequality we have: $|X-\mathbb EX+\mathbb EX|\leqslant|X-\mathbb EX|+|\mathbb EX|$? Where did the minus sign in $-|X-\mathbb EX|$ come from?
Jun
30
comment Proving that $\Pr[|X| > T\sqrt{n}/2] \geq \Pr[|X-\mathbb EX| < T\sqrt{n}/2]$
Could you kindly elaborate the first inequality? I don't seem to be able to know how it was deduced.
Jun
30
asked Proving that $\Pr[|X| > T\sqrt{n}/2] \geq \Pr[|X-\mathbb EX| < T\sqrt{n}/2]$
Jun
26
comment Does this probability distribution have a name?
Hey, thanks for the answer. Would you kindly provide some explanation of how you reached this conclusion?
Mar
5
accepted The smallest normal set containing a subset $X$, what is a “normal set”?
Feb
22
awarded  Announcer
Jun
8
awarded  Constituent
Jun
8
awarded  Caucus