I finally found a Squirtle and caught it, but then the app froze and I lost it.  However, my Pokedex still shows a picture of a Squirtle, but I don't actually have it. This isn't the first time this has happened to me, is there anything I can do to recover the Pokemon that this happened to? Any way I can fix this? Just an annoying bug?
Sometimes when attempting to capture a Pokémon, the mon will be wiggling in the ball and the app will disconnect from servers (for mobile connection reasons or whatever), and I'll be left staring at a ball containing Schrodinger's Pokémon. I'll sigh at the loss (or in the case of a 500CP Pinsir, scream in agony), and reload the game.  I thought maybe encounters in general couldn't handle disconnecting well, but sometimes immediately after a Pokémon broke free of a ball, the little icon in the upper left would appear, the game would stall for a bit, but it would recover.   Is there anything I can do to avoid this? Should I be waiting it out, instead of reloading?  (and before anyone says it, yes I have filed a bug report)
What are irreducibles in the ring $R$ of entire functions? If I run history, I can see my latest executed commands.  But if I do tail -f $HISTFILE or tail -f ~/.bash_history, they do not get listed.  Does the file get locked, is there a temporary location or something similar? Why does a minimal prime ideal consist of zerodivisors? Let $A$ be a commutative ring. Suppose $P \subset A$ is a minimal prime ideal. Then it is a theorem that $P$ consists of zero-divisors.  This can be proved using localization, when $A$ is noetherian: $A_P$ is local artinian, so every element of $PA_P$ is nilpotent.  Hence every element of $P$ is a zero-divisor. (As Matt E has observed, when $A$ is nonnoetherian, one can still use a similar argument: $PA_P$ is the only prime in $A_P$, hence is the radical of $A_P$ by elementary commutative algebra.)  Can this be proved without using localization? Proof of greatest integer theorem: floor function is well-defined I have to prove that  $$\forall x \in \mathbb{R},\exists !\,n \in \mathbb{Z} \text{ s.t. }n \leq x &lt; n+1\;.$$  where $\exists !\,n $ means there exists a unique (exactly one) $n$. I'm done with proving that there are at least one integers for the solution. I couldn't prove the &quot;uniqueness&quot; of the solution, and so I looked up the internet, and here's what I found:  Let $\hspace{2mm}n,m \in \mathbb{Z} \text{ s.t. }n \leq x &lt; n+1$ and $m \leq x &lt; m+1$. Since $n \leq x \text{ and } -(m+1) &lt; -x$, by adding both, $n + (-m-1) &lt; (x-x) = 0$. And (some steps here) likewise, $(x-x) &lt; n+m+1$.  Now, can I add up inequalities like that, even when the book is about real analysis (and in which assumptions are supposed to be really minimal)? Or should I also prove those addition of inequalities? Thank you :D I can't use the terminal while gedit command is running I recently installed 12.04. When I try to edit a file with gedit, I can't use the terminal until I close the editing file or I have to open a new terminal. But I think I didn't have this problem with 11.04, however I'm not sure. Is there anyway to avoid this and to use same terminal while editing files. Does photon have size measurement because of its particle nature Does photon have size measurement because of its particle nature like electron's 3.86*10^-13m etc.. A man enters a comatose author's dreams to wake him up How to duplicate sys.stdout to a log file? Edit: Since it appears that there's either no solution, or I'm doing something so non-standard that nobody knows - I'll revise my question to also ask: What is the best way to accomplish logging when a python app is making a lot of system calls?  My app has two modes.  In interactive mode, I want all output to go to the screen as well as to a log file, including output from any system calls.  In daemon mode, all output goes to the log.  Daemon mode works great using os.dup2().  I can't find a way to "tee" all output to a log in interactive mode, without modifying each and every system call.    In other words, I want the functionality of the command line 'tee' for any output generated by a python app, including system call output.  To clarify:   To redirect all output I do something like this, and it works great:  # open our log file so = se = open("%s.log" % self.name, 'w', 0)  # re-open stdout without buffering sys.stdout = os.fdopen(sys.stdout.fileno(), 'w', 0)  # redirect stdout and stderr to the log file opened above os.dup2(so.fileno(), sys.stdout.fileno()) os.dup2(se.fileno(), sys.stderr.fileno())   The nice thing about this is that it requires no special print calls from the rest of the code.  