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To estimate the time for brute-force attack we need to compute keyspace size divided by hash rate, where the hash rate (hash/second) varies depending on the computer's capabilities.  The website  shows the hash rate for different computers.  Can you explain the following two confusing phenomena?   The hash rate decreases as the message size increases. Core i3 has a higher hash rate than Core i7 for SHAKE128. Do downvotes on Meta truly mean something different? Pick an urn and pick two balls from the urn There are 5 urns, the nth urn contains n − 1 red balls and 5 − n green balls (in total 4 ineach urn). We pick an urn at random and take 2 balls out of it without replacement.  (a) What is the probability the second ball is green?  (b) What is the probability the second ball is green given the ﬁrst is green?  (c) What is the probability we picked either the ﬁrst urn or second urn given both balls are green?  for (a), I did:   P(2nd ball is green) = 5*[{P(i th urn picked) * P(first ball drawn red/ ith urn picked) P(2nd ball drwan green/first ball red)}+ {P(i th urn picked) * P(first ball drawn green/ ith urn picked) P(2nd ball drwan green/first ball green)}]  = 5*[[{1/5 * (n-1)/4 * (5-n)/3 } + {1/5 * (5-n)/4 * (4-n)/3}] = 5* [{(n-1)(5-n)/60} + {(5-n)(4-n)/ 60}]  =[(n-1)(5-n) + (5-n)(4-n)]/12 = (5-n)(n-1+4-n)/12 = (5-n)*3/12 = (5-n)/4  And for b): 5*{P(i th urn picked) * P(first ball drawn green/ ith urn picked) P(2nd ball drwan green/first ball green)} = 5*{1/5 * (5-n)/4 * (4-n)/3} =(4-n)(5-n)/12  c), P(1st urn or 2nd urn/ both are green) = [ [P(1st or 2nd]* (4-n)(5-n)/6 ] / [{P(1st or 2nd]* (4-n)(5-n)/6} + {{P(3rd or 4th]* (4-n)(5-n)/6}] = {(4-n)(5-n)/15} /  2*{(4-n)(5-n)/15} = 1/2  But I am really uncertain about my solutions. If anyone has different solution or easier way to get the answer please let me know! Thanks a lot! What is the fastest way to get from Courmayeur/Aosta to Milan? Why with two classes, LDA gives only one dimension? 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? All of my Firefox add-ons have been disabled suddenly, how can I re-enable them? All of my Firefox extensions were spontaneously and simultaneously disabled (converted to Legacy status) a few moments ago.  How can I re-enable them? Firefox 66.0.3 (64-bit) When saving a .qgs file in QGIS, another appears. Is this a backup? When saving a project file (.qgs), which when executed is automatically loaded with qgis.bat, there's another .qgs file which appears in the same directory of the original project file, although this one is not loaded with the qgis.bat file:    This is just out of curiosity but is the latter designed to be a backup of the original .qgs file? Sum of a power series $n x^n$ I would like to know: How come that  $$\sum_{n=1}^\infty n x^n=\frac{x}{(x-1)^2}$$  Why isn't it infinity? Which is faster in Python: x**.5 or math.sqrt(x)? I've been wondering this for some time. As the title say, which is faster, the actual function or simply raising to the half power?  UPDATE  This is not a matter of premature optimization. This is simply a question of how the underlying code actually works. What is the theory of how Python code works?  I sent Guido van Rossum an email cause I really wanted to know the differences in these methods.  My email:     There are at least 3 ways to do a square root in Python: math.sqrt, the   '**' operator and pow(x,.5).  I'm just curious as to the differences in   the implementation of each of these.  When it comes to efficiency which   is better?   His response:     pow and ** are equivalent; math.sqrt doesn't work for complex numbers,   and links to the C  sqrt() function. As to which one is   faster, I have no idea... what are all the open subgroups of $(\mathbb{R},+)$ What "exactly" happens when someone turns to the Dark Side? I had this "almost bonus" question on the final in Group Theory recently: prove that there is no such group $G$ which would satisfy Aut$(G)\cong \mathbb{Z}_n$, where $n$ is an odd integer. I don't have much certainty if this proof is OK, and one's opinion would be appreciated.  