« first day (2290 days earlier)   

8:45 AM
@e-sushi Nice one: "seven of the ten most common vulnerabilities seen across federal agency networks at the issuance of this directive would be addressed through complying with the required actions in this directive related to web security" xD
 
 
1 hour later…
9:47 AM
@Lery “Governmental InfoSec” ;)
 
10:19 AM
In case anyone is interested in grabbing a micro-bounty…
6
Q: Homogenous vs heterogeneous unbalanced Feistel networks?

e-sushiUnbalanced Feistel networks can be homogenous (F-function identical in each round), or they can be heterogeneous (F-function not always identical in each round). The advantage of heterogeneous UFNs is, that its internal properties change each round, making it more difficult to find specific char...

 
 
8 hours later…
6:15 PM
Hello @SEJPM
Can I ask you something about a proof?
I am reading about quadratic residues...
 
@Evinda surely you can ask, whether I will be able to help is a different question though ;)
 
I haven't understood the proof from the part: The $\frac{1}{2}(p-1)$ elements.. till the end.
Why do we pick i to belong in the interval [0,p-1) ? Do you have an idea? @SEJPM
 
@Evinda because you want to talk about all squares
and all squares have the form $g^{2i}=(g^i)^2$
and if you were to go bigger than $1/2(p-1)$ you'd repeat
 
6:30 PM
How do we get that the number of the $g^{2i}$s, $0 \leq i<p-1$ is $\frac{1}{2}(p-1)$ ? @SEJPM
 
@Evinda because $g^k$ generates $p-1$ elements and exactly half of them are captured by introducing the $2$
(note also that you are doing math in $\mathbb Z_{p-1}^*$ with $i$ and $k$ with $p-1$ even
 
6:54 PM
First of all do we know that when we have $g^{2i}$ that this will be a quadratic residue modulo p, and if we consider $g^{2i+1}$ that it will be not? @SEJPM
 
@Evinda well, $x$ is a quadratic residue if there is some $y$ such that $y^2=x\bmod p$, right?
so if we pick $g^{2i}$ we know it will be a quadratic residue because $(g^{i})^2=g^{2i}$
 
Yes, and if we consider $g^{2i+1}=(g^{i})^2 g$ this will be not a quadtratic residue, since g cannot be written as $g^{2k}$, for some k, right? @SEJPM
 
@Evinda 2i+1 is odd and thus you can't find any (integer) k such that 2k=2i+1 modulo an even number (p-1)
@Evinda yes, because $g^{2k}=g\implies g^{2k-1}=1$ and $2k-1\neq p-1$ because p is odd
 
I see... And we see that for $i \in [\frac{1}{2}(p-1),p)$ and for $j=i-\frac{p-1}{2} \in [1, \frac{p-1}{2})$, it holds that $g^{2i}=g^{2j}$, so there are in total $\frac{p-1}{2}$ quadratic residues, right? @SEJPM
 
7:19 PM
yes
 
I see, thank you :) @SEJPM
 
:)
 

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