In his  of the Solitaire keystream generator algorithm, Bruce Schneier warns not to use the same key for two different messages:     The first rule of an output-feedback mode stream cipher, any of them,   is that you should never use the same key to encrypt two different   messages. Repeat after me: NEVER USE THE SAME KEY TO ENCRYPT TWO   DIFFERENT MESSAGES. If you do, you completely break the security of   the system. Here's why: if you have two ciphertext streams, A+K and   B+K, and you subtract one from the other, you get (A+K)-(B+K) =   A+K-B-K = A-B. That's two plaintext streams combined with each other   with no key involved, and is very easy to break. Trust me on this one:   you might not be able to recover A and B from A-B, but a professional   cryptanalyst can. This is vitally important: never use the same key to   encrypt two different messages.   Intuitively, this makes sense, but as he said, I'm not able to recover A and B.  What sort of attack is he talking about?
My question might appear the same as the question , but actually I did read all the answers and none of them helped me with the details I need.  I am new to cryptography and my problem is with two time pad attacks on OTP. The problem I had in my course was that I have 10 ciphertexts encrypted with the same key $K$. I am then given another ciphertext that I should decrypt.  I know that XOR-ing two ciphers gives me the XOR of their original messages.  My question is what is the correct thing to do after that?  I tried to take 3 ciphertexts $C_1, C_2$ and $C_3$.  Then get $S_1 = C_1 \oplus C_2  \oplus $' ', also get $S_2 = C_1 \oplus C_3 \oplus$ ' '.  After that I compared all corresponding characters in $S_1$ and $S_2$,  and if $S_1[i] = S_2[i]$ then I calculate $S_1[i] \oplus C_2[i]$ to get $K[i]$.  I tried this on paper before coding and it worked, but I might be missing something.  Is this the right approach? Why does it work?
I've been asked to show that:  $n(n+1)(2n+1) \equiv 0 \pmod 6$  I found in a previous question that:  $n(n+1)$ was divisible by $2$ and resulted in an even number e.g  $n(n+1) \equiv 0 \pmod 2$  so I figured I needed to find:  $(2n+1) \equiv 0 \pmod 3$ in order to complete  $n(n+1)(2n+1) \equiv 0 \pmod 6$  but I am unsure on how to find $(2n+1) \equiv 0 \pmod 3$  Is this the right way to find the mod 6 and if so could you tell me how I could find  $(2n+1) \equiv 0 \pmod 3?$ The qtree package seems to facilitate the drawing of simple trees only. Is there a way to incorporate:   Multiple roots &amp; Nodes with multiple parents   in a tree in LaTeX? hyperlink name with biblatex authoryear (biblatex 1.4b) Need Google Geocoding to return City name in English Is temperature a Lorentz invariant in relativity? Will having multiple domains improve my seo? Lets say I have a domain already, for example www.automobile4u.com (not mine), with a website fully running and all.  The title of my "Website" says:    &lt;title&gt;Used cars - buy and sell your used cars here&lt;/title&gt;   Also, lets say I have fully SEO the website so when people searching for the term buy used cars, I end up on the second or first page.  Now, I want to end up higher, so I go to the google adwords page where you can check how many searches are made on specific terms.  Lets say the term "used cars" has 20 million searches each month.  Here comes the question, could I just go and buy that domain with the search terms adress, in this case www.usedcars.com and make a redirect to my original page, and this way when people search for "used cars", my newly bought domain name comes up redirecting people to my original website (www.automobile4u.com)?  The reason I believe this benefits me is because it seems search engines first of all check website adresses matching the search, so the query "used cars" would automatically bring www.usedcars.com to the first result right?  What are the downsides for this?  I already know about google spiders not liking redirects, but there are many methods of redirecting...  