CBC-MAC padded message difference If an attacker had a padded message and its corresponding CBC-MAC tag, how would they construct another message which produces the same tag for the same key as the original? The padding scheme used in this scenario is to add 0's to the end of the message until is it a full block.   What about when the attacker has an non-padded message? How would they accomplish a similar attack and what conditions are needed for the message for the attack?
CBC-MAC Padding weaknesses I'm trying to wrap my head around how different padding methods affect the strength of CBC-MAC's. Suppose a message is split into different blocks, m1, m2, m3 etc. and the last block is padded with 0's to fill it if needed.  If the message is intercepted, why can an attacker easily create another message with the same MAC tag for the same key as the message that was intercepted?  If another message is intercepted that does not contain any padding (so it completely fills the last block), how does this make it more difficult for the attacker to create a tag that matches? Does the message need to meet any conditions for the attack to work?  If this padding method is not secure, what is a better method to secure against these attacks? Would it be more secure if the padding was just random? Not sure how that would work.
Prove that if $|G| = p^n$ then $G$ has a subgroup of order $p^m$ for all $0\le m &lt;n.$   Since $G$ is of prime-power order I know $|Z(G)| \ne e$ so there is an $a\in Z(G)$ with order $p$ such that $p \mid |Z(G)|$. Now, the subgroup generated by is normal since it's a subgroup of the center. How can I get this normal subgroup to be less than $n$? I would like to see a full how-to guide on how to install Ubuntu. Software recommendations How can I send single file to multiple remote sites at the same time? Can scp be used to send a single file to multiple remote servers at the same time? If so, how? If not, what's the alternative? Finding minimum bounding extent of given pixel value within raster? Imaging new hard drive in Windows 7 laptop? I'm upgrading my Windows 7 laptop to a new hard drive.  How can I clone or image the old hard drive over to the new hard drive, so that it looks like nothing has changed (other than the new hard drive) to my operating system and applications?  Can I use what is built into Windows 7 Backup or should I use something else? I have a dictionary of values read from two fields in a database: a string field and a numeric field. The string field is unique, so that is the key of the dictionary.  I can sort on the keys, but how can I sort based on the values?  Note: I have read Stack Overflow question here  and probably could change my code to have a list of dictionaries, but since I do not really need a list of dictionaries I wanted to know if there is a simpler solution to sort either in ascending or descending order. A question about $f(x)\equiv 0$ Alternate proofs (other than diagonalization and topological nested sets) for uncountability of the reals? I recently started studying set theory and am having quite a bit of difficulty accepting Cantor's diagonal proof for the uncountability of the reals. I also saw a topological proof via nested sets for uncountability which still does not satisfy me completely, given that just like the diagonal it relies on a never ending process. In fact, the nested sets proof sounds very much like the diagonalization proof to me.  Do all proofs of the uncountability of the reals involve diagonalization? Are there any other proofs I can look at to understand? I couldn't find any on searching stack exchange. Thanks. Embedded Prolog Interpreter/Compiler for Java Xcode 6.1 won't let me create a new project I'm running Xcode 6.1 on OS X Yosemite and I'm unable to create a new project, as you can see below it just says 'loading'.  I've removed the product and org name for the purpose of this image. I have tried uninstalling but that didn't work. I am new to Xcode and coding in general so I would appreciate it if any solution could be given in lay terms. GroupBy a list of associations with their keys Show inline math as if it were display math (and vice versa) I'm very familiar with how symbols display differently in inline mode (with $ ... $) vs. how display math shows them (with \[ ... \] or $$ ... $$).  Two examples would be with limits  $\lim_{n\rightarrow \infty}f(x)$   vs.  \[ \lim_{n\rightarrow \infty}f(x) \]   and with sums  $\sum_{n=1}^{x} n^2$   vs.  \[ \sum_{n=1}^{x} n^2 \]   My Question is:     How do you display these things in inline mode (with all the nice inline formatting) as if they were in display mode ?   The inverse -- how to display things in display mode as if they were in inline mode -- is also of interest. $(a_n+a_{n+1})$ is convergent implies $(a_n/n)$ converges to $0$ Let $(a_n)$ be a sequence such that $(a_n+a_{n+1})$ is convergent. Prove that $(a_n/n)$ converges to $0$.   $(a_n+a_{n+1})$ convergent to $L$ means that for all $\epsilon$, there exists $N$ such that for all $n\geq N$, $|L-a_n-a_{n+1}|&lt;\epsilon$, or in other words $L-\epsilon&lt;a_n+a_{n+1}&lt;L+\epsilon$. How to proceed from here? Context  In one of his most popular books , Terry Pratchett makes an entropy joke:     Knowledge equals Power, which equals Energy, which equals Mass   Pratchett is a fantasy comedian and every third phrase in his book is a joke, therefore there is no good reason to believe it.  Pratchett uses that madness to make up that a huge library has a tremendous gravitational push.  The question  I work with computers and mostly with encryption.  My work colleagues believe Terry Pratchett's statement because of entropy.  On the other hand, I believe, it is incorrect since entropy of information is a different entropy than the one used in thermodynamics.  Am I correct?  And if so, why do use the same name (entropy) to mean two different things?  Also, what would be a good way to explain that these two "entropies" are different things to non-scientists (i.e. people without a chemistry or physics background)? How to typeset some text including math content in sans serif Why are hand-built wheels considered to be better than machine-built wheels? Why are hand-built wheels considered to be better than machine-built wheels? I would think that a machine-built wheel would be better and cheaper, since machines generally do a more precise job than humans. Can you use Wish to duplicate a spell at a higher level? Wish allows you to duplicate the effects of any spell L8 or lower. Does that include a lower-level spell, cast at L8? Say, as a Wizard, casting Heal at L8.  I'm leaning towards yes, but it wouldn't be the first time I assume the wrong answer... Can a hotel kick you out if you let an unregistered guest share a room with you? We are going to an event and because of this event all hotels are fully booked. Now a friend wants to join us, but can't find accommodation. We are okay with him crashing in our hotel room, either on the couch or on a self-inflatable mattress. We have done this in the past where we secretly sneaked the "guest" in.   It is not that we don't want to pay the tourist tax, it is that when ask and the hotel says no, we can't sneak him in, because we drew attention to ourself. In many hotels you can just order another bed, but there are also a lot that don't have that service.   I don't have any ethical problems with my behavior of secretly hosting a guest. I did pay for the room. I don't like being sneaky about it, since again I do pay for the hotel room. Also if I have to choose between sneaking someone in the room to crash on the couch or let him/her sleep on bench in the park. I opt for the first.   Am I just worrying to much and will being honest about it in the end be acceptable to most if not all hotels? Why is $R\cos{a} = mg$ in circular motion compared and not $R = mg\cos{a}$? Normally, if an object of mass $m$ is inclined to the horizontal at an angle $b$, we set the reaction force of the object on the inclined plane as $R = mg\cos{b}$ (if we resolve the force of gravity so the line of action coming out of the plane is perpendicular to it).  However in circular motion*. it's assumed that $R\cos{b} = mg$. In the example above, one would have to do this in order to arrive to the correct answer, instead of $R = mg\cos{b}$. Using $R = mg\cos{b}$ seems natural enough, as I am resolving vertically, however, both equations would produce two different values for $R$. Why is this?   To show what I mean: If we set the reaction force in this question as $mg\cos{a}$, then the centripetal force will be $mg\cos{b}\cos(\pi/2-b) = mg\cos{b}\sin{b} = \frac{1}{2}mg\sin(2b)$   Whereas If we use $R\cos{b} = mg$, $R = mg\sec{b}$ and the centripetal force will be $mg\sec{b}\sin{b} = mg\tan{b}$. This will end up with two different values for the radius of the circular motion, and hence two different final answers.  *In the circular motion questions I've seen in my mechanics module