I am trying to add NOPASSWD option for existing user who belongs to the sudo group and have admin rights. I added this line in sudoers file:    max     ALL=(ALL:ALL) NOPASSWD: ALL   But it still ask me for password when i try to execute command with sudo.
I have a NOPASSWD line in /etc/sudoers (edited with visudo)  gatoatigrado    ALL=(ALL) NOPASSWD: /bin/set-slow-cpufreq   However, the output is,  gatoatigrado@coral:~&gt; sudo -n /bin/set-slow-cpufreq sudo: sorry, a password is required to run sudo   This kind of command works on an OpenSuSE machine, but not on Ubuntu 11.10. What am I doing wrong?  Note: I cannot find any relevant system log messages, e.g. via tail -f /var/log/syslog.  edit  Here is /etc/sudoers.  Defaults    env_reset  # things I've tried copying from an opensuse machine Defaults always_set_home Defaults env_keep = "LANG LC_ADDRESS LC_CTYPE LC_COLLATE LC_IDENTIFICATION LC_MEASUREMENT LC_MESSAGES LC_MONETARY LC_NAME LC_NUMERIC LC_PAPER LC_TELEPHONE LC_TIME LC_ALL LANGUAGE LINGUAS XDG_SESSION_COOKIE"  root    ALL=(ALL:ALL) ALL gatoatigrado ALL=NOPASSWD: /bin/set-slow-cpufreq %admin ALL=(ALL) ALL %sudo   ALL=(ALL:ALL) ALL
In the sentence below, are the usages of apostrophes correct?     In research about parent stereotypes, the correlation between parents’ beliefs about their early elementary age children’s’ mathematics abilities and the children’s actual abilities increased as children aged. Word for low level noise among restless crowd, as sign of confusion or complaint? Dictionary gives me "drone" and "murmur" but I'm not sure if that's the word I am thinking of.  I want to use it to describe a group of people on strike who hear that their demands are not met and the CEO has resigned.  They're not shouting, but they're not whispering either.  They're restless and confused.  I want to say that a certain something (murmur?) was heard among the crowds, to indicate their state. How to plot and visualize a single linear vector in 3D? Does Google use the Meta Description Tag for Description of Page? How are private keys handed to users in Identity-based Encryption? Why does Gandhi want to nuke me? Different civilizations obviously have different priorities. For example, Gandhi always seems to rush nuke technology and the Mongols will most likely start invading and spread like mad. Are these variables public knowledge? Can you tell by who you are playing what strategy they will adopt during the game? Does the universe have a center? If the big bang was the birth of everything, and the big bang was an event in the sense that it had a location and a time (time 0), wouldn't that mean that our universe has a center?  Where was the big bang? What is there now? Are we moving away from the center still? Are these even valid questions? I tried to add a new repository to software sources, but "add source" button is unavailable. I can't click on it.  I also tried to add a new software source from the terminal, but it didn't work.  sudo‬‬ ‫‪add-apt-repository‬‬ ‫‪ppa:n-muench/burg‬‬   s‫‪udo‬‬: command not found     I also tried these commands:  ‫‪sudo‬‬ ‫‪add-apt-repository‬‬ ‫‪deb‬‬ ‫‪http://ppa.launchpad.net/n-muench/burg/ubuntu‬‬ natty‬‬ ‫‪main‬‬   sudo‬‬: command not found Include reviews made outside of /review in progress towards the badge The recent changes to /review has gotten everyone interested in cleaning up the muck. At present, only posts "reviewed" (by clicking on the review link) and actions taken (up/down/close/edit/flag) from within /review count towards the badge.   However, it is entirely possible that one catches a poor question outside of /review (e.g., on the front page/tag page, etc.) and "reviews" it and takes some action on it. However, these actions do not count for the badge, at present. In all likelihood, this question would've also been in the /review queue and if one were fortuitous enough to have reviewed it there, these actions would've counted towards the badge.   Would it be possible to include reviews made outside of /review to count if the question was also in the review queue? Note that this shouldn't apply if the question dropped out of the queue (either by inaction or after it was edited and improved by previous reviews). Setting-up SSH-tunnel for VPN? I have a laptop running Ubuntu.  I also rent a VPS running Ubuntu.  I've bought a paid VPN-service, which I use on my laptop.  I was wondering how I can set-up a SSH-tunnel or something similar, from my laptop to the VPS I'm renting, so I then can run VPN (from my laptop) "through" it (to the VPS), to the VPN-server.  The reason I wish to do so, is that a place I often work blocks my VPN-service, but not my SSHing to my VPS...  While I have been using Linux for a while, I'm a novice when it comes to setting-up virtual network-devices, advanced routing, and so on... so please spoonfeed me (using small words and colorful pictures ;-)  I have been googling and looking at various answers on the StackExchange-network, but not seems to be exactly what I'm after.  SSH-tunneling seemed promising, but since I don't have/know an exact domain and port for my VPN, I couldn't figure out how to use it...  