The code also runs some shell commands, so it's nice not having to deal with each of their output individually as well.  Simply, I want to do the same, except duplicating instead of redirecting.  At first thought, I thought that simply reversing the dup2's should work.  Why doesn't it?  Here's my test:   import os, sys  ### my broken solution: so = se = open("a.log", 'w', 0) sys.stdout = os.fdopen(sys.stdout.fileno(), 'w', 0)  os.dup2(sys.stdout.fileno(), so.fileno()) os.dup2(sys.stderr.fileno(), se.fileno()) ###  print("foo bar")  os.spawnve("P_WAIT", "/bin/ls", ["/bin/ls"], {}) os.execve("/bin/ls", ["/bin/ls"], os.environ)   The file "a.log" should be identical to what was displayed on the screen. Function with same name but different signature in derived class MWE for externalize + automatic recompilation after tikz figure modification How do I clear Bash's cache of paths to executables? When I execute a program without specifying the full path to the executable, and Bash must search the directories in $PATH to find the binary, it seems that Bash remembers the path in some sort of cache. For example, I installed a build of Subversion from source to /usr/local, then typed svnsync help at the Bash prompt. Bash located the binary /usr/local/bin/svnsync for "svnsync" and executed it. Then when I deleted the installation of Subversion in /usr/local and re-ran svnsync help, Bash responds:  bash: /usr/local/bin/svnsync: No such file or directory   But, when I start a new instance of Bash, it finds and executes /usr/bin/svnsync.  How do I clear the cache of paths to executables? Polynomial rings in infinitely indeterminants How can we prove that (the set of all polynomials in infinitely non commutative indeterminants over $k$) is a ring, where $k$ is a ring with unity?   Adding two polynomials is a polynomial, but for commutativity  How can we show the following: $ax_1+bx_2=bx_2+ax_1$ (commutativity of addition) If we have finite number of variable it is easy to prove it, (we take the $n-1$ variables...), but what about the infinite number of variables?  Can someone give me a hint? Purpose of memory alignment Admittedly I don't get it. Say you have a memory with a memory word of length of 1 byte. Why can't you access a 4 byte long variable in a single memory access on an unaligned address(i.e. not divisible by 4), as it's the case with aligned addresses? I frequently cross this issue, and always have to google for an answer. Does anyone have a permanent fix for BADSIG errors from apt-get?     W: GPG error:    lucid   Release: The following signatures were   invalid: BADSIG 54422A4B98AB5139   Oracle Corporation (VirtualBox archive   signing key) Variables losing their values Properties of p-adic valuations Let $n$ be a integer not equal to zero, $p$ a prime and $$|n|_p = p^{-\operatorname{ord}_p(n)}; |0|_p=0$$  Proposition: $|xy|_p=|x|_p|y|_p$  We want to show: $\operatorname{ord}_p(ab)=\operatorname{ord}_{p}(a)+\operatorname{ord}_p(b)$ So if we suppose there is a largest integral $v$ such that $p^{v}|(ab)$, then we can split up v such that $p^{v_1}|a,p^{v_2}|b$. If we suppose that there exist $v_1,v_2$ with them both being the largest integers such that $p^{v_1}|a, p^{v_2}|b$,  with $v_1+v_2= : v $ then $p^v|ab$. (I don't believe this to be a formal proof but rather my best own "idea" I can come up with...)  The script also mentions that the padic valuations' sums are bounded: $|x+y|_p\le \sup\{|x|_p,|y|_p\}$, how to show this?   Edit:   According to Robert Israel and Bill Dubuques answers, the first proposition is proven, if I understood correctly:   Given $ord(a)= v$ and $ord(b) = u$ which is the same as $a:= p^{v}m, b:= p^{u}n$ with $gcd(m,p)=gcd(n,p)=1$, because $ab=p^{u+v}mn$ with $gcd(mn,p)=1$ it follows that $ord(ab)=ord(a)+ord(b)$ and so $|xy|_{p}=p^{-ord(xy)}=p^{-(ord(x)+ord(y))}=|x|_{p}|y|_{p}$  Now to show: $|x+y|_{p}\le sup\{|x|_{p},|y|_{p} \}$  and the second can be tackled like this: set $a:=p^{v}m. b:=p^{u}n$, then checking two cases $1:(u=v)$ and $2:(v&lt;u)$: $(a+b)=p^{v}m+p^{u}n= p^{v}(m+p^{u-v}n)$ so in case of (u=v) it follows that  $ord(a+b)= u = sup\{ord(a),ord(b)\}$  and in case of $(v&lt;u)$ because the biggest v is $v:= u-1$: $ord(a+b)\le p^{u-1}(m+pn) \le p^{u}(m+n)= sup\{ord(a),ord(b)\}$  It seems that I did something wrong in both cases... How to fix Permission Denied (Public key) error? While trying to communicate between my desktop Ubuntu 12.10 and Fedora 19 desktop, I am facing the error of Permission Denied (Public key). I am able to login through ssh from fedora to Ubuntu, but not vice-versa. I even performed telnet on both machines via port 22 and both machines are listening, but not through ssh. What to do? Any idea?  