Here's my attempt:     Suppose that there exists such a group $G$ that satisfies the above condition. Since $\mathbb{Z}_n$ is cyclic, Aut$(G)$ is also cyclic and $G$ is abelian, which implies that Inn$(G)\cong \{e\}$. Thus Aut$(G)=$Out$(G)$. (I don't think this fact is important here though). Now, since an automorphism sends a generator to a generator, and since each automorphism is completely determined by such mapping, |Aut($G$)| must be of factorial order. But an integer factorial is always even. However, $|\mathbb{Z}_n|$ is odd, which is a contradiction. Therefore, no such $G$ exists.   I'm afraid that my proof is not very nice, but at least a genuine attempt was made. What I would like is a method to convert a double to a string which rounds using the half-up method - i.e. if the decimal to be rounded is 5, it always rounds up to the next number. This is the standard method of rounding most people expect in most situations.  I also would like only significant digits to be displayed - i.e. there should not be any trailing zeroes.  I know one method of doing this is to use the String.format method:  String.format("%.5g%n", 0.912385);   returns:  0.91239   which is great, however it always displays numbers with 5 decimal places even if they are not significant:   String.format("%.5g%n", 0.912300);   returns:  0.91230   Another method is to use the DecimalFormatter:  DecimalFormat df = new DecimalFormat("#.#####"); df.format(0.912385);   returns:  0.91238   However as you can see this uses half-even rounding. That is it will round down if the previous digit is even. What I'd like is this:  0.912385 -&gt; 0.91239 0.912300 -&gt; 0.9123   What is the best way to achieve this in Java? How can I prove that all rational numbers are either terminating decimal or repeating decimal numerals? I am trying to figure out how to prove that all rational numbers are either terminating decimal or repeating decimal numerals, but I am having a great difficulty in doing so. Any help will be greatly appreciated. Primary ideals of $\mathbb{Z}$ Let $R=\mathbb{Z}$ I must prove that      In $\mathbb{Z}$ the primary ideals s are precisely the ideals $(p^k)$ where $p$ is a prime number and $n\ge 1$.   $$$$ The definition of primary ideal.     Definition. An ideal $I$ of the ring $R$ is called primary if for all $a,b\in R$ $$(ab\in I\quad\text{and}\quad a\notin I)\Rightarrow b^n\in I, \text{for some positive integer $n$}.$$   Therefore I must prove that      $$(n)\;\text{is primary in}\;\mathbb{Z}\iff n=p^k, k\ge1,p\ge2,\;\text{where}\;p\;\text{is a prime}.$$   Proof ($\Leftarrow$) We suppose that $n=p^k$ for same prime $p$ and $k\ge 1$. Let $a,b\in\mathbb{Z}$ such that $ab\in(p^k)$ and $a\notin (p^k)$. Then $p^k\nmid a$, but $p^k \mid ab$ I no longer know how to proceed.  $(\Rightarrow)$ Let $(n)$ a primary ideal and we suppose for absurd that $n$ is not a power of a prime, then? how could I proceed?  Now, let $ab\in (p^k)$ with $a\notin(p^k)$, then $p^k\mid ab.$ Can I conclude that $p^k\mid b$?   How can I set up the other implication? Thanks How do I remove an incomplete or broken snap installation of nextcloud on a Raspberry Pi? What is the value of $\arctan(1/2)+\arctan(1/5)+\arctan(1/8)$? Determine app sending SMS Premium messages I normally have "Block Premium Messaging" set up on the phones in my family. This is a verizon content block the prevents you (or family members) from sending messages to premium services that charge.  Recently, I had to replace a phone that broke and when the new phone was activated the blocks I had set up were cleared. On my bill was a $9.99 charge for a text to one of these premium services.   I know nobody sent the SMS to the service to sign up for this. I suspect some application on the phone sent the SMS message. Is there a way I can find out which app on my phone is sending SMS messages to premium services? Can I view all the apps that have permission to send SMS messages?  -- update -- well, I ran RL Permissions and no downloaded apps have the Send SMS permissions. I must have already deleted the offending app or something else is going on. Find $\min(\operatorname{trace}(AA^T))$ for invertible $A_{n\times n}$ For invertible $A_{n\times n}$ find $\min(\operatorname{trace}(AA^T))$  (a) $0$ (b) $1$ (c) $n$ (d) $n^2$  Clearly for $A=I$, it is $n$, and I am unable to get any lower values, but how do I prove it.