I guess I may need to set-up some virtual network-device on my laptop (which ends on the VPS), and connect VPN through it - but how to do this, sadly goes over my head. I'm playing around with fdisk and I tried to create a partition on an empty USB drive. Here's how that went: Command (m for help): n Partition number (1-128, default 1): 1 First sector (34-61187102, default 2048): 2048 Last sector, +/-sectors or +/-size{K,M,G,T,P} (2048-61187102, default 61187102): +30000000           Created a new partition 1 of type 'Linux filesystem' and of size 14,3 GiB.  so just one partition with 30 million sectors. However, when I check how many sectors it contains I get:  I'm checking with a different program and I see again that additional sector:  Sorry for the pettiness :) I'm learning. Which way is up? (electric outlet) Using "${a:-b}" for variable assignment in scripts How can I recover my account? I tried to log into Minecraft today to find that my password isn't working. I tried to change it, but I can't login to my email because I forgot the answer to my security question. I tried changing the email to a different one, but that didn't work for some reason (it kept saying "invalid email"). I don't have my transation ID because I forgot which credit card I used.  Does this mean that I have to buy a new Minecraft account? Understanding the -exec option of `find` Here is the problem: From any IP address not belonging to your mail server:  telnet me.myemailserver.com 25    helo me.someserver.com mail from: &lt;yourusername@mydomain.com&gt; rcpt to: &lt;yourusername@mydomain.com&gt; data This is spam.  Buy my stuff. .   I'm using Postfix.  I'm having a problem finding a solution to requiring SMTP-AUTH for email claiming to be from mydomain.com.  Googling around, this guy has identified the same problem (where I cut-n-paste with some modifications) the above example from:   This link   gets close to a solution but it has a side effect of requiring SMTP-AUTH for mail not from mydomain.com.  For mail not claiming to be from mydomain.com, I would do the usual RBL and Spam filtering.  In short, I want to reject mail to local domains (mydomain.com) from outside/unauthenticated clients claiming to be from local domains (mydomain.com).  This is what I tried: I've tried both permit and reject as the default.  Here is exact excerpt from my main.cf:  smtpd_recipient_restrictions = reject_unauth_pipelining,                            permit_sasl_authenticated,                            check_recipient_access pgsql:/etc/postfix/pgsql-recipient.cf,                            reject_unauthenticated_sender_login_mismatch,                            reject_unauth_destination,                            reject_unlisted_recipient,                            check_sender_access pgsql:/etc/postfix/pgsql-sender.cf,                            reject_unlisted_sender,                            reject_invalid_hostname,                            reject_non_fqdn_hostname,                            reject_non_fqdn_sender,                            reject_non_fqdn_recipient,                            reject_unknown_sender_domain,                            reject_unknown_recipient_domain,                            reject_rbl_client cbl.abuseat.org,                            reject_rbl_client sbl.spamhaus.org,                            reject_rbl_client sbl-xbl.spamhaus.org,                            reject_rbl_client bl.spamcop.net,                            reject_rbl_client dnsbl.njabl.org,                            reject_rbl_client blackholes.wirehub.net,                            reject_rbl_client relays.mail-abuse.org,                            reject_rbl_client dialups.mail-abuse.org,                            reject_rbl_client blackholes.mail-abuse.org,                            reject_rhsbl_sender dsn.rfc-ignorant.org,                            (reject and permit both tried here) Cannot log into Area 51 with OpenID Interesting topics to research in mathematical physics for undergraduates We login to applications almost every day. The password is stored in our fingers, as an automated process (if they have used it long enough). When users realize they have entered a wrong character (before hitting login-button) they erase the whole password, and retype the entire password – instead of just the last wrong character. Question is why?         &ndash; Wireframes created with extending convergence radius using one boundary point Show that if $f(z)=\sum\limits_{n\geq0}a_nz^n$ is analytic in $\{z\in\mathbb{C}:|z|&lt;1\}\cup\{1\}$ and $\forall n\geq0:a_n\geq0$ then the radius of convergence of the power series is strictly larger than $1$.  My approach:   Let $D=\{z\in\mathbb{C}:|z|\leq1\}$.  We have that $f(1)$ converges, so $\sum\limits_{n\geq 0} a_n&lt;\infty$. Moreover since the $a_n$'s are all positive we have that $|a_n|=a_n$, therefore $\sum\limits_{n\geq 0} |a_n|=\sum\limits_{n\geq 0} a_n&lt;\infty$.  Let $f_n(z)=a_nz^n$, then $\forall z\in D, n\geq 0$ we have:  $$|f_n(z)|=|a_nz^n|=|a_n|\cdot|z^n|=a_n\cdot |z|^n\leq a_n\cdot 1^n\leq a_n$$  Then, applying Weierstrass M-test we get that the power series is convergent in $D$.  Now, I need to "find" an open ball $B$ such that $D\subset B$ and the power series is convergent in it.  Edit  $f$ is analytic in $1$, so there is an open set $U$ that contains $1$ such that   $$f(z)=\sum\limits_{n\geq 0} b_n(z-1)^n$$  Does it mean that the series is convergent in $U$?   I can see that we have that the two series coincide in $D\cap U$.