Is it possible to compute these probabilities concerning a 6-digit password using theory of Markov chains?
6-digit password - a special decoding method
Is the total variation function uniform continuous or continuous? Question mark or bold citation key instead of citation number Is this sentence acceptable?     You’re welcome, have a nice day ahead. What is aperture, and how does it affect my photographs? How does aperture affect my photographs?  Why should I care about the aperture with which a photo was taken? Xcode - How to fix 'NSUnknownKeyException', reason: … this class is not key value coding-compliant for the key X" error? Disable chat migration notification if one of the users has insufficient rep? Simple request, really.  Can we add a detection for the     Please avoid extended discussions in   comments. Would you like to   automatically move this discussion to   chat?   prompt, so that if one of the users in the "extended discussion" has under 20 rep it doesn't appear?  I think a lot of the users this apply to will have low rep and it may take some back and forth to get good info out of them.  New suggestion: how about if a sufficiently high-rep user is clicking on the link, the low-rep user gets a field promotion sufficient to chat? Low-rep users are particularly prone to post questions to lead in this direction of needing to move to chat. Why does "for good" mean "forever"? A very recent and similar  was closed asks what "for good" means.  While general reference can answer the question, I became curious as to the etymology of the idiom.   around got me nowhere, which is unusual.  The closest I found was a thread in :     OED   for good (and all): as a valid conclusion; hence, as a final act, finally.    15.. Parl. Byrdes Aij, Than desyred al the Byrdes great and smal to mewe the hauke for good and all. 1603 in Crt. &amp; Times Jas. I (1849) I. 25 D’Auval.. is gone for good and all. 1687 CONGREVE Old Bach. I. i, Ay, you may take him for good-and-all if you will. 1711 SWIFT Jrnl. to Stella 4 July, This day I left Chelsea for good, (that’s a genteel phrase). a 1732 T. BOSTON Crook in Lot (1805) 37 He was obliged for good and all to leave his country. 1850 J. H. NEWMAN Diffic. Anglic. 324 Throw off, for good and all, the illusions of your intellect. 1882 W. E. FORSTER Let. To Gladstone 10 Apr. in T. W. Reid Life (1888) II. viii. 421 This morning we released Parnell—not for good, but on parole.   However, this only traces some historical usages, and suggests that for good and all used to be a more popular version of the current idiom.  Is there anything more substantial?  Why did "for good" come to mean "forever"? Auto incrementing field based on groups within feature class? I am needing to auto increment a field based on groups within a feature class. I have 8 plots within a given polygon and I need to assign them an ID from 1-8 for each set of plots within each polygon. The polygon would have its own unique ID number to be used to group the plots.  I assume it would be an alteration of this:  rec=0 def autoIncrement():  global rec  pStart = 1   pInterval = 1   if (rec == 0):    rec = pStart   else:    rec = rec + pInterval   return rec Can you lookup a domains past nameserver & mx record history? What is the difference between if(CONST==variable) or if(variable==CONST)? I am lost.   So far...  If $\gcd (a,b) = 1$, by Bezout's Formula  $ax + by = 1$  If $c|(a+b)$, then $cf = a+b$  Then,  $a (x-y) + cfy = 1$   $b(yx) + cfx = 1$  Am I on the right track? Any suggestions? I have a process I can't kill with kill -9 &lt;pid&gt;. What's the problem in such a case, especially since I am the owner of that process. I thought nothing could evade that kill option. I'm working on a 2nd order passive low pass filter, consisting of two passive low pass filters chained together.         &ndash; Schematic created using   Let \$ H(s) = H_1(s)H_2(s) \$ where \$ H_1(s) \$ and \$ H_2(s) \$ are the transfer functions for each separate filter stage.  Then \$ |H(s)| = |H_1(s)||H_2(s)| \$  Knowing the magnitude of a passive low pass filter,   $$|H(s)| = \dfrac1{\sqrt{ (\omega R_1C_1)^2 + 1} } \times \dfrac1{\sqrt{ (\omega R_2C_2)^2 + 1} } = \dfrac1{\sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}} $$  Then trying to find the cutoff frequency:  $$ \left(\dfrac1{\sqrt{2}}\right)^2  = \dfrac1{\sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}} $$ $$ 2  = \sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)} $$ $$ 4  = ((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1) $$ $$ 4  = (\omega R_1C_1)^2(\omega R_2C_2)^2 + (\omega R_1C_1)^2 + (\omega R_2C_2)^2 + 1 $$  And I'm stuck. Research on the web tells me \$ \omega_c = \dfrac1{\sqrt{R_1C_1R_2C_2}} \$, but I can't find why? Can anybody show my the derivation to find this? Which graphics cards are supported by the new AMDGPU driver in Ubuntu 16.04? Windows 7 stuck on CLASSPNP.SYS I'm running Win 7 x86 on a Lenovo Thinkpad Z61m in dual boot mode with Win XP, both OSes are running on separate partitions. I have a faulty battery and if left unplugged my laptop shuts usually down within 10 minutes. Last night, it shut down due to this, but after plugging in AC power and starting back up I'm stuck at the "Starting Windows" screen, I'm unable to boot into safe mode or into recovery mode(XP boots fine).  I believe I had installed some Win 7 updates the same/previous day. From one of the boot options I was able to see that Windows 7 is getting stuck on CLASSPNP.SYS  How can I go about fixing this?  I do have a complete Win 7 Restore CD that I created after setting up my machine, I'm unable to boot into this either.  Please help! Font size setting is missed in iPhone application Can you add a setting for choosing font size in iPhone mobile app?   It's hard to read questions. How do I implement a retro-style password-based "savegame" system? Impossible to hide navigation bars in Safari iOS 7 for iPhone/iPod touch A geometric reason why the square of the focal length of a hyperbola is equal to the sum of the squares of the axes? When I teach conics, I give a simple geometric argument for the Pythagorean relation for ellipses. (By "Pythagorean relation" I mean the fact that the square of the focal length $c$ is the difference of the squares of the semimajor and semiminor axes, i.e. $c^2=|a^2-b^2|$ for the ellipse defined by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.) The picture argument goes as follows: just draw the obvious isosceles triangle from the two foci to one of the two points at the end of the minor axis. We see two right triangles; in each, one leg is the focal length $c$ and the other leg is the semiminor axis, while the hypotenuse is the semimajor axis. The result follows from the Pythagorean theorem applied to this right triangle.  But when it comes to the Pythagorean relation for hyperbolas, I can't think of an equally convincing picture to draw. Instead I have to give a rather inelegant algebraic argument for the relation $c^2=a^2+b^2$ (for the hyperbola defined by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, with focal length $c$), directly from the definition of the hyperbola. The messy details are below.*  Now, obviously there is an asymmetry between the argument for the ellipse and the argument for the hyperbola: the ellipse argument is almost trivial and doesn't involve working straight from the locus definition, but the hyperbola argument is tedious and does require working straight from the definition. So, my question:     Is there an equally persuasive picture I could draw to convince high   school students of the Pythagorean relation for hyperbolas -- without having to get technical with points at infinity, say?   The natural triangle to draw is the right triangle with legs from the center to a vertex and from that vertex to an asymptote, which has legs of length $a$ and $b$; but then the problem becomes why the length along the asymptote should be equal to $c$. I doubt there is a purely (Euclidean) geometrical reason because this triangle lies on the asymptote, not on the hyperbola itself; I suspect some sort of limiting procedure will be required.    [*] The tedious algebraic details:     The foci lie at $(\pm c,0)$.   Say $a&gt;b$. Then hyperbola is the locus of points $(x,y)$ such that       $$\left|\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right|=2a$$      Assume for simplicity the expression between the absolute value bars   is positive. (The algebra is even more tedious if we don't make such a   simplifying assumption.) Then moving the right root to the other side   and squaring gives      $$(x+c)^2+y^2=4a^2+(x-c)^2+y^2+4a\sqrt{(x-c)^2+y^2}$$      Simplifying gives      $$xc-a^2=a\sqrt{(x-c)^2+y^2}$$      and upon squaring again we have      $$x^2c^2-2xca^2+a^4=a^2(x^2-2xc+c^2+y^2)=a^2x^2-2xca^2+c^2a^2+a^2y^2$$      which reduces to      $$x^2(c^2-a^2)-y^2a^2=a^2(c^2-a^2)$$      Defining $b^2:=c^2-a^2$ brings the equation into the form   $x^2/a^2+y^2/b^2=1$, and the Pythagorean relation follows from the   definition of $b$. Why is there a difference between ping "localhost" and ping "local IP address"? Using cmd and ping on Windows gave me the following results:   Pinging "localhost":       Pinging "192.168.0.10" (local IP address):     Aren't both situations exactly the same?   I mean, I'm pinging the same interface, the same machine and the same address. Why do I get such different results?  EDIT: Here is my ipconfig /all screen: