In the real world, it's best to use at least two or more algorithms to encrypt data.  From what I know, RSA acts as a wrapper to transfer data safely across web systems or API(s), or acts as a wrapper for symmetric keys such as AES.  If computational power and time are not the main factor (it has reached the secure standard), why is there (from what I know), no one using RSA to really encrypt data instead of just wrapping them up (as stated by the two usages)?  What I meant was really using it alongside other encryption algorithms such as AES.  Wouldn't the stored data be safer in this way (considering that the network is secure from both outsider and insider attack)?
RSA is not designed to be used on long blocks of plaintext like a block cipher, but I need to use it to send a large (encrypted) message.  How can I do this?
What is the difference between IQueryable and IEnumerable? What is the difference between IQueryable&lt;T&gt; and IEnumerable&lt;T&gt;?    See also  that overlaps with this question. Lines too long due to text inside custom command not being hyphenated I'm having some problems with lines that turn out to be too long. There are correct hyphenation rules for the word but they aren't being applied.  Let me first give the sentence that is too long:  An isomorphism that maps a graph to itself is called an \concept{automorphism}.   The command concept is defined as follows  \newcommand{\concept}[1]{\index{#1}\marginpar{\raggedright\textbf{#1}}\textbf{#1}}   The problem seems to be that words inside \concept aren't being hyphenated. How can I fix this? If I type   An isomorphism that maps a graph to itself is called an \textbf{automorphism}.   then the word is hyphenated perfectly. 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Unable to correct problems, you have held broken packages After upgrading from 10.04 to 12.04 I am trying to install different packages.  For instance ia32-libs and skype (4.0).  When trying to install these, I am getting the 'Unable to correct problems, you have held broken packages' error message.   Output of commands:  sudo apt-get install -f Reading package lists... Done Building dependency tree        Reading state information... Done 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded.   After running this:  sudo dpkg --configure -a foo@foo:~$ sudo apt-get install -f Reading package lists... Done Building dependency tree        Reading state information... Done 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded. Programmatical Setting of 'Store Relative Pathways' for multiple MXDs ArcGIS 10.1 I am looking to move large number of mxds to a different location on the same drive. 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Thanks in advance, Magda      import arcpy, os  folderPath = r"C:\My Folder Path\...\...\..."  for filename in os.listdir(folderPath):     fullpath = os.path.join(folderPath, filename)     if os.path.isfile(fullpath):         basename, extension = os.path.splitext(fullpath)         if extension.lower() == ".mxd":             mxd = arcpy.mapping.MapDocument(fullpath)             mxd.relativePaths = 'True'         mxd.save() del mxd Heroku and Twisted In case of shell-builtins (eg type itself):  $ type type type is a shell builtin  $ which type &lt;Doesn't return anything since it's a shell builtin, silently exits&gt;   In case of commands (normally) (eg python):  $ type python python is /usr/bin/python  $ which python /usr/bin/python   In case of which (which is a command located at /usr/bin/which)  $ type which which is hashed (/usr/bin/which) $ which which /usr/bin/which   Why does type which say that which is hashed? 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In the course of testing things out, I've also found that calling a function defined with pgfmathdeclarefunction can also change the positioning of the graph...  \documentclass{article}  \usepackage{xparse}  \usepackage{pgfplots} \pgfplotsset{compat=1.10}  \begin{document}  % Example of a macro which does math outside of pgf     \ExplSyntaxOn     \DeclareExpandableDocumentCommand \doubleMacro { m }%      {%       \fp_eval:n { #1*2 }%Math performed by a package other than pgfplots      }%     \ExplSyntaxOff    %Demo: macro does work in regular text \doubleMacro{1.1}.  % Defining custom pgf functions for plotting      \pgfmathdeclarefunction{mathDoubleMacro}{1}{%         \pgfmathparse{\doubleMacro{#1}} %Calls macro which performs math done by another package     }      \pgfmathdeclarefunction{mathDoublePgf}{1}{%         \pgfmathparse{#1*2} %Uses pgf to do the same thing as an external macro could     }  %Generic plot with function given directly to \addplot \begin{center} \begin{tikzpicture}     \begin{axis}         \addplot[domain=0:1] {2*x};     \end{axis} \end{tikzpicture} \end{center}  %Plot using first macro: complains about 'Unknown fp word Y.' \begin{center} \begin{tikzpicture}     \begin{axis}         %\addplot[domain=0:1] {mathDoubleMacro(x)};%Commented out because it gives problems     \end{axis} \end{tikzpicture} \end{center}  %Plot using second macro: works, but graph is displaced to the right of the expected position \begin{center} \begin{tikzpicture}     \begin{axis}         \addplot[domain=0:1] {mathDoublePgf(x)};     \end{axis} \end{tikzpicture} \end{center}  \end{document}   If I define mathDoubleMacro as calling a non-existent macro, then the errors show which x-values which the addplot is trying to evaluate y at:    0Y0.0e0  1Y4.1666e-2 1Y8.3332e-2 ...  1Y9.9998401e-1   What are these nY's all about, and why does \addplot[domain=0:1] {mathDoublePgf(x)} give a graph with different positioning? What is the difference between Reduce and Solve? Can I use my laptop as a second monitor? Deep cloning objects How did Sauron remain in control of the Nazgûl? The One Ring gave Sauron control of the Nine, yet when it was taken from him he still possessed the power to control them (or, at least he regained it towards the end of the third age).  Why is this?   mentions that he drew the Nine to him, but not how. Cases environment compresses fractions? what should I do to stop the cases environment compressing the first line of the following? \begin{equation} \alpha_{mn} = \begin{cases} \frac{16\mu^2}{mn\pi^2\left[\pi^2(m^2/a^2 + n^2/b^2) - \mu^2\right]} &amp;\mbox{for } m \mbox{ and } n \mbox{ odd;}  \\ 0 &amp;\mbox{for } m \mbox{ or } n \mbox{ even;} \end{cases} \\[1ex] \end{equation}  I'm not sure to do tex on this site, but if anyone tries it they see the first case is compressed because it's a fraction. Given two increasing continuous functions $f,g$ prove that $(b-a) \int^b_a f(x)g(x) dx > \int^b_a f(x) dx \int^b_a g(x) dx$ Given two monotonically increasing continuous functions $f,g$ prove that  $$(b-a) \int^b_a f(x)g(x) dx &gt; \int^b_a f(x) dx \int^b_a g(x) dx,\; b&gt;a$$  what I have tried:  Let $$h(x) = (x-a)\int^x_a f(t)g(t) dt - \int^x_a f(t)dt\int^x_a g(t)dt $$ and trying to figure out that $h'&gt;0$ using MVT to reform $h'(x)$ as   $$h'(x) = (x-a)(-f(x)g(\xi_1) - g(x)f(\xi_2) + f(x)g(x) + f(\xi_3)g(\xi_3))$$ where $\xi_{1,2,3} \in (a,x)$ (That method does not seem to work)