Is this a good idea generally? Are enemies with immunity to the Poisoned condition unaffected by Contagion? After the , the spell  now inflicts the  condition upon a successful melee spell attack. The relevant spell text is below, with updated portions highlighted in bold:  Your touch inflicts disease. Make a melee spell attack against a creature within your reach. On a hit, the target is poisoned. At the end of each of the poisoned target's turns, the target must make a Constitution saving throw. If the target succeeds on three of these saves, it is no longer poisoned, and the spell ends. If the target fails three of these saves, the target is no longer poisoned, but choose one of the diseases below. The target is subjected to the chosen disease for the spell's duration. Since this spell induces a natural disease in its target, any effect that removes a disease or otherwise ameliorates a disease's effects apply to it.  The wording of the spell leads me to conclude that this spell still inflicts a disease, which has the immediate effect of poisoning the target. Therefore, even if a target has poison immunity, they will still make Constitution saves as designated to see if the disease takes effect. However, the line, &quot;At the end of each of the poisoned target's turns...&quot; also leads me to think that an enemy with poison-immunity cannot be the Poisoned target, therefore disallowing the remaining sequence of events. Are targets with immunity to the Poisoned condition therefore immune to contagion? The Help Vampire problem Differences between strong and weak in Objective-C Generic List - moving an item within the list So I have a generic list, and an oldIndex and a newIndex value.  I want to move the item at oldIndex, to newIndex...as simply as possible.  Any suggestions?  Note  The item should be end up between the items at (newIndex - 1) and newIndex before it was removed. I have a PDF with 100 pages. How can I make a new PDF that displays 4 pages on one?  I know that I can do it with Chrome by printing the PDF to new PDF and change this in the options. But I want to make it for a lot of files and put it as a script. How to share worlds between platforms with Minecraft Better Together I have read that with the new "Better Together Update" (BTU) it is now possible to .  However, I am having trouble finding details of how this is supposed to work to determine if it would do what I am hoping it would and therefore whether it is worth my time/money.  I have some saved worlds on Minecraft for XBox One, all done in creative mode.  I am able to import them into what is now just called "Minecraft" which as of the BTU is what is being used to mean the cross-platform version.   Can I now play this same world on my Windows 10 PC?   If I go to the Minecraft web site it offers to sell me the "Java Edition" with the addendum "includes edition for Windows 10".   But from what I have read, anything that has the word "edition" in it isn't the cross platform version.   It's very unclear to me what I would actually be buying here.  If it is possible to open the same world on different platforms, exactly how would I go about doing this?  On the XBox I have to sign in with an outlook.com e-mail address to access my worlds, or my kids' sign-ins to access their worlds, is it the same from the PC? And where would my world be saved?   I have read that XBox saves will be in the cloud, but PC saves will be local, so how can I get a cloud save onto my PC?  And then how can I get it back to the cloud when I am done so I can play it on the console again?  Will there be two copies floating around, making it possible for me to make a backup on my PC?  How will this interact with world editors?  At some point will I be able to start a game on XBox, transfer it to the PC, open it in a world editor, make some changes, and then transfer it back to the XBox? Let $A =$ {$(x,y,z) \in \mathbb{R}^3 |$  $x,y,z \geq 0$ and $x+y+z \leq 1$}.   Compute the integral $$\int_{A} x^m y^n z^p (1-x-y-z)^q dx dy dz$$  using new variables $u = x+y+z$, $v=\frac{y+z}{x+y+z}$, $w=\frac{z}{y+z}$.  Also, although it was not stated in the problem, I'm assuming that $m,n,p,q \in \mathbb{Z}$.   Since $f(x,y,z)=(x+y+z, \frac{y+z}{x+y+z}, \frac{z}{y+z}) = (u,v,w)$, $f^{-1}(u,v,w) = (u(1-v), uv(1-w), uvw)$  Thus J $= \begin{vmatrix} 1-v &amp; -u &amp; 0\\  v(1-w) &amp; u(1-w) &amp; -uv\\  vw &amp; uw &amp; uv\\ \end{vmatrix} = u^2v$  Skipping the part where I find the new limits of integration, and skipping some algebraic simplification we have:  $$\int_{A} x^m y^n z^p (1-x-y-z)^q dx dy dz$$   $$= \int_{0}^1 \int_{0}^1 \int_{0}^1 u^{m+n+p+2}(1-u)^q v^{n+p+1}(1-v)^mw^p(1-w)^n du dv dw$$  However, I have no  idea where to go from here. Any help would be much appreciated. I've just answered .  After accepting the answer, the OP then changed his question to another problem - which I know I shouldn't have answered (I should have requested as new question) but I updated my answer anyway.  The OP has now edited his question to remove the essence of either the original or new problem, and instead has a statement saying that the problem has been fixed.  What is the correct thing to do here?  Roll the question back so it's in its original state? Roll it back to the "new problem" state? Leave it and let him continue to get downvoted and/or closed by people who didn't see the original question? Can we model gravitation as a repulsive force? This question is actually related to my earlier question ("").  The fact that objects move a lot in the universe and that the universe is expanding, can imply that gravity is a repulsive force that increases with distance.. so the farthest objects repel us more.  This can still explain several existing observations, e.g., why does the apple fall?   Motion is the result of such repulsion. Two objects unlucky enough not to be moving relative to each other get squished due to the repulsion of the rest of the universe  around them. The earth repels the apple less than the stars so it is pushed towards the earth.   Furthermore, it can explain the expanding universe without the need for dark energy.  This could be demonstrated in a thought experiment. If we take a lot of same-charge particles (with small mass) such as electrons and lock them in a large box at a low enough temperature. The mutual repulsion of the particles may cause similar motion as if due to gravitational attraction.   Another experiment would be to measure the slight changes in our weight during day and night when the sun and earth align (if their masses are large enough to detect the feeble change in repulsion).   [EDIT: the question in the original form may not have been clear. It is "can we model".. with a yes/no answer and why (not). If downvoting, please justify. Wrong hyperref with repeated counter I'm having an interesting problem with hyperref.  Apparently, when a counter is reset by \setcounter{mycounter}{0} then hyperref always links to the first possible instance of the counter and not the one where the respective label sits.   Check this MWE out:   % !TEX TS-program = xelatexmk \documentclass{article} \usepackage[paperwidth=6in,paperheight=3in]{geometry} \usepackage{hyperref} \hypersetup{colorlinks=true, allcolors=blue, } \newcounter{mycounter}\setcounter{mycounter}{0} \newcommand\countme{\refstepcounter{mycounter}\themycounter}  \begin{document}  First line \countme{} has some interesting text.  The second line \countme \label{onelabel} has \textbf{more} interesting text.  As we saw at line \ref{onelabel} ...  \newpage  A new page brings new numbers \setcounter{mycounter}{0}   A very different first line \countme{} has other interesting text.  The second line \countme \label{otherlabel} has \textbf{different} interesting text.  As we saw at line \ref{otherlabel} ... \end{document}   As you can see in the screenshot below, the pop up preview of the \ref on the second page doesn't refer to the second line with \label{otherlabel} but to the one on the first page (also has the tooltip "Go to page 1" -- in TeXShop).   Am I not understanding something fundamental about hyperref?      The context for all this is an environment that resets my counter to 0 at begin.  And all subsequent \refs to labels in a second or third iteration of the environment point to the very first instance of that environment, not the correct one where the label sits. Unlock by fingerprint only Moderncv package - cventry date width In a tennis tournement of 128 players, elimination system, what is the probability that two twins will meet at some point? How to set programmatically multiple arguments for views contextual filters How to set programmatically multiple arguments for views contextual filters?  I can do it with one argument, it works great, but not with multiple arguments.  I want to select all nodes created by a list of people, but views only takes first in array into consideration:  $array = array(3250,32); $view = views_get_view('myview'); $view-&gt;set_display("myblock"); $view-&gt;set_arguments($array); $view-&gt;pre_execute(); $view-&gt;execute(); $content = $view-&gt;render();    Am I doing something wrong?