My sshd_config file is as follows:  # Package generated configuration file # See the sshd_config(5) manpage for details  # What ports, IPs and protocols we listen for Port 22 # Use these options to restrict which interfaces/protocols sshd will bind to #ListenAddress :: #ListenAddress 0.0.0.0 Protocol 2 # HostKeys for protocol version 2 HostKey /etc/ssh/ssh_host_rsa_key HostKey /etc/ssh/ssh_host_dsa_key HostKey /etc/ssh/ssh_host_ecdsa_key #Privilege Separation is turned on for security UsePrivilegeSeparation yes  # Lifetime and size of ephemeral version 1 server key KeyRegenerationInterval 3600 ServerKeyBits 768  # Logging SyslogFacility AUTH LogLevel INFO  # Authentication: LoginGraceTime 120 PermitRootLogin yes StrictModes yes  RSAAuthentication yes PubkeyAuthentication yes #AuthorizedKeysFile    %h/.ssh/authorized_keys  # Don't read the user's ~/.rhosts and ~/.shosts files IgnoreRhosts yes # For this to work you will also need host keys in /etc/ssh_known_hosts RhostsRSAAuthentication no # similar for protocol version 2 HostbasedAuthentication no # Uncomment if you don't trust ~/.ssh/known_hosts for RhostsRSAAuthentication #IgnoreUserKnownHosts yes  # To enable empty passwords, change to yes (NOT RECOMMENDED) PermitEmptyPasswords no  # Change to yes to enable challenge-response passwords (beware issues with # some PAM modules and threads) ChallengeResponseAuthentication no  # Change to no to disable tunnelled clear text passwords PasswordAuthentication yes  # Kerberos options #KerberosAuthentication no #KerberosGetAFSToken no #KerberosOrLocalPasswd yes #KerberosTicketCleanup yes  # GSSAPI options GSSAPIAuthentication no GSSAPICleanupCredentials yes  X11Forwarding yes X11DisplayOffset 10 PrintMotd no PrintLastLog yes TCPKeepAlive yes #UseLogin no  #MaxStartups 10:30:60 #Banner /etc/issue.net  # Allow client to pass locale environment variables AcceptEnv LANG LC_*  Subsystem sftp /usr/lib/openssh/sftp-server  # Set this to 'yes' to enable PAM authentication, account processing, # and session processing. If this is enabled, PAM authentication will # be allowed through the ChallengeResponseAuthentication and # PasswordAuthentication.  Depending on your PAM configuration, # PAM authentication via ChallengeResponseAuthentication may bypass # the setting of "PermitRootLogin without-password". # If you just want the PAM account and session checks to run without # PAM authentication, then enable this but set PasswordAuthentication # and ChallengeResponseAuthentication to 'no'. UsePAM yes   This is sshd_config of ubuntu system. A deck of 52 cards is shuffled, and we deal a bridge of 13 cards. Let $X$ be the number of aces and $Y$ be the number of spades. Show that $X$, $Y$ are uncorrelated.  Here is what I did:  $Cov(X,Y) = E[XY]-E[X]E[Y]$   uncorrelated means $Cov(X,Y) = 0$, hence  $E[XY]=E[X]E[Y]$   $E[X] =  \sum_{k=0}^{k=4} k \frac{\dbinom{4}{k} \dbinom{48}{13-k}}{\dbinom{52}{13}} $   $E[Y] =  \sum_{k=0}^{k=13} k \frac{\dbinom{13}{k} \dbinom{39}{13-k}}{\dbinom{52}{13}} $   Are the summations above correct? and how do I calculate $E[XY]$? What is the best way to remove duplicate rows from a fairly large SQL Server table (i.e. 300,000+ rows)?  The rows, of course, will not be perfect duplicates because of the existence of the RowID identity field.  MyTable  RowID int not null identity(1,1) primary key, Col1 varchar(20) not null, Col2 varchar(2048) not null, Col3 tinyint not null Mutineers force captain to record instructions to spaceship's computer but he leaves out "please" The crew mutinies on a spaceship whose captain controls it with verbal orders. They test whether they can control the ship with recordings of the captain's voice with an order to increase orbital speed (currently orbiting a planet). It works, so they force the captain to record all the possible orders they may need, and execute him. Only then do they realize that the ship obeys only orders couched in polite language, and only the test recording was like this, but the other recordings are not. They will die once the ship's faster orbit somehow causes it to run out of fuel, or something like that. I seem to remember the ship's &quot;computer&quot; following the captain around on robotic legs. I read the story circa 1980, but it might have been published